GSEB Class 12 Statistics Solutions Chapter 3 Normal Distribution Solutions

Exercise 3

Section A

Answer the following questions by selecting a correct option from the given options:

 

Question 1. Which of the following is probability density function for normal variable X with mean \(\mu\) and standard deviation \(\sigma\)?
(a) \(f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)}\), - \(\infty\) < X < \(\infty\)
(b) \(f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-\left(\frac{x-\mu}{\sigma}\right)^{2}}\) - \(\infty\) < x < \(\infty\)
(c) \(f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^{2}}\) – \(\infty\) < X < \(\infty\)
(d) \(f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^{2}}\), 0 < x < \(\infty\)
Answer: (c) \(f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^{2}}\) – \(\infty\) < X < \(\infty\)}
In simple words: The correct formula shows how the probability changes for a normal variable, where the values can range from negative infinity to positive infinity.

🎯 Exam Tip: Remember the exact form of the probability density function for a normal distribution, including the exponent and the range of X, as it is a fundamental concept.

 

Question 2. For a normal variable X with mean \(\mu\) and standard deviation \(\sigma\), which of the following is standard normal variable Z for it?
(a) \(Z = \frac{x-\sigma}{\mu}\)
(b) \(Z = \frac{\sigma-x}{\mu}\)
(c) \(Z = \frac{\mu-x}{\sigma}\)
(d) \(Z = \frac{x-\mu}{\sigma}\)
Answer: (d) \(Z = \frac{x-\mu}{\sigma}\)}
In simple words: To change a normal variable X into a standard normal variable Z, you subtract the mean (\(\mu\)) and then divide by the standard deviation (\(\sigma\)).

🎯 Exam Tip: Know the Z-score formula perfectly, as it is used to standardize any normal distribution, allowing for comparisons.

 

Question 3. Which of the following is probability density function for standard normal variable?
(a) \(f(z) = e^{-\frac{1}{2} z^{2}}\), -\(\infty\) < Z < \(\infty\)
(b) \(f(z) = \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} z^{2}}\), -\(\infty\) < z < \(\infty\)
(c) \(f(z) = \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} z^{2}}\), 0 < z < \(\infty\)
(d) \(f(z) = \frac{1}{\sqrt{2 \pi}}e^{-z^{2}}\), – \(\infty\) < z < \(\infty\)
Answer: (b) \(f(z) = \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} z^{2}}\), -\(\infty\) < z < \(\infty\)}
In simple words: The standard normal variable Z has a specific probability function, which includes `1` over `sqrt(2π)` and `e` raised to the power of negative half of Z squared, covering all real numbers.

🎯 Exam Tip: Distinguish between the probability density function for a general normal variable and the specific form for a standard normal variable (Z-score).

 

Question 4. Which of the following are mean and variance of standard normal variable ?
(a) Mean = 0, Variance = 1
(b) Mean = 1, Variance = 0
(c) Mean = 0, Variance = 0
(d) Mean = 1, Variance = 1
Answer: (a) Mean = 0, Variance = 1}
In simple words: A standard normal variable always has a mean of zero and a variance (and thus standard deviation) of one.

🎯 Exam Tip: Remember these key properties of the standard normal distribution (mean=0, variance=1) as they simplify calculations and interpretations.

 

Question 5. What is the total area under normal curve among the following?
(a) - 1
(b) 0
(c) 1
(d) 0.5
Answer: (c) 1}
In simple words: For any probability distribution, the total area under its curve must always be one, representing 100% of all possible outcomes.

🎯 Exam Tip: This is a fundamental property of all probability distributions; the total area under the curve always sums to 1.

 

Question 6. What is the area under the normal curve to the right hand side of perpendicular line at X = \(\mu\)?
(a) 0
(b) 0.5
(c) 1
(d) - 0.5
Answer: (b) 0.5}
In simple words: The normal curve is perfectly symmetrical around its mean, so exactly half of the area is on the right side of the mean.

🎯 Exam Tip: The symmetry of the normal curve around its mean (\(\mu\)) means the area on either side of the mean is 0.5.

 

Question 7. In normal distribution, usually which limits include 99 % of the observations?
(a) \(\mu\) + 1.96\(\sigma\)
(b) \(\mu\) \(\pm\) 2\(\sigma\)
(c) \(\mu\) \(\pm\) 3\(\sigma\)
(d) \(\mu\) \(\pm\) 2.575\(\sigma\)
Answer: (d) \(\mu\) \(\pm\) 2.575\(\sigma\)}
In simple words: For a normal distribution, about 99% of the data falls within 2.575 standard deviations from the mean.

🎯 Exam Tip: Memorize the approximate percentages of observations within specific standard deviation ranges from the mean (e.g., 68%, 95%, 99%).

 

Question 8. In normal distribution, usually what percentage of the observations are included in the limits \(\mu\) \(\pm\) \(\sigma\)?
(a) 34.13%
(b) 95.45 %
(c) 68.26 %
(d) 50%
Answer: (c) 68.26%}
In simple words: Approximately 68.26% of all data points in a normal distribution lie within one standard deviation away from the average value.

🎯 Exam Tip: This is a key part of the empirical rule (68-95-99.7 rule) for normal distributions, which is frequently tested.

 

Question 9. Which of the following is approximate value of mean deviation for normal variable?
(a) \(\frac{4}{5} \sigma\)
(b) \(\frac{4}{5} \mu\)
(c) \(\frac{2}{3} \sigma\)
(d) \(\frac{2}{3} \mu\)
Answer: (a) \(\frac{4}{5} \sigma\)}
In simple words: The mean deviation for a normal variable is roughly four-fifths of its standard deviation.

🎯 Exam Tip: Understand the relationships between different measures of dispersion (mean deviation, quartile deviation) and the standard deviation in a normal distribution.

 

Question 10. Which of the following is approximate value of quartile deviation for standard normal variable ?
(a) \(\frac{2}{3} \sigma\)
(b) \(\frac{2}{3}\)
(c) \(\frac{4}{5} \sigma\)
(d) \(\frac{4}{5}\)
Answer: (b) \(\frac{2}{3}\)}
In simple words: For a standard normal variable, the quartile deviation is approximately two-thirds.

🎯 Exam Tip: Recall the specific relationships for standard normal variables; here, \(\sigma\) is 1, so the formula simplifies.

 

Question 11. Mean and the first quartile for a normal distribution are 11 and 3 respectively. Which of the following is the value of the third quartile ?
(a) 8
(b) 14
(c) 19
(d) 10
Answer: (c) 19}
In simple words: In a normal distribution, the mean is exactly in the middle of the first and third quartiles. If the mean is 11 and the first quartile is 3, the third quartile must be 19 to keep 11 as the midpoint.

🎯 Exam Tip: For a symmetric distribution like the normal distribution, the mean, median, and mode are equal, and the quartiles are equidistant from the median.

 

Question 12. For a normal distribution, approximate value of mean deviation is 20. Which of the following is the value of quartile deviation?
(a) \(\frac{25}{3}\)
(b) \(\frac{32}{3}\)
(c) 24
(d) \(\frac{50}{3}\)
Answer: (d) \(\frac{50}{3}\)}
In simple words: If the mean deviation is 20, we can find the standard deviation first. Since mean deviation is about `4/5` of standard deviation, `σ` is `25`. Then, quartile deviation is about `2/3` of standard deviation, which gives `50/3`.

🎯 Exam Tip: Remember the approximate relationships between mean deviation (\(\approx \frac{4}{5}\sigma\)) and quartile deviation (\(\approx \frac{2}{3}\sigma\)) in a normal distribution to solve problems efficiently.

 

Question 13. In usual notation of normal distribution, x = 25; \(\mu\) = 20 and \(\sigma\) = 5, then which of the following is the value of standard normal variable ?
(a) 1
(b) - 1
(d) \(\frac{10}{3}\)
Answer: (a) 1}
In simple words: To find the standard normal variable (Z-score), subtract the mean from the given value and divide by the standard deviation. Here, Z = (25 - 20) / 5 = 5 / 5 = 1.

🎯 Exam Tip: Practice calculating Z-scores accurately, as it is a core skill for transforming raw data into standard units for normal distribution analysis.

 

Question 14. Mean of a normal variable X is 50. If the value of Z-score is - 2.5 for x = 25, then which of the following is a variance of the distribution ?
(a) 10
(b) 100
(c) 50
(d) 25
Answer: (b) 100}
In simple words: We know Z = (X - mean) / standard deviation. Putting in the given values, -2.5 = (25 - 50) / \(\sigma\), so -2.5 = -25 / \(\sigma\). This means \(\sigma\) = 10. The variance is \(\sigma\) squared, which is 100.

🎯 Exam Tip: Work backwards from the Z-score formula to find missing parameters like standard deviation or mean. Remember that variance is the square of the standard deviation.

 

Question 15. If the distribution of normal variable is shown as N (20, 4), then which of the following intervals includes 99.73 % of observations ?
(a) (18, 22)
(b) (16, 24)
(c) (14, 26)
(d) (12, 28)
Answer: (c) (14, 26)}
In simple words: For N(20, 4), the mean is 20 and variance is 4, so standard deviation is 2. The 99.73% of observations fall within `\(\mu\) \(\pm\) 3\(\sigma\)` range. So, the interval is `20 \(\pm\) 3(2)`, which means `20 \(\pm\) 6`, resulting in (14, 26).

🎯 Exam Tip: The 68-95-99.7 empirical rule is crucial. 99.73% of observations lie within three standard deviations of the mean. N(mean, variance) means mean is the first number and standard deviation is the square root of the second.

 

Section B

Answer the following questions in one sentence:

 

Question 1. Give the values of the constants used in probability density function of normal variable.
Answer: In probability density function of normal variable the values of the constants are \(\pi\) = 3.1416 and e = 2.7183.
In simple words: The constants used are pi (\(\pi\)) which is about 3.1416 and Euler's number (e) which is about 2.7183.

🎯 Exam Tip: These are universal mathematical constants; know their approximate values for context, even if not required for direct calculation in simple questions.

 

Question. What is the probability that a continuous random variable takes definite value?
Answer: The probability that a continuous random variable takes definite value is zero (0).
In simple words: For a continuous variable, the chance of it being any *exact* single number is zero because there are infinitely many possible numbers.

🎯 Exam Tip: For continuous probability distributions, probability is calculated over intervals, not exact points, as the probability of any single point is infinitesimally small.

 

Question 3. What is the shape of normal curve ?
Answer: The shape of normal curve is completely bell-shaped.
In simple words: A normal curve looks like a symmetrical bell.

🎯 Exam Tip: The term "bell-shaped" is a standard and important descriptor for the normal distribution's curve.

 

Question 4. What is the skewness of normal distribution ?
Answer: The skewness of normal distribution is zero.
In simple words: A normal distribution has no skewness because it is perfectly balanced on both sides.

🎯 Exam Tip: Zero skewness is a direct result of the normal distribution's perfect symmetry.

 

Question 5. "Standard score is independent of unit of measurement.” Is this statement true or false?
Answer: “Standard score is independent of unit of measurement.” This statement is true.
In simple words: A standard score does not depend on the units used (like cm or inches) because it shows how many standard deviations a value is from the mean.

🎯 Exam Tip: The unit-free nature of the Z-score is what allows it to be used for comparing data from different distributions.

 

Question 6. For which value of standard normal variable, the standard normal curve is symmetric on both the sides?
Answer: For Z = 0, the standard normal curve is symmetric on both the sides.
In simple words: The standard normal curve is symmetrical around the point where Z is zero.

🎯 Exam Tip: The mean of the standard normal distribution is 0, which also serves as its point of symmetry.

 

Question 7. Which value of normal variable divides the area of normal curve in two equal parts ?
Answer: The value of normal variable X = \(\mu\) mean divides the area of normal curve in two equal parts.
In simple words: The mean (\(\mu\)) of a normal variable splits the normal curve's area into two equal halves.

🎯 Exam Tip: In a normal distribution, the mean, median, and mode are all equal and divide the curve's area into two equal halves.

 

Question 8. What percentage of area is covered under the normal curve within the range \(\mu\) – 2\(\sigma\) to \(\mu\) + 2\(\sigma\)?
Answer: 95.45% of area is covered under the normal curve within the range \(\mu\) – 2\(\sigma\) to \(\mu\) + 2\(\sigma\).
In simple words: About 95.45% of data falls within two standard deviations of the mean in a normal distribution.

🎯 Exam Tip: This is another critical component of the empirical rule. Know that approximately 95% of data lies within \(\pm\)2 standard deviations.

 

Question 9. Mean of a normal distribution is 13.25 and its standard deviation is 10. Estimate the value of its third quartile.
Answer: Here, \(\mu\) = 13.25; \(\sigma\) = 10
Now, Q3 = \(\mu\) + 0.675 \(\sigma\)
\(\implies\) Q3 = 13.25 + 0.675(10)
= 13.25 + 6.75
= 20
Hence, Q3 = 20
In simple words: To find the third quartile, add `0.675` times the standard deviation to the mean. So, `13.25 + 0.675 * 10` gives `13.25 + 6.75`, which is `20`.

🎯 Exam Tip: For a normal distribution, the quartiles can be estimated using the mean and standard deviation with specific constants (Q1 = \(\mu\) - 0.675\(\sigma\), Q3 = \(\mu\) + 0.675\(\sigma\)).

 

Question 10. For a normal distribution having mean 10 and standard deviation 6, estimate the value of quartile deviation.
Answer: Here, \(\mu\) = 10; \(\sigma\) = 6
Estimated value of quartile deviation:
In normal distribution, Quartile deviation \(\approx \frac{2}{3} \sigma\)
Putting, \(\sigma\) = 6
Quartile deviation \(\approx \frac{2}{3} \times 6\)
\(\implies\) Quartile deviation \(\approx 4\)
Hence, the estimated value of quartile deviation obtained is 4.
In simple words: To estimate the quartile deviation, we multiply two-thirds by the standard deviation. With a standard deviation of 6, it comes out to 4.

🎯 Exam Tip: Remember the approximate relationship between quartile deviation and standard deviation (\(QD \approx \frac{2}{3}\sigma\)) for normal distributions.

 

Question 11. The approximate value of mean deviation for a normal distribution is 8. Find the value of its standard deviation.
Answer: Here, mean deviation = 8
In normal distribution, mean deviation \(\approx \frac{4}{5} \sigma\)
\(\implies\) 8 \(\approx \frac{4}{5} \sigma\)
\(\implies \sigma \approx \frac{8 \times 5}{4}\)
\(\implies \sigma \approx 10\)
Hence, the standard deviation obtained is 10.
In simple words: If the mean deviation is 8, and knowing it's about `4/5` of the standard deviation, we can calculate that the standard deviation is 10.

🎯 Exam Tip: Understand how to use the approximate relationship \(MD \approx \frac{4}{5}\sigma\) to find \(\sigma\) when mean deviation is given.

 

Question 12. For a normal distribution, the estimated value of quartile deviation is 12. Find the value of its standard deviation.
Answer: Here, quartile deviation = 12
In normal distribution,
Estimated value of quartile deviation \(\approx \frac{2}{3} \sigma\)
\(\implies\) 12 \(\approx \frac{2}{3} \sigma\)
\(\implies \sigma \approx \frac{12 \times 3}{2}\)
\(\implies \sigma \approx 18\)
Hence, the value of standard deviation obtained is 18.
In simple words: Given that quartile deviation is 12, and it is approximately `2/3` of the standard deviation, we can find the standard deviation by multiplying 12 by `3/2`, which equals 18.

🎯 Exam Tip: Apply the formula for quartile deviation in terms of standard deviation (\(QD \approx \frac{2}{3}\sigma\)) to calculate \(\sigma\) when QD is known.

 

Question 13. For a probability distribution of standard normal variable, state the estimated limits for the middle 50% observations.
Answer: For a probability distribution of standard normal variable, the estimated limits for the middle 50% observations is (Q1, Q3) = (- 0.675, 0.675).
In simple words: For a standard normal distribution, the middle 50% of values are found between Z = -0.675 (first quartile) and Z = 0.675 (third quartile).

🎯 Exam Tip: The first and third quartiles for a standard normal distribution are symmetrical around the mean (0) and are approximately \(\pm\)0.675.

 

Question 14. The extreme quartiles of normal distribution are 20 and 30. Find its mean.
Answer: Here, Q\(_1\) = 20; Q\(_3\) = 30
\(\implies M = \frac{Q_{3}+Q_{1}}{2} = \frac{30+20}{2} = \frac{50}{2} = 25\)
In normal distribution, Mean = Median = Mode
\(\implies\) Mean \(\mu\) = 25
In simple words: The mean of a normal distribution is the midpoint between its first and third quartiles. So, (20 + 30) / 2 gives a mean of 25.

🎯 Exam Tip: In a perfectly symmetrical distribution like the normal distribution, the mean is exactly halfway between any two symmetrically placed percentiles, such as the quartiles.

 

Question 15. The monthly expense of a group of persons follows normal distribution with mean Rs. 10,000 and standard deviation Rs. 1000. A student has obtained a Z-score = 1 for randomly selected person having monthly expense more than Rs. 11,000. Is this calculation of Z-score true? Give reason.
Answer: Here, \(\mu\) = 10,000; \(\sigma\) = 1000
For x > 11000, \(Z > \frac{x-\mu}{\sigma} \implies Z > \frac{11000-10000}{1000} \implies Z > 1\)
A student has obtained a Z-score = 1. Hence, this calculation of Z-score is not true.
Reason: For given data, we must have Z \(\ge\) 1.
In simple words: If someone's expense is *more than* Rs. 11,000, their Z-score must be *greater than* 1. A Z-score of exactly 1 means their expense is exactly Rs. 11,000, not more than it.

🎯 Exam Tip: Pay close attention to inequality signs (\(<\), \(>\), \(\le\), \(\ge\)) when interpreting Z-scores and probabilities. A Z-score of 1 means the value is *exactly* one standard deviation above the mean.

 

Question 16. The age of a group of persons follows normal distribution with mean 45 years and standard deviation 10 years. Calculate Z-score for a randomly selected person having age 60 years.
Answer: Here, \(\mu\) = 45; \(\sigma\) = 10, x = 60
\(\implies Z = \frac{x-\mu}{\sigma} = \frac{60-45}{10} = 1.5\)
Hence, Z-score = 1.5
In simple words: To find the Z-score for a 60-year-old, subtract the average age (45) from 60, then divide by the standard deviation (10). This gives a Z-score of 1.5.

🎯 Exam Tip: Ensure correct substitution of values into the Z-score formula, paying attention to the order of operations for accurate results.

 

Question 17. Marks obtained by students of a school in Economics subject follows normal distribution with mean \(\mu\) and standard deviation \(\sigma\). The value of standard score that a randomly selected student obtained 60 marks is 1. If the variance of variable is 100 (marks)\(_2\), then find average marks.
Answer: Here, x = 60; Z = 1; \(\sigma^2\) = 100 \(\implies \sigma\) = 10
\(\implies Z = \frac{x-\mu}{\sigma}\)
\(\implies 1 = \frac{60-\mu}{10}\)
\(\implies 10 = 60 – \mu\)
\(\implies \mu = 60 – 10\)
\(\implies \mu = 50\)
Hence, the average marks obtained is 50.
In simple words: We know the Z-score is 1 for 60 marks, and the standard deviation is 10 (because variance is 100). Using the Z-score formula `1 = (60 - mean) / 10`, we find that the average marks (mean) are 50.

🎯 Exam Tip: Remember to calculate standard deviation from variance by taking the square root. Use the Z-score formula to solve for any unknown variable if others are provided.

 

Section C

Answer the following questions as required:

 

Question 1. Define probability density function of continuous random variable.
Answer: A function for obtaining probability that a continuous random variable assumes value between specified interval is called probability density function of continuous random variable. It is denoted by f(x).
In simple words: A probability density function (PDF) is a special function that helps us find the probability of a continuous variable falling within a certain range. We write it as `f(x)`.

🎯 Exam Tip: The key concept of a PDF for continuous variables is that it determines probabilities over intervals, not at single points.

 

Question 2. Write the conditions for probability density function for continuous variable.
Answer: The conditions for probability density function for continuous variable are as follows:

• The probability that the value of random variable X lies within the specified interval must be non-negative (positive).
• The total probability that the random variable X assumes any value in the specified interval must be 1.

In simple words: For a probability density function, two things must be true: first, all probabilities must be positive or zero; second, the total probability for all possible values must add up to one.

🎯 Exam Tip: These two conditions (non-negativity and total area of 1) are fundamental requirements for any valid probability density function.

 

Question 3. How is the normal curve drawn ?
Answer: The normal curve is drawn by considering different values of normal random variable X and its respective values of probability density function f(x).
In simple words: We draw a normal curve by plotting many different values of a normal variable (X) against their corresponding probability density function values `f(x)`.

🎯 Exam Tip: The curve is a graphical representation of the probability density function, showing the distribution of data points.

 

Question 4. Define probability density function for normal variable.
Answer: Probability density function for normal variable is defined as follows ;
\(f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^{2}}\), – \(\infty\) < X < \(\infty\)
Where, x = Value of normal variable X
\(\mu\) = Mean of normal distribution
\(\sigma\) = Standard deviation of normal distribution
\(\pi\) = Constant = 3.1416
e = Constant = 2.7183
In simple words: This special formula `f(x)` tells us the probability density for any value `x` in a normal distribution. It uses the mean (`\(\mu\)`) and standard deviation (`\(\sigma\)`) to show how likely different values are.

🎯 Exam Tip: Recognize and understand each component of the normal probability density function, including the constants and parameters, and the range of X.

 

Question 5. What is the shape of standard normal curve ? To which value of variable it is symmetric ?
Answer: The shape of standard normal curve is complete bell-shaped. It is symmetric about the value of variable Z = 0.
In simple words: The standard normal curve looks like a perfect bell and is symmetrical around the point where the Z-value is zero.

🎯 Exam Tip: The standard normal curve is a special case of the normal curve with mean 0 and standard deviation 1, and its symmetry at Z=0 is a defining characteristic.

 

Question 6. Define standard normal variable and write its probability density function.
Answer: If X is normal variable with mean = \(\mu\) and standard deviation = \(\sigma\), then the random variable \(Z = \frac{\mathrm{X}-\mu}{\sigma}\) is called the standard normal variable. It is free from units of measurement.
The probability density function of Z is as follows :
\(f(z) = \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} z^{2}}\), - \(\infty\) < z < \(\infty\)
In simple words: A standard normal variable, called Z, shows how many standard deviations a value is from its mean. It doesn't have units. Its probability function is a bell-shaped curve that is always the same, centered at zero.

🎯 Exam Tip: Understand that the standard normal variable (Z-score) transforms any normal variable into a standardized scale, which simplifies probability calculations using standard tables.

 

Question 7. A normal variable X has the probability density function as,
\(f(x) = \text{constant } x e^{-\frac{1}{50}(x-10)^{2}}\), - \(\infty\) < x < \(\infty\)
Find the first quartile from this information.
Answer: \(f(x) = \text{constant } x e^{-\frac{1}{50}(x-10)^{2}}\), - \(\infty\) < x < \(\infty\)
\(= \text{constant } x e^{-\frac{1}{2}\left(\frac{x-10}{5}\right)^{2}}\)
Hence, \(\mu\) = 10, \(\sigma\) = 5
First Quartile Q1:
25 % of observations of the distribution is less than Q1.
\(\implies P [X \le Q1] = \frac{25}{100} = 0.25\)
Now, \(z_1 = \frac{\mathrm{Q}_{1}-\mu}{\sigma} = \frac{Q_{1}-10}{5}\)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सामान्य वितरण वक्र को दिखाता है, जिसमें पहला चतुर्थक (Q1) दर्शाया गया है। Q1 वह बिंदु है जिसके बाईं ओर कुल क्षेत्रफल का 25% (0.25) होता है, जो यह बताता है कि 25% अवलोकन इस मान से कम हैं।

P[Z \(\le\) z1] = 0.25
\(= P[-\infty < Z \le 0] – P[z_1 \le Z \le 0]\)
\(= P[0 \le Z < \infty] – P[0 \le Z \le Z1]\)
\(\implies P[0 \le Z \le Z1] = 0.5000 – 0.2500\)
\(= 0.2500\)
From the table, corresponding to probability 0.2486, the value of z\(_1\) = -0.67 and corresponding to probability 0.2518, the value of z\(_1\) = -0.68
\(\implies\) For probability 0.25
\(z_1 = \frac{(-0.67)+(-0.68)}{2} = - 0.675\)
Now, \(z_1 = \frac{Q_{1}-10}{5}\)
\(\implies -0.675 = \frac{Q_{1}-10}{5}\)
\(\implies - 0.675 \times 5 = Q1 - 10\)
\(\implies -3.375 = Q1 - 10\)
\(\implies 10 – 3.375 = Q1\)
\(\implies Q_1 = 6.625 \approx 6.63\)
Hence, first quartile obtained is 6.83.
In simple words: From the given function, the mean is 10 and standard deviation is 5. To find the first quartile (Q1), which is the point where 25% of observations are less than it, we use the Z-score table. We find that a Z-score of -0.675 corresponds to 25% of the area to its left. Then, we use the Z-score formula to solve for Q1, getting approximately 6.63.

🎯 Exam Tip: Be careful to identify the correct mean and standard deviation from the probability density function. Use the Z-table accurately for finding corresponding Z-scores for given probabilities, and calculate Q1 using the Z-score formula.

 

Question 8. The extreme quartiles of a normal variable are 10 and 30. Find its mean deviation.
Answer: Here, Q\(_1\) = 10; Q\(_3\) = 30
\(\implies\) Quartile deviation \( = \frac{Q_{3}-Q_{1}}{2} = \frac{30-10}{2} = 10\)
Now, Quartile deviation \( = \frac{2}{3} \sigma\)
\(\implies 10 = \frac{2}{3} \sigma\)
\(\implies \sigma = 10 \times \frac{3}{2} = 15\)
Mean deviation \( = \frac{4}{5} \sigma\)
\(\implies\) Mean deviation \( = \frac{4}{5} \times 15 = 12\)
Hence, the mean deviation obtained is 12.
In simple words: First, calculate the quartile deviation from Q1 and Q3, which is 10. Then, use the relationship that quartile deviation is about `2/3` of the standard deviation to find the standard deviation, which is 15. Finally, use the relationship that mean deviation is about `4/5` of the standard deviation to get a mean deviation of 12.

🎯 Exam Tip: This question tests your knowledge of the relationships between quartiles, quartile deviation, standard deviation, and mean deviation in a normal distribution. Remember all the approximate formulas.

 

Question 9. For a normal variable, mean deviation is 48 and its third quartile is 120. Estimate its first quartile.
Answer: Here, Mean deviation = 48; Q\(_3\) = 120
Now, mean deviation \( = \frac{4}{5} \sigma\)
\(\implies 48 = \frac{4}{5} \sigma\)
\(\implies \sigma = 48 \times \frac{5}{4} \approx 60\)
Quartile deviation \(\approx \frac{2}{3} \sigma\)
\(\implies\) Quartile deviation \(\approx \frac{2}{3} \times 60\)
\(\implies \frac{Q_{3}-Q_{1}}{2} \approx 40\)
\(\implies Q3 – Q1 \approx 80\)
Putting, Q3 = 120
\(\implies 120 - Q1 \approx 80\)
\(\implies Q_1 \approx 120 – 80 \approx 40\)
Hence, the first quartile obtained is 40.
In simple words: First, use the mean deviation (48) to find the standard deviation (60). Then, calculate the quartile deviation (40) from the standard deviation. Since the quartile deviation is half the difference between Q3 and Q1, we can find Q1 by subtracting the quartile deviation from Q3, which gives 40.

🎯 Exam Tip: This problem requires a multi-step calculation. First, find \(\sigma\) from MD, then QD from \(\sigma\), and finally Q1 using QD and Q3. Make sure to clearly lay out each step.

 

Question 10. A normal variable X has the probability density function as,
\(f(x) = \frac{1}{10 \sqrt{2 \pi}} e^{-\frac{1}{2}\left(\frac{x-100}{10}\right)^{2}}\), - \(\infty\) < X < \(\infty\)
For this distribution, obtain the limits which include middle 68.26% of the observations.
Answer: \(f (x) = \frac{1}{10 \sqrt{2 \pi}} e^{-\frac{1}{2}\left(\frac{x-100}{10}\right)^{2}}\), - \(\infty\) < X < \(\infty\)
\(\implies \mu\) = 100, \(\sigma\) = 10
Now, the limits that include middle 68.26 % of the observations = \(\mu\) \(\pm\) \(\sigma\)
\(= (\mu – \sigma)\) to \((\mu+ \sigma)\)
\(= (100 – 10)\) to \((100 + 10)\)
\(= 90\) to 110
Hence, the limits that include middle 68.26% of the observations obtained is 90 to 110.
In simple words: From the given function, the mean is 100 and standard deviation is 10. The middle 68.26% of observations fall within one standard deviation from the mean, so the range is `100 \(\pm\) 10`, which is from 90 to 110.

🎯 Exam Tip: Identify the mean and standard deviation correctly from the given PDF. Recall that approximately 68.26% of observations in a normal distribution lie within \(\pm\)1 standard deviation of the mean.

 

Question 11. The probability that the value of standard normal variable lies between 0 and Z-score (Z1) is 0.4950. Find the possible values of Z-score.
Answer: Here, P [0 \(\le\) Z \(\le\) z\(_1\)) = 0.4950
From the table of area under the standard normal curve, corresponding to probability 0.4949 (which is near to 0.4950) the value of Z-score = 2.57. Also corresponding to probability 0.4951 (which is also near to 0.4950), the value of Z-score = 2.58. .
\(\implies\) Corresponding to probability 0.4950, the value of Z-score \( = \frac{2.57+2.58}{2} = 2.575\)
Hence, the possible value of Z-score obtained is 2.575.
In simple words: We are looking for the Z-score where the area under the standard normal curve from 0 to Z is 0.4950. By checking the Z-table, we find that a Z-score of 2.575 gives this area.

🎯 Exam Tip: Use the standard normal distribution table (Z-table) to find the Z-score corresponding to a given probability. If the exact probability is not found, interpolate between the closest values.

 

Section D

Answer the following questions as required:

 

Question 1. Define normal distribution and state the characteristics of normal curve.
Answer: The distribution of a normal variable X is called Normal distribution. It is denoted by N [\(\mu\), \(\sigma^2\)].
The characteristics of normal curve:

• It is completely bell-shaped.
• It is asymptotic to X-axis, i.e., the tails of the curve never touch X-axis.
• The total area of the region bounded by normal curve and X-axis = Total probability 1.
• It is symmetric about the sides of mean \(\mu\) of normal variate. So the area of the region of normal curve on both the sides of X = \(\mu\) is equal to 0.5.
• P [\(\mu \le X \le a\)] is the area of region bounded by X-axis and two perpendicular lines at X = \(\mu\) and X = a.
• To find the area under the normal curve, the normal variable is transformed to standard normal variable Z and readymade table is used.

In simple words: Normal distribution describes how many values are spread out around a central value, shown as `N[\(\mu\), \(\sigma^2\)]`. Its curve is bell-shaped and never touches the X-axis. It is symmetrical around the mean, with half the area on each side, and the total area under the curve is 1. We use Z-scores and tables to find areas under this curve.

🎯 Exam Tip: Be able to both define normal distribution and list its key characteristics, such as shape, symmetry, asymptotic behavior, and total area under the curve.

 

Question 2. State the properties of normal distribution.
Answer: The normal distribution possesses the following properties:
1. The normal distribution is a probability distribution of continuous random variable.
2. \(\mu\) and \(\sigma\) are its parameters.
3. The graph of density function of this distribution is continuous graph of bell-shape.
4. The normal curve is symmetric about X = \(\mu\).
5. The total area under normal curve is 1 and the area of the region of normal curve on both sides of X = \(\mu\) is equal and 0.5.
6. The mean, median and mode of the distribution are equal, i.e., \(\mu\) = M = M\(_0\).
7. The skewness of this distribution is zero.
8. The first quartile Q\(_1\) and third quartile Q\(_3\) are equidistant from the second quartile Q\(_2\) in this distribution, i.e., Q\(_3\) – Q\(_2\) = Q\(_2\) – Q\(_1\) and \(M = \frac{Q_{3}+Q_{1}}{2}\)
9. In the form of \(\mu\) and \(\sigma\) the estimated values of Q\(_1\) and Q\(_3\) are obtained as follows:
Q\(_1 \approx \mu\) – 0.675\(\sigma\) and Q\(_3 \approx \mu\) + 0.675\(\sigma\)
10. Two tails of the curve of normal distribution are asymptotic to X-axis, i.e., they never touch X-axis.
11. In this distribution,

• Quartile deviation \(\approx \frac{2}{3}\sigma\)
• Mean deviation \(\approx \frac{4}{5}\sigma\)

12. The important areas under normal curve are as follows :

• The area under the normal curve between \(\mu\) \(\pm\) \(\sigma\) = 0.6826

In simple words: Normal distribution is for continuous data, defined by its mean (`\(\mu\)`) and standard deviation (`\(\sigma\)`). Its graph is a smooth, bell-shaped curve, symmetrical around the mean. The total area under the curve is 1, with 0.5 on each side of the mean. The mean, median, and mode are all the same, and it has zero skewness. Quartiles are equally spaced around the median. Its tails get very close to the X-axis but never touch. Also, specific percentages of data lie within certain standard deviation ranges, like 68.26% within \(\pm\)\(\sigma\).

🎯 Exam Tip: This is a comprehensive question. Be prepared to list and explain multiple properties of the normal distribution, including its shape, measures of central tendency, dispersion relationships, and asymptotic behavior.

Question 3. State the properties of standard normal distribution.

Answer:The standard normal distribution has several key characteristics:
(1) It is a distribution for a continuous random variable.
(2) The curve representing this distribution is perfectly bell-shaped.
(3) The curve's tails get very close to the X-axis but never actually touch it; they are asymptotic.
(4) The mean, variance, and standard deviation of this distribution are all 0 and 1 respectively.
(5) For this distribution, the mean, median, and mode values are all zero.
(6) The estimated values for the first quartile (\(Q_1\)) and third quartile (\(Q_3\)) are approximately \(Q_1 \approx -0.675\) and \(Q_3 \approx 0.675\).
(7) The curve of this distribution is perfectly symmetrical around \(Z = 0\), meaning its skewness is zero.
(8) The total area under the curve bounded by the X-axis, with \(Z = 0\) at its center, is 0.5 on both sides.
(9) In this distribution, the approximate values are:
• Quartile deviation \( \approx \frac{2}{3} \)
• Mean deviation \( \approx \frac{4}{5} \)
(10) Key areas under the standard normal curve are:
• The area between \(z = \pm 1\) is 0.6826
• The area between \(z = \pm 2\) is 0.9545
• The area between \(z = \pm 3\) is 0.9973
• The area between \(z = \pm 1.96\) is 0.95
• The area between \(z = \pm 2.575\) is 0.99In simple words: The standard normal distribution has a bell-shaped curve that is perfectly balanced around zero. Its average is zero, and its spread is one. The curve never touches the horizontal line, and its total area is one.

🎯 Exam Tip: Remember the mean, variance, and standard deviation values (0, 1, 1 respectively) for standard normal distribution, and the property of symmetry around Z=0, as these are frequently tested.

Question 4. A normal variable X has a mean of 50 and a variance of 9. Find the probability that:
(i) The value of normal variable X lies between 50 and 53.
(ii) The value of normal variable X lies between 47 and 53.

Answer:Here, the mean (\(\mu\)) is 50, and the variance (\(\sigma^2\)) is 9.
Therefore, the standard deviation (\(\sigma\)) is \( \sqrt{9} = 3 \).
(i) We need to find the probability \( P [50 \le X \le 53] \).
First, convert the X-values to Z-scores:
For \( x_1 = 50 \), the Z-score is \( Z_1 = \frac{x_1 - \mu}{\sigma} = \frac{50 - 50}{3} = 0 \).
For \( x_2 = 53 \), the Z-score is \( Z_2 = \frac{53 - 50}{3} = \frac{3}{3} = 1 \).
So, \( P [50 \le X \le 53] = P [0 \le Z \le 1] \).
From the standard normal distribution table, the area for \( P [0 \le Z \le 1] \) is 0.3413.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सामान्य वितरण वक्र को दर्शाता है। इसमें, Z-स्कोर 0 से 1 तक के क्षेत्र को छायांकित किया गया है, जो X के 50 से 53 के बीच होने की संभावना को दर्शाता है। यह क्षेत्र वक्र के कुल क्षेत्र का एक हिस्सा है।
(ii) We need to find the probability \( P [47 \le X \le 53] \).
First, convert the X-values to Z-scores:
For \( x_1 = 47 \), the Z-score is \( Z_1 = \frac{x_1 - \mu}{\sigma} = \frac{47 - 50}{3} = \frac{-3}{3} = -1 \).
For \( x_2 = 53 \), the Z-score is \( Z_2 = \frac{53 - 50}{3} = \frac{3}{3} = 1 \).
So, \( P [47 \le X \le 53] = P [-1 \le Z \le 1] \).
Since the normal distribution is symmetric around 0, \( P [-1 \le Z \le 1] = 2 \times P [0 \le Z \le 1] \).
Using the value from part (i), \( 2 \times 0.3413 = 0.6826 \).In simple words: First, convert the given numbers into Z-scores using the mean and standard deviation. Then, use the Z-scores to find the area under the normal curve, which tells you the probability. For symmetric ranges, you can double the probability from 0 to the positive Z-score.

🎯 Exam Tip: Always convert X-values to Z-scores when dealing with normal distribution probabilities. Remember the standard normal table provides areas from 0 to Z, and utilize the symmetry property (\(P[-Z \le Z \le Z] = 2 \times P[0 \le Z \le Z]\)) to simplify calculations.

Question 5. If X is a normal variable with a mean of 100 and a standard deviation of 15, then find the percentage of observations that are:
(i) More than 85.
(ii) Less than 130.

Answer:Here, the mean (\(\mu\)) is 100, and the standard deviation (\(\sigma\)) is 15.
(i) We want to find the percentage of observations having a value more than 85, i.e., \( P[X \ge 85] \).
First, convert \( x = 85 \) to a Z-score:
\( Z = \frac{x - \mu}{\sigma} = \frac{85 - 100}{15} = \frac{-15}{15} = -1 \).
So, we need to find \( P[Z \ge -1] \).
This probability can be split into two parts: \( P[-1 \le Z \le 0] + P[0 \le Z \le \infty] \).
Due to symmetry, \( P[-1 \le Z \le 0] = P[0 \le Z \le 1] \).
From the standard normal table, \( P[0 \le Z \le 1] = 0.3413 \).
Also, \( P[0 \le Z \le \infty] \) represents half of the total area under the curve, which is 0.5000.
So, \( P[Z \ge -1] = 0.3413 + 0.5000 = 0.8413 \).
To express this as a percentage, multiply by 100: \( 0.8413 \times 100 = 84.13\% \).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सामान्य वितरण वक्र को दिखाता है। इसमें Z-स्कोर -1 से शुरू होकर दाहिनी ओर अनंत तक के क्षेत्र को छायांकित किया गया है। यह 85 से अधिक मानों की संभावना को दर्शाता है, जिसमें वक्र का अधिकांश हिस्सा शामिल है।
(ii) We want to find the percentage of observations having a value less than 130, i.e., \( P[X \le 130] \).
First, convert \( x = 130 \) to a Z-score:
\( Z = \frac{x - \mu}{\sigma} = \frac{130 - 100}{15} = \frac{30}{15} = 2 \).
So, we need to find \( P[Z \le 2] \).
This probability can be split into two parts: \( P[-\infty \le Z \le 0] + P[0 \le Z \le 2] \).
\( P[-\infty \le Z \le 0] \) represents half of the total area under the curve, which is 0.5000.
From the standard normal table, \( P[0 \le Z \le 2] = 0.4772 \).
So, \( P[Z \le 2] = 0.5000 + 0.4772 = 0.9772 \).
To express this as a percentage, multiply by 100: \( 0.9772 \times 100 = 97.72\% \).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सामान्य वितरण वक्र को प्रदर्शित करता है। इसमें Z-स्कोर अनंत से शुरू होकर 2 तक के क्षेत्र को छायांकित किया गया है। यह 130 से कम मानों की संभावना को दर्शाता है, जिसमें वक्र का एक बहुत बड़ा हिस्सा कवर होता है।In simple words: To find the percentage, first change the given X-values to Z-scores using the formula. Then, look up the area for these Z-scores in a standard normal table. Add 0.5 for areas covering half the curve, and convert the final probability to a percentage by multiplying by 100.

🎯 Exam Tip: When calculating probabilities for \(P[X \ge x]\) or \(P[X \le x]\), remember to consider the area from \(-\infty\) or to \(+\infty\) respectively, which is 0.5, and add or subtract areas from the Z-table accordingly.

Question 6. The weights of 500 randomly selected adult persons from a city region follow a normal distribution. The average weight of these persons is 55 kg, and their variance is 100 kg\(^2\). Based on their weight, persons are categorized as:
(1) Persons weighing more than 70 kg are in the fat persons group.
(2) Persons weighing between 50 kg and 60 kg are in the healthy persons group.
(3) Persons weighing less than 35 kg are in the physically weak person's group.
Using this information, estimate the number of fat persons, healthy persons, and physically weak persons in that area.

Answer:Here, the total number of persons (\(N\)) is 500.
The mean weight (\(\mu\)) is 55 kg.
The variance (\(\sigma^2\)) is 100 kg\(^2\), so the standard deviation (\(\sigma\)) is \( \sqrt{100} = 10 \) kg.
X represents the average weight in kg.
(i) **Number of fat persons (weight more than 70 kg):**
We need to find \( P[X \ge 70] \).
Convert \( x_1 = 70 \) kg to a Z-score: \( Z_1 = \frac{x_1 - \mu}{\sigma} = \frac{70 - 55}{10} = \frac{15}{10} = 1.5 \).
So, we need to find \( P[Z \ge 1.5] \).
\( P[Z \ge 1.5] = P[0 \le Z < \infty] - P[0 \le Z \le 1.5] \).
From the standard normal table, \( P[0 \le Z \le 1.5] = 0.4332 \).
Since \( P[0 \le Z < \infty] = 0.5000 \),
\( P[Z \ge 1.5] = 0.5000 - 0.4332 = 0.0668 \).
The estimated number of fat persons is \( N \times P[Z \ge 1.5] = 500 \times 0.0668 = 33.4 \).
Approximately, there are 33 fat persons.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सामान्य वितरण वक्र को दर्शाता है। Z-स्कोर 1.5 से दाहिनी ओर अनंत तक का क्षेत्र छायांकित है, जो 70 किलोग्राम से अधिक वजन वाले व्यक्तियों की संभावना को प्रदर्शित करता है।
(ii) **Number of healthy persons (weight between 50 kg to 60 kg):**
We need to find \( P[50 \le X \le 60] \).
Convert \( x_1 = 50 \) kg to a Z-score: \( Z_1 = \frac{x_1 - \mu}{\sigma} = \frac{50 - 55}{10} = \frac{-5}{10} = -0.5 \).
Convert \( x_2 = 60 \) kg to a Z-score: \( Z_2 = \frac{x_2 - \mu}{\sigma} = \frac{60 - 55}{10} = \frac{5}{10} = 0.5 \).
So, we need to find \( P[-0.5 \le Z \le 0.5] \).
Due to symmetry, \( P[-0.5 \le Z \le 0.5] = P[-0.5 \le Z \le 0] + P[0 \le Z \le 0.5] = 2 \times P[0 \le Z \le 0.5] \).
From the standard normal table, \( P[0 \le Z \le 0.5] = 0.1915 \).
So, \( P[-0.5 \le Z \le 0.5] = 2 \times 0.1915 = 0.3830 \).
The estimated number of healthy persons is \( N \times P[-0.5 \le Z \le 0.5] = 500 \times 0.3830 = 191.5 \).
Approximately, there are 192 healthy persons.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सामान्य वितरण वक्र को दर्शाता है। Z-स्कोर -0.5 से 0.5 के बीच का क्षेत्र छायांकित है, जो 50 से 60 किलोग्राम के बीच वजन वाले स्वस्थ व्यक्तियों की संभावना को प्रदर्शित करता है।
(iii) **Number of physically weak persons (weight less than 35 kg):**
We need to find \( P[X \le 35] \).
Convert \( x_1 = 35 \) kg to a Z-score: \( Z_1 = \frac{x_1 - \mu}{\sigma} = \frac{35 - 55}{10} = \frac{-20}{10} = -2 \).
So, we need to find \( P[Z \le -2] \).
\( P[Z \le -2] = P[-\infty < Z \le 0] - P[-2 \le Z \le 0] \).
Due to symmetry, \( P[-2 \le Z \le 0] = P[0 \le Z \le 2] \).
From the standard normal table, \( P[0 \le Z \le 2] = 0.4772 \).
Since \( P[-\infty < Z \le 0] = 0.5000 \),
\( P[Z \le -2] = 0.5000 - 0.4772 = 0.0228 \).
The estimated number of physically weak persons is \( N \times P[Z \le -2] = 500 \times 0.0228 = 11.4 \).
Approximately, there are 11 physically weak persons.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सामान्य वितरण वक्र को दिखाता है। Z-स्कोर अनंत से -2 तक का क्षेत्र छायांकित है, जो 35 किलोग्राम से कम वजन वाले शारीरिक रूप से कमजोर व्यक्तियों की संभावना को प्रदर्शित करता है।In simple words: To estimate the number of people in each group, first calculate the Z-score for the given weight limits using the mean and standard deviation. Then, find the probability (area under the curve) for those Z-score ranges from a standard table. Finally, multiply this probability by the total number of people to get the estimated count for each group.

🎯 Exam Tip: When calculating counts from probabilities, remember to multiply the probability by the total number of observations (N). Pay attention to whether the question asks for "more than", "less than", or "between" specific values to correctly set up your Z-score ranges and probability calculations.

Question 7. If the probabilities for the value of standard normal variable Z are as under, then estimate the value of Z-score (\(Z_1\)):
(i) Area to the left of \(Z = z_1\) is 0.9928.
(ii) Area to the right of \(Z = z_1\) is 0.0250.

Answer:(i) **Area to the left of \(Z = z_1\) is 0.9928.**
This means \( P[Z \le z_1] = 0.9928 \).
Since the area to the left is greater than 0.5, \( z_1 \) must be a positive value.
We can write \( P[Z \le z_1] = P[-\infty < Z \le 0] + P[0 \le Z \le z_1] \).
So, \( 0.9928 = 0.5000 + P[0 \le Z \le z_1] \).
\( P[0 \le Z \le z_1] = 0.9928 - 0.5000 = 0.4928 \).
From the standard normal table:
- For probability 0.4927, \( z_1 = 2.44 \).
- For probability 0.4929, \( z_1 = 2.45 \).
To estimate \( z_1 \) for 0.4928, we can take the average: \( z_1 = \frac{2.44 + 2.45}{2} = 2.445 \).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सामान्य वितरण वक्र को दर्शाता है। इसमें \(z_1\) तक बाईं ओर के पूरे क्षेत्र को छायांकित किया गया है, जो 0.9928 की संभावना को प्रदर्शित करता है। \(z_1\) का मान धनात्मक है क्योंकि कुल क्षेत्र 0.5 से अधिक है।
(ii) **Area to the right of \(Z = z_1\) is 0.0250.**
This means \( P[Z \ge z_1] = 0.0250 \).
Since the area to the right is less than 0.5, \( z_1 \) must be a positive value.
We can write \( P[Z \ge z_1] = P[0 \le Z < \infty] - P[0 \le Z \le z_1] \).
So, \( 0.0250 = 0.5000 - P[0 \le Z \le z_1] \).
\( P[0 \le Z \le z_1] = 0.5000 - 0.0250 = 0.4750 \).
From the standard normal table, for probability 0.4750, \( z_1 = 1.96 \).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सामान्य वितरण वक्र को दर्शाता है। इसमें \(z_1\) से दाहिनी ओर के छोटे क्षेत्र को छायांकित किया गया है, जो 0.0250 की संभावना को प्रदर्शित करता है। \(z_1\) का मान धनात्मक है क्योंकि यह क्षेत्र 0.5 से कम है।In simple words: To find the Z-score, if the given area is to the left and more than 0.5, or to the right and less than 0.5, subtract 0.5 from the area to get the value for 0 to Z. Then, find the Z-score in the normal distribution table that matches this probability. If the area is to the left and less than 0.5, or to the right and more than 0.5, you will need to adjust your calculation to find the probability from 0 to Z and then assign the correct negative Z-score.

🎯 Exam Tip: Always sketch the normal curve to visualize the given area (left or right of \(z_1\)) and whether \(z_1\) should be positive or negative. This helps correctly set up the equation using the 0.5 area property and Z-table values.

Question 8. If Z is a standard normal variable, then estimate the value of Z-score (\(Z_x\)) such that the following conditions are met:
(i) Area to the left of \(Z = z_1\) is 0.15.
(ii) Area to the right of \(Z = z_1\) is 0.75.

Answer:(i) **Area to the left of \(Z = z_1\) is 0.15.**
This means \( P[Z \le z_1] = 0.15 \).
Since the area to the left is less than 0.5, \( z_1 \) must be a negative value.
We can write \( P[Z \le z_1] = P[-\infty < Z \le 0] - P[z_1 \le Z \le 0] \).
So, \( 0.15 = 0.5000 - P[z_1 \le Z \le 0] \).
\( P[z_1 \le Z \le 0] = 0.5000 - 0.15 = 0.3500 \).
Due to symmetry, \( P[z_1 \le Z \le 0] = P[0 \le Z \le |z_1|] \). So, \( P[0 \le Z \le |z_1|] = 0.3500 \).
From the standard normal table:
- For probability 0.3485, \( Z = 1.03 \).
- For probability 0.3508, \( Z = 1.04 \).
To estimate \( |z_1| \) for 0.3500, we can take the average: \( |z_1| = \frac{1.03 + 1.04}{2} = 1.035 \).
Since \( z_1 \) is negative, \( z_1 = -1.035 \).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक मानक सामान्य वितरण वक्र को दर्शाता है। \(z_1\) तक बाईं ओर का क्षेत्र छायांकित है, जो 0.15 की संभावना दिखाता है। चूंकि यह क्षेत्र 0.5 से कम है, \(z_1\) का मान नकारात्मक है, जो वक्र के बाईं ओर स्थित है।
(ii) **Area to the right of \(Z = z_1\) is 0.75.**
This means \( P[Z \ge z_1] = 0.75 \).
Since the area to the right is greater than 0.5, \( z_1 \) must be a negative value.
We can write \( P[Z \ge z_1] = P[z_1 \le Z \le 0] + P[0 \le Z < \infty] \).
So, \( 0.75 = P[z_1 \le Z \le 0] + 0.5000 \).
\( P[z_1 \le Z \le 0] = 0.75 - 0.5000 = 0.2500 \).
Due to symmetry, \( P[z_1 \le Z \le 0] = P[0 \le Z \le |z_1|] \). So, \( P[0 \le Z \le |z_1|] = 0.2500 \).
From the standard normal table:
- For probability 0.2486, \( Z = 0.67 \).
- For probability 0.2518, \( Z = 0.68 \).
To estimate \( |z_1| \) for 0.2500, we can take the average: \( |z_1| = \frac{0.67 + 0.68}{2} = 0.675 \).
Since \( z_1 \) is negative, \( z_1 = -0.675 \).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक मानक सामान्य वितरण वक्र को प्रदर्शित करता है। \(z_1\) से दाहिनी ओर का क्षेत्र छायांकित है, जो 0.75 की संभावना दर्शाता है। चूंकि यह क्षेत्र 0.5 से अधिक है, \(z_1\) का मान नकारात्मक है, जो वक्र के बाईं ओर स्थित है।In simple words: When finding a Z-score from an area, first determine if the Z-score is positive or negative based on whether the area is more or less than 0.5. Then, calculate the area from 0 to Z, and use the normal distribution table to find the corresponding Z-score.

🎯 Exam Tip: Pay close attention to whether the area is to the "left" or "right" and its magnitude (greater or less than 0.5). This dictates the sign of the Z-score and how you use the symmetry of the normal curve with the Z-table.

Question 9. If Z is a standard normal variable and \(z_1\) represents the Z-score, then estimate the value of \(z_1\) such that the following conditions are satisfied:
(i) \( P[-2 \le Z \le z_1] = 0.2857 \)
(ii) \( P[z_1 \le Z \le 1.75] = 0.10 \)

Answer:(i) We are given \( P[-2 \le Z \le z_1] = 0.2857 \).
From the normal distribution table, \( P[0 \le Z \le 2] = 0.4772 \).
We know that \( P[-2 \le Z \le z_1] = P[-2 \le Z \le 0] - P[z_1 \le Z \le 0] \).
So, \( 0.2857 = 0.4772 - P[z_1 \le Z \le 0] \).
\( P[z_1 \le Z \le 0] = 0.4772 - 0.2857 = 0.1915 \).
From the standard normal table, for a probability of 0.1915, the corresponding Z-score is approximately 0.50.
Since the area \( P[z_1 \le Z \le 0] \) is positive and \( z_1 \) is to the left of 0, \( z_1 \) must be negative.
Therefore, \( z_1 = -0.5 \).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक मानक सामान्य वितरण वक्र को दर्शाता है। इसमें Z-स्कोर -2 से \(z_1\) तक का क्षेत्र छायांकित है, जिसकी संभावना 0.2857 है। \(z_1\) का मान नकारात्मक है और -2 के दाईं ओर स्थित है।
(ii) We are given \( P[z_1 \le Z \le 1.75] = 0.10 \).
From the normal distribution table, \( P[0 \le Z \le 1.75] = 0.4599 \).
Since \( P[z_1 \le Z \le 1.75] = 0.10 \) is positive and less than \( P[0 \le Z \le 1.75] \), \( z_1 \) must be positive and to the left of 1.75, or negative. Let's assume \( z_1 \) is positive.
If \( z_1 \) is positive and less than 1.75, then \( P[z_1 \le Z \le 1.75] = P[0 \le Z \le 1.75] - P[0 \le Z \le z_1] \).
So, \( 0.10 = 0.4599 - P[0 \le Z \le z_1] \).
\( P[0 \le Z \le z_1] = 0.4599 - 0.10 = 0.3599 \).
From the standard normal table, for a probability of 0.3599, the corresponding Z-score is approximately 1.08.
Therefore, \( z_1 = 1.08 \).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक मानक सामान्य वितरण वक्र को दिखाता है। इसमें Z-स्कोर \(z_1\) से 1.75 तक का क्षेत्र छायांकित है, जिसकी संभावना 0.10 है। \(z_1\) का मान धनात्मक है और 1.75 के बाईं ओर स्थित है।In simple words: To find \(z_1\) when given a probability range, use the Z-table to find the area for known Z-scores. Then, rearrange the probability equation to solve for the area corresponding to \(z_1\). Finally, use the Z-table again to find \(z_1\) for that calculated area, remembering to check the sign.

🎯 Exam Tip: Always subtract areas from the origin (0) when dealing with probabilities between two Z-scores, or between a Z-score and infinity/negative infinity. Use the standard normal table to find the Z-score for the calculated probability or vice-versa.

Question 10. The monthly production of units in a factory follows a normal distribution with mean \(\mu\) and standard deviation \(\sigma\). The Z-scores corresponding to the production of 2400 units and 1800 units are 1 and -0.5 respectively. Find the mean and standard deviation of this distribution.

Answer:Let \(X_1 = 2400\) units and \(X_2 = 1800\) units.
The corresponding Z-scores are \(z_1 = 1\) and \(z_2 = -0.5\).
The formula for Z-score is \( Z = \frac{X - \mu}{\sigma} \).
Using the first data point:
\( 1 = \frac{2400 - \mu}{\sigma} \)
Multiplying both sides by \(\sigma\), we get:
\( \sigma = 2400 - \mu \) --- (Equation 1)
Using the second data point:
\( -0.5 = \frac{1800 - \mu}{\sigma} \)
Multiplying both sides by \(\sigma\), we get:
\( -0.5\sigma = 1800 - \mu \) --- (Equation 2)
Now, we subtract Equation 2 from Equation 1:
\( (\sigma) - (-0.5\sigma) = (2400 - \mu) - (1800 - \mu) \)
\( \sigma + 0.5\sigma = 2400 - \mu - 1800 + \mu \)
\( 1.5\sigma = 600 \)
\( \sigma = \frac{600}{1.5} \)
\( \sigma = 400 \)
Now, substitute the value of \(\sigma\) into Equation 1 to find \(\mu\):
\( 400 = 2400 - \mu \)
\( \mu = 2400 - 400 \)
\( \mu = 2000 \)
Thus, the mean of the distribution is 2000 units and the standard deviation is 400 units.In simple words: We used the given Z-scores and production units to form two simple math equations. By solving these two equations together, we found that the average production (mean) is 2000 units, and the spread (standard deviation) is 400 units.

🎯 Exam Tip: When given two data points (X-values) and their corresponding Z-scores, set up two simultaneous equations using the Z-score formula. Solving these equations is a standard method to find the unknown mean (\(\mu\)) and standard deviation (\(\sigma\)) of the distribution.

Question 11. A normal variable X has the following probability density function: \(f(x) = \frac{1}{6 \sqrt{2 \pi}} e^{-\frac{1}{72}(x-100)^{2}}, - \infty < x < \infty\).
For this distribution, obtain the estimated limits for the exact middle 50% of the observations.

Answer:The given probability density function is \(f(x) = \frac{1}{6 \sqrt{2 \pi}} e^{-\frac{1}{72}(x-100)^{2}}\).
This can be rewritten as \(f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^{2}}\).
Comparing the given function with the standard form:
\( \frac{1}{\sigma \sqrt{2 \pi}} = \frac{1}{6 \sqrt{2 \pi}} \implies \sigma = 6 \).
\( \frac{(x-\mu)^2}{2\sigma^2} = \frac{(x-100)^2}{72} \).
Using \(\sigma = 6\), \( 2\sigma^2 = 2(6^2) = 2(36) = 72 \).
So, \(\mu = 100\) and \(\sigma = 6\).
The limits for the exact middle 50% of observations are the first quartile (\(Q_1\)) and the third quartile (\(Q_3\)).
\(Q_1\) is the value below which 25% of observations lie, and \(Q_3\) is the value below which 75% of observations lie (or above which 25% of observations lie).
So, \( P[X \le Q_1] = 0.25 \) and \( P[X \ge Q_3] = 0.25 \).
For \( P[X \le Q_1] = 0.25 \):
Convert to Z-score: \( z_1 = \frac{Q_1 - \mu}{\sigma} = \frac{Q_1 - 100}{6} \).
\( P[Z \le z_1] = 0.25 \). Since 0.25 < 0.5, \(z_1\) is negative.
We find \( P[z_1 \le Z \le 0] = P[-\infty < Z \le 0] - P[-\infty < Z \le z_1] = 0.50 - 0.25 = 0.25 \).
From the standard normal table, the Z-score for an area of 0.25 from 0 is approximately 0.675.
Since \(z_1\) is negative, \( z_1 = -0.675 \).
Now, \( -0.675 = \frac{Q_1 - 100}{6} \).
\( -0.675 \times 6 = Q_1 - 100 \).
\( -4.05 = Q_1 - 100 \).
\( Q_1 = 100 - 4.05 = 95.95 \).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सामान्य वितरण वक्र को दर्शाता है। इसमें \(z_1\) के बाईं ओर का 25% क्षेत्र और \(z_2\) के दाहिनी ओर का 25% क्षेत्र छायांकित है। बीच का बिना छायांकित क्षेत्र 50% मध्य अवलोकन को दर्शाता है, जिसमें \(Q_1\) और \(Q_3\) उनकी सीमाएँ हैं।
For \( P[X \ge Q_3] = 0.25 \):
Convert to Z-score: \( z_2 = \frac{Q_3 - \mu}{\sigma} = \frac{Q_3 - 100}{6} \).
\( P[Z \ge z_2] = 0.25 \). Since 0.25 < 0.5, \(z_2\) is positive.
We find \( P[0 \le Z \le z_2] = P[0 \le Z < \infty] - P[Z \ge z_2] = 0.50 - 0.25 = 0.25 \).
From the standard normal table, the Z-score for an area of 0.25 from 0 is approximately 0.675.
Since \(z_2\) is positive, \( z_2 = 0.675 \).
Now, \( 0.675 = \frac{Q_3 - 100}{6} \).
\( 0.675 \times 6 = Q_3 - 100 \).
\( 4.05 = Q_3 - 100 \).
\( Q_3 = 100 + 4.05 = 104.05 \).
Thus, the estimated limits for the exact middle 50% of the observations are from 95.95 to 104.05.In simple words: We first found the mean and standard deviation from the given function. Then, we calculated the first and third quartiles, which mark the boundaries for the middle 50% of the data. To do this, we found the Z-scores that correspond to the 25% and 75% marks of the data.

🎯 Exam Tip: To find quartiles in a normal distribution, remember that the middle 50% is bounded by Q1 and Q3. These correspond to Z-scores of approximately -0.675 and +0.675, respectively, which are crucial values to memorize or derive quickly from the Z-table for \(P[0 \le Z \le 0.675] \approx 0.25\).

Question 12. For a normal distribution, the first quartile is 35 and its third quartile is 65. Estimate the limits that include exactly the middle 95.45% of the observations.

Answer:Given: First quartile \(Q_1 = 35\) and third quartile \(Q_3 = 65\).
For a normal distribution, the mean (\(\mu\)) is also the median, which is the midpoint of \(Q_1\) and \(Q_3\).
\( \mu = \frac{Q_1 + Q_3}{2} = \frac{35 + 65}{2} = \frac{100}{2} = 50 \).
The quartile deviation (\(QD\)) is given by \( QD = \frac{Q_3 - Q_1}{2} = \frac{65 - 35}{2} = \frac{30}{2} = 15 \).
For a normal distribution, \( QD \approx \frac{2}{3}\sigma \).
So, \( 15 = \frac{2}{3}\sigma \).
Solving for \(\sigma\): \( \sigma = \frac{15 \times 3}{2} = \frac{45}{2} = 22.5 \).
We need to find the limits that include exactly the middle 95.45% of the observations.
This corresponds to the range \( \mu \pm 2\sigma \).
Limits are \( (\mu - 2\sigma) \) to \( (\mu + 2\sigma) \).
\( \mu - 2\sigma = 50 - 2(22.5) = 50 - 45 = 5 \).
\( \mu + 2\sigma = 50 + 2(22.5) = 50 + 45 = 95 \).
Therefore, the limits that include exactly the middle 95.45% of the observations are (5, 95).In simple words: We first used the given quartiles to find the average (mean) and the spread (standard deviation) of the data. Then, we used the rule that about 95.45% of data in a normal distribution falls within two standard deviations from the mean to find the specific range of values.

🎯 Exam Tip: Remember the empirical rule for normal distributions: approximately 68.27% of data falls within \( \mu \pm \sigma \), 95.45% within \( \mu \pm 2\sigma \), and 99.73% within \( \mu \pm 3\sigma \). This shortcut is vital for quick calculations of such percentage-based limits.

Question 13. For a normal distribution, the third quartile and quartile deviations are 36 and 24 respectively. Find the mean of the distribution.

Answer:Given: Third quartile \(Q_3 = 36\).
Given: Mean deviation (\(MD\)) = 24.
For a normal distribution, \( MD \approx \frac{4}{5}\sigma \).
So, \( 24 = \frac{4}{5}\sigma \).
Solving for \(\sigma\): \( \sigma = \frac{24 \times 5}{4} = 6 \times 5 = 30 \).
For a normal distribution, the third quartile \(Q_3\) is related to the mean (\(\mu\)) and standard deviation (\(\sigma\)) by the formula:
\( Q_3 = \mu + 0.675\sigma \).
We are given \(Q_3 = 36\) and we found \(\sigma = 30\).
So, \( 36 = \mu + 0.675(30) \).
\( 36 = \mu + 20.25 \).
\( \mu = 36 - 20.25 \).
\( \mu = 15.75 \).
The mean of the distribution is 15.75.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सामान्य वितरण वक्र को दर्शाता है। इसमें \(z_1\) के दाहिनी ओर का 25% क्षेत्र छायांकित है, जो \(Q_3\) के ऊपर के अवलोकनों को दिखाता है। यह Z-स्कोर \(z_1\) का उपयोग करके वितरण के माध्य (\(\mu\)) को खोजने में मदद करता है।In simple words: We used the given mean deviation to first calculate the standard deviation. Then, knowing the standard deviation and the value of the third quartile, we used the formula linking the third quartile to the mean and standard deviation to find the mean of the distribution.

🎯 Exam Tip: Memorize the approximate relationships between mean deviation (MD), quartile deviation (QD), and standard deviation (\(\sigma\)) for normal distributions: \(MD \approx \frac{4}{5}\sigma\) and \(QD \approx \frac{2}{3}\sigma\). Also, remember the formulas for quartiles: \(Q_1 = \mu - 0.675\sigma\) and \(Q_3 = \mu + 0.675\sigma\).

Question 14. A normal variable X has a mean of 200 and a variance of 100.
(i) Estimate the values of extreme quartiles.
(ii) Find the approximate value of quartile deviation.
(iii) Find the approximate value of mean deviation.

Answer:Here, the mean (\(\mu\)) is 200.
The variance (\(\sigma^2\)) is 100, so the standard deviation (\(\sigma\)) is \( \sqrt{100} = 10 \).
(i) **Estimate the values of extreme quartiles (\(Q_1\) and \(Q_3\)):**
For a normal distribution, the quartiles are given by:
\( Q_1 = \mu - 0.675\sigma \)
\( Q_3 = \mu + 0.675\sigma \)
Substitute \(\mu = 200\) and \(\sigma = 10\):
\( Q_1 = 200 - 0.675(10) = 200 - 6.75 = 193.25 \).
\( Q_3 = 200 + 0.675(10) = 200 + 6.75 = 206.75 \).
The extreme quartiles are 193.25 and 206.75.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सामान्य वितरण वक्र को दिखाता है। इसमें \(z_1\) के बाईं ओर का 25% क्षेत्र और \(z_2\) के दाहिनी ओर का 25% क्षेत्र छायांकित है। बीच का बिना छायांकित क्षेत्र मध्य अवलोकनों को दर्शाता है, जहाँ \(Q_1\) और \(Q_3\) उनकी सीमाएँ हैं।
(ii) **Find the approximate value of quartile deviation:**
For a normal distribution, \( QD \approx \frac{2}{3}\sigma \).
Substitute \(\sigma = 10\):
\( QD = \frac{2}{3}(10) = \frac{20}{3} \approx 6.67 \).
The approximate quartile deviation is 6.67.
(iii) **Find the approximate value of mean deviation:**
For a normal distribution, \( MD \approx \frac{4}{5}\sigma \).
Substitute \(\sigma = 10\):
\( MD = \frac{4}{5}(10) = 4 \times 2 = 8 \).
The approximate mean deviation is 8.In simple words: We used the average (mean) and spread (standard deviation) to find the first and third quartiles, which are the "extreme" quartiles. Then, we applied simple formulas to estimate the quartile deviation and mean deviation based on the standard deviation.

🎯 Exam Tip: Remember the specific formulas for quartiles (\(Q_1 = \mu - 0.675\sigma\), \(Q_3 = \mu + 0.675\sigma\)) and the approximate relationships for quartile deviation (\(\approx \frac{2}{3}\sigma\)) and mean deviation (\(\approx \frac{4}{5}\sigma\)) in a normal distribution. These are essential for solving problems quickly.

Section E

Solve the Following:

Question 1. An amount of purchase made by a customer in a mall in a city follows a normal distribution with a mean of Rs. 800 and a standard deviation of Rs. 200. If a customer is selected randomly, find the probabilities for the following events:
(1) The amount of purchase made by him is between Rs. 850 and Rs. 1200.
(2) The amount of purchase made by him is between Rs. 600 and Rs. 750.

Answer:Here, the mean (\(\mu\)) is Rs. 800, and the standard deviation (\(\sigma\)) is Rs. 200.
X represents the amount of purchase in Rs.
(1) **Amount of purchase made by the customer is between Rs. 850 and Rs. 1200:**
We need to find \( P[850 \le X \le 1200] \).
Convert the X-values to Z-scores:
For \( x_1 = 850 \): \( Z_1 = \frac{x_1 - \mu}{\sigma} = \frac{850 - 800}{200} = \frac{50}{200} = 0.25 \).
For \( x_2 = 1200 \): \( Z_2 = \frac{x_2 - \mu}{\sigma} = \frac{1200 - 800}{200} = \frac{400}{200} = 2.00 \).
So, we need to find \( P[0.25 \le Z \le 2.00] \).
This can be calculated as \( P[0 \le Z \le 2.00] - P[0 \le Z \le 0.25] \).
From the standard normal table:
\( P[0 \le Z \le 2.00] = 0.4772 \).
\( P[0 \le Z \le 0.25] = 0.0987 \).
Therefore, \( P[0.25 \le Z \le 2.00] = 0.4772 - 0.0987 = 0.3785 \).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सामान्य वितरण वक्र को दर्शाता है। Z-स्कोर 0.25 से 2.00 के बीच का क्षेत्र छायांकित है, जो 850 रुपये से 1200 रुपये के बीच खरीदारी की संभावना को प्रदर्शित करता है। यह क्षेत्र वक्र के माध्य के दाहिनी ओर स्थित है।
(2) **Amount of purchase made by the customer is between Rs. 600 and Rs. 750:**
We need to find \( P[600 \le X \le 750] \).
Convert the X-values to Z-scores:
For \( x_1 = 600 \): \( Z_1 = \frac{x_1 - \mu}{\sigma} = \frac{600 - 800}{200} = \frac{-200}{200} = -1.00 \).
For \( x_2 = 750 \): \( Z_2 = \frac{x_2 - \mu}{\sigma} = \frac{750 - 800}{200} = \frac{-50}{200} = -0.25 \).
So, we need to find \( P[-1.00 \le Z \le -0.25] \).
This can be calculated as \( P[-1.00 \le Z \le 0] - P[-0.25 \le Z \le 0] \).
Due to symmetry, \( P[-1.00 \le Z \le 0] = P[0 \le Z \le 1.00] \) and \( P[-0.25 \le Z \le 0] = P[0 \le Z \le 0.25] \).
From the standard normal table:
\( P[0 \le Z \le 1.00] = 0.3413 \).
\( P[0 \le Z \le 0.25] = 0.0987 \).
Therefore, \( P[-1.00 \le Z \le -0.25] = 0.3413 - 0.0987 = 0.2426 \).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सामान्य वितरण वक्र को दर्शाता है। Z-स्कोर -1.00 से -0.25 के बीच का क्षेत्र छायांकित है, जो 600 रुपये से 750 रुपये के बीच खरीदारी की संभावना को प्रदर्शित करता है। यह क्षेत्र वक्र के माध्य के बाईं ओर स्थित है।In simple words: To find the probability for a range, convert the range limits to Z-scores. Then, use the standard normal table to find the areas from the mean (Z=0) to these Z-scores. Subtract the smaller area from the larger one to get the probability for the specified range.

🎯 Exam Tip: Always draw a rough sketch of the normal curve to visualize the Z-score range. This helps prevent errors when adding or subtracting areas from the Z-table, especially for ranges crossing the mean or entirely on one side of it.

 

(1) Persons having weight more than 70 kg is in the fat persons group:
Answer:To find the number of fat persons, we check for weights over 70 kg. Here, the mean weight (μ) is 55 kg and the standard deviation (σ) is 10 kg. For x₁ = 70 kg, the Z-score is calculated as: \[ Z_1 = \frac{x_1 - \mu}{\sigma} = \frac{70 - 55}{10} = \frac{15}{10} = 1.5 \] So, the probability that a person's weight is more than 70 kg is P[Z ≥ 1.5]. From the standard normal distribution table, P[0 ≤ Z < ∞] is 0.5000 and P[0 ≤ Z ≤ 1.5] is 0.4332. Thus, P[Z ≥ 1.5] = P[0 ≤ Z < ∞] - P[0 ≤ Z ≤ 1.5] = 0.5000 - 0.4332 = 0.0668. The total number of persons (N) is 500. The estimated number of fat persons = N × P[Z ≥ 1.5] = 500 × 0.0668 = 33.4. Approximately, there are 33 fat persons.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सामान्य वितरण वक्र दिखाता है। छायांकित क्षेत्र Z = 1.5 से अधिक के क्षेत्र को दर्शाता है, जो उन व्यक्तियों का प्रतिनिधित्व करता है जिनका वजन 70 किलो से अधिक है और जिन्हें "मोटे व्यक्ति" के रूप में वर्गीकृत किया गया है।In simple words: We calculated how many people weigh more than 70 kg using a special score called the Z-score. Out of 500 people, about 33 people are considered fat.

🎯 Exam Tip: Remember to clearly state the mean and standard deviation. Calculations for Z-score and probability from the Z-table are key. Make sure to multiply the probability by the total number of individuals for the final count.

 

(2) Persons having weight between 50 kg to 60 kg is in the healthy persons group:
Answer:To find the number of healthy persons, we look for weights between 50 kg and 60 kg. Here, the mean weight (μ) is 55 kg and the standard deviation (σ) is 10 kg. For x₁ = 50 kg, the Z-score is: \[ Z_1 = \frac{x_1 - \mu}{\sigma} = \frac{50 - 55}{10} = \frac{-5}{10} = -0.5 \] For x₂ = 60 kg, the Z-score is: \[ Z_2 = \frac{x_2 - \mu}{\sigma} = \frac{60 - 55}{10} = \frac{5}{10} = 0.5 \] So, the probability for healthy persons is P[-0.5 ≤ Z ≤ 0.5]. From the standard normal distribution table, P[0 ≤ Z ≤ 0.5] is 0.1915. Since the normal curve is symmetric, P[-0.5 ≤ Z ≤ 0.5] = P[-0.5 ≤ Z ≤ 0] + P[0 ≤ Z ≤ 0.5] = 2 × P[0 ≤ Z ≤ 0.5] = 2 × 0.1915 = 0.3830. The total number of persons (N) is 500. The estimated number of healthy persons = N × P[-0.5 ≤ Z ≤ 0.5] = 500 × 0.3830 = 191.5. Approximately, there are 192 healthy persons.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सामान्य वितरण वक्र दिखाता है, जिसमें Z = -0.5 और Z = 0.5 के बीच का क्षेत्र छायांकित है। यह क्षेत्र उन व्यक्तियों का प्रतिनिधित्व करता है जिनका वजन 50 किलोग्राम से 60 किलोग्राम के बीच है, जिन्हें "स्वस्थ व्यक्ति" के रूप में वर्गीकृत किया गया है।In simple words: We found how many people weigh between 50 kg and 60 kg. Using Z-scores, we calculated that about 192 out of 500 people are in the healthy weight group.

🎯 Exam Tip: When finding probabilities between two Z-scores that are symmetric around the mean (e.g., -Z and +Z), you can double the probability from the mean to one side (e.g., 2 * P[0 ≤ Z ≤ Z]). Ensure your Z-score calculations are precise.

 

(3) Person having weight less than 35 kg is in the physically weak person's group.
Answer:To estimate the number of physically weak persons, we need to find the probability of a person having a weight less than 35 kg. Here, the mean weight (μ) is 55 kg and the standard deviation (σ) is 10 kg. For x₁ = 35 kg, the Z-score is: \[ Z_1 = \frac{x_1 - \mu}{\sigma} = \frac{35 - 55}{10} = \frac{-20}{10} = -2 \] The probability of a person having weight less than 35 kg is P[Z ≤ -2]. Using the standard normal distribution table, P[Z ≤ -2] = P[-∞ < Z ≤ 0] - P[-2 ≤ Z ≤ 0]. This is also equal to P[0 ≤ Z < ∞] - P[0 ≤ Z ≤ 2] = 0.5000 - 0.4772 = 0.0228. The total number of persons (N) is 500. The estimated number of physically weak persons = N × P[Z ≤ -2] = 500 × 0.0228 = 11.4. Approximately, there are 11 physically weak persons.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सामान्य वितरण वक्र के Z = -2 से बाईं ओर के क्षेत्र को दर्शाता है। यह छायांकित क्षेत्र उन व्यक्तियों का प्रतिनिधित्व करता है जिनका वजन 35 किलोग्राम से कम है, जिन्हें "शारीरिक रूप से कमजोर व्यक्ति" के रूप में वर्गीकृत किया गया है।In simple words: We calculated how many people weigh less than 35 kg. Using the Z-score, we found that about 11 out of 500 people are in the physically weak group.

🎯 Exam Tip: For probabilities like P[Z ≤ -a], use the symmetry of the normal curve: P[Z ≤ -a] = P[Z ≥ a]. Then, calculate it as P[0 ≤ Z < ∞] - P[0 ≤ Z ≤ a]. Be careful with negative Z-scores and their corresponding areas.

 

Question 3. The average monthly expense of students residing in university hostel is Rs. 2000 and its standard deviation is Rs. 500. If the monthly expense of a student follows normal distribution, then
Answer:Here, the mean monthly expense (μ) is Rs. 2000, and the standard deviation (σ) is Rs. 500. Let X be the average monthly expense.
(1) Find percentage of students having expense between Rs. 750 and Rs. 1250. For x₁ = 750, the Z-score is: \[ Z_1 = \frac{x_1 - \mu}{\sigma} = \frac{750 - 2000}{500} = \frac{-1250}{500} = -2.5 \] For x₂ = 1250, the Z-score is: \[ Z_2 = \frac{x_2 - \mu}{\sigma} = \frac{1250 - 2000}{500} = \frac{-750}{500} = -1.5 \] The probability that expenses are between Rs. 750 and Rs. 1250 is P[-2.5 ≤ Z ≤ -1.5]. This can be calculated as P[-2.5 ≤ Z ≤ 0] - P[-1.5 ≤ Z ≤ 0]. Due to symmetry, this is equal to P[0 ≤ Z ≤ 2.5] - P[0 ≤ Z ≤ 1.5]. From the standard normal distribution table: P[0 ≤ Z ≤ 2.5] = 0.4938 P[0 ≤ Z ≤ 1.5] = 0.4332 So, P[-2.5 ≤ Z ≤ -1.5] = 0.4938 - 0.4332 = 0.0606. The percentage of students with expenses between Rs. 750 and Rs. 1250 is 0.0606 × 100 = 6.06%.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सामान्य वितरण वक्र दिखाता है, जिसमें Z = -2.5 और Z = -1.5 के बीच का क्षेत्र छायांकित है। यह क्षेत्र उन छात्रों का प्रतिनिधित्व करता है जिनका मासिक खर्च Rs. 750 से Rs. 1250 के बीच है।In simple words: We want to find the percentage of students whose monthly expense is between Rs. 750 and Rs. 1250. We convert these amounts to Z-scores, which are -2.5 and -1.5. Then, we find the area between these Z-scores under the normal curve, which is 6.06%.

🎯 Exam Tip: When calculating the probability between two negative Z-scores, subtract the smaller area from the larger area (e.g., P[Z₁ ≤ Z ≤ Z₂] = P[0 ≤ Z ≤ |Z₁|] - P[0 ≤ Z ≤ |Z₂|] if both Z₁ and Z₂ are negative). Be precise with Z-score calculations and table lookups.

 

(2) Find percentage of students having expense more than Rs. 1800.
Answer:To find the percentage of students with expenses more than Rs. 1800: Here, the mean expense (μ) is Rs. 2000, and the standard deviation (σ) is Rs. 500. For x₁ = 1800, the Z-score is: \[ Z_1 = \frac{x_1 - \mu}{\sigma} = \frac{1800 - 2000}{500} = \frac{-200}{500} = -0.4 \] The probability that expenses are more than Rs. 1800 is P[Z ≥ -0.4]. This can be calculated as P[-0.4 ≤ Z ≤ 0] + P[0 ≤ Z < ∞]. Due to symmetry, P[-0.4 ≤ Z ≤ 0] is equal to P[0 ≤ Z ≤ 0.4]. From the standard normal distribution table, P[0 ≤ Z ≤ 0.4] = 0.1554. So, P[Z ≥ -0.4] = 0.1554 + 0.5000 = 0.6554. The percentage of students with expenses more than Rs. 1800 is 0.6554 × 100 = 65.54%.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सामान्य वितरण वक्र के Z = -0.4 से दाईं ओर के क्षेत्र को दर्शाता है। यह छायांकित क्षेत्र उन छात्रों का प्रतिनिधित्व करता है जिनका मासिक खर्च Rs. 1800 से अधिक है।In simple words: We want to find the percentage of students spending more than Rs. 1800. The Z-score for Rs. 1800 is -0.4. The total percentage of students spending more than this amount is 65.54%.

🎯 Exam Tip: For probabilities like P[Z ≥ -a], remember to add the area from -a to 0 (which is P[0 ≤ Z ≤ a] due to symmetry) to the area from 0 to infinity (0.5). Always draw a quick sketch to visualize the area.

 

(3) The percentage of students having expense less than Rs. 2400:
Answer:To find the percentage of students with expenses less than Rs. 2400: Here, the mean expense (μ) is Rs. 2000, and the standard deviation (σ) is Rs. 500. For x₁ = 2400, the Z-score is: \[ Z_1 = \frac{x_1 - \mu}{\sigma} = \frac{2400 - 2000}{500} = \frac{400}{500} = 0.8 \] The probability that expenses are less than Rs. 2400 is P[Z ≤ 0.8]. This can be calculated as P[-∞ < Z ≤ 0] + P[0 ≤ Z ≤ 0.8]. From the standard normal distribution table, P[0 ≤ Z ≤ 0.8] = 0.2881. So, P[Z ≤ 0.8] = 0.5000 + 0.2881 = 0.7881. The percentage of students with expenses less than Rs. 2400 is 0.7881 × 100 = 78.81%.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सामान्य वितरण वक्र दिखाता है, जिसमें Z = 0.8 से बाईं ओर का क्षेत्र छायांकित है। यह क्षेत्र उन छात्रों का प्रतिनिधित्व करता है जिनका मासिक खर्च Rs. 2400 से कम है।In simple words: We need to find the percentage of students whose monthly expense is less than Rs. 2400. The Z-score for Rs. 2400 is 0.8. We calculate the area to the left of this Z-score, which is 78.81%.

🎯 Exam Tip: For probabilities like P[Z ≤ a] (where 'a' is positive), add the area from -∞ to 0 (0.5) to the area from 0 to a (P[0 ≤ Z ≤ a]). Visualizing the area on the normal curve helps prevent mistakes.

 

Question 4. The monthly average salary of workers working in a production house is Rs. 10000 and its standard deviation is Rs. 2000. By assuming that the monthly salary of a worker follows normal distribution, estimate the maximum salary of 20% of the workers having lowest salary. Also estimate the minimum salary of 10 % of the workers having highest salary.
Answer:Here, X represents the monthly salary. The mean salary (μ) is Rs. 10,000, and the standard deviation (σ) is Rs. 2000. Maximum salary of 20 % workers having lowest salary: Let x₁ be the maximum salary for the 20% of workers with the lowest salaries. This means P[X ≤ x₁] = 20/100 = 0.2. We need to find the Z-score (Z₁) corresponding to this probability. Since P[X ≤ x₁] = 0.2, this means the area to the left of Z₁ is 0.2. So, P[Z₁ ≤ Z ≤ 0] = P[-∞ < Z ≤ 0] - P[Z ≤ Z₁] = 0.5000 - 0.2000 = 0.3000. From the Z-table, the area 0.2995 corresponds to Z = -0.84, and 0.3023 corresponds to Z = -0.85. For an area of 0.3000, Z₁ is approximately -0.84 (since 0.2995 is closer). Using the Z-score formula: \[ Z_1 = \frac{x_1 - \mu}{\sigma} \] \[ -0.84 = \frac{x_1 - 10000}{2000} \] \[ -0.84 \times 2000 = x_1 - 10000 \] \[ -1680 = x_1 - 10000 \] \[ x_1 = 10000 - 1680 \]
\[ x_1 = 8320 \] Therefore, the maximum salary for 20% of workers having the lowest salary is Rs. 8320.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सामान्य वितरण वक्र के Z1 से बाईं ओर के 0.2 क्षेत्र को दर्शाता है। यह क्षेत्र उन 20% श्रमिकों का प्रतिनिधित्व करता है जिनका वेतन सबसे कम है।In simple words: We found the highest salary for the 20% of workers who earn the least. We used Z-scores to figure this out. The highest salary for these workers is Rs. 8320.

🎯 Exam Tip: When finding a value (like salary) corresponding to a given percentile, first find the Z-score for that percentile using the normal distribution table. Then, use the Z-score formula to solve for the unknown value. Pay attention to whether the area is to the left or right of the mean.

 

The minimum salary of 10% of workers having maximum salary:
Answer:Let x₂ be the minimum salary for the 10% of workers with the highest salaries. This means P[X ≥ x₂] = 10/100 = 0.1. We need to find the Z-score (Z₂) corresponding to this probability. Since P[X ≥ x₂] = 0.1, this means the area to the right of Z₂ is 0.1. So, P[0 ≤ Z ≤ Z₂] = P[0 ≤ Z < ∞] - P[Z ≥ Z₂] = 0.5000 - 0.1000 = 0.4000. From the Z-table, the area 0.3997 corresponds to Z = 1.28, and 0.4015 corresponds to Z = 1.29. For an area of 0.4000, Z₂ is approximately 1.28 (since 0.3997 is closer). Using the Z-score formula: \[ Z_2 = \frac{x_2 - \mu}{\sigma} \] \[ 1.28 = \frac{x_2 - 10000}{2000} \] \[ 1.28 \times 2000 = x_2 - 10000 \] \[ 2560 = x_2 - 10000 \] \[ x_2 = 10000 + 2560 \]
\[ x_2 = 12560 \] Therefore, the minimum salary for 10% of workers having the highest salary is Rs. 12560.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सामान्य वितरण वक्र के Z2 से दाईं ओर के 0.1 क्षेत्र को दर्शाता है। यह क्षेत्र उन 10% श्रमिकों का प्रतिनिधित्व करता है जिनका वेतन सबसे अधिक है।In simple words: We found the lowest salary for the 10% of workers who earn the most. We used Z-scores for this. The lowest salary for these workers is Rs. 12560.

🎯 Exam Tip: For probabilities like P[X ≥ x], first calculate the area from 0 to Z (0.5 minus the given probability). Then find the Z-score from the table and use it to find x. Accuracy in Z-score lookup is vital.

 

Question 5. A normal distribution has mean 52 and variance 64. Obtain estimated limits which include exactly middle 25 % of the observations.
Answer:Here, the mean (μ) is 52. The variance (σ²) is 64, so the standard deviation (σ) is \( \sqrt{64} = 8 \). We need to find the limits (x₁ and x₂) that include the exactly middle 25% of the observations. This means we are looking for the central 25% of the data. This implies that 25%/2 = 12.5% of the observations lie on each side of the mean. So, P[x₁ ≤ X ≤ μ] = 0.125 and P[μ ≤ X ≤ x₂] = 0.125. Therefore, P[Z₁ ≤ Z ≤ 0] = 0.125 and P[0 ≤ Z ≤ Z₂] = 0.125. From the Z-table, for an area of 0.125: Area 0.1217 corresponds to Z = 0.31. Area 0.1255 corresponds to Z = 0.32. So, for an area of 0.125, Z is between 0.31 and 0.32. We can take the average Z = (0.31 + 0.32) / 2 = 0.315. Thus, Z₁ = -0.315 (since it's to the left of the mean) and Z₂ = 0.315 (since it's to the right of the mean). For Z₁ = -0.315: \[ Z_1 = \frac{x_1 - \mu}{\sigma} \] \[ -0.315 = \frac{x_1 - 52}{8} \] \[ -0.315 \times 8 = x_1 - 52 \] \[ -2.52 = x_1 - 52 \] \[ x_1 = 52 - 2.52 \]
\[ x_1 = 49.48 \] For Z₂ = 0.315: \[ Z_2 = \frac{x_2 - \mu}{\sigma} \] \[ 0.315 = \frac{x_2 - 52}{8} \] \[ 0.315 \times 8 = x_2 - 52 \] \[ 2.52 = x_2 - 52 \] \[ x_2 = 52 + 2.52 \]
\[ x_2 = 54.52 \] Hence, the limits that include exactly the middle 25% of the observations are 49.48 to 54.52.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सामान्य वितरण वक्र दिखाता है जिसमें Z1 और Z2 के बीच का केंद्रीय क्षेत्र छायांकित है। यह मध्य 25% अवलोकनों की सीमा को दर्शाता है, जिसमें Z1 नकारात्मक और Z2 सकारात्मक Z-स्कोर है, जो माध्य के चारों ओर सममित हैं।In simple words: We needed to find the range of values that includes the middle 25% of the data. With a mean of 52 and a standard deviation of 8, we found that this range is from 49.48 to 54.52.

🎯 Exam Tip: For central percentage problems, divide the percentage by 2 to find the area on each side of the mean. Use the Z-table to find the corresponding Z-score, then convert the Z-scores back to X values. Remember that Z-scores to the left of the mean are negative.

 

Question 6. In a big showroom of electronic items, on an average 52 electronic units are sold every week and its variance is 9 (unit)². Sale of electronic items follows normal distribution. The probability that the sale of electronic items during a week out of 50 weeks is from x₁ units to 61 units is 0.1574. Estimate the value of x₁. Also estimate the number of weeks during which the sale of electronic items is more than 55 units.
Answer:Here, the mean (μ) is 52. The variance (σ²) is 9, so the standard deviation (σ) is \( \sqrt{9} = 3 \). The total number of weeks (N) is 50. First, estimate the value of x₁: Given, P[x₁ ≤ X ≤ 61] = 0.1574. For x₂ = 61, the Z-score is: \[ Z_2 = \frac{x_2 - \mu}{\sigma} = \frac{61 - 52}{3} = \frac{9}{3} = 3 \] We know P[x₁ ≤ X ≤ 61] = P[Z₁ ≤ Z ≤ Z₂] = P[Z₁ ≤ Z ≤ 3] = 0.1574. From the Z-table, P[0 ≤ Z ≤ 3] = 0.4987. Since Z₁ is to the left of the mean (because P[Z₁ ≤ Z ≤ 3] is less than P[0 ≤ Z ≤ 3]), we have: P[Z₁ ≤ Z ≤ 3] = P[0 ≤ Z ≤ 3] - P[0 ≤ Z ≤ Z₁] So, 0.1574 = 0.4987 - P[0 ≤ Z ≤ Z₁]. P[0 ≤ Z ≤ Z₁] = 0.4987 - 0.1574 = 0.3413. From the Z-table, for an area of 0.3413, the Z-score is 1.00. Since Z₁ is to the left of the mean, Z₁ = -1.00. Now, use the Z-score formula to find x₁: \[ Z_1 = \frac{x_1 - \mu}{\sigma} \] \[ -1 = \frac{x_1 - 52}{3} \] \[ -1 \times 3 = x_1 - 52 \] \[ -3 = x_1 - 52 \] \[ x_1 = 52 - 3 \]
\[ x_1 = 49 \] Therefore, the value of x₁ is 49 units. Next, estimate the number of weeks when sales are more than 55 units: We need to find P[X ≥ 55]. For x₁ = 55, the Z-score is: \[ Z_1 = \frac{x_1 - \mu}{\sigma} = \frac{55 - 52}{3} = \frac{3}{3} = 1 \] The probability P[X ≥ 55] = P[Z ≥ 1]. This can be calculated as P[0 ≤ Z < ∞] - P[0 ≤ Z ≤ 1]. From the Z-table, P[0 ≤ Z ≤ 1] = 0.3413. So, P[Z ≥ 1] = 0.5000 - 0.3413 = 0.1587. The number of weeks when sales are more than 55 units = N × P[Z ≥ 1] = 50 × 0.1587 = 7.935. Approximately, this is 8 weeks.
ℹ️ चित्र व्याख्या (Diagram Explanation): पहला चित्र Z1 से Z2=3 के बीच का छायांकित क्षेत्र दिखाता है, जो 0.1574 की संभावना को दर्शाता है। दूसरा चित्र Z=1 से दाईं ओर का छायांकित क्षेत्र दिखाता है, जो उन हफ्तों का प्रतिनिधित्व करता है जब बिक्री 55 इकाइयों से अधिक थी।In simple words: The average weekly sale is 52 units with a spread of 3 units. First, we found that for a probability of 0.1574 between two sales figures, the lower sales figure (x1) is 49 units. Second, we calculated that sales will be more than 55 units in about 8 out of 50 weeks.

🎯 Exam Tip: Break down complex problems into smaller, manageable parts. For finding an unknown x-value within a given probability range, first convert the known x-value to a Z-score, then use the given probability to find the Z-score of the unknown x-value, and finally solve for x. Always consider both sides of the mean.

 

Question 7. It is known that on an average a person spends 61 minutes in a painting exhibition. If this time is normally distributed and 20 % of the persons spent less than 30 minutes in the exhibition then find variance of the distribution. Also determine the probability that a person spends more than 90 minutes in the exhibition.
Answer:Here, the mean time spent (μ) is 61 minutes. Given, 20% of persons spent less than 30 minutes, so P[X ≤ 30] = 20/100 = 0.2. First, find the variance (σ²): For x₁ = 30, we have P[X ≤ 30] = 0.2. This means the area to the left of Z₁ (corresponding to x₁=30) is 0.2. So, P[Z₁ ≤ Z ≤ 0] = P[-∞ < Z ≤ 0] - P[Z ≤ Z₁] = 0.5000 - 0.2000 = 0.3000. From the Z-table, the area 0.2995 corresponds to Z = 0.84. Since Z₁ is to the left of the mean, Z₁ = -0.84. Now, use the Z-score formula to find the standard deviation (σ): \[ Z_1 = \frac{x_1 - \mu}{\sigma} \] \[ -0.84 = \frac{30 - 61}{\sigma} \] \[ -0.84 = \frac{-31}{\sigma} \] \[ \sigma = \frac{-31}{-0.84} \]
\[ \sigma \approx 36.9 \text{ (approximately)} \] The variance (σ²) = (36.9)² = 1361.61. Next, determine the probability that a person spends more than 90 minutes: We need to find P[X ≥ 90]. For x₂ = 90, the Z-score is: \[ Z_2 = \frac{x_2 - \mu}{\sigma} = \frac{90 - 61}{36.9} = \frac{29}{36.9} \approx 0.7859 \approx 0.79 \] The probability P[X ≥ 90] = P[Z ≥ 0.79]. This can be calculated as P[0 ≤ Z < ∞] - P[0 ≤ Z ≤ 0.79]. From the Z-table, P[0 ≤ Z ≤ 0.79] = 0.2852. So, P[Z ≥ 0.79] = 0.5000 - 0.2852 = 0.2148.
ℹ️ चित्र व्याख्या (Diagram Explanation): पहला चित्र Z1 से बाईं ओर के 0.2 क्षेत्र को दर्शाता है, जिसका उपयोग मानक विचलन ज्ञात करने के लिए किया गया था। दूसरा चित्र Z2=0.79 से दाईं ओर के क्षेत्र को दर्शाता है, जो उन व्यक्तियों का प्रतिनिधित्व करता है जिन्होंने 90 मिनट से अधिक समय बिताया।In simple words: People spend an average of 61 minutes at the exhibition. 20% spend less than 30 minutes. From this, we figured out the spread (variance) of the time spent. Then, we calculated that the chance of someone spending more than 90 minutes is 0.2148.

🎯 Exam Tip: To find variance (or standard deviation) from a given percentile, first find the Z-score for that percentile. Then, use the Z-score formula to solve for σ. When dealing with probabilities, clearly identify the area under the curve (left or right tail, or central) and use the Z-table accordingly.

 

Question 8. If the diameter of pipes produced by a company manufacturing pipes is 20 mm to 22 mm, then it is accepted by specified group of customers. The standard deviation of produced pipes is 4 mm and it is known that 70 % of the pipes produced in the unit have diameter more than 19.05 mm. Find the average diameter of the produced pipes. Also find the percentage of pipes rejected by specified group of customers. Note: The diameter of the produced pipes follows normal distribution.
Answer:Here, the standard deviation (σ) is 4 mm. Given, 70% of pipes have a diameter more than 19.05 mm, so P[X ≥ 19.05] = 70/100 = 0.7. First, find the average diameter (μ): For x₁ = 19.05, we have P[X ≥ 19.05] = 0.7. This means the area to the right of Z₁ (corresponding to x₁=19.05) is 0.7. Since the area is greater than 0.5, Z₁ must be to the left of the mean (negative). So, P[Z₁ ≤ Z ≤ 0] = P[Z ≥ Z₁] - P[0 ≤ Z < ∞] = 0.7000 - 0.5000 = 0.2000. From the Z-table, the area 0.1985 corresponds to Z = 0.52, and 0.2019 corresponds to Z = 0.53. For an area of 0.2000, Z is between 0.52 and 0.53. We can take the average Z = (0.52 + 0.53) / 2 = 0.525. Since Z₁ is to the left of the mean, Z₁ = -0.525. Now, use the Z-score formula to find the mean (μ): \[ Z_1 = \frac{x_1 - \mu}{\sigma} \] \[ -0.525 = \frac{19.05 - \mu}{4} \] \[ -0.525 \times 4 = 19.05 - \mu \] \[ -2.1 = 19.05 - \mu \] \[ \mu = 19.05 + 2.1 \]
\[ \mu = 21.15 \text{ mm} \] Therefore, the average diameter of the produced pipes is 21.15 mm. Next, find the percentage of pipes rejected: Pipes are accepted if their diameter is between 20 mm and 22 mm. So, pipes are rejected if their diameter is not in this range. The probability that pipes are accepted is P[20 ≤ X ≤ 22]. For x₁ = 20, the Z-score is: \[ Z_1 = \frac{x_1 - \mu}{\sigma} = \frac{20 - 21.15}{4} = \frac{-1.15}{4} = -0.2875 \approx -0.29 \] For x₂ = 22, the Z-score is: \[ Z_2 = \frac{x_2 - \mu}{\sigma} = \frac{22 - 21.15}{4} = \frac{0.85}{4} = 0.2125 \approx 0.21 \] The probability P[20 ≤ X ≤ 22] = P[-0.29 ≤ Z ≤ 0.21]. This can be calculated as P[-0.29 ≤ Z ≤ 0] + P[0 ≤ Z ≤ 0.21]. Due to symmetry, P[-0.29 ≤ Z ≤ 0] = P[0 ≤ Z ≤ 0.29]. From the Z-table: P[0 ≤ Z ≤ 0.29] = 0.1141 P[0 ≤ Z ≤ 0.21] = 0.0832 So, P[-0.29 ≤ Z ≤ 0.21] = 0.1141 + 0.0832 = 0.1973. The probability that pipes are accepted is 0.1973. The probability that pipes are rejected = 1 - P[accepted] = 1 - 0.1973 = 0.8027. The percentage of rejected pipes = 0.8027 × 100 = 80.27%.
ℹ️ चित्र व्याख्या (Diagram Explanation): पहला चित्र Z1 से दाईं ओर के 0.7 क्षेत्र को दर्शाता है, जिसका उपयोग माध्य व्यास ज्ञात करने के लिए किया गया था। दूसरा चित्र Z=-0.29 और Z=0.21 के बीच के छायांकित क्षेत्र को दर्शाता है, जो स्वीकार्य व्यास वाली पाइपों की संभावना को दर्शाता है।In simple words: The average pipe diameter is 21.15 mm with a standard deviation of 4 mm. We found this mean because 70% of pipes are over 19.05 mm. Pipes are accepted if their diameter is between 20 mm and 22 mm. Only 19.73% of pipes are accepted, meaning 80.27% are rejected.

🎯 Exam Tip: When given a probability to find a parameter (like mean or standard deviation), first identify the Z-score, then solve the Z-score formula for the unknown parameter. For rejection/acceptance rates, calculate the probability of acceptance and subtract from 1 for the rejection probability.

 

Question 9. A normal variable X has mean 400 and variance 900. FInd the fourth decile and 90th percentile of this distribution and also interpret the values.
Answer:Here, the mean (μ) is 400. The variance (σ²) is 900, so the standard deviation (σ) is \( \sqrt{900} = 30 \). Fourth Decile (D₄): The fourth decile (D₄) is the value below which 40% of the observations fall. So, P[X ≤ D₄] = 40/100 = 0.4. This means the area to the left of Z₁ (corresponding to D₄) is 0.4. Since the area is less than 0.5, Z₁ must be to the left of the mean (negative). So, P[Z₁ ≤ Z ≤ 0] = P[-∞ < Z ≤ 0] - P[Z ≤ Z₁] = 0.5000 - 0.4000 = 0.1000. From the Z-table, the area 0.0987 corresponds to Z = 0.25, and 0.1026 corresponds to Z = 0.26. For an area of 0.1000, Z is between 0.25 and 0.26. We can take the average Z = (0.25 + 0.26) / 2 = 0.255. Since Z₁ is to the left of the mean, Z₁ = -0.255. Now, use the Z-score formula to find D₄: \[ Z_1 = \frac{D_4 - \mu}{\sigma} \] \[ -0.255 = \frac{D_4 - 400}{30} \] \[ -0.255 \times 30 = D_4 - 400 \] \[ -7.65 = D_4 - 400 \] \[ D_4 = 400 - 7.65 \]
\[ D_4 = 392.35 \] The fourth decile (D₄) is 392.35. Interpretation: 40% of the observations in this distribution are less than 392.35.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सामान्य वितरण वक्र के Z1 से बाईं ओर के 0.4 क्षेत्र को दर्शाता है, जो चौथे दशमक (D4) का प्रतिनिधित्व करता है। Z1 माध्य के बाईं ओर नकारात्मक है।In simple words: The average is 400, and the spread is 30. The fourth decile is 392.35, meaning 40% of the data values are below 392.35.

🎯 Exam Tip: Deciles and percentiles are specific points in a distribution. To find them, first determine the proportion of data below that point, convert it to a Z-score, and then use the Z-score formula to find the actual value (X). Interpretation is key for these questions.

 

90th Percentile (P₉₀):
Answer:The 90th percentile (P₉₀) is the value below which 90% of the observations fall. So, P[X ≤ P₉₀] = 90/100 = 0.9. This means the area to the left of Z₂ (corresponding to P₉₀) is 0.9. Since the area is greater than 0.5, Z₂ must be to the right of the mean (positive). So, P[0 ≤ Z ≤ Z₂] = P[Z ≤ Z₂] - P[-∞ < Z ≤ 0] = 0.9000 - 0.5000 = 0.4000. From the Z-table, the area 0.3997 corresponds to Z = 1.28, and 0.4015 corresponds to Z = 1.29. For an area of 0.4000, Z is between 1.28 and 1.29. We can take Z₂ = 1.28 (since 0.3997 is closer to 0.4). Now, use the Z-score formula to find P₉₀: \[ Z_2 = \frac{P_{90} - \mu}{\sigma} \] \[ 1.28 = \frac{P_{90} - 400}{30} \] \[ 1.28 \times 30 = P_{90} - 400 \] \[ 38.4 = P_{90} - 400 \] \[ P_{90} = 400 + 38.4 \]
\[ P_{90} = 438.4 \] The 90th percentile (P₉₀) is 438.4. Interpretation: 90% of the observations in this distribution are less than 438.4.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सामान्य वितरण वक्र के Z2 से बाईं ओर के 0.9 क्षेत्र को दर्शाता है, जो 90वें प्रतिशतक (P90) का प्रतिनिधित्व करता है। Z2 माध्य के दाईं ओर सकारात्मक है।In simple words: With an average of 400 and a spread of 30, the 90th percentile is 438.4. This means 90% of the data values are below 438.4.

🎯 Exam Tip: Distinguish between deciles and percentiles. Both indicate a point in the distribution. Ensure the correct area is used for Z-score lookup (area to the left for percentiles) and remember the sign of the Z-score (positive for values above the mean, negative for values below).

 

Question 10. A normal variable X has the following density function:
\( f(x) = \frac{1}{50 \sqrt{2 \pi}} e^{-\frac{1}{2}\left(\frac{x-150}{50}\right)^2}, - \infty < x < \infty \)
For this distribution,
(1) If P[x₁ ≤ X ≤ 250] = 0.4772, then estimate x₁
Answer:From the given probability density function, we can identify the mean (μ) and standard deviation (σ). The general form of a normal PDF is: \( f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2} \) Comparing this with the given function: \( f(x) = \frac{1}{50 \sqrt{2 \pi}} e^{-\frac{1}{2}\left(\frac{x-150}{50}\right)^2} \) We find that the mean (μ) is 150 and the standard deviation (σ) is 50. (1) To estimate x₁ when P[x₁ ≤ X ≤ 250] = 0.4772: For x₂ = 250, the Z-score is: \[ Z_2 = \frac{x_2 - \mu}{\sigma} = \frac{250 - 150}{50} = \frac{100}{50} = 2 \] We are given P[x₁ ≤ X ≤ 250] = P[Z₁ ≤ Z ≤ Z₂] = P[Z₁ ≤ Z ≤ 2] = 0.4772. From the Z-table, P[0 ≤ Z ≤ 2] = 0.4772. Since P[Z₁ ≤ Z ≤ 2] = P[0 ≤ Z ≤ 2] = 0.4772, this implies that Z₁ must be 0 (meaning x₁ is the mean). If Z₁ = 0, then: \[ Z_1 = \frac{x_1 - \mu}{\sigma} \] \[ 0 = \frac{x_1 - 150}{50} \] \[ 0 = x_1 - 150 \]
\[ x_1 = 150 \] Therefore, the estimated value of x₁ is 150.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सामान्य वितरण वक्र को दर्शाता है। Z=0 (माध्य) और Z=2 के बीच का क्षेत्र छायांकित है, जो 0.4772 की संभावना को दर्शाता है। यह दिखाता है कि x1 माध्य के बराबर है।In simple words: The average value is 150 with a spread of 50. We were given a probability for values between x1 and 250. Since the probability matched the area from the mean to 250, it means x1 must be 150, which is the mean.

🎯 Exam Tip: Always extract μ and σ directly from the probability density function if it's provided. When a given probability P[x₁ ≤ X ≤ x₂] matches P[μ ≤ X ≤ x₂], it directly implies that x₁ = μ. This is a common shortcut if you recognize it.

 

(2) If P[75 ≤ X ≤ x₂] = 0.3539, then estimate x₂.
Answer:Here, the mean (μ) is 150 and the standard deviation (σ) is 50. To estimate x₂ when P[75 ≤ X ≤ x₂] = 0.3539: For x₁ = 75, the Z-score is: \[ Z_1 = \frac{x_1 - \mu}{\sigma} = \frac{75 - 150}{50} = \frac{-75}{50} = -1.5 \] We are given P[75 ≤ X ≤ x₂] = P[Z₁ ≤ Z ≤ Z₂] = P[-1.5 ≤ Z ≤ Z₂] = 0.3539. From the Z-table, P[0 ≤ Z ≤ 1.5] = 0.4332. Since P[-1.5 ≤ Z ≤ Z₂] is less than P[0 ≤ Z ≤ 1.5], Z₂ must be to the left of 0 (negative). So, P[-1.5 ≤ Z ≤ Z₂] = P[-1.5 ≤ Z ≤ 0] - P[Z₂ ≤ Z ≤ 0]. Due to symmetry, P[-1.5 ≤ Z ≤ 0] = P[0 ≤ Z ≤ 1.5] = 0.4332. So, 0.3539 = 0.4332 - P[Z₂ ≤ Z ≤ 0]. P[Z₂ ≤ Z ≤ 0] = 0.4332 - 0.3539 = 0.0793. From the Z-table, for an area of 0.0793, the closest Z-score is 0.20 (P[0 ≤ Z ≤ 0.20] = 0.0793). Since Z₂ is to the left of the mean, Z₂ = -0.20. Now, use the Z-score formula to find x₂: \[ Z_2 = \frac{x_2 - \mu}{\sigma} \] \[ -0.20 = \frac{x_2 - 150}{50} \] \[ -0.20 \times 50 = x_2 - 150 \] \[ -10 = x_2 - 150 \] \[ x_2 = 150 - 10 \]
\[ x_2 = 140 \] Therefore, the estimated value of x₂ is 140.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सामान्य वितरण वक्र के Z1=-1.5 और Z2=-0.2 के बीच के छायांकित क्षेत्र को दर्शाता है, जो 0.3539 की संभावना को दर्शाता है। दोनों Z-स्कोर माध्य के बाईं ओर हैं।In simple words: We want to find x2, given that the probability of values between 75 and x2 is 0.3539. We first found the Z-score for 75, which is -1.5. Then, by calculating the required area and finding its Z-score from the table, we determined x2 to be 140.

🎯 Exam Tip: When finding an unknown value (x₂) within a probability range involving a negative Z-score (Z₁), visualize the area carefully. Determine if Z₂ will be positive or negative. Calculate the area from Z₁ to 0, subtract the given probability, and then find the Z-score for the remaining area. Always check the sign of Z₂ based on your visualization.

 

Question 1.
(2) Find the percentage of children getting marks between 70 and 90.
(3) Find the minimum score of most intelligent 50 children.

Answer:

(2) To find the percentage of children with marks between 70 and 90, we use the given data: total children \(N = 500\), mean marks `\(\mu = 68\)`, and standard deviation `\(\sigma = 22\)`.
First, calculate the Z-scores for the mark boundaries:
For \(x_1 = 70\), \(Z_1 = \frac{x_1 - \mu}{\sigma} = \frac{70-68}{22} = \frac{2}{22} \approx 0.09\).
For \(x_2 = 90\), \(Z_2 = \frac{x_2 - \mu}{\sigma} = \frac{90-68}{22} = \frac{22}{22} = 1\).
The probability of marks between 70 and 90 corresponds to \(P[0.09 \le Z \le 1]\).
This probability is calculated as \(P[0 \le Z \le 1] - P[0 \le Z \le 0.09]\).
From the standard normal distribution table, \(P[0 \le Z \le 1] = 0.3413\) and \(P[0 \le Z \le 0.09] = 0.0359\).
So, the probability is \(0.3413 - 0.0359 = 0.3054\).
To express this as a percentage, multiply by 100: \(0.3054 \times 100 = 30.54\%\).

ℹ️ चित्र व्याख्या (Diagram Explanation): एक सामान्य वितरण वक्र दिखाया गया है जिसका केंद्र Z=0 पर है। इसमें Z=0.09 और Z=1 के बीच का क्षेत्र छायांकित है, जो छात्रों के उन अंकों का प्रतिशत दर्शाता है जो 70 और 90 के बीच आते हैं।

(3) To find the minimum score of the 50 most intelligent children out of 500, we consider the probability of a student being in this group. This is \(P[X \ge x_1] = \frac{50}{500} = 0.1\).
We are given `\(\mu = 68\)` and `\(\sigma = 22\)`. We need to find the Z-score \(Z_1\) such that \(P[Z \ge Z_1] = 0.1\).
This can be rewritten as \(P[0 \le Z \le Z_1] = P[0 \le Z < \infty] - P[Z \ge Z_1] = 0.5 - 0.1 = 0.4\).
From the standard normal distribution table, the Z-score corresponding to an area of 0.4 is approximately \(Z_1 = 1.28\) (using interpolation for 0.3997).
Now, use the Z-score formula: \(Z_1 = \frac{x_1 - \mu}{\sigma}\).
So, \(1.28 = \frac{x_1 - 68}{22}\).
Multiply both sides by 22: \(1.28 \times 22 = x_1 - 68 \implies 28.16 = x_1 - 68\).
Solve for \(x_1\): \(x_1 = 68 + 28.16 = 96.16\).
Therefore, the minimum score for the most intelligent 50 children is approximately 96 marks.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सामान्य वितरण वक्र को दर्शाता है जहाँ Z=0 केंद्र में है। वक्र के दाहिने तरफ Z1 से अनंत तक का क्षेत्र छायांकित है, जो 0.1 की प्रायिकता को प्रदर्शित करता है, जो शीर्ष 50 सबसे बुद्धिमान छात्रों के न्यूनतम अंक को दर्शाता है।

In simple words: For marks between 70 and 90, we change them to Z-scores, find the area in the Z-table, and multiply by 100 to get 30.54%. For the top 50 smartest children, we find the Z-score for the top 10% of students (1.28) and use it to calculate their minimum score (96.16 marks).

🎯 Exam Tip: Remember to clearly define the Z-scores and use the correct Z-table values for areas to the left or right of the mean. Always convert probabilities to percentages by multiplying by 100 when asked for a percentage.

 

Question 2. Age of 500 employees working in a private company follows normal distribution with mean 40 years and standard deviation 6 years. The company wants to reduce its staff by 25 % in the following manner: (1) To retrench 5 % of the employees having minimum age. (2) After retrenching 5 % of employees having minimum age, next 10 % of the employees are to be transferred to another company. (3) To retire 10% of employees having maximum age. From this information, find the age of employees who are to be retrenched, transferred and retired from the company.

Answer:Here, the total number of employees is \(N = 500\), the mean age is `\(\mu = 40\)` years, and the standard deviation is `\(\sigma = 6\)` years. Let \(X\) represent the age of an employee.

(1) To retrench 5% of employees with minimum age:
We need to find the age \(x_1\) such that \(P[X \le x_1] = \frac{5}{100} = 0.05\).
We convert this probability to a Z-score. Since 0.05 is less than 0.5, \(Z_1\) will be negative.
\(P[Z_1 \le Z \le 0] = P[-\infty < Z \le 0] - P[Z \le Z_1] = 0.5000 - 0.05 = 0.4500\).
From the standard normal distribution table, the Z-score for an area of 0.4500 is approximately \(Z_1 = -1.645\) (interpolating between -1.64 for 0.4495 and -1.65 for 0.4505).
Now, use the Z-score formula: \(Z_1 = \frac{x_1 - \mu}{\sigma}\).
So, \(-1.645 = \frac{x_1 - 40}{6}\).
Multiply both sides by 6: \(-1.645 \times 6 = x_1 - 40 \implies -9.87 = x_1 - 40\).
Solve for \(x_1\): \(x_1 = 40 - 9.87 = 30.13\) years.
Therefore, employees aged 30.13 years or less are to be retrenched.

ℹ️ चित्र व्याख्या (Diagram Explanation): एक सामान्य वितरण वक्र दिखाया गया है जिसका केंद्र Z=0 पर है। वक्र के बाएं तरफ Z1 तक का क्षेत्र छायांकित है, जो 0.05 की प्रायिकता को दर्शाता है, जो न्यूनतम आयु वाले 5% कर्मचारियों की आयु सीमा को प्रदर्शित करता है।

(2) After retrenching the initial 5%, the next 10% of employees are to be transferred. This means we are looking for employees whose age is above the retrenched group but within the next 10% from the lowest end. So, the cumulative percentage from the minimum age side is \(5\% + 10\% = 15\%\).
We need to find the age \(x_2\) such that \(P[X \le x_2] = \frac{15}{100} = 0.15\).
As before, we convert this probability to a Z-score. Since 0.15 is less than 0.5, \(Z_2\) will be negative.
\(P[Z_2 \le Z \le 0] = P[-\infty < Z \le 0] - P[Z \le Z_2] = 0.5000 - 0.15 = 0.3500\).
From the standard normal distribution table, the Z-score for an area of 0.3500 is approximately \(Z_2 = -1.035\) (interpolating between -1.03 for 0.3485 and -1.04 for 0.3508).
Now, use the Z-score formula: \(Z_2 = \frac{x_2 - \mu}{\sigma}\).
So, \(-1.035 = \frac{x_2 - 40}{6}\).
Multiply both sides by 6: \(-1.035 \times 6 = x_2 - 40 \implies -6.21 = x_2 - 40\).
Solve for \(x_2\): \(x_2 = 40 - 6.21 = 33.79\) years.
Therefore, employees with ages between 30.13 years (from part 1) and 33.79 years are to be transferred to another company.

ℹ️ चित्र व्याख्या (Diagram Explanation): एक सामान्य वितरण वक्र दिखाया गया है जिसका केंद्र Z=0 पर है। वक्र के बाएं तरफ Z2 तक का क्षेत्र छायांकित है, जो 0.15 की प्रायिकता को दर्शाता है, जो स्थानांतरित होने वाले 10% कर्मचारियों की आयु सीमा को प्रदर्शित करता है, इसमें न्यूनतम आयु वाले 5% कर्मचारी भी शामिल हैं।

(3) To retire 10% of employees having maximum age:
We need to find the age \(x_1\) (using \(x_1\) again as in the source) such that \(P[X \ge x_1] = \frac{10}{100} = 0.1\).
We convert this probability to a Z-score. Since we're looking at the upper tail, \(Z_1\) will be positive.
\(P[0 \le Z \le Z_1] = P[0 \le Z < \infty] - P[Z \ge Z_1] = 0.5000 - 0.1 = 0.4000\).
From the standard normal distribution table, the Z-score for an area of 0.4000 is approximately \(Z_1 = 1.28\) (using interpolation for 0.3997).
Now, use the Z-score formula: \(Z_1 = \frac{x_1 - \mu}{\sigma}\).
So, \(1.28 = \frac{x_1 - 40}{6}\).
Multiply both sides by 6: \(1.28 \times 6 = x_1 - 40 \implies 7.68 = x_1 - 40\).
Solve for \(x_1\): \(x_1 = 40 + 7.68 = 47.68\) years.
Therefore, employees aged 47.68 years or more are to be retired.

ℹ️ चित्र व्याख्या (Diagram Explanation): एक सामान्य वितरण वक्र दिखाया गया है जिसका केंद्र Z=0 पर है। वक्र के दाहिने तरफ Z1 से अनंत तक का क्षेत्र छायांकित है, जो 0.1 की प्रायिकता को दर्शाता है, जो अधिकतम आयु वाले 10% कर्मचारियों की आयु सीमा को प्रदर्शित करता है।

In simple words: For retrenchment (lowest 5% age), the age limit is 30.13 years. For transfer (next 10% age), the age limit is 33.79 years. For retirement (highest 10% age), the age limit is 47.68 years.

🎯 Exam Tip: When dealing with multiple percentage cut-offs, carefully determine if they are cumulative from one end or represent distinct intervals. Always use the correct Z-score sign (positive for above mean, negative for below mean) corresponding to the probability area.

 

Question 3. An entrance test of 200 marks is conducted for higher studies. 20,000 students remain present in the examination and the marks obtained by them follows normal distribution with mean 120 and standard deviation 20. The rules for the result are as under:
(1) Students who acquire less than 40 per cent marks are failed.
(2) An additional test is conducted for the students acquiring marks between 40 per cent and 48 per cent.
(3) Students who acquire marks between 48 per cent and 75 per cent are called for personal interview.
(4) Students who acquire marks more than 75 per cent get direct admission for the higher studies.
Find approximate number of students who:
(i) failed In test,
(ii) appeared for additional 100 marks test,
(iii) appeared for personal interview and
(iv) got direct admission for the higher studies.

Answer:Here, total number of students \(N = 20000\), mean marks `\(\mu = 120\)`, standard deviation `\(\sigma = 20\)`, and total marks for the test = 200. Let \(X\) represent the marks obtained by students.

(i) Number of students who failed the test:
Students fail if they score less than 40% marks. 40% of 200 marks is \(\frac{40}{100} \times 200 = 80\) marks.
So, we need to find \(P[X \le 80]\).
First, calculate the Z-score for \(x_1 = 80\): \(Z_1 = \frac{x_1 - \mu}{\sigma} = \frac{80 - 120}{20} = \frac{-40}{20} = -2\).
We need to find \(P[Z \le -2]\).
This probability is calculated as \(P[-\infty < Z \le 0] - P[-2 \le Z \le 0]\) or \(0.5 - P[0 \le Z \le 2]\).
From the standard normal distribution table, \(P[0 \le Z \le 2] = 0.4772\).
So, \(P[Z \le -2] = 0.5000 - 0.4772 = 0.0228\).
The approximate number of students who failed is \(N \times P[Z \le -2] = 20000 \times 0.0228 = 456\) students.

ℹ️ चित्र व्याख्या (Diagram Explanation): एक सामान्य वितरण वक्र दिखाया गया है जिसका केंद्र Z=0 पर है। वक्र के बाएं तरफ Z=-2 तक का क्षेत्र छायांकित है, जो 0.0228 की प्रायिकता को दर्शाता है, जो उन छात्रों की संख्या को प्रदर्शित करता है जो परीक्षा में अनुत्तीर्ण हुए।

(ii) Number of students who appeared for an additional test:
Students appear for an additional test if their marks are between 40% and 48%.
40% of 200 marks = 80 marks.
48% of 200 marks = \(\frac{48}{100} \times 200 = 96\) marks.
So, we need to find \(P[80 \le X \le 96]\).
Calculate the Z-scores for these boundaries:
For \(x_1 = 80\), \(Z_1 = \frac{80 - 120}{20} = -2\).
For \(x_2 = 96\), \(Z_2 = \frac{96 - 120}{20} = \frac{-24}{20} = -1.2\).
We need to find \(P[-2 \le Z \le -1.2]\).
This probability is calculated as \(P[-2 \le Z \le 0] - P[-1.2 \le Z \le 0]\) or \(P[0 \le Z \le 2] - P[0 \le Z \le 1.2]\).
From the standard normal distribution table, \(P[0 \le Z \le 2] = 0.4772\) and \(P[0 \le Z \le 1.2] = 0.3849\).
So, the probability is \(0.4772 - 0.3849 = 0.0923\).
The approximate number of students who appeared for the additional test is \(N \times 0.0923 = 20000 \times 0.0923 = 1846\) students.

ℹ️ चित्र व्याख्या (Diagram Explanation): एक सामान्य वितरण वक्र दिखाया गया है जिसका केंद्र Z=0 पर है। वक्र के बाएं तरफ Z=-2 और Z=-1.2 के बीच का क्षेत्र छायांकित है, जो 0.0923 की प्रायिकता को दर्शाता है, जो उन छात्रों की संख्या को प्रदर्शित करता है जो अतिरिक्त परीक्षा के लिए उपस्थित हुए।

(iii) Number of students who appeared for a personal interview:
Students are called for a personal interview if their marks are between 48% and 75%.
48% of 200 marks = 96 marks.
75% of 200 marks = \(\frac{75}{100} \times 200 = 150\) marks.
So, we need to find \(P[96 \le X \le 150]\).
Calculate the Z-scores for these boundaries:
For \(x_1 = 96\), \(Z_1 = \frac{96 - 120}{20} = -1.2\).
For \(x_2 = 150\), \(Z_2 = \frac{150 - 120}{20} = 1.5\).
We need to find \(P[-1.2 \le Z \le 1.5]\).
This probability is calculated as \(P[-1.2 \le Z \le 0] + P[0 \le Z \le 1.5]\) or \(P[0 \le Z \le 1.2] + P[0 \le Z \le 1.5]\).
From the standard normal distribution table, \(P[0 \le Z \le 1.2] = 0.3849\) and \(P[0 \le Z \le 1.5] = 0.4332\).
So, the probability is \(0.3849 + 0.4332 = 0.8181\).
The approximate number of students called for a personal interview is \(N \times 0.8181 = 20000 \times 0.8181 = 16362\) students.

ℹ️ चित्र व्याख्या (Diagram Explanation): एक सामान्य वितरण वक्र दिखाया गया है जिसका केंद्र Z=0 पर है। वक्र पर Z=-1.2 और Z=1.5 के बीच का क्षेत्र छायांकित है, जो 0.8181 की प्रायिकता को दर्शाता है, जो उन छात्रों की संख्या को प्रदर्शित करता है जिन्हें व्यक्तिगत साक्षात्कार के लिए बुलाया गया।

(iv) Number of students who got direct admission:
Students get direct admission if their marks are more than 75%.
75% of 200 marks = 150 marks.
So, we need to find \(P[X \ge 150]\).
Calculate the Z-score for \(x_1 = 150\): \(Z_1 = \frac{150 - 120}{20} = 1.5\).
We need to find \(P[Z \ge 1.5]\).
This probability is calculated as \(P[0 \le Z < \infty] - P[0 \le Z \le 1.5]\) or \(0.5 - P[0 \le Z \le 1.5]\).
From the standard normal distribution table, \(P[0 \le Z \le 1.5] = 0.4332\).
So, \(P[Z \ge 1.5] = 0.5000 - 0.4332 = 0.0668\).
The approximate number of students who got direct admission is \(N \times 0.0668 = 20000 \times 0.0668 = 1336\) students.

ℹ️ चित्र व्याख्या (Diagram Explanation): एक सामान्य वितरण वक्र दिखाया गया है जिसका केंद्र Z=0 पर है। वक्र के दाहिने तरफ Z=1.5 से अनंत तक का क्षेत्र छायांकित है, जो 0.0668 की प्रायिकता को दर्शाता है, जो उन छात्रों की संख्या को प्रदर्शित करता है जिन्हें सीधे प्रवेश मिला।

In simple words: Out of 20,000 students, 456 failed (less than 80 marks). 1846 students got between 80-96 marks for an extra test. 16362 students scored between 96-150 marks and were called for interviews. 1336 students scored over 150 marks and got direct admission.

🎯 Exam Tip: Convert percentage marks to actual marks before calculating Z-scores. Be careful with Z-score calculations for different intervals (less than, between, more than) and use the Z-table correctly for positive and negative Z-values by utilizing symmetry.

 

Question 4. The monthly income of a group of persons follows normal distribution with mean Rs. 20000 and standard deviation Rs. 5000. If the minimum monthly income of 50 richest persons is Rs. 31625, then how many persons are in the group ? Also, what is the maximum income of 50 persons having lowest income?

Answer:Here, the mean monthly income is `\(\mu = \text{Rs. } 20000\)`, and the standard deviation is `\(\sigma = \text{Rs. } 5000\)`.

First, find the total number of persons (\(N\)) in the group.
We are given that the minimum monthly income of the 50 richest persons is Rs. 31625. This means that 50 persons have an income of Rs. 31625 or more.
So, \(P[X \ge 31625] = \frac{50}{N}\).
Calculate the Z-score for \(x_2 = 31625\): \(Z_2 = \frac{x_2 - \mu}{\sigma} = \frac{31625 - 20000}{5000} = \frac{11625}{5000} = 2.325 \approx 2.33\).
We need to find \(P[Z \ge 2.33]\).
This probability is calculated as \(P[0 \le Z < \infty] - P[0 \le Z \le 2.33]\) or \(0.5 - P[0 \le Z \le 2.33]\).
From the standard normal distribution table, \(P[0 \le Z \le 2.33] = 0.4901\).
So, \(P[Z \ge 2.33] = 0.5000 - 0.4901 = 0.0099\).
Now, we have \(\frac{50}{N} = 0.0099\).
Solving for \(N\): \(N = \frac{50}{0.0099} \approx 5050.5\). Rounding up, the total number of persons in the group is approximately 5051.

ℹ️ चित्र व्याख्या (Diagram Explanation): एक सामान्य वितरण वक्र दिखाया गया है जिसका केंद्र Z=0 पर है। वक्र के दाहिने तरफ Z=2.33 से अनंत तक का क्षेत्र छायांकित है, जो 0.0099 की प्रायिकता को दर्शाता है, जो उन 50 सबसे अमीर व्यक्तियों की न्यूनतम आय से संबंधित है।

Next, find the maximum income of the 50 persons having the lowest income.
We found the total number of persons \(N = 5051\).
We need to find the income \(x_1\) such that \(P[X \le x_1] = \frac{50}{5051} \approx 0.0099\).
We convert this probability to a Z-score. Since 0.0099 is very small and less than 0.5, \(Z_1\) will be negative.
\(P[Z_1 \le Z \le 0] = P[-\infty < Z \le 0] - P[Z \le Z_1] = 0.5000 - 0.0099 = 0.4901\).
From the standard normal distribution table, the Z-score for an area of 0.4901 is approximately \(Z_1 = -2.33\) (close to the Z-score for 0.4901, which is 2.33).
Now, use the Z-score formula: \(Z_1 = \frac{x_1 - \mu}{\sigma}\).
So, \(-2.33 = \frac{x_1 - 20000}{5000}\).
Multiply both sides by 5000: \(-2.33 \times 5000 = x_1 - 20000 \implies -11650 = x_1 - 20000\).
Solve for \(x_1\): \(x_1 = 20000 - 11650 = 8350\).
Therefore, the maximum income of the 50 persons with the lowest income is Rs. 8350.

ℹ️ चित्र व्याख्या (Diagram Explanation): एक सामान्य वितरण वक्र दिखाया गया है जिसका केंद्र Z=0 पर है। वक्र के बाएं तरफ Z=-2.33 तक का क्षेत्र छायांकित है, जो 0.0099 की प्रायिकता को दर्शाता है, जो उन 50 सबसे कम आय वाले व्यक्तियों की अधिकतम आय को प्रदर्शित करता है।

In simple words: If the 50 richest people earn at least Rs. 31625, then there are about 5051 people in total. For the 50 people with the lowest incomes, their highest income is Rs. 8350.

🎯 Exam Tip: When finding the total number of items from a proportion, remember to divide the number of items by their calculated probability. Ensure Z-scores are correctly interpreted for upper and lower tails of the distribution.

 

Question 5. Analysis of result of 12th standard students of a school is as under:
Pass with distinction: 15% of total students
Pass without distinction: 75% of total students
Fail: 10% of total students
For passing the examination, minimum 40 % of the total marks and for distinction minimum 80% marks are required. If the percentage of result of the students follows normal distribution then find mean and standard deviation and by using it determine the percentage marks for which 75 % of the students have less than that percentage marks.

Answer:Let `\(\mu\)` be the mean percentage of marks and `\(\sigma\)` be the standard deviation of percentage marks.

From the given information:
1. 10% of students fail, meaning they score less than 40% marks.
So, \(P[X < 40] = 0.10\).
For \(x_1 = 40\), we find the corresponding Z-score \(Z_1\). Since \(P[Z \le Z_1] = 0.10\) (which is less than 0.5), \(Z_1\) must be negative.
\(P[Z_1 \le Z \le 0] = P[-\infty < Z \le 0] - P[Z \le Z_1] = 0.5000 - 0.10 = 0.4000\).
From the standard normal distribution table, the Z-score for an area of 0.4000 is approximately \(Z_1 = -1.28\).
Using the Z-score formula: \(Z_1 = \frac{x_1 - \mu}{\sigma} \implies -1.28 = \frac{40 - \mu}{\sigma}\).
This gives us the equation: \(-1.28\sigma = 40 - \mu\) (Equation 1).

ℹ️ चित्र व्याख्या (Diagram Explanation): एक सामान्य वितरण वक्र दिखाया गया है जिसका केंद्र Z=0 पर है। वक्र के बाएं तरफ Z1 तक का क्षेत्र छायांकित है, जो 0.1 की प्रायिकता को दर्शाता है, जो 40% से कम अंक प्राप्त करने वाले छात्रों के प्रतिशत को प्रदर्शित करता है।

2. 15% of students pass with distinction, meaning they score more than 80% marks.
So, \(P[X > 80] = 0.15\).
For \(x_2 = 80\), we find the corresponding Z-score \(Z_2\). Since \(P[Z \ge Z_2] = 0.15\) (which is less than 0.5 for the upper tail), \(Z_2\) must be positive.
\(P[0 \le Z \le Z_2] = P[0 \le Z < \infty] - P[Z \ge Z_2] = 0.5000 - 0.15 = 0.3500\).
From the standard normal distribution table, the Z-score for an area of 0.3500 is approximately \(Z_2 = 1.035\) (interpolating between 1.03 for 0.3485 and 1.04 for 0.3508).
Using the Z-score formula: \(Z_2 = \frac{x_2 - \mu}{\sigma} \implies 1.035 = \frac{80 - \mu}{\sigma}\).
This gives us the equation: \(1.035\sigma = 80 - \mu\) (Equation 2).

ℹ️ चित्र व्याख्या (Diagram Explanation): एक सामान्य वितरण वक्र दिखाया गया है जिसका केंद्र Z=0 पर है। वक्र के दाहिने तरफ Z2 से अनंत तक का क्षेत्र छायांकित है, जो 0.15 की प्रायिकता को दर्शाता है, जो 80% से अधिक अंक प्राप्त करने वाले छात्रों के प्रतिशत को प्रदर्शित करता है।

Now, solve Equation 1 and Equation 2 simultaneously to find `\(\mu\)` and `\(\sigma\)`:
\(-1.28\sigma = 40 - \mu\) (1)
\(1.035\sigma = 80 - \mu\) (2)
Subtract Equation (1) from Equation (2):
\((1.035\sigma) - (-1.28\sigma) = (80 - \mu) - (40 - \mu)\)
\(1.035\sigma + 1.28\sigma = 80 - 40\)
\(2.315\sigma = 40\)

\( \implies \sigma = \frac{40}{2.315} \approx 17.278 \approx 17.28\).
Substitute `\(\sigma = 17.28\)` into Equation (1):
\(-1.28 \times 17.28 = 40 - \mu\)
\(-22.1184 = 40 - \mu\)
\( \implies \mu = 40 + 22.1184 = 62.1184 \approx 62.12\).
Hence, the mean percentage of marks `\(\mu\)` is 62.12% and the standard deviation `\(\sigma\)` is 17.28%.

Next, determine the percentage marks for which 75% of the students have less than that.
We need to find the mark \(x_1\) such that \(P[X \le x_1] = \frac{75}{100} = 0.75\).
We convert this probability to a Z-score. Since \(P[Z \le Z_1] = 0.75\) (which is greater than 0.5), \(Z_1\) will be positive.
\(P[0 \le Z \le Z_1] = P[Z \le Z_1] - P[-\infty < Z \le 0] = 0.75 - 0.50 = 0.25\).
From the standard normal distribution table, the Z-score for an area of 0.2500 is approximately \(Z_1 = 0.675\) (interpolating between 0.67 for 0.2486 and 0.68 for 0.2518).
Now, use the Z-score formula: \(Z_1 = \frac{x_1 - \mu}{\sigma}\).
Using `\(\mu = 62.12\)` and `\(\sigma = 17.28\)`:
\(0.675 = \frac{x_1 - 62.12}{17.28}\).
Multiply both sides by 17.28: \(0.675 \times 17.28 = x_1 - 62.12 \implies 11.664 = x_1 - 62.12\).
Solve for \(x_1\): \(x_1 = 62.12 + 11.664 = 73.784 \approx 73.78\).
Therefore, 75% of students have less than 73.78 percentage marks.

ℹ️ चित्र व्याख्या (Diagram Explanation): एक सामान्य वितरण वक्र दिखाया गया है जिसका केंद्र Z=0 पर है। वक्र के बाएं तरफ Z1 तक का क्षेत्र छायांकित है, जो 0.75 की प्रायिकता को दर्शाता है, जो उन 75% छात्रों के प्रतिशत अंकों को प्रदर्शित करता है जिनके अंक इस सीमा से कम हैं।

In simple words: Based on student results, the average mark is 62.12% with a spread of 17.28%. We found this using the fail and distinction percentages. Also, 75% of students scored less than 73.78%.

🎯 Exam Tip: When given two probability points, set up two simultaneous equations involving `\(\mu\)` and `\(\sigma\)` to solve for these parameters. Always ensure you use the correct area for a given Z-score (or vice versa) from the standard normal table.

 

Question 6. The monthly bill amount of regular customers of a provision store follows normal distribution. If 7.78 % customers have monthly bill amount less than Rs. 3590 and 94.52% customers have bill amount less than Rs. 5100, then determine the parameters of the normal distribution. Also determine the interval for monthly bill amount of exactly middle 95% customers.

Answer:Let \(X\) be the monthly bill amount of customers.

First, determine the mean `\(\mu\)` and standard deviation `\(\sigma\)` of the normal distribution.
1. 7.78% of customers have a monthly bill amount less than Rs. 3590.
So, \(P[X \le 3590] = \frac{7.78}{100} = 0.0778\).
For \(x_1 = 3590\), we find the corresponding Z-score \(Z_1\). Since \(P[Z \le Z_1] = 0.0778\) (less than 0.5), \(Z_1\) must be negative.
\(P[Z_1 \le Z \le 0] = P[-\infty < Z \le 0] - P[Z \le Z_1] = 0.5000 - 0.0778 = 0.4222\).
From the standard normal distribution table, the Z-score for an area of 0.4222 is approximately \(Z_1 = -1.42\).
Using the Z-score formula: \(Z_1 = \frac{x_1 - \mu}{\sigma} \implies -1.42 = \frac{3590 - \mu}{\sigma}\).
This gives us the equation: \(-1.42\sigma = 3590 - \mu\) (Equation 1).

ℹ️ चित्र व्याख्या (Diagram Explanation): एक सामान्य वितरण वक्र दिखाया गया है जिसका केंद्र Z=0 पर है। वक्र के बाएं तरफ Z1 तक का क्षेत्र छायांकित है, जो 0.0778 की प्रायिकता को दर्शाता है, जो Rs. 3590 से कम बिल वाले ग्राहकों के प्रतिशत को प्रदर्शित करता है।

2. 94.52% of customers have a bill amount less than Rs. 5100.
So, \(P[X \le 5100] = \frac{94.52}{100} = 0.9452\).
For \(x_2 = 5100\), we find the corresponding Z-score \(Z_2\). Since \(P[Z \le Z_2] = 0.9452\) (greater than 0.5), \(Z_2\) must be positive.
\(P[0 \le Z \le Z_2] = P[Z \le Z_2] - P[-\infty < Z \le 0] = 0.9452 - 0.5000 = 0.4452\).
From the standard normal distribution table, the Z-score for an area of 0.4452 is approximately \(Z_2 = 1.60\).
Using the Z-score formula: \(Z_2 = \frac{x_2 - \mu}{\sigma} \implies 1.60 = \frac{5100 - \mu}{\sigma}\).
This gives us the equation: \(1.60\sigma = 5100 - \mu\) (Equation 2).

ℹ️ चित्र व्याख्या (Diagram Explanation): एक सामान्य वितरण वक्र दिखाया गया है जिसका केंद्र Z=0 पर है। वक्र के बाएं तरफ Z2 तक का क्षेत्र छायांकित है, जो 0.9452 की प्रायिकता को दर्शाता है, जो Rs. 5100 से कम बिल वाले ग्राहकों के प्रतिशत को प्रदर्शित करता है।

Now, solve Equation 1 and Equation 2 simultaneously:
\(-1.42\sigma = 3590 - \mu\) (1)
\(1.60\sigma = 5100 - \mu\) (2)
Subtract Equation (1) from Equation (2):
\((1.60\sigma) - (-1.42\sigma) = (5100 - \mu) - (3590 - \mu)\)
\(1.60\sigma + 1.42\sigma = 5100 - 3590\)
\(3.02\sigma = 1510\)

\( \implies \sigma = \frac{1510}{3.02} \approx 500\).
Substitute `\(\sigma = 500\)` into Equation (2):
\(1.60 \times 500 = 5100 - \mu\)
\(800 = 5100 - \mu\)
\( \implies \mu = 5100 - 800 = 4300\).
Hence, the mean monthly bill amount `\(\mu\)` is Rs. 4300 and the standard deviation `\(\sigma\)` is Rs. 500.

Next, determine the interval for the monthly bill amount of exactly the middle 95% customers.
For the middle 95% of observations in a normal distribution, the interval is given by `\(\mu \pm 1.96\sigma\)`.
Using `\(\mu = 4300\)` and `\(\sigma = 500\)`:
Interval = \(4300 \pm (1.96 \times 500)\)
Interval = \(4300 \pm 980\)
Lower limit: \(4300 - 980 = 3320\).
Upper limit: \(4300 + 980 = 5280\).
The interval for the monthly bill amount of the middle 95% customers is (Rs. 3320, Rs. 5280).

In simple words: We found that the average monthly bill is Rs. 4300 with a standard deviation of Rs. 500. This was calculated using the percentages of customers with bills less than certain amounts. The middle 95% of customer bills fall between Rs. 3320 and Rs. 5280.

🎯 Exam Tip: When given two cumulative probabilities, construct two Z-score equations and solve them simultaneously for `\(\mu\)` and `\(\sigma\)` using algebraic methods. Remember that the middle 95% interval for a normal distribution is always `\(\mu \pm 1.96\sigma\)`.

 

Question 7. A normal variable X has following probability density function:
\(f(x) = \frac{1}{50 \sqrt{2 \pi}} e^{-\frac{1}{2}\left(\frac{x-75}{50}\right)^{2}}, - \infty < x < \infty\)
From this, answer the following questions:
(1) If \(P[60 \le X \le x_2] = 0.5670\), then find \(x_2\).
(2) If \(P[x_1 \le X \le 125] = 0.3979\), then find \(x_1\).
(3) Find \(P[|X - 50| \le 10]\).

Answer:Comparing the given probability density function with the general form \(f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^{2}}\), we can identify:
Mean `\(\mu = 75\)`
Standard deviation `\(\sigma = 50\)`

(1) Find \(x_2\) if \(P[60 \le X \le x_2] = 0.5670\).
First, convert the lower bound \(x_1 = 60\) to a Z-score:
\(Z_1 = \frac{x_1 - \mu}{\sigma} = \frac{60 - 75}{50} = \frac{-15}{50} = -0.3\).
So, we have \(P[-0.3 \le Z \le Z_2] = 0.5670\).
This probability can be expressed as \(P[-0.3 \le Z \le 0] + P[0 \le Z \le Z_2]\).
\(P[0 \le Z \le 0.3]\) (due to symmetry) is found from the Z-table as 0.1179.
So, \(0.1179 + P[0 \le Z \le Z_2] = 0.5670\).
\(P[0 \le Z \le Z_2] = 0.5670 - 0.1179 = 0.4491\).
From the standard normal distribution table, the Z-score corresponding to an area of 0.4491 is approximately \(Z_2 = 1.635\) (interpolating between 1.63 for 0.4484 and 1.64 for 0.4495).
Now, use the Z-score formula to find \(x_2\): \(Z_2 = \frac{x_2 - \mu}{\sigma}\).
\(1.635 = \frac{x_2 - 75}{50}\).
Multiply both sides by 50: \(1.635 \times 50 = x_2 - 75 \implies 81.75 = x_2 - 75\).
Solve for \(x_2\): \(x_2 = 75 + 81.75 = 156.75 \approx 157\).
Therefore, \(x_2\) is approximately 157.

ℹ️ चित्र व्याख्या (Diagram Explanation): एक सामान्य वितरण वक्र दिखाया गया है जिसका केंद्र Z=0 पर है। वक्र पर Z=-0.3 और Z2 के बीच का क्षेत्र छायांकित है, जो 0.5670 की प्रायिकता को दर्शाता है। यह क्षेत्र 60 और x2 के बीच X के मानों के लिए है।

(2) Find \(x_1\) if \(P[x_1 \le X \le 125] = 0.3979\).
First, convert the upper bound \(x_2 = 125\) to a Z-score:
\(Z_2 = \frac{x_2 - \mu}{\sigma} = \frac{125 - 75}{50} = \frac{50}{50} = 1\).
So, we have \(P[Z_1 \le Z \le 1] = 0.3979\).
This probability can be expressed as \(P[Z_1 \le Z \le 0] + P[0 \le Z \le 1]\).
From the standard normal distribution table, \(P[0 \le Z \le 1] = 0.3413\).
So, \(P[Z_1 \le Z \le 0] + 0.3413 = 0.3979\).
\(P[Z_1 \le Z \le 0] = 0.3979 - 0.3413 = 0.0566\).
Since this area is to the left of the mean, \(Z_1\) must be negative.
From the standard normal distribution table, the Z-score corresponding to an area of 0.0566 is approximately \(Z_1 = -0.145\) (interpolating between -0.14 for 0.0557 and -0.15 for 0.0596).
Now, use the Z-score formula to find \(x_1\): \(Z_1 = \frac{x_1 - \mu}{\sigma}\).
\(-0.145 = \frac{x_1 - 75}{50}\).
Multiply both sides by 50: \(-0.145 \times 50 = x_1 - 75 \implies -7.25 = x_1 - 75\).
Solve for \(x_1\): \(x_1 = 75 - 7.25 = 67.75 \approx 68\).
Therefore, \(x_1\) is approximately 68.

ℹ️ चित्र व्याख्या (Diagram Explanation): एक सामान्य वितरण वक्र दिखाया गया है जिसका केंद्र Z=0 पर है। वक्र पर Z1 और Z=1 के बीच का क्षेत्र छायांकित है, जो 0.3979 की प्रायिकता को दर्शाता है। यह क्षेत्र x1 और 125 के बीच X के मानों के लिए है।

(3) Find \(P[|X - 50| \le 10]\).
The inequality \(|X - 50| \le 10\) means \(-10 \le X - 50 \le 10\).
Adding 50 to all parts: \(50 - 10 \le X \le 50 + 10 \implies 40 \le X \le 60\).
So, we need to find \(P[40 \le X \le 60]\).
Convert the boundaries to Z-scores using `\(\mu = 75\)` and `\(\sigma = 50\)`:
For \(x_1 = 40\), \(Z_1 = \frac{40 - 75}{50} = \frac{-35}{50} = -0.7\).
For \(x_2 = 60\), \(Z_2 = \frac{60 - 75}{50} = \frac{-15}{50} = -0.3\).
We need to find \(P[-0.7 \le Z \le -0.3]\).
This probability is calculated as \(P[-0.7 \le Z \le 0] - P[-0.3 \le Z \le 0]\) or \(P[0 \le Z \le 0.7] - P[0 \le Z \le 0.3]\).
From the standard normal distribution table, \(P[0 \le Z \le 0.7] = 0.2580\) and \(P[0 \le Z \le 0.3] = 0.1179\).
So, the probability is \(0.2580 - 0.1179 = 0.1401\).

ℹ️ चित्र व्याख्या (Diagram Explanation): एक सामान्य वितरण वक्र दिखाया गया है जिसका केंद्र Z=0 पर है। वक्र पर Z=-0.7 और Z=-0.3 के बीच का क्षेत्र छायांकित है, जो 0.1401 की प्रायिकता को दर्शाता है। यह क्षेत्र 40 और 60 के बीच X के मानों के लिए है।

In simple words: Given the distribution's mean (75) and standard deviation (50), we found that for a probability of 0.5670 between 60 and some value, that value is about 157. For a probability of 0.3979 between some value and 125, that value is about 68. Lastly, the chance that X is between 40 and 60 (which is \(|X - 50| \le 10\)) is 0.1401.

🎯 Exam Tip: Carefully extract `\(\mu\)` and `\(\sigma\)` from the PDF's formula. When solving for an unknown boundary (\(x_1\) or \(x_2\)), always find the Z-score first using the area, then use the Z-score formula to solve for the unknown \(x\) value. Remember how to convert absolute value inequalities into ranges.

Question 7.
(3) Find \(P[|X - 50| \le 10]\)
Answer:To find \(P[|X - 50| \le 10]\), we understand this means the value of X is between \(50 - 10\) and \(50 + 10\). So, we need to find \(P[40 \le X \le 60]\). From the probability density function given earlier in the question (on page 56, \(f(x) = \frac{1}{50 \sqrt{2 \pi}} e^{-\frac{1}{2}(\frac{x-75}{50})^2}\)), we know the mean \( \mu = 75 \) and standard deviation \( \sigma = 50 \). First, convert the X-values to Z-scores: For \(x_1 = 40\): \(Z_1 = \frac{x_1 - \mu}{\sigma} = \frac{40 - 75}{50} = \frac{-35}{50} = -0.7\) For \(x_2 = 60\): \(Z_2 = \frac{x_2 - \mu}{\sigma} = \frac{60 - 75}{50} = \frac{-15}{50} = -0.3\) Now, we need to find \(P[-0.7 \le Z \le -0.3]\). This can be calculated as \(P[-0.7 \le Z \le 0] - P[-0.3 \le Z \le 0]\). Because the normal distribution is symmetric, \(P[-0.7 \le Z \le 0] = P[0 \le Z \le 0.7]\) and \(P[-0.3 \le Z \le 0] = P[0 \le Z \le 0.3]\). Using a standard normal distribution table: \(P[0 \le Z \le 0.7] = 0.2580\) \(P[0 \le Z \le 0.3] = 0.1179\) So, \(P[-0.7 \le Z \le -0.3] = 0.2580 - 0.1179 = 0.1401\).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक मानक सामान्य वितरण वक्र (Standard Normal Distribution Curve) को दर्शाता है। इसमें माध्य (mean) Z=0 पर है। चित्र में -0.7 और -0.3 के बीच के क्षेत्र को छायांकित किया गया है, जो इस श्रेणी के Z-स्कोर के लिए प्रायिकता (probability) को दर्शाता है।In simple words: This part asks for the chance that a value X is within 10 units of 50. We change the X values into Z-scores and then look up these Z-scores in a special table to find the probability.

🎯 Exam Tip: Remember to always convert X values to Z-scores using the given mean and standard deviation before calculating probabilities for normal distributions. Pay attention to the direction of inequalities (less than, greater than, or between) when using the Z-table.

 

Question 8.
A normal variable X has the following probability density function:
\( f(x) = \text{constant} \times e^{-\frac{1}{200}(x-50)^2}, -\infty < x < \infty \)
From this distribution, answer the following questions:
(1) Find median.
(2) Find estimated values of the extreme quartiles.
(3) Find approximate value of quartile deviation.
(4) Find approximate value of mean deviation.
Answer:The general form of a normal probability density function is \( f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2} \). By comparing the given function \( f(x) = \text{constant} \times e^{-\frac{1}{200}(x-50)^2} \) with the general form, we can identify the parameters: \(-\frac{1}{2}(\frac{x-\mu}{\sigma})^2 = -\frac{1}{200}(x-50)^2\) \( \implies \frac{(x-\mu)^2}{\sigma^2} = \frac{(x-50)^2}{100} \) Therefore, the mean \( \mu = 50 \) and the variance \( \sigma^2 = 100 \), which means the standard deviation \( \sigma = \sqrt{100} = 10 \). (1) Median: In a normal distribution, the mean, median, and mode are all equal. So, Median \( = \mu = 50 \). (2) Estimated values of the extreme quartiles (Q1 and Q3): The first quartile (Q1) is the value below which 25% of observations fall. The third quartile (Q3) is the value below which 75% of observations fall, or above which 25% of observations fall. For Q1: We need \( P[X \le Q_1] = 0.25 \). In terms of Z-scores, \( P[Z \le z_1] = 0.25 \). This means the area from \( -\infty \) to \( z_1 \) is 0.25. Since the total area to the left of 0 is 0.5, \( P[z_1 \le Z \le 0] = 0.50 - 0.25 = 0.25 \). From the standard normal distribution table, the Z-score corresponding to an area of 0.25 (between 0 and Z) is approximately 0.675. Since it's to the left of the mean, \( z_1 = -0.675 \). Using the formula \( z_1 = \frac{Q_1 - \mu}{\sigma} \): \( -0.675 = \frac{Q_1 - 50}{10} \) \( -0.675 \times 10 = Q_1 - 50 \) \( -6.75 = Q_1 - 50 \) \( Q_1 = 50 - 6.75 = 43.25 \) For Q3: We need \( P[X \ge Q_3] = 0.25 \). In terms of Z-scores, \( P[Z \ge z_2] = 0.25 \). This means the area from \( z_2 \) to \( \infty \) is 0.25. Since the total area to the right of 0 is 0.5, \( P[0 \le Z \le z_2] = 0.50 - 0.25 = 0.25 \). From the standard normal distribution table, the Z-score corresponding to an area of 0.25 (between 0 and Z) is approximately 0.675. Since it's to the right of the mean, \( z_2 = 0.675 \). Using the formula \( z_2 = \frac{Q_3 - \mu}{\sigma} \): \( 0.675 = \frac{Q_3 - 50}{10} \) \( 0.675 \times 10 = Q_3 - 50 \) \( 6.75 = Q_3 - 50 \) \( Q_3 = 50 + 6.75 = 56.75 \) So, the extreme quartiles are Q1 = 43.25 and Q3 = 56.75.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक मानक सामान्य वितरण वक्र को दर्शाता है जिसका माध्य Z=0 पर है। इसमें \( Z_1 \) और \( Z_2 \) के Z-स्कोर के क्षेत्रों को छायांकित किया गया है, जो क्रमशः निचले 25% और ऊपरी 25% अवलोकनों के लिए चतुर्थक (quartiles) को दर्शाते हैं। (3) Approximate value of quartile deviation: For a normal distribution, Quartile Deviation (QD) is approximately \( \frac{2}{3}\sigma \). QD \( \approx \frac{2}{3} \times 10 = \frac{20}{3} \approx 6.67 \). (4) Approximate value of mean deviation: For a normal distribution, Mean Deviation (MD) is approximately \( \frac{4}{5}\sigma \). MD \( \approx \frac{4}{5} \times 10 = 8 \).In simple words: First, we figure out the average and spread from the given math rule. Then, we find the middle point (median), the points that mark the bottom 25% and top 25% (quartiles), and how much numbers usually spread out from the average using special formulas.

🎯 Exam Tip: Remember that for a normal distribution, the mean, median, and mode are identical. Also, learn the standard approximations for quartile deviation (\( \frac{2}{3}\sigma \)) and mean deviation (\( \frac{4}{5}\sigma \)) as they are frequently tested and simplify calculations.