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The mean and standard deviation of 20 items is found to be 10 and 2 respectively. At the time of checking it was found that one item 8 was incorrect. If it is replaced by 12, then the mean and variance are respectively

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10.2, 4.01

10.1, 3.69

10.2, 3.96

10.2, 3.76

answer is C.

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Detailed Solution

$\begin{array}{l} \bar{x} = \frac{10 \times 20 - 8 + 12}{20} = 12 .2; \\ σ^{2} = 4 = \frac{\sum x_{1}^{2}}{20} - 10^{2} = 2080 Correct \\ \sum x_{1}^{2} = 2080 - 64 + 144 = 2160 \Rightarrow σ^{2} = 3 .96 \end{array}$

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