Question 1

Let two lines of regression be x+y+11=0 and 2x+3y+4=0 for some data. What is the value of correlation coefficient between x and y ?

Accepted Answer

Let the regression line of y on x be 2x+3y+4=0. 3y = -2x - 4 y = - 2 3 x - 4 3 The regression coefficient of y on x is b_ yx = - 2 3 . Let the regression line of x on y be x+y+11=0. x = -y - 11 The regression coefficient of x on y is b_ xy = -1. The square of the correlation coefficient is given by r^2 = b_ yx b_ xy . r^2 = (- 2 3 ) (-1) = 2 3 Since r^2 1, our assumption for the regression lines is correct. Furthermore, since both regression coefficients b_ yx and b_ xy are negative, the correla

Question 2

Passage: The frequency distribution of marks obtained by 100 students in a certain subject is given below : Marks 0-10 10-20 20-30 30-40 No. of students 10 20 f 40 Question: What is the standard devia

Accepted Answer

The total number of students is 100. 10 + 20 + f + 40 = 100 f = 30 The mid-points of the class intervals are x_i = 5, 15, 25, 35. The mean x is given by: x = f_i x_i f_i = 10(5) + 20(15) + 30(25) + 40(35) 100 x = 50 + 300 + 750 + 1400 100 = 2500 100 = 25 The variance ^2 is given by: ^2 = f_i (x_i - x )^2 f_i ^2 = 10(5 - 25)^2 + 20(15 - 25)^2 + 30(25 - 25)^2 + 40(35 - 25)^2 100 ^2 = 10(-20)^2 + 20(-10)^2 + 30(0)^2 + 40(10)^2 100 ^2 = 10(400) + 20(100) + 0 + 40(100) 100 ^2 = 4000 + 2000 + 0 + 4000

Question 3

The standard deviation of 100 observations is 10. If 5 is multiplied to each of the observations, then what is the new standard deviation ?

Accepted Answer

Given the standard deviation of 100 observations is = 10. If each observation in a dataset is multiplied by a constant k, the new standard deviation becomes |k| . Here, k = 5. New standard deviation = 5 10 = 50. Answer: 50

Question 4

Passage: The frequency distribution of marks obtained by 100 students in a certain subject is given below : Marks 0-10 10-20 20-30 30-40 No. of students 10 20 f 40 Question: What is the mean deviation

Accepted Answer

Total number of students is 100. 10 + 20 + f + 40 = 100 f = 30 The mid-points x_i of the classes are 5, 15, 25, 35. The arithmetic mean x = f_i x_i f_i x = 10 5 + 20 15 + 30 25 + 40 35 100 x = 50 + 300 + 750 + 1400 100 = 2500 100 = 25 The absolute deviations |x_i - x | are |5 - 25| = 20, |15 - 25| = 10, |25 - 25| = 0, and |35 - 25| = 10. Mean deviation about the mean = f_i |x_i - x | f_i = 10 20 + 20 10 + 30 0 + 40 10 100 = 200 + 200 + 0 + 400 100 = 800 100 = 8 Answer: 8

Question 5

If the mean of 20 observations, namely x_1,x_2,x_3, ,x_ 20 is 1 414, then what is the value of _ i=1 ^ 20 100(2x_i+4) ?

Accepted Answer

Given the mean of 20 observations is 1.414. 1 20 _ i=1 ^ 20 x_i = 1.414 _ i=1 ^ 20 x_i = 20 1.414 = 28.28 We need to find the value of _ i=1 ^ 20 100(2x_i + 4). _ i=1 ^ 20 100(2x_i + 4) = 100 ( 2 _ i=1 ^ 20 x_i + _ i=1 ^ 20 4 ) = 100 ( 2(28.28) + 20 4 ) = 100 (56.56 + 80) = 100 (136.56) = 13656

Question 6

If the random variable X has mean 3 and standard deviation 5, then what is the variance of the random variable Y=2X-5 ?

Accepted Answer

Given the standard deviation of X is 5. The variance of X is Var(X) = 5^2 = 25. Using the property of variance, Var(aX + b) = a^2 Var(X). For the random variable Y = 2X - 5, we have a = 2 and b = -5. Var(Y) = 2^2 Var(X) = 4 25 = 100. Answer: 100

Question 7

The standard deviation of 100 observations is 10. If 20 is added to each observation, then what will be the new standard deviation?

Accepted Answer

The standard deviation of a set of observations is independent of the change of origin. If a constant is added to or subtracted from each observation, the standard deviation remains unchanged. Given that the original standard deviation is 10. When 20 is added to each observation, the new standard deviation will still be 10. Answer: 10

Question 8

Consider the following statements: I. Mean and variance have the same unit of measurement. II. Mean deviation and standard deviation have the same unit of measurement. Which of the statements given ab

Accepted Answer

Let the unit of observation x_i be u. The mean is given by x = 1 n x_i, which has the unit u. The variance is given by ^2 = 1 n (x_i - x )^2, which has the unit u^2. Thus, mean and variance have different units of measurement. Statement I is incorrect. The mean deviation is given by 1 n |x_i - x |, which has the unit u. The standard deviation is given by = 1 n (x_i - x )^2 , which has the unit u^2 = u. Thus, mean deviation and standard deviation have the same unit of measurement. Statement II is

Question 9

Consider the following for the four (04) items that follow: The frequency distribution of height of students of a class is given below: Height (in cm) Number of Students 160–162 12 162–164 15 164–166

Accepted Answer

The cumulative frequencies for the given classes are: 160-162: 12 162-164: 12 + 15 = 27 164-166: 27 + 24 = 51 166-168: 51 + 13 = 64 Total number of students, N = 64 N 2 = 64 2 = 32 The cumulative frequency just greater than 32 is 51, which corresponds to the class 164-166. Thus, the median class is 164-166. Lower limit of the median class, L = 164 Cumulative frequency of the class preceding the median class, cf = 27 Frequency of the median class, f = 24 Class width, h = 2 Using the formula for m

Question 10

Consider the following for the four (04) items that follow: The frequency distribution of height of students of a class is given below: Height (in cm) Number of Students 160–162 12 162–164 15 164–166

Accepted Answer

The given frequency distribution consists of continuous class intervals (160-162, 162-164, 164-166, 166-168) representing a continuous variable (height). For continuous grouped frequency distributions, the most appropriate graphical representation is a histogram, as it displays the frequencies of continuous data using adjacent rectangular bars. Bar charts, percentage bar charts, and pie charts are generally used for categorical or discrete data. Answer: histogram

Question 11

Consider the following for the two (02) items that follow: The sum and the sum of squares of the observations corresponding to length X (in cm) and weight Y (in gm) of 50 tropical tubers are given as

Accepted Answer

Given n = 50, X = 200, Y = 250, X^2 = 900, and Y^2 = 1400. The variance of X is calculated as: Var (X) = X^2 n - ( X n )^2 Var (X) = 900 50 - ( 200 50 )^2 Var (X) = 18 - (4)^2 = 18 - 16 = 2 The variance of Y is calculated as: Var (Y) = Y^2 n - ( Y n )^2 Var (Y) = 1400 50 - ( 250 50 )^2 Var (Y) = 28 - (5)^2 = 28 - 25 = 3 Comparing the variances, we get Var (X) Answer: Variance (X) < Variance (Y)

Question 12

The arithmetic mean of 100 observations is 50. If 5 is subtracted from each observation and then divided by 20, then what is the new arithmetic mean?

Accepted Answer

Let the original observations be x_1, x_2, , x_ 100 . The original arithmetic mean is x = 50. The new observations are given by y_i = x_i - 5 20 . Using the property of arithmetic mean, if each observation is subjected to a linear transformation y_i = a x_i + b, the new mean is y = a x + b. Therefore, the new mean is y = x - 5 20 . Substituting x = 50, we get y = 50 - 5 20 = 45 20 = 2.25. Answer: 2.25

Question 13

The standard deviation of 100 observations is 10. If 5 is added to each observation and then divided by 20, then what will be the new standard deviation?

Accepted Answer

Let the original observations be x_1, x_2, , x_ 100 . The standard deviation of these observations is _x = 10. The new observations are given by y_i = x_i + 5 20 = 1 20 x_i + 1 4 . If a new variable is defined as y_i = ax_i + b, its standard deviation is given by _y = |a| _x. Substituting a = 1 20 and _x = 10: _y = | 1 20 | 10 = 0.5 The new standard deviation is 0.5. Answer: 0.5

Question 14

Consider the following for the two (02) items that follow: The sum and the sum of squares of the observations corresponding to length X (in cm) and weight Y (in gm) of 50 tropical tubers are given as

Accepted Answer

Mean of X, X = X n = 200 50 = 4 Mean of Y, Y = Y n = 250 50 = 5 Variance of X, _X^2 = X^2 n - ( X )^2 = 900 50 - 4^2 = 18 - 16 = 2 Standard deviation of X, _X = 2 Variance of Y, _Y^2 = Y^2 n - ( Y )^2 = 1400 50 - 5^2 = 28 - 25 = 3 Standard deviation of Y, _Y = 3 Coefficient of variation of X, CV_X = _X X 100 = 2 4 100 = 25 2 35.35\% Coefficient of variation of Y, CV_Y = _Y Y 100 = 3 5 100 = 20 3 34.64\% Since 35.35\% > 34.64\%, the coefficient of variation of X is strictly more than the coeffici

Question 15

Consider the following for the four (04) items that follow: The frequency distribution of height of students of a class is given below: Height (in cm) Number of Students 160–162 12 162–164 15 164–166

Accepted Answer

The height which occurs most frequently corresponds to the mode of the given frequency distribution. The maximum frequency is 24, which corresponds to the class interval 164-166. Thus, the modal class is 164-166. The lower limit of the modal class, l = 164 Frequency of the modal class, f_1 = 24 Frequency of the class preceding the modal class, f_0 = 15 Frequency of the class succeeding the modal class, f_2 = 13 Class size, h = 2 The mode is calculated using the formula: Mode = l + ( f_1 - f_0 2f

Question 16

The frequency distribution of the marks obtained by students in a Science examination is given below: Marks 5–15 15–25 25–35 35–45 Number of students 20 30 30 20 What is the arithmetic mean?

Accepted Answer

The class marks (midpoints) x_i for the given intervals are 10, 20, 30, 40. The corresponding frequencies f_i are 20, 30, 30, 20. Total number of students, N = f_i = 20 + 30 + 30 + 20 = 100 Calculating the sum of f_i x_i: f_i x_i = (20 10) + (30 20) + (30 30) + (20 40) f_i x_i = 200 + 600 + 900 + 800 = 2500 Arithmetic mean x = f_i x_i N = 2500 100 = 25 Answer: 25

Question 17

What is the arithmetic mean of 8^2, 9^2, 10^2, , 15^2?

Accepted Answer

The given sequence is 8^2, 9^2, 10^2, , 15^2. Number of terms, n = 15 - 8 + 1 = 8. Sum of the series S = _ k=8 ^ 15 k^2 = _ k=1 ^ 15 k^2 - _ k=1 ^ 7 k^2. Using the formula _ k=1 ^ n k^2 = n(n+1)(2n+1) 6 , _ k=1 ^ 15 k^2 = 15 16 31 6 = 1240 _ k=1 ^ 7 k^2 = 7 8 15 6 = 140 S = 1240 - 140 = 1100 Arithmetic mean = S n = 1100 8 = 137.5 Answer: 137.5

Question 18

Consider the following for the four (04) items that follow: The frequency distribution of height of students of a class is given below: Height (in cm) Number of Students 160–162 12 162–164 15 164–166

Accepted Answer

The cumulative frequencies for the given class intervals are calculated as follows: Height 162 cm: 12 Height 164 cm: 12 + 15 = 27 Height 166 cm: 27 + 24 = 51 To find the number of students with height less than or equal to 165 cm, we use linear interpolation within the class interval 164–166. The number of students in the interval 164–166 is 24. Assuming a uniform distribution of frequencies within the class interval, the number of students with height between 164 cm and 165 cm is: 165 - 164 166

Statistics — Previous Year Questions & Solutions