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
The correct answer is : 58
\(\bar{X_c}=\) mean of c = \(\frac{(5-3)+(12+2)}{2}=8\)
\(\sigma^2_{12}\)=variance of c
\(= \frac{n_1(\sigma_1^2-d_1^2)+n_1(\sigma_2^2+d_2^2)}{n_1+n_2}\)
\(d_1=\bar{x_{12}-\bar{x_1}}\)
\(d_2=\bar{x_{12}-\bar{x_2}}\)
\(n_1=5,\sigma_1^2=8,d_1=8-2=6\)
\(n_2=5,\sigma_2^2=20,d_2=8-14=-6\)
\(\sigma^2_{12}=\frac{5(8+36)+5(20+36)}{10}=50\)
\(\sigma^2_{12}+\bar{x_c}=50+8=58\)
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