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Q.

Let sets A and B have 5 elements each. Let 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_________.

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a

36

b

38

c

32

d

40

answer is D.

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Detailed Solution

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

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, $C = {C_{1}, C_{2}, ........ C_{10}}$

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

$\bar{C} = \frac{10 + 50}{10} = 6 .$

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