BITSAT Mathematics Question (2024) — Solution

Step-by-step solution

Let the observations be x_1, x_2, , x_ 20 and x be their mean. Given that, variance =5 and n= 20. We know that, Variance ( ^2 )= 1 n _ i=1 ^ 20 (x_i- x )^2 i.e. 5= 1 20 _ i=1 ^ 20 (x_i- x )^2 or _ i=1 ^ 20 (x_i- x )^2=100 .....(i) If each observation is multiplied by 2 and the new resulting observations are y_i, then y_i=2 x_i i.e., x_i= 1 2 y_i Therefore, y = 1 n _ i =1 ^ 20 y _ i = 1 20 _ i =1 ^ 20 2 x _ i =2 1 20 _ i =1 ^ 20 x _ i i.e., y =2 x or x = 1 2 y . On substituting the values of x_i and x in eq. (i), we get _ i=1 ^ 20 ( 1 2 y_i- 1 2 y )^2=100 i.e., _ i=1 ^ 20 (y_i- y )^2=400 Thus, the variance of new observations = 1 20 400=20=2^2 5.