Samacheer Kalvi Class 10 Maths Solutions Chapter 8 Statistics and Probability Exercise 8.5

Multiple Choice Questions.

Question 1. Which of the following is not a measure of dispersion?
(a) Range
(b) Standard deviation
(c) Arithmetic mean
(d) Variance
Answer: (c) Arithmetic mean
In simple words: Measures of dispersion tell us how spread out a data set is. Range, standard deviation, and variance all show how spread out the numbers are. Arithmetic mean, however, tells us the average value, not the spread.

🎯 Exam Tip: Remember that measures of dispersion quantify the variability or scatter of data, while measures of central tendency (like the mean) describe the center of the data.

Question 2. The range of the data 8, 8, 8, 8, 8... 8 is
(a) 0
(b) 1
(c) 8
(d) 3
Answer: (a) 0
Hint: The largest value in the data is 8. The smallest value in the data is 8.
Solution: Range \( R = L - S = 8 - 8 = 0 \)
In simple words: To find the range, you subtract the smallest number from the largest number in a set of data. Since all the numbers here are the same (8), there is no difference between the largest and smallest, so the range is 0.

🎯 Exam Tip: The range is the simplest measure of dispersion and is found by subtracting the minimum value from the maximum value in a dataset.

Question 3. The sum of all deviations of the data from its mean is ____
(a) always positive
(b) always negative
(c) zero
(d) None of the options
Answer: (c) zero
In simple words: If you take each number in a data set and subtract the average (mean) from it, then add all these differences together, the total will always be zero. This is a basic property of the arithmetic mean.

🎯 Exam Tip: Understanding this property is crucial for grasping concepts like variance and standard deviation, where squared deviations are used to avoid the sum being zero.

Question 4. The mean of 100 observations is 40 and their standard deviation is 3. The sum of squares of all deviations is
(a) 40000
(b) 160900
(c) 160000
(d) 30000
Answer: (b) 160900
Hint: The formula for standard deviation \( \sigma \) is \( \sigma = \sqrt { \frac { \Sigma x ^ { 2 } }{ n } - \left( \frac { \Sigma x }{ n } \right) ^ { 2 } } \).
Solution: We are given the mean \( \left( \frac { \Sigma x }{ n } \right) = 40 \) and standard deviation \( \sigma = 3 \).
We know the formula for standard deviation is \( \sigma = \sqrt { \frac { \Sigma x ^ { 2 } }{ n } - \left( \text {mean} \right) ^ { 2 } } \).
So, \( 3 = \sqrt { \frac { \Sigma x ^ { 2 } }{ 100 } - \left( 40 \right) ^ { 2 } } \)
Squaring both sides: \( 3 ^ { 2 } = \frac { \Sigma x ^ { 2 } }{ 100 } - 40 ^ { 2 } \)
\( 9 = \frac { \Sigma x ^ { 2 } }{ 100 } - 1600 \)
Now, we add 1600 to both sides: \( 9 + 1600 = \frac { \Sigma x ^ { 2 } }{ 100 } \)
\( 1609 = \frac { \Sigma x ^ { 2 } }{ 100 } \)
Finally, we multiply both sides by 100 to find \( \Sigma x ^ { 2 } \):
\( \implies \Sigma x ^ { 2 } = 1609 \times 100 \)
\( \implies \Sigma x ^ { 2 } = 160900 \)
The sum of squares of all deviations is \( 160900 \).
In simple words: We used the formula for standard deviation, which connects the sum of squares, the number of observations, and the mean. By plugging in the given values for mean and standard deviation, we can solve for the sum of squares. This shows how spread out the numbers are from their average.

🎯 Exam Tip: Always remember the formula for standard deviation and variance, as they are often rearranged to find other components like the sum of squares.

Question 5. Variance of first 20 natural numbers is ____
(a) 32.25
(b) 44.25
(c) 33.25
(d) 30
Answer: (c) 33.25
Hint: The formula for the variance of the first \( n \) natural numbers is \( \frac { n ^ { 2 } - 1 }{ 12 } \).
Solution: Here, \( n = 20 \).
Variance \( = \frac { n ^ { 2 } - 1 }{ 12 } \)
\( = \frac { 20 ^ { 2 } - 1 }{ 12 } \)
\( = \frac { 400 - 1 }{ 12 } \)
\( = \frac { 399 }{ 12 } \)
\( = 33.25 \)
In simple words: The variance for the first 'n' natural numbers can be found using a specific formula. We just need to put the number 20 into this formula to get the answer. This formula helps us quickly find how spread out the numbers are.

🎯 Exam Tip: Memorize direct formulas for common sequences like the first 'n' natural numbers, as they save significant time in calculations.

Question 6. The standard deviation of a data Is 3. If each value is multiplied by 5 then the new variance is
(a) 3
(b) 15
(c) 5
(d) 225
Answer: (d) 225
Solution: Given, standard deviation \( \sigma = 3 \).
If each value in the data is multiplied by 5, then the new standard deviation also gets multiplied by 5.
New standard deviation \( = 5 \times \sigma = 5 \times 3 = 15 \).
Variance is the square of the standard deviation.
So, new variance \( = \left( \text {new standard deviation} \right) ^ { 2 } \)
\( = 15 ^ { 2 } = 225 \).
In simple words: When every number in a data set is multiplied by a constant, the standard deviation also gets multiplied by that same constant. Since variance is simply the square of the standard deviation, we square the new standard deviation to find the new variance. This makes the spread much larger.

🎯 Exam Tip: Multiplying each data value by a constant 'k' multiplies the standard deviation by 'k' and the variance by 'k²'. Adding a constant to each data value does not change the standard deviation or variance.

Question 7. If the standard deviation of x, y, z is p then the standard deviation of 3x + 5, 3y + 5, 3z + 5 is ____
(a) 3p + 5
(b) 3p
(c) p + 5
(d) 9p + 15
Answer: (b) 3p
Hint: When each value in a dataset is multiplied by a constant, the standard deviation is also multiplied by that constant. When a constant is added or subtracted from each value, the standard deviation remains unchanged. In this case, each value is multiplied by 3 and then 5 is added.
Solution: Given, the standard deviation of \( x, y, z \) is \( p \).
When each value is multiplied by 3, the standard deviation becomes \( 3p \).
When 5 is added to each value, the standard deviation does not change.
Therefore, the standard deviation of \( 3x + 5, 3y + 5, 3z + 5 \) is \( 3p \).
In simple words: If you multiply all numbers in a data set by 3, the standard deviation also becomes 3 times bigger. But if you then add 5 to all those numbers, the standard deviation does not change at all. So, the standard deviation will be \( 3p \).

🎯 Exam Tip: Remember that adding or subtracting a constant shifts the data without changing its spread, while multiplying or dividing scales the spread proportionally.

Question 8. If the mean and coefficient of variation of a data are 4 and 87.5% then the standard deviation is
(a) 3.5
(b) 3
(c) 4.5
(d) 2.5
Answer: (a) 3.5
Hint: The formula for coefficient of variation (C.V) is \( \frac { \sigma }{ \bar { x } } \times 100 \), where \( \sigma \) is the standard deviation and \( \bar { x } \) is the mean.
Solution: Given, mean \( \bar { x } = 4 \)
Coefficient of variation \( \text {C.V} = 87.5\% \)
Using the formula for coefficient of variation:
\( \text {C.V} = \frac { \sigma }{ \bar { x } } \times 100 \)
We plug in the given values:
\( 87.5 = \frac { \sigma }{ 4 } \times 100 \)
To find \( \sigma \), we first divide 87.5 by 100:
\( \frac { 87.5 }{ 100 } = \frac { \sigma }{ 4 } \)
\( 0.875 = \frac { \sigma }{ 4 } \)
Now, we multiply both sides by 4:
\( \implies \sigma = 0.875 \times 4 \)
\( \implies \sigma = 3.5 \)
The standard deviation is 3.5.
In simple words: The coefficient of variation shows how much the standard deviation is compared to the mean, usually as a percentage. We are given the mean and the coefficient of variation. By using the formula, we can work backward to find the standard deviation, which tells us the spread of the data.

🎯 Exam Tip: Always remember the Coefficient of Variation (C.V) formula, which is a useful relative measure of dispersion, especially when comparing datasets with different means.

Question 9. Which of the following is incorrect?
(a) P(A) > 1
(b) \( 0 \leq P(A) \leq 1 \)
(c) \( P(\Phi) = 0 \)
(d) \( P(A) + P(\overline{A}) = 1 \)
Answer: (a) P(A) > 1
Hint: The probability of any event must be between 0 and 1, inclusive.
Solution: We know that the probability of any event A, denoted as P(A), must always be a value between 0 and 1. It cannot be less than 0 (negative) or greater than 1.
Option (b) \( 0 \leq P(A) \leq 1 \) is correct.
Option (c) \( P(\Phi) = 0 \) is correct, as the probability of an impossible event (empty set \( \Phi \)) is 0.
Option (d) \( P(A) + P(\overline{A}) = 1 \) is correct, as the sum of the probability of an event and its complement is always 1.
Therefore, P(A) > 1 is incorrect.
In simple words: Probability is always a number from 0 to 1. If an event is impossible, its probability is 0. If an event is certain, its probability is 1. The chance of something happening plus the chance of it not happening always adds up to 1. So, a probability cannot be more than 1.

🎯 Exam Tip: A fundamental rule of probability is that the likelihood of any event must fall within the range of 0 (impossible) to 1 (certainty).

Question 10. The probability a red marble selected at random from a jar containing p red, q blue and r green marbles is ____
(a) \( \frac { q }{ p+q+r } \)
(b) \( \frac { p }{ p+q+r } \)
(c) \( \frac { p+q }{ p+q+r } \)
(d) \( \frac { p+r }{ p+q+r } \)
Answer: (b) \( \frac { p }{ p+q+r } \)
Hint: Probability is calculated as the number of favorable outcomes divided by the total number of possible outcomes.
Solution: Total number of marbles in the jar (sample space) \( = p \text { (red) } + q \text { (blue) } + r \text { (green) } = p+q+r \).
Let A be the event of getting a red marble.
The number of red marbles (favorable outcomes) \( n(A) = p \).
The probability of getting a red marble \( P(A) = \frac { \text {Number of red marbles} }{ \text {Total number of marbles} } \)
\( \implies P(A) = \frac { p }{ p+q+r } \)
In simple words: To find the chance of picking a red marble, you just count how many red marbles there are and divide that by the total number of all marbles in the jar. This gives you the probability.

🎯 Exam Tip: Clearly identify the number of favorable outcomes and the total number of possible outcomes before applying the probability formula.

Question 11. A page is selected at random from a book. The probability that the digit at units place of the page number chosen is less than 7 is ____
(a) \( \frac { 3 }{ 10 } \)
(b) \( \frac { 7 }{ 10 } \)
(c) \( \frac { 3 }{ 9 } \)
(d) \( \frac { 7 }{ 9 } \)
Answer: (b) \( \frac { 7 }{ 10 } \)
Hint: The digits that can appear in the units place are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. There are 10 possible digits.
Solution: In any page number, the digit at the units place can be any digit from 0 to 9.
So, the total number of possible outcomes for the units digit \( n(S) = 10 \) (i.e., {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}).
Let A be the event that the digit at the units place is less than 7.
The digits less than 7 are 0, 1, 2, 3, 4, 5, 6.
So, the number of favorable outcomes \( n(A) = 7 \).
The probability \( P(A) = \frac { n(A) }{ n(S) } = \frac { 7 }{ 10 } \).
In simple words: The last digit of a page number can be any number from 0 to 9. We want to find the chance that this last digit is smaller than 7. We count how many digits are smaller than 7 (0 through 6) and divide that by the total possible digits (0 through 9).

🎯 Exam Tip: When dealing with digits, ensure you include 0 in your count of possible outcomes if it's a valid option.

Question 12. The probability of getting a job for a person is \( \frac { x }{ 3 } \). If the probability of not getting the job is \( \frac { 2 }{ 3 } \) then the value of x is.
(a) 2
(b) 1
(c) 3
(d) 1.5
Answer: (b) 1
Hint: The sum of the probability of an event happening and the probability of it not happening is always 1.
Solution: Let P(Job) be the probability of getting a job and P(Not Job) be the probability of not getting a job.
We are given \( P(\text {Job}) = \frac { x }{ 3 } \) and \( P(\text {Not Job}) = \frac { 2 }{ 3 } \).
We know that \( P(\text {Job}) + P(\text {Not Job}) = 1 \).
So, \( \frac { x }{ 3 } + \frac { 2 }{ 3 } = 1 \)
Combine the fractions on the left side: \( \frac { x + 2 }{ 3 } = 1 \)
Multiply both sides by 3: \( x + 2 = 3 \)
Subtract 2 from both sides: \( \implies x = 3 - 2 \)
\( \implies x = 1 \)
The value of \( x \) is 1.
In simple words: The chance of getting a job and the chance of not getting a job must add up to 1 (or 100%). We know the chance of not getting the job is \( \frac { 2 }{ 3 } \). So, the chance of getting the job must be \( 1 - \frac { 2 }{ 3 } = \frac { 1 }{ 3 } \). This means \( x \) must be 1.

🎯 Exam Tip: Always use the complement rule \( P(A) + P(\overline{A}) = 1 \) to find missing probabilities when an event has two mutually exclusive outcomes.

Question 13. Kamalam went to play a lucky draw contest. 135 tickets of the lucky draw were sold. If the probability of Kamalam winning is \( \frac { 1 }{ 9 } \), then the number of tickets bought by Kamalam is ____
(a) 5
(b) 10
(c) 15
(d) 20
Answer: (c) 15
Hint: Probability is the ratio of favorable outcomes to the total possible outcomes.
Solution: Total number of tickets sold \( n(S) = 135 \).
Let A be the event that Kamalam wins. Let \( n(A) \) be the number of tickets Kamalam bought.
The probability of Kamalam winning is given as \( P(A) = \frac { 1 }{ 9 } \).
We use the probability formula: \( P(A) = \frac { n(A) }{ n(S) } \)
\( \frac { 1 }{ 9 } = \frac { n(A) }{ 135 } \)
To find \( n(A) \), multiply both sides by 135:
\( \implies n(A) = \frac { 1 }{ 9 } \times 135 \)
\( \implies n(A) = 15 \)
Kamalam bought 15 tickets.
In simple words: The chance of winning a lucky draw is found by dividing the number of tickets you bought by the total number of tickets sold. We know the total tickets and the winning chance, so we can multiply these to find out how many tickets Kamalam bought.

🎯 Exam Tip: Be careful with units and ensure that the "favorable outcomes" (tickets bought by Kamalam) are consistent with the "total outcomes" (total tickets sold).

Question 14. If a letter is chosen at random from the English alphabets {a, b, ..., z}, then the probability that the letter chosen precedes x.
(a) \( \frac { 12 }{ 13 } \)
(b) \( \frac { 1 }{ 13 } \)
(c) \( \frac { 23 }{ 26 } \)
(d) \( \frac { 3 }{ 26 } \)
Answer: (c) \( \frac { 23 }{ 26 } \)
Hint: There are 26 letters in the English alphabet.
Solution: Total number of letters in the English alphabet \( n(S) = 26 \).
Let A denote the event that the letter chosen precedes 'x'.
The letters that precede 'x' are a, b, c, ..., w. (This means all letters from 'a' up to, but not including, 'x').
The position of 'x' in the alphabet is the 24th letter.
So, the number of letters before 'x' is \( 24 - 1 = 23 \).
Therefore, the number of favorable outcomes \( n(A) = 23 \).
The probability \( P(A) = \frac { n(A) }{ n(S) } = \frac { 23 }{ 26 } \).
In simple words: The English alphabet has 26 letters. We want to pick a letter that comes before 'x'. There are 23 letters from 'a' to 'w'. So, the chance of picking one of these 23 letters out of the total 26 letters is \( \frac { 23 }{ 26 } \).

🎯 Exam Tip: Always count carefully, especially when determining the number of elements in a set based on alphabetical order, ensuring you include or exclude boundaries correctly.

Question 15. A purse contains 10 notes of Rs. 2000, 15 notes of Rs. 500, and 25 notes of Rs. 200. One note is drawn at random. What is the probability that the note is either a Rs. 500 note or Rs. 200 note?
(a) \( \frac { 1 }{ 5 } \)
(b) \( \frac { 2 }{ 10 } \)
(c) \( \frac { 3 }{ 10 } \)
(d) \( \frac { 4 }{ 5 } \)
Answer: (d) \( \frac { 4 }{ 5 } \)
Hint: For mutually exclusive events, the probability of either one happening is the sum of their individual probabilities.
Solution: Number of Rs. 2000 notes = 10
Number of Rs. 500 notes = 15
Number of Rs. 200 notes = 25
Total number of notes in the purse (sample space) \( n(S) = 10 + 15 + 25 = 50 \).
Let A be the event of getting a Rs. 500 note.
Number of Rs. 500 notes \( n(A) = 15 \).
Probability of getting a Rs. 500 note \( P(A) = \frac { n(A) }{ n(S) } = \frac { 15 }{ 50 } \).
Let B be the event of getting a Rs. 200 note.
Number of Rs. 200 notes \( n(B) = 25 \).
Probability of getting a Rs. 200 note \( P(B) = \frac { n(B) }{ n(S) } = \frac { 25 }{ 50 } \).
Since choosing a Rs. 500 note and choosing a Rs. 200 note are mutually exclusive events (they cannot happen at the same time), the probability that the note is either a Rs. 500 note or a Rs. 200 note is the sum of their individual probabilities:
\( P(A \text { or } B) = P(A) + P(B) \)
\( = \frac { 15 }{ 50 } + \frac { 25 }{ 50 } \)
\( = \frac { 15 + 25 }{ 50 } \)
\( = \frac { 40 }{ 50 } \)
\( = \frac { 4 }{ 5 } \)
The probability is \( \frac { 4 }{ 5 } \).
In simple words: First, count all the notes to find the total. Then, find the chance of picking a Rs. 500 note and the chance of picking a Rs. 200 note separately. Because you can't pick both at once, just add these two chances together to get the probability of picking either one.

🎯 Exam Tip: For "either/or" probability questions, determine if the events are mutually exclusive. If they are, simply add their individual probabilities.