Let sets A and B have 5 elements each. Let th | vedclass

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(B) Let $A = \{a_1, a_2, a_3, a_4, a_5\}$ and $B = \{b_1, b_2, b_3, b_4, b_5\}$.Given,$\overline{A} = 5 \implies \sum a_i = 25$ and $\overline{B} = 8

Vedclass · Accepted Answer

$38$

Vedclass · Answer

(B) Let $A = \{a_1, a_2, a_3, a_4, a_5\}$ and $B = \{b_1, b_2, b_3, b_4, b_5\}$.Given,$\overline{A} = 5 \implies \sum a_i = 25$ and $\overline{B} = 8 \implies \sum b_i = 40$.Variance $\sigma_A^2 = 12 \implies \frac{\sum a_i^2}{5} - 5^2 = 12 \implies \sum a_i^2 = 5(37) = 185$.Variance $\sigma_B^2 = 20 \implies \frac{\sum b_i^2}{5} - 8^2 = 20 \implies \sum b_i^2 = 5(84) = 420$.Set $C$ consists of $a_i - 3$ and $b_i + 2$ for $i=1$ to $5$.Mean of $C$,$\overline{C} = \frac{\sum (a_i - 3) + \sum (b_i + 2)}{10} = \frac{(25 - 15) + (40 + 10)}{10} = \frac{60}{10} = 6$.Variance of $C$,$\sigma_C^2 = \frac{\sum (a_i - 3)^2 + \sum (b_i + 2)^2}{10} - (\overline{C})^2$.$\sum (a_i - 3)^2 = \sum a_i^2 - 6\sum a_i + 45 = 185 - 6(25) + 45 = 80$.$\sum (b_i + 2)^2 = \sum b_i^2 + 4\sum b_i + 20 = 420 + 4(40) + 20 = 600$.$\sigma_C^2 = \frac{80 + 600}{10} - 6^2 = 68 - 36 = 32$.Sum of mean and variance $= 6 + 32 = 38$.

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