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Q.

The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12. If the new mean of the marks is 10.2, then their new variance is equal to:

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a

3.96

b

4.04

c

3.92

d

4.08

answer is B.

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Detailed Solution

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New mean $= (\frac{n \times 10 - 8 + 12}{n}) = 10.2$
$\begin{array}{l} \Rightarrow 4 = (0.2) n ∴ n = 20 \\ σ^{2} = \frac{\sum x^{2}}{n} - {\bar{x}}^{2} = 4 \Rightarrow \sum x^{2} = 2080 \\ New \sum x_{i}^{2} = (2080 + 12^{2} - 8^{2}) = 2160 \\ New σ^{2} = (\frac{2160}{20} - (10.2)^{2}) = 108 - 104.04 = 3.96 \end{array}$

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