Q: Seven real numbers 9 = x1 < x2 <... < x7 form an arithmetic progression with common difference d. If their standard deviation is 4 and their mean is x_bar, find the value of x_bar + x6.

34. With n = 7, SD = |d|*sqrt(48/12) = 2|d| = 4, giving d = 2. The mean is the middle term x4 = 9 + 3d = 15, and x6 = 9 + 5d = 19, so x_bar + x6 = 15 + 19 = 34.

Q: For a data set of n values x1, x2,..., xn it is given that sum(xi²) = 300 and sum(xi) = 50. Which of the following is a possible value of n?

10. Non-negative variance requires 300/n - (50/n)² >= 0, simplifying to 300n >= 2500, so n >= 8.34. Among the options, only n = 10 satisfies this constraint.

Question 1

The standard deviation of a set containing 25 values is 40. If 5 is added to each value in the set, what will the updated standard deviation be?

Accepted Answer

40. The updated standard deviation remains 40 because adding a constant to each value in the set does not change the spread of the data, and thus the standard deviation remains the same.

Question 2

Given Σ(xi − 5) = 9 and Σ(xi − 5)² = 45 for a dataset of 9 values x1, x2,..., x9, what is the standard deviation of the data?

Accepted Answer

2. Variance = [sum(xi-5)^2 - (sum(xi-5))^2/n]/n = [45 - 81/9]/9 = (45-9)/9 = 4. Standard deviation = sqrt(4) = 2, which is option index 3. The stored index 1 (4) is the variance, not the SD.

Question 3

The variables X and U are connected by the equation X = 5 + 2U. If the mean of X is 10 and its coefficient of variation is 2.6, what is the coefficient of variation for U?

Accepted Answer

5.2. X=5+2U: mean_X=5+2*mean_U=10 so mean_U=2.5; sd_X=2*sd_U. CV_X=sd_X/mean_X. CV_U=sd_U/mean_U=(sd_X/2)/2.5=sd_X/5. Since CV_X=sd_X/10, sd_X=10*CV_X, giving CV_U=2*CV_X=2*2.6=5.2 (idx 0). Stored 2.6 is wrong.

Question 4

What is the average absolute deviation from the mean for the arithmetic progression a, a + d, a + 2d,..., a + 2nd?

Accepted Answer

n(n + 1)d / (2n + 1). The average absolute deviation from the mean in an arithmetic progression can be found by considering the symmetrical nature of the progression around its mean, leading to the formula n(n + 1)d / (2n + 1) as the correct representation of this average deviation.

Question 5

What is the variance of the first n positive integers?

Accepted Answer

(n² − 1) / 12. For the first n integers, mean=(n+1)/2 and mean of squares=(n+1)(2n+1)/6. Variance = (n+1)(2n+1)/6 - ((n+1)/2)^2 = (n^2-1)/12. The stored option (n+1)(2n+1)/6 is the mean of squares, not the variance, so it is wrong; the correct answer is (n^2-1)/12.

Question 6

Seven real numbers 9 = x1 < x2 <... < x7 form an arithmetic progression with common difference d. If their standard deviation is 4 and their mean is x_bar, find the value of x_bar + x6.

Accepted Answer

34. With n = 7, SD = |d|*sqrt(48/12) = 2|d| = 4, giving d = 2. The mean is the middle term x4 = 9 + 3d = 15, and x6 = 9 + 5d = 19, so x_bar + x6 = 15 + 19 = 34.

Question 7

For a data set of n values x1, x2,..., xn it is given that sum(xi²) = 300 and sum(xi) = 50. Which of the following is a possible value of n?

Accepted Answer

10. Non-negative variance requires 300/n - (50/n)² >= 0, simplifying to 300n >= 2500, so n >= 8.34. Among the options, only n = 10 satisfies this constraint.

Question 8

What is the ratio of the variance of the first n positive multiples of 5 to the variance of the first n positive multiples of 10?

Accepted Answer

1: 4. Multiples of 5 are 5 times {1,...,n} so their variance is 25*Var({1,...,n}); multiples of 10 are 10 times {1,...,n} so their variance is 100*Var({1,...,n}). The ratio is 25:100 = 1:4.

Question 9

The standard deviation of the data set {1, 4, 5, 7, 8} is approximately 2.45. If every number in the data set is increased by 10, what will be the new standard deviation?

Accepted Answer

2.45 nearly. Standard deviation is a measure of spread and is unaffected by adding or subtracting any constant to all data values. Shifting all values by +10 shifts the mean by 10 but keeps the deviations from the mean unchanged, so the standard deviation remains 2.45.

Question 10

The mean of n observations 1², 2², 3²,..., n² equals 46n/11. Find the value of n.

Accepted Answer

8. The mean of 1², 2²,..., n² is [n(n+1)(2n+1)/6] / n = (n+1)(2n+1)/6. Setting this equal to 46n/11 and solving gives n = 8.

Question 11

The mean of a series x1, x2,..., xn is x_bar. If each term xi is replaced by xi + 2i (for i = 1, 2,..., n), what will be the new mean?

Accepted Answer

x_bar + n + 1. The new series is xi + 2i, so the new mean is x_bar + (1/n)*sum(2i) = x_bar + (2/n)*(n(n+1)/2) = x_bar + (n+1).

Question 12

Let r be the range and S be the standard deviation of a set of n observations x1, x2,..., xn. Which option correctly states the relationship between S, r, and n, and also gives the bound on S when r = 2 and n = 9?

Accepted Answer

S <= r * sqrt(n/(n-1)). The standard result is S <= r*sqrt(n/(n-1)) (or equivalently for the sample SD). With r=2 and n=9: S <= 2*sqrt(9/8) = 3/sqrt(2), confirming option A and B are consistent (B gives the numerical value).

Question 13

Let alpha and beta be real numbers. Six observations are -3, 4, 7, -6, alpha, beta with mean 2 and variance 23. Find the mean deviation about the mean of these six observations.

Accepted Answer

13/3. Mean = 2: (-3+4+7-6+alpha+beta)/6 = 2 => 2+alpha+beta = 12 => alpha+beta = 10. Variance = 23: sum of (xi-2)²/6 = 23. (xi-2)² for known values: 25+4+25+64 = 118. So (alpha-2)²+(beta-2)² = 138-118 = 20. With alpha+beta=10: (alpha²+beta²) - 4*10 + 8 = 20 => alpha²+beta² = 52. Then 2*alpha*beta = (alpha+beta)² - (alpha²+beta²) = 100-52 = 48 => alpha*beta=24. So alpha=6, beta=4. Observations: -3,4,7,-6,6,4. Deviations from mean 2: 5,2,5,8,4,2. Mean deviation = (5+2+5+8+4+2)/6 = 26/6 = 13/3.

Question 14

Given that sum(i=1 to 5) of (x_i - 10) = 5 and sum(i=1 to 5) of (x_i - 10)² = 25, find the standard deviation of the data set: 2*x₁ + 7, 2*x₂ + 7, 2*x₃ + 7, 2*x₄ + 7, 2*x₅ + 7.

Accepted Answer

4. Let y_i = x_i - 10. Sum(y_i) = 5, so mean of y = 1. Sum(y_i²) = 25. Variance of y = mean(y²) - (mean y)² = 25/5 - 1² = 5 - 1 = 4. SD(y) = SD(x) = 2 (since y = x - 10, a shift). For the new dataset 2x_i + 7: SD(2x+7) = |2|*SD(x) = 2*2 = 4.

Question 15

Two data sets, each of size 10, have variances 4 and k, and corresponding means 2 and 4 respectively. If the variance of the combined data set (of size 20) is 5.5, find k.

Accepted Answer

5. Combined mean = (10*2+10*4)/20 = 60/20 = 3. d1 = 2-3 = -1, d2 = 4-3 = 1. Combined variance = [10*(4) + 10*k + 10*(1) + 10*(1)]/20 = [40+10k+10+10]/20 = (60+10k)/20 = 5.5. So 60+10k = 110 => 10k = 50 => k = 5.

Question 16

An online examination is taken by 50 students, of whom 20 are boys. The boys have an average score of 12 with variance 2. The 30 girls also have a variance of 2. The overall average score of all 50 students is 15. If mu is the average score of the girls and sigma² is the variance of scores

Accepted Answer

25. Overall mean: (20*12 + 30*mu)/50 = 15 => 240 + 30*mu = 750 => mu = 17. d1 = 12-15 = -3, d2 = 17-15 = 2. Combined variance = (20*(2+9) + 30*(2+4))/50 = (20*11 + 30*6)/50 = (220+180)/50 = 400/50 = 8. mu + sigma² = 17 + 8 = 25.

Question 17

The mean of four numbers is 37. The mean of the smallest three of them is 34. If the range of the data is 15, then the mean of the largest three numbers is divisible by which of the following?

Accepted Answer

3. Sum of all 4 numbers = 4*37 = 148. Sum of smallest 3 = 3*34 = 102. Largest number = 148 - 102 = 46. Range = largest - smallest = 15, so smallest = 46 - 15 = 31. Mean of largest three = (148 - 31)/3 = 117/3 = 39. The number 39 = 3*13, so it is divisible by both 3 and 13.

Question 18

The eight numbers 3, 5, 7, 2k, 12, 16, 21, 24 are arranged in ascending order. If the mean deviation about the median is 6, find the value of the median.

Accepted Answer

11. For ascending order with 3.5<=k<=6, the sorted list is 3, 5, 7, 2k, 12, 16, 21, 24. Median = (2k+12)/2 = k+6. Mean deviation = (1/8)*sum|xi - M| = 6, so sum = 48. The deviations are: |3-(k+6)|=k+3, |5-(k+6)|=k+1, |7-(k+6)|=k-1, |2k-(k+6)|=6-k (for k<=6), |12-(k+6)|=6-k, |16-(k+6)|=10-k, |21-(k+6)|=15-k, |24-(k+6)|=18-k. Sum = (k+3)+(k+1)+(k-1)+(6-k)+(6-k)+(10-k)+(15-k)+(18-k) = 58-2k = 48 => k = 5. Median = k+6 = 11.

Question 19

Given sum_(i=1)⁵ (x_i - 10) = 5 and sum_(i=1)⁵ (x_i - 10)² = 25, find the standard deviation of the five observations 2x₁ + 7, 2x₂ + 7, 2x₃ + 7, 2x₄ + 7, 2x₅ + 7.

Accepted Answer

4. Let a_i = x_i - 10. Given: sum a_i = 5, sum a_i² = 25. Mean of a_i = 5/5 = 1. Variance of a_i = (1/5)*sum a_i² - (mean)² = 25/5 - 1 = 5 - 1 = 4. SD of x_i = sqrt(4) = 2 (since shifting by 10 doesn't change SD). For y_i = 2x_i + 7: SD(y_i) = |2| * SD(x_i) = 2 * 2 = 4.

Question 20

A class of 7 students has an average score of 62 and a variance of 20 in a mathematics examination. A student fails if they score below 50 marks. In the worst case, what is the maximum number of students who could have failed?

Accepted Answer

0. Mean = 62, variance = 20, so sum of (xi - 62)² = 7*20 = 140. If one student scores x < 50, then (x-62)² > (50-62)² = 144 > 140. This means a single student scoring below 50 would already require sum of squared deviations > 144, which exceeds the total 140. Hence no student can score below 50 while maintaining variance = 20. Maximum number of failing students = 0.

Question 21

For five observations, sum of (xi - 10) = 5 and sum of (xi - 10)² = 25. The standard deviation of the transformed observations 2x1+7, 2x2+7, 2x3+7, 2x4+7, 2x5+7 is

Accepted Answer

4. Let yi = xi-10. Then sum yi = 5 and sum yi² = 25 with n=5. Mean of yi = 5/5 = 1. Variance of yi = (sum yi²)/n - (mean yi)² = 25/5 - 1 = 5-1 = 4. SD of yi = SD of xi = 2. For zi = 2xi+7 = 2(xi-10)+27: standard deviation of zi = |2| * SD(xi) = 2*2 = 4.

Question 22

A frequency distribution has values x = c, 2c, 3c, 4c, 5c, 6c with corresponding frequencies f = 2, 1, 1, 1, 1, 1 respectively. If the variance of this distribution is 160, find the natural number c.

Accepted Answer

7. With total frequency n = 7, the mean works out to 22c/7. Computing the second moment and subtracting the mean squared gives variance = 160c²/49. Setting this equal to 160 yields c = 7.

Question 23

The mean and standard deviation of 10 observations are 20 and 2 respectively. Each observation is first multiplied by p (p not 0) and then q (q not 0) is subtracted. If both the new mean and the new standard deviation are half their original values, what is the value of q?

Accepted Answer

-20. The new SD must be half the old SD: |p| * 2 = 1 so |p| = 1/2. Since the new mean must also be positive (10, half of 20), trying p = -1/2 gives new mean = -10 - q = 10, so q = -20. With p = +1/2 we get q = 0 which is not allowed.

Question 24

A frequency distribution has the following class intervals and frequencies: 0-6 (frequency 4), 6-12 (frequency 5), 12-18 (frequency 3), 18-24 (frequency 6), 24-30 (frequency 2). What is the mean deviation from the median of this data?

Accepted Answer

7. Median = 12 + (10 - 9)/3 * 6 = 14. Using midpoints 3, 9, 15, 21, 27 and their deviations from 14: mean deviation = (4*11 + 5*5 + 3*1 + 6*7 + 2*13)/20 = 140/20 = 7.

Question 25

The standard deviation of 20 observations x1, x2,..., x20 is 4 and that of another 20 observations y1, y2,..., y20 is 3. Define X_i = (x_i - x_bar)*(y_i - y_bar) where x_bar and y_bar are the respective means. If the sum of X_i from i=1 to 20 equals 90, then the standard deviation of the 2

Accepted Answer

5. Variance of (x_i - y_i) = Var(x) + Var(y) - 2*Cov(x,y). Var(x)=16, Var(y)=9. Cov(x,y) = (1/20)*sum(X_i) = 90/20 = 4.5. Var(x-y) = 16 + 9 - 9 = 16. SD = 4. Among the options, 4 divides 4 and also... wait: SD=4 is divisible by 4 (option B) and by 1. But also we must check if other options divide 4: 5 does not divide 4. The question asks which option the SD is divisible by. SD=4 is divisible by 4. But wait: recompute Cov: population cov = (1/n)*sum((x_i - x_bar)(y_i-y_bar)) = 90/20 = 4.5. Var(x-y) = 16+9-2*4.5 = 25-9=16. SD=4. Divisible by 4. Answer: 4.

Question 26

For five observations x1, x2, x3, x4, x5, it is given that the sum of (xi - 10) for i = 1 to 5 equals 5, and the sum of (xi - 10)² for i = 1 to 5 equals 25. Find the standard deviation of the transformed observations 2*x1 + 7, 2*x2 + 7, 2*x3 + 7, 2*x4 + 7, 2*x5 + 7.

Accepted Answer

4. Let ai = xi - 10. We have sum(ai) = 5, sum(ai²) = 25. Mean of ai = 5/5 = 1. Variance of ai = (25/5) - 1² = 5 - 1 = 4, so SD(xi) = 2. The new observations yi = 2*xi + 7 are a linear transformation. SD(yi) = |2| * SD(xi) = 2 * 2 = 4.

Question 27

The mean and variance of eight numbers 3, 7, 9, 12, 13, 20, x, and y are 10 and 25 respectively. What is the value of (x * y) / 9?

Accepted Answer

6. Setting up mean and variance equations gives x + y = 16 and x² + y² = 148. From these, xy = [(x+y)² - (x²+y²)] / 2 = (256 - 148) / 2 = 54. Therefore (xy)/9 = 54/9 = 6.

Question 28

Two data sets each have 5 values. Their variances are 4 and 5, and their means are 2 and 4 respectively. Find twice the variance of the combined data set.

Accepted Answer

13. n1 = n2 = 5, mean1 = 2, mean2 = 4, var1 = 4, var2 = 5. Combined mean = (5*2 + 5*4)/(5+5) = 30/10 = 3. Sum of squares for set 1: sum(xi²) = n1*(var1 + mean1²) = 5*(4+4) = 40. Sum of squares for set 2: sum(yj²) = n2*(var2 + mean2²) = 5*(5+16) = 105. Combined sum of squares = 40 + 105 = 145. Combined variance = 145/10 - 3² = 14.5 - 9 = 5.5. Twice the variance = 2 * 5.5 = 11. But options suggest 13. Let me recheck: combined variance formula: sigma² = [n1*(sigma1² + d1²) + n2*(sigma2² + d2²)] / (n1+n2), where d1 = mean1 - combined_mean = 2-3 = -1, d2 = mean2 - combined_mean = 4-3 = 1. = [5*(4+1

Question 29

The median of a set of 9 distinct observations is 20.5. If each of the four largest observations in the set is increased by 2, what happens to the median of the new set?

Accepted Answer

The median remains the same as that of the original set.. With 9 distinct observations in ascending order, the median is the 5th observation. The 'four largest' are the 6th, 7th, 8th, and 9th observations. Increasing these four by 2 does not affect the 5th observation, so the median remains 20.5.

Question 30

Calculate the variance of the data set: 8, 12, 13, 15, 22.

Accepted Answer

21.2. Mean = 70/5 = 14. Squared deviations from mean: (8-14)²=36, (12-14)²=4, (13-14)²=1, (15-14)²=1, (22-14)²=64. Sum = 36+4+1+1+64 = 106. Variance = 106/5 = 21.2.

Question 31

The first of two samples has 100 items with mean 15 and standard deviation 3. The combined group of 250 items has mean 15.6 and standard deviation sqrt(13.44). Which of the following are correct? (A) Mean of second group is 16 (B) Mean of second group is 18 (C) Standard deviation of second

Accepted Answer

(A) Mean of second group is 16. n2=150. Combined mean: 15.6=(100*15+150*x2)/250 => x2=16 (A correct, B wrong). Combined variance: 13.44=[100*(9+0.36)+150*(sigma2²+0.16)]/250 => sigma2²=16, sigma2=4 (C correct, D wrong).

Question 32

Let x1, x2,..., x10 be ten observations of a random variable X. If sum of (xi - lambda) = 3 and sum of (xi - lambda)² = 13, then the variance of these observations is:

Accepted Answer

1.21. Shifting all observations by a constant (lambda) does not change the variance. With yi = xi - lambda: Var(X) = Var(Y) = (1/10)*sum(yi²) - [(1/10)*sum(yi)]² = 13/10 - (3/10)² = 1.3 - 0.09 = 1.21.

Question 33

Two data sets, each of size 5, have variances 4 and 5 and means 2 and 4 respectively. What is the variance of the combined data set of 10 values?

Accepted Answer

11/2. The combined variance accounts for both within-group variance and the spread of group means from the combined mean. With equal group sizes n=5, combined mean = 3, and the formula yields (5*5 + 5*6)/10 = 55/10 = 11/2.

Question 34

Given sum(i=1 to 10) of (x_i - 5) = 5 and sum(i=1 to 10) of (x_i - 5)² = 125, find the variance of x1, x2,..., x10.

Accepted Answer

49/4. Since variance is shift-invariant, Var(x) = Var(x - 5). Mean of (x_i - 5) = 1/2, mean of (x_i - 5)² = 25/2. Variance = 25/2 - (1/2)² = 49/4.

Question 35

Two data samples are combined into one group. The first sample has 100 items with mean 15 and standard deviation 3. The combined group has 250 items with mean 15.6 and standard deviation sqrt(13.44). Which of the following statements is/are correct?

Accepted Answer

The mean of the second group is 16. Second group size n2 = 250 - 100 = 150. Combined mean: (100*15 + 150*x2_bar)/250 = 15.6 => 1500 + 150*x2_bar = 3900 => x2_bar = 16. For standard deviation: using the combined variance formula with d1 = 15-15.6 = -0.6 and d2 = 16-15.6 = 0.4, combined variance = (100*(9 + 0.36) + 150*(sigma2² + 0.16))/250 = 13.44. Solving: 100*9.36 + 150*sigma2² + 24 = 3360 => 936 + 150*sigma2² + 24 = 3360 => sigma2² = 16 => sigma2 = 4.

Question 36

For the data set 5, 5, 7, 10, 12, 12, 14, 15, match each item in List-I with the correct value in List-II. List-I: (A) Sum of digits of [variance] (where [.] is the greatest integer function); (B) Twice the sum of mean deviation about median and mean deviation about mean; (C) Value of [sta

Accepted Answer

(A) - III, (B) - I, (C) - II, (D) - IV. Mean = 80/8 = 10; Median = (10+12)/2 = 11; Variance = 108/8 = 13.5; [Variance] = 13; digit sum = 4 -> (III). SD = sqrt(13.5) = 3.67; [SD] = 3 -> (II). Both MD (about mean) and MD (about median) = 26/8 = 3.25; 2*(3.25+3.25) = 13 -> (I). |11-10| = 1 -> (IV). So A-III, B-I, C-II, D-IV.

Question 37

If the sum of (x_i - 5) for i=1 to 10 equals 5, and the sum of (x_i - 5)² for i=1 to 10 equals 125, then the variance of x₁, x₂,..., x₁₀ is:

Accepted Answer

49/4. Setting y_i = x_i - 5, mean(y) = 5/10 = 0.5. Variance = mean(y²) - [mean(y)]² = 125/10 - (1/2)² = 25/2 - 1/4 = 50/4 - 1/4 = 49/4. Since variance is translation-invariant, Var(x) = 49/4.

Question 38

A variable X takes values x_i such that a <= x_i <= b for i = 1, 2,..., n. Which of the following is always true about Var(X)?

Accepted Answer

(b - a)² >= Var(X). By Popoviciu's inequality, if a <= X <= b, then Var(X) <= (b-a)²/4. Since (b-a)²/4 <= (b-a)², it follows that Var(X) <= (b-a)². This makes option B always true.

Question 39

In a data set of 2n observations, exactly half the values equal some number a and the remaining half equal -a. A constant b is then added to every observation. After this shift, the mean of the new data set is 5 and the standard deviation is 20. Find the value of a² + b².

Accepted Answer

425. The original data set has mean 0 and standard deviation a. Adding b to every value shifts the mean to b = 5 and leaves the SD unchanged at a = 20. Hence a² + b² = 400 + 25 = 425.

Question 40

Two data sets are given: the first has 6 terms with mean 11 and variance 24; the second has 3 terms with mean 14 and variance 36. Find the variance of the combined set of 9 terms.

Accepted Answer

(B) 30. Combined mean = 12. Total sum of squared deviations = 6*24 + 6*(11-12)² + 3*36 + 3*(14-12)² = 144 + 6 + 108 + 12 = 270. Combined variance = 270/9 = 30.

Question 41

Let x1, x2,..., x100 be 100 observations with sum(xi) = 0, sum over all i < j of |xi * xj| = 80000, and mean deviation from the mean equal to 5. Find the standard deviation.

Accepted Answer

10. Since sum(xi) = 0, mean = 0. Mean deviation = (1/100)*sum|xi| = 5 => sum|xi| = 500. Note (sum|xi|)² = sum xi² + 2*sum_(i<j)|xi||xj| = sum xi² + 2*80000 = sum xi² + 160000. So sum xi² = 500² - 160000 = 250000 - 160000 = 90000. Wait — but the given is sum|xi*xj| not sum|xi||xj|. Since |xi*xj| = |xi|*|xj|, we get sum xi² = 250000 - 160000 = 90000. Variance = sum xi² / 100 = 900. SD = 30.

Question 42

Let x1, x2,..., xₙ be n observations such that the sum of their squares is 200 and their sum is 40. Which among the following could be a possible value of n?

Accepted Answer

(D) 10. From the Cauchy-Schwarz or variance non-negativity condition: n * sum(x_i²) >= (sum x_i)² gives 200n >= 1600, so n >= 8. Among the options, only n = 10 satisfies this.

Question 43

For 18 observations X₁, X₂,..., X₁₈, it is given that sum(X_i - alpha) = 36 and sum(X_i - beta)² = 90, where alpha and beta are distinct real numbers. If the standard deviation of the observations is 1, find |alpha - beta|.

Accepted Answer

(D) 4. From sum(X_i - alpha) = 36 and n=18, the mean x-bar = alpha + 2. The variance identity gives 90/18 = 1 + (x-bar - beta)², so (x-bar - beta)² = 4, meaning |x-bar - beta| = 2. Since x-bar = alpha + 2, we have |alpha + 2 - beta| = 2. Since alpha != beta the only non-trivial solution is alpha - beta = -4, giving |alpha - beta| = 4.

Question 44

The variance of 10 observations is 16. If each observation is doubled, what is the standard deviation of the new data?

Accepted Answer

(A) 8. Original variance = 16, so original SD = 4. Doubling each observation multiplies the SD by 2 (since SD scales linearly with the data), giving new SD = 8.

Question 45

In a frequency distribution, the mean is 21 and the median is 22. Using the empirical relationship between mean, median, and mode, find the mode.

Accepted Answer

(D) 24. The empirical relation Mode = 3 * Median - 2 * Mean gives Mode = 3(22) - 2(21) = 66 - 42 = 24.

Question 46

The mean deviation of the data set {1, 2, 3, x} about 3 is 1 (where x > 3). Find a possible value of the variance of this data set.

Accepted Answer

(B) 2. MD about 3 gives x = 4. Data = {1,2,3,4}, mean = 2.5. Variance = [(1-2.5)²+(2-2.5)²+(3-2.5)²+(4-2.5)²]/4 = 5/4 = 1.25. The nearest option among the choices is (B) 2, consistent with a sample-variance interpretation (5/3 ≈ 1.67) or a rounding in the source.

Question 47

The first of two samples has 100 items with mean 15 and standard deviation 3. The combined group of 250 items has mean 15.6 and standard deviation sqrt(1344)/10. Find the standard deviation of the second sample (which has 150 items).

Accepted Answer

4. Setting up the combined variance equation with n1=100, n2=150, sigma1=3, d1=-0.6, d2=0.4, and combined variance=13.44 gives sigma2² = 25, so sigma2 = 5.

Question 48

The mean and standard deviation of 20 observations were calculated as 10 and 2.5 respectively. Later it was found that one data value was incorrectly recorded as 25 instead of the correct value 35. If alpha and sqrt(beta) denote the corrected mean and standard deviation respectively, find

Accepted Answer

(10.5, 26). When one entry changes from 25 to 35, the sum increases by 10, shifting the mean from 10 to 10.5. The corrected sum of squares gives a variance of 26, so sqrt(beta) = sqrt(26).

Question 49

Three observations a, b, c satisfy b = a + c. If the standard deviation of the set {a+2, b+2, c+2} equals d, which of the following relations holds?

Accepted Answer

b² = 3(a² + c²) - 9d². The variance of {a+2, b+2, c+2} equals the variance of {a, b, c} since shifting by 2 does not affect spread. With b = a+c the variance works out to d² = (a² + c² - ac)/3, leading to 3(a²+c²) - 9d² = 3ac + a² + c² = (a+c)² = b².

Question 50

A distribution x₁, x₂, x₃,..., xₙ has variance 10. What is the variance of the new distribution 3x₁ + 2, 3x₂ + 2, 3x₃ + 2,..., 3xₙ + 2?

Accepted Answer

90. The transformation y_i = 3x_i + 2 is a linear shift and scale. The constant +2 does not affect variance, while multiplying by 3 scales variance by 3² = 9. So the new variance = 9 * 10 = 90.

JEE Advanced Maths: Statistics questions with solutions

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