Let sets A and B have 5 elements each. Let the mean of the elements in sets A and B be 5 and 8 respectively and the variance of the elements in sets A and B be 12 and 20 respectively. A new set C of 10 elements is formed by subtracting 3 from each element

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Solution

Q.
Let sets A and B have 5 elements each. Let the mean of the elements in sets A and B be 5 and 8 respectively and the variance of the elements in sets A and B be 12 and 20 respectively. A new set C of 10 elements is formed by subtracting 3 from each element of A and adding 2 to each element of B. Then the sum of the mean and variance of the elements of C is ______. [2023]

1 32

2 38

3 40

4 36

Ans.
(2)

$A = {a_{1}, a_{2}, a_{3}, a_{4}, a_{5}},$    $B = {b_{1}, b_{2}, b_{3}, b_{4}, b_{5}}$

Given, $\sum_{i = 1}^{5} a_{i} = 25,$    $\sum_{i = 1}^{5} b_{i} = 40$

$\frac{\sum_{i = 1}^{5} a_{i}^{2}}{5} - {(\frac{\sum_{i = 1}^{5} a_{i}}{5})}^{2} = 12,$    $\frac{\sum_{i = 1}^{5} b_{i}^{2}}{5} - {(\frac{\sum_{i = 1}^{5} b_{i}}{5})}^{2} = 20$

$\Rightarrow \sum_{i = 1}^{5} a_{i}^{2} = 185,$    $\sum_{i = 1}^{5} b_{i}^{2} = 420$

Now, $C = {C_{1}, C_{2}, \dots, C_{10}}$

Such that $C_{i} =$ ${\begin{matrix} a_{i} - 3, & i = 1, 2, 3, 4, 5 \\ b_{i} + 2, & i = 6, 7, 8, 9, 10 \end{matrix}$

$∴$    $Mean of C, \bar{C} = \frac{(\sum a_{i} - 15) + (\sum b_{i} + 10)}{10} = \frac{10 + 50}{10} = 6$

$∴$    $σ^{2} = \frac{\sum_{i = 1}^{10} C_{i}^{2}}{10} - {(\bar{C})}^{2} = \frac{\sum {(a_{i} - 3)}^{2} + \sum {(b_{i} + 2)}^{2}}{10} - 6^{2}$

$= \frac{\sum a_{i}^{2} + \sum b_{i}^{2} - 6 \sum a_{i} + 4 \sum b_{i} + 65}{10} - 36 = 32$

$∴$ $Mean + Variance = \bar{C} + σ^{2} = 6 + 32 = 38$

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