Samacheer Kalvi Class 12 Business Maths Solutions Chapter 7 Probability Distributions Exercise 7.4

Choose The Correct Answer

Question 1. Normal distribution was invented by
(a) Laplace
(b) De-Moivre
(c) Gauss
(d) All of the options
Answer: (b) De-Moivre
In simple words: The Normal distribution, a very common bell-shaped curve in statistics, was first thought of by Abraham de Moivre. He laid the groundwork for understanding this important concept.

🎯 Exam Tip: Remember key figures in statistics and their contributions, especially for foundational concepts like the normal distribution, as these are frequently tested for basic knowledge.

Question 2. If X ~ N(9, 81) the standard normal variate Z will be
(a) Z = \( \frac { X-81 }{9} \)
(b) Z = \( \frac { X-9 }{81} \)
(c) Z = \( \frac { X-9 }{9} \)
(d) Z = \( \frac { 9-X }{9} \)
Answer: (c) Z = \( \frac { X-9 }{9} \)
In simple words: To find the standard normal variate Z, we subtract the mean (which is 9) from X and then divide by the standard deviation. Since the variance is 81, the standard deviation is its square root, which is 9. This calculation helps us compare different normal distributions.

🎯 Exam Tip: Always remember that in the notation N(\( \mu \), \( \sigma^2 \)), the second parameter is the variance (\( \sigma^2 \)), not the standard deviation (\( \sigma \)). Take the square root to find \( \sigma \) before calculating Z.

Question 3. If Z is a standard normal variate, the proportion of items lying between Z = -0.5 and Z = -3.0 is
(a) 0.4987
(b) 0.1915
(c) 0.3072
(d) 0.3098
Answer: (c) 0.3072
In simple words: We are looking for the area under the normal curve between two negative Z-scores. We find this by subtracting the area from Z=0 to Z=0.5 from the area from Z=0 to Z=3.0, because the normal distribution is symmetrical.

🎯 Exam Tip: When calculating areas between negative Z-scores, use the symmetry of the normal distribution: \( P(-b < Z < -a) = P(a < Z < b) = P(0 < Z < b) - P(0 < Z < a) \).

Question 4. If X ~ Ν(\( \mu \), \( \sigma^2 \)), the maximum probability at the point of inflexion of normal distribution
(a) \( (\frac { 1 }{\sqrt{2\pi}})e^{\frac { -1 }{2}} \)
(b) \( (\frac { 1 }{\sqrt{2\pi}})e^{\frac { 1 }{2}} \)
(c) \( (\frac { 1 }{\sigma\sqrt{2\pi}})e^{\frac { -1 }{2}} \)
(d) \( (\frac { 1 }{\sigma\sqrt{2\pi}}) \)
Answer: (c) \( (\frac { 1 }{\sigma\sqrt{2\pi}})e^{\frac { -1 }{2}} \)
In simple words: The points where the curve changes its bending direction are called points of inflexion. At these special points in a normal distribution, the probability density is given by a specific formula that depends on the standard deviation (\( \sigma \)) and the mathematical constant 'e'.

🎯 Exam Tip: Understand that the maximum probability density of a normal distribution is at its mean (\( \frac{1}{\sigma\sqrt{2\pi}} \)), and the points of inflection are at \( \mu \pm \sigma \), where the probability density function value is \( \frac{1}{\sigma\sqrt{2\pi}} e^{-1/2} \).

Question 5. In a parametric distribution the mean is equal to variance is
(a) binomial
(b) normal
(c) poisson
(d) All of the options
Answer: (c) poisson
In simple words: In a Poisson distribution, which models rare events, the average number of events (mean) is always the same as how spread out the events are (variance). This is a unique property of this type of distribution.

🎯 Exam Tip: Remember the defining characteristics of common distributions. For Poisson, the mean (\( \lambda \)) is equal to the variance (\( \lambda \)). For Binomial, mean is \( np \) and variance is \( npq \).

Question 6. In a manufacturing company, the average number of defectives is 1%. The probability that the sample of 100 toys there will be 3 defectives is
(a) 0.0613
(b) 0.613
(c) 0.00613
(d) 0.3913
Answer: (a) 0.0613
In simple words: We can use a Poisson distribution to find the chance of exactly 3 defective toys in a sample of 100, given that 1% are usually defective. We first calculate the average number of defectives expected in this sample.

🎯 Exam Tip: Use the Poisson distribution when dealing with probabilities of a certain number of events occurring in a fixed interval or space, especially when the events are rare and the average rate (\( \lambda \)) is known. Remember \( \lambda = np \).

Question 7. The parameters of the normal distribution f(x) \( = \frac { 1 }{\sqrt{72\pi}} e^{\frac{-(x-10)^2}{72}} \) are
(a) (10, 6)
(b) (10, 36)
(c) (6, 10)
(d) (36, 10)
Answer: (b) (10, 36)
In simple words: By comparing the given formula to the general formula for a normal distribution, we can find its mean (\( \mu \)) and variance (\( \sigma^2 \)). The \( (x-10)^2 \) part tells us the mean is 10, and the 72 in the exponent helps us find the variance.

🎯 Exam Tip: Memorize the standard form of the normal probability density function: \( f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2} \). This allows direct identification of mean \( \mu \) and standard deviation \( \sigma \) (and thus variance \( \sigma^2 \)).

Question 8. A manufacturer produces switches and experiences that 2 per cent switches are defective. The probability that in a box of 50 switches, there are at most two defective is:
(a) 2.5 e\(^{-1}\)
(b) e\(^{-1}\)
(c) 2 e\(^{-1}\)
(d) None of the options
Answer: (a) 2.5 e\(^{-1}\)
In simple words: We calculate the probability of having 0, 1, or 2 defective switches using the Poisson distribution. First, find the average number of defectives, which is \( \lambda = np = 50 \times 0.02 = 1 \). Then sum the probabilities for each case.

🎯 Exam Tip: "At most two defective" means the number of defectives can be 0, 1, or 2. Sum the individual probabilities \( P(X=0) + P(X=1) + P(X=2) \) for the Poisson distribution to get the final answer.

Question 9. An experiment succeeds twice as often as it fails. The chance that in the next six trials, there shall be at least four successes is
(a) 240/729
(b) 489/729
(c) 496/729
(d) 251/729
Answer: (c) 496/729
In simple words: First, we figure out the probability of success (p) and failure (q) from the given information. Then, using the binomial distribution formula for 6 trials, we add up the probabilities of getting 4, 5, or 6 successes to find the chance of at least four successes.

🎯 Exam Tip: When \( p=2q \), remember \( p+q=1 \), so \( 2q+q=1 \implies 3q=1 \implies q=1/3 \) and \( p=2/3 \). "At least four successes" means \( P(X \ge 4) = P(X=4) + P(X=5) + P(X=6) \).

Question 10. If for a binomial distribution b(n, p) mean = 4 and variance = 4/3, the probability, P(X ≥ 5) is equal to:
(a) \( (2/3)^6 \)
(b) \( (2/3)^5(1/3) \)
(c) \( (1/3)^6 \)
(d) \( 4(2/3)^6 \)
Answer: (d) \( 4(2/3)^6 \)
In simple words: We use the given mean (\( np=4 \)) and variance (\( npq=4/3 \)) to find 'n', 'p', and 'q' for the binomial distribution. Once we have these values, we can calculate the probability of getting at least 5 successes in 'n' trials.

🎯 Exam Tip: For binomial distribution, always remember the relationships: Mean = \( np \) and Variance = \( npq \). Use these to find \( p, q, n \) before calculating any probabilities.

Question 11. The average percentage of failure in a certain examination is 40. The probability that out of a group of 6 candidates atleast 4 passed in the examination are:
(a) 0.5443
(b) 0.4543
(c) 0.5543
(d) 0.4573
Answer: (a) 0.5443
In simple words: If the failure rate is 40%, then the success rate (passing) is 60%. For a group of 6 candidates, we need to find the chance that 4, 5, or all 6 of them pass. We use the binomial distribution for this.

🎯 Exam Tip: Clearly define success and failure probabilities. If 40% fail, then \( q = 0.4 \), so \( p = 1 - 0.4 = 0.6 \). "At least 4 passed" means calculating \( P(X=4) + P(X=5) + P(X=6) \).

Question 12. Forty percent of the passengers who fly on a certain route do not check in any luggage. The planes on this route seat 15 passengers. For a full flight, what is the mean of the number of passengers who do not check in any luggage?
(a) 6.00
(b) 6.45
(c) 7.20
(d) 7.50
Answer: (a) 6.00
In simple words: If 40% of passengers don't check luggage on a plane with 15 seats, we can find the average number of such passengers. We multiply the total number of passengers by the percentage who don't check luggage.

🎯 Exam Tip: The mean (average) for a binomial distribution is simply \( np \), where \( n \) is the number of trials (passengers) and \( p \) is the probability of success (not checking luggage).

Question 13. Which of the following statements is/are true regarding the normal distribution curve?
(a) it is symmetrical and bell shaped curve
(b) it is asymptotic in that each end approaches the horizontal axis but never reaches it
(c) its mean, median and mode are located at the same point
(d) All of the options
Answer: (d) All of the options
In simple words: The normal distribution curve is perfectly balanced, looks like a bell, stretches out infinitely without ever quite touching the x-axis, and its middle-most value (median), most common value (mode), and average value (mean) are all exactly the same. All these points describe its shape.

🎯 Exam Tip: Understanding the characteristics of the normal distribution (bell-shaped, symmetrical, asymptotic, mean=median=mode) is fundamental for solving problems related to it.

Question 14. Which of the following cannot generate a Poisson distribution?
(a) The number of telephone calls received in a ten-minute interval
(b) The number of customers arriving at a petrol station
(c) The number of bacteria found in a cubic feet of soil
(d) The number of misprints per page
Answer: (b) The number of customers arriving at a petrol station
In simple words: A Poisson distribution usually models events that happen rarely and independently over a fixed time or space. While events like calls, bacteria, or misprints fit this well, customer arrivals at a petrol station might not always fit the "rare and independent" assumption, especially if there are many external factors or burst arrivals.

🎯 Exam Tip: Poisson distribution is suitable for counting discrete events occurring in a fixed interval of time or space, provided events occur independently and at a constant average rate, and are relatively rare. Options (a), (c), (d) are classic examples of Poisson processes.

Question 15. X is normally distributed with a mean of 70 and a standard deviation of 10. What is the probability that X is between 72 and 84?
(a) 0.683
(b) 0.954
(c) 0.271
(d) 0.340
Answer: (d) 0.340
In simple words: To find the probability, we convert the X values (72 and 84) into Z-scores using the given mean and standard deviation. Then, we find the area under the standard normal curve between these two Z-scores using a Z-table.

🎯 Exam Tip: Always convert the raw data points (X values) into Z-scores using \( Z = \frac{X - \mu}{\sigma} \) before using a standard normal distribution table to find probabilities or areas.

Question 16. The starting annual salaries of newly qualified chartered accountants (CA's) in South Africa follow a normal distribution with a mean of Rs 180,000 and a standard deviation of Rs 10,000. What is the probability that a randomly selected CA will earn between Rs 165,000 and Rs 175,000 per annum?
(a) 0.819
(b) 0.242
(c) 0.286
(d) 0.533
Answer: (b) 0.242
In simple words: We first change the salary amounts into Z-scores using the given mean and standard deviation. Since both Z-scores are negative, we find the area between them on the normal curve by using symmetry and subtracting the smaller area from the larger one, referencing a standard normal distribution table.

🎯 Exam Tip: When calculating probabilities for an interval like \( P(a < X < b) \), convert both 'a' and 'b' to Z-scores, and then use the Z-table. Remember that \( P(-b < Z < -a) = P(a < Z < b) \).

Question 17. In a large statistics class the heights of the students are normally distributed with a mean of 172 cm and a variance of 25 cm. What proportion of students are between 165 cm and 181 cm in height?
(a) 0.954
(b) 0.601
(c) 0.718
(d) 0.883
Answer: (d) 0.883
In simple words: We convert the height values into Z-scores using the mean and standard deviation (which is the square root of the variance). Then, we find the area under the normal curve between these two Z-scores. Since one Z-score is negative and the other is positive, we add the areas from the middle (mean) to each Z-score.

🎯 Exam Tip: Always correctly calculate the standard deviation from the given variance. For probabilities between a negative and a positive Z-score, add the two areas from the mean to each Z-score: \( P(-a < Z < b) = P(0 < Z < a) + P(0 < Z < b) \).

Question 18. A statistical analysis of long-distance telephone calls indicates that the length of these calls is normally distributed with a mean of 240 seconds and a standard deviation of 40 seconds. What proportion of calls lasts less than 180 seconds?
(b) 0.094
(c) 0933
(d) 0.067
Answer: (d) 0.067
In simple words: We change the 180 seconds into a Z-score using the given mean and standard deviation. Then, we find the area under the normal curve to the left of this Z-score. Because the Z-score is negative, we use the symmetry of the curve to find the area to the right of the positive equivalent, and subtract it from 0.5.

🎯 Exam Tip: To find \( P(X < a) \) when 'a' is below the mean (resulting in a negative Z-score), you calculate \( P(Z < -z_0) \). Due to symmetry, this is equal to \( P(Z > z_0) = 0.5 - P(0 < Z < z_0) \).

Question 19. Cape town is estimated to have 21% of homes whose owners subscribe to the satellite service, DSTV. If a random sample of your home in taken, what is the probability that all four home subscribe to DSTV?
(a) 0.2100
(b) 0.5000
(c) 0.8791
(d) 0.0019
Answer: (d) 0.0019
In simple words: If there's a 21% chance a single home subscribes, then for four homes to *all* subscribe, we multiply this probability by itself four times. This is because each home's subscription is an independent event.

🎯 Exam Tip: For independent events, the probability of all events occurring is the product of their individual probabilities. Here, \( P(\text{all four subscribe}) = p \times p \times p \times p = p^4 \).

Question 20. Using the standard normal table, the sum of the probabilities to the right of z = 2.18 and to the left of z = -1.75 is:
(a) 0.4854
(b) 0.4599
(c) 0.0146
(d) 0.0547
Answer: (d) 0.0547
The problem asks for the sum of the area to the right of \( z = 2.18 \) and the area to the left of \( z = -1.75 \).
Area to the right of \( z = 2.18 \) is \( P(Z > 2.18) = 0.5 - P(0 < Z < 2.18) \). From the standard normal table, \( P(0 < Z < 2.18) = 0.4854 \). So, \( 0.5 - 0.4854 = 0.0146 \).
Area to the left of \( z = -1.75 \) is \( P(Z < -1.75) \). By symmetry, \( P(Z < -1.75) = P(Z > 1.75) = 0.5 - P(0 < Z < 1.75) \). From the table, \( P(0 < Z < 1.75) = 0.4599 \). So, \( 0.5 - 0.4599 = 0.0401 \).
The total sum is \( 0.0146 + 0.0401 = 0.0547 \). This type of problem helps understand how to use z-tables for different parts of a normal distribution.
In simple words: We add up two small areas from the "tails" of the normal curve. One area is very far to the right, and the other is very far to the left.

🎯 Exam Tip: Remember that the standard normal table typically gives areas from 0 to z. For areas in the tails or between two z-scores, you'll need to subtract or add these table values from 0.5 or each other.

Question 21. A brand of inkjet printers is normally distributed with a mean of 1,500 hours and a standard deviation of 200 hours. What proportion of printers fails before 1000 hours?
(a) 0.0062
(b) 0.0668
(c) 0.8413
(d) 0.0228
Answer: (a) 0.0062
Given mean \( \mu = 1500 \) hours and standard deviation \( \sigma = 200 \) hours. We want to find the proportion of printers that fail before \( X = 1000 \) hours, which is \( P(X < 1000) \).
First, we convert \( X = 1000 \) to a z-score:
\( z = \frac { X - \mu }{ \sigma } = \frac { 1000 - 1500 }{ 200 } = \frac { -500 }{ 200 } = -2.5 \)
So, we need to find \( P(Z < -2.5) \). Due to the symmetry of the normal distribution, \( P(Z < -2.5) = P(Z > 2.5) \).
The area to the right of \( z = 2.5 \) is \( 0.5 - P(0 < Z < 2.5) \).
From the standard normal table, \( P(0 < Z < 2.5) = 0.4938 \).
Therefore, \( P(Z < -2.5) = 0.5 - 0.4938 = 0.0062 \). This small probability shows that it is very unlikely for a printer to fail so early.
In simple words: Only a very tiny number of printers are expected to stop working before reaching 1000 hours of use.

🎯 Exam Tip: When finding probabilities for values less than the mean, the z-score will be negative. Always remember to use the symmetry of the normal curve to find the corresponding positive z-score area if your table only provides areas from 0 to z.

Question 22. The weights of newborn human babies are normally distributed with a mean of 3.2 kg and a standard deviation of 1.1 kg. What is the probability that a randomly selected newborn baby weighs less than 2.0 kg?
(a) 0.138
(b) 0.428
(c) 0.766
Answer: (a) 0.138
Given mean \( \mu = 3.2 \) kg and standard deviation \( \sigma = 1.1 \) kg. We need to find the probability that a newborn baby weighs less than \( X = 2.0 \) kg, which is \( P(X < 2.0) \).
First, calculate the z-score for \( X = 2.0 \):
\( z = \frac { X - \mu }{ \sigma } = \frac { 2.0 - 3.2 }{ 1.1 } = \frac { -1.2 }{ 1.1 } \approx -1.09 \)
So, we need to find \( P(Z < -1.09) \). Using the symmetry of the normal distribution, \( P(Z < -1.09) = P(Z > 1.09) \).
The area to the right of \( z = 1.09 \) is \( 0.5 - P(0 < Z < 1.09) \).
From the standard normal table, \( P(0 < Z < 1.09) \approx 0.3621 \).
Therefore, \( P(Z < -1.09) = 0.5 - 0.3621 = 0.1379 \), which rounds to \( 0.138 \). This means about 13.8% of newborn babies weigh less than 2.0 kg.
In simple words: We calculate how far 2.0 kg is from the average weight in terms of standard deviations. Then we use a table to find the chance that a baby's weight is below that point.

🎯 Exam Tip: Always draw a quick sketch of the normal curve to visualize the area you need to find. This helps avoid common errors when dealing with positive and negative z-scores or areas on either side of the mean.

Question 23. Monthly expenditure on their credit cards, by credit card holders from a certain bank, follows a normal distribution with a mean of Rs 1,295.00 and a standard deviation of Rs 750.00. What proportion of credit card holders spend more than Rs 1,500.00 on their credit cards per month?
(a) 0.487
(b) 0.392
(c) 0.500
(d) 0.791
Answer: (b) 0.392
Given mean \( \mu = Rs 1295 \) and standard deviation \( \sigma = Rs 750 \). We need to find the proportion of holders who spend more than \( X = Rs 1500 \), which is \( P(X > 1500) \).
First, convert \( X = 1500 \) to a z-score:
\( z = \frac { X - \mu }{ \sigma } = \frac { 1500 - 1295 }{ 750 } = \frac { 205 }{ 750 } \approx 0.273 \)
So, we need to find \( P(Z > 0.273) \). This area is found by subtracting the area from 0 to z from 0.5.
\( P(Z > 0.273) = 0.5 - P(0 < Z < 0.273) \).
From the standard normal table, \( P(0 < Z < 0.273) \approx 0.1064 \).
Therefore, \( P(Z > 0.273) = 0.5 - 0.1064 = 0.3936 \), which rounds to \( 0.392 \). This means about 39.2% of credit card holders spend more than Rs 1,500 monthly.
In simple words: We find the z-score for Rs 1500 and then use the normal distribution table to figure out the percentage of people who spend more than that amount.

🎯 Exam Tip: Pay attention to whether the question asks for "less than," "greater than," or "between" a value, as this determines how you use the 0.5 value and table lookups.

Question 24. Let z be a standard normal variable. If the area to the right of z is 0.8413, then the value of z must be:
(a) 1.00
(b) -1.00
(c) 0.00
(d) -0.41
Answer: (b) -1.00
We are given that the area to the right of z is \( P(Z > z) = 0.8413 \).
Since this probability is greater than 0.5, the z-value must be negative, meaning it is to the left of the mean (z=0).
We can write \( P(Z > z) = P(Z > 0) + P(z < Z < 0) \).
Since \( P(Z > 0) = 0.5 \), we have \( 0.8413 = 0.5 + P(z < Z < 0) \).

\( \implies P(z < Z < 0) = 0.8413 - 0.5 = 0.3413 \)
By symmetry, the area \( P(z < Z < 0) \) is equal to \( P(0 < Z < |z|) \).
So, we look up \( 0.3413 \) in the standard normal table to find the corresponding z-value, which is approximately \( 1.00 \).
Therefore, \( |z| = 1.00 \). Since we determined that z must be negative, \( z = -1.00 \). The normal distribution is symmetrical around its mean, so areas on the left side are mirror images of those on the right.
In simple words: Since a large part of the curve (more than half) is to the right of z, z must be a negative number. We find that negative z-score using a table.

🎯 Exam Tip: If the area to the right of z is greater than 0.5, then z must be a negative value. If the area to the right of z is less than 0.5, then z must be a positive value.

Question 25. If the area to the left of a value of z (z has a standard normal distribution) is 0.0793, what is the value of z?
(a) -1.41
(b) 1.41
(c) -2.25
(d) 2.25
Answer: (a) -1.41
We are given that the area to the left of z is \( P(Z < z) = 0.0793 \).
Since this probability is less than 0.5, the z-value must be negative, meaning it is in the left tail of the normal distribution.
We can write \( P(Z < z) = P(Z < 0) - P(z < Z < 0) \).
Since \( P(Z < 0) = 0.5 \), we have \( 0.0793 = 0.5 - P(z < Z < 0) \).

\( \implies P(z < Z < 0) = 0.5 - 0.0793 = 0.4207 \)
By symmetry, the area \( P(z < Z < 0) \) is equal to \( P(0 < Z < |z|) \).
So, we look up \( 0.4207 \) in the standard normal table to find the corresponding z-value, which is approximately \( 1.41 \).
Therefore, \( |z| = 1.41 \). Since z must be negative, \( z = -1.41 \). This calculation helps in understanding how to reverse-lookup z-scores from given probabilities.
In simple words: A very small area (less than half) is to the left of z, so z must be a negative number far to the left. We find this negative z-score using the normal distribution table.

🎯 Exam Tip: If the area to the left of z is less than 0.5, z must be negative. If the area to the left of z is greater than 0.5, z must be positive. This visual check is crucial for determining the sign of your z-score.

Question 26. If the area to the right of z (z has a standard normal distribution) is 0.8508, what is the value of z?
(a) -0.48
(b) 0.48
(c) -1.04
(d) 1.04
Answer: (c) -1.04
We are given that the area to the right of z is \( P(Z > z) = 0.8508 \).
Since this probability is greater than 0.5, the z-value must be negative, meaning it lies to the left of the mean (z=0).
We can express \( P(Z > z) \) as the sum of the area to the right of 0 and the area between z and 0: \( P(Z > z) = P(Z > 0) + P(z < Z < 0) \).

\( \implies 0.8508 = 0.5 + P(z < Z < 0) \)

\( \implies P(z < Z < 0) = 0.8508 - 0.5 = 0.3508 \)
By symmetry, the area \( P(z < Z < 0) \) is equal to \( P(0 < Z < |z|) \).
Looking up \( 0.3508 \) in the standard normal table, we find that the corresponding z-value is approximately \( 1.04 \).
Therefore, \( |z| = 1.04 \). Since z must be negative, \( z = -1.04 \). Understanding the implications of areas greater than 0.5 is key here.
In simple words: Because over half of the curve is to the right of z, z has to be a negative number. We find this specific negative z-score using the normal distribution table.

🎯 Exam Tip: When the given probability represents an area in a normal distribution, visually determining if the z-score should be positive or negative based on whether the area is greater or less than 0.5 is your first step.

Question 27. If P(Z > z) = 0.5832, what is the value of z (z has a standard normal distribution)?
(a) -0.48
(b) 0.48
(c) 1.04
(d) -0.21
Answer: (d) -0.21
We are given that the area to the right of z is \( P(Z > z) = 0.5832 \).
Since this probability is greater than 0.5, the z-value must be negative, meaning z is to the left of the mean (z=0).
We can write \( P(Z > z) = P(Z > 0) + P(z < Z < 0) \).
Since \( P(Z > 0) = 0.5 \), we have \( 0.5832 = 0.5 + P(z < Z < 0) \).

\( \implies P(z < Z < 0) = 0.5832 - 0.5 = 0.0832 \)
By symmetry, the area \( P(z < Z < 0) \) is equal to \( P(0 < Z < |z|) \).
Looking up \( 0.0832 \) in the standard normal table, the corresponding z-value is approximately \( 0.21 \).
Therefore, \( |z| = 0.21 \). Since z must be negative, \( z = -0.21 \). This calculation demonstrates how to find a specific negative z-score when given an area to its right.
In simple words: Since a bit more than half of the curve is to the right of z, z must be a negative number, but not too far from zero. We find this z-score using the normal distribution table.

🎯 Exam Tip: A small z-value (close to 0) means the probability is close to 0.5. A larger z-value (further from 0) indicates a probability closer to 0 or 1, depending on the tail.

Question 28. In a binomial distribution, the probability of success is twice as that of failure. Then out of 4 trials, the probability of no success is
(a) 16/81
(b) 1/16
(c) 2/27
(d) 1/81
Answer: (d) 1/81
Let \( p \) be the probability of success and \( q \) be the probability of failure.
We are given that the probability of success is twice that of failure, so \( p = 2q \).
Also, in any probability distribution, the sum of probabilities of success and failure is 1: \( p + q = 1 \).
Substitute \( p = 2q \) into the second equation:
\( 2q + q = 1 \)
\( 3q = 1 \)
\( q = \frac { 1 }{ 3 } \)
Now, find \( p \): \( p = 2q = 2 \times \frac { 1 }{ 3 } = \frac { 2 }{ 3 } \).
We have \( n = 4 \) trials and we need to find the probability of no success, which means \( x = 0 \).
The binomial probability formula is \( P(X = x) = nC_x p^x q^{n-x} \).
For \( x = 0 \):
\( P(X = 0) = 4C_0 \left( \frac { 2 }{ 3 } \right)^0 \left( \frac { 1 }{ 3 } \right)^{4-0} \)
We know that \( 4C_0 = 1 \) and any non-zero number raised to the power of 0 is 1, so \( \left( \frac { 2 }{ 3 } \right)^0 = 1 \).

\( \implies P(X = 0) = 1 \times 1 \times \left( \frac { 1 }{ 3 } \right)^4 \)

\( \implies P(X = 0) = \frac { 1^4 }{ 3^4 } = \frac { 1 }{ 81 } \). This result demonstrates how to apply the binomial probability formula after calculating individual probabilities of success and failure.
In simple words: We first find the exact chances for success and failure. Then, we use a special formula for "binomial distribution" to calculate the chance of having zero successes in four tries.

🎯 Exam Tip: Always define your 'p' and 'q' clearly in binomial distribution problems. Remember that 'p' is the probability of success and 'q' is the probability of failure, and their sum must always be 1.