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Q.

The mean and variance of the marks obtained by the students in a test are 10 and 4 alter marks of one student increased from 8 to 12. If new mean of marks B 10.2 then new varience

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a

3.96

b

4.04

c

3.92

d

4.08

answer is C.

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Detailed Solution

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$\begin{array}{l} \sum_{i = 1}^{n} x_{i} = 10 n \Rightarrow \sum_{i = 1}^{n} (x_{i} - 8 + 12) = (10.2) n \Rightarrow n = 20 \\ N o w \frac{\sum_{i = 1}^{20} {x_{i}}^{2}}{20} - 10^{2} = 4 \Rightarrow \sum x_{i}^{2} = 2080 \Rightarrow \frac{\sum x_{i}^{2} - 8^{2} + 12^{2}}{20} = {(10.2)}^{2} \\ \Rightarrow 108 - 104.04 = 3.96 \end{array}$

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