Question

Consider two sets A and B. Set A has 5 elements whose mean & variance are 5 and 8 respectively. Set B has also 5 elements whose mean & variance are 12 & 20 respectively. A new set C is formed by subtracting 3 from each element of set A and by adding 2 to each element of set B. The sum of mean & variance of the set C is

Answer 1

The correct answer is : 58

$\bar{X_c}=$ mean of c = $\frac{(5-3)+(12+2)}{2}=8$

$\sigma^2_{12}$=variance of c

$= \frac{n_1(\sigma_1^2-d_1^2)+n_1(\sigma_2^2+d_2^2)}{n_1+n_2}$

$d_1=\bar{x_{12}-\bar{x_1}}$

$d_2=\bar{x_{12}-\bar{x_2}}$

$n_1=5,\sigma_1^2=8,d_1=8-2=6$

$n_2=5,\sigma_2^2=20,d_2=8-14=-6$

$\sigma^2_{12}=\frac{5(8+36)+5(20+36)}{10}=50$

$\sigma^2_{12}+\bar{x_c}=50+8=58$