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Q.

The mean and variance of 20 observations are found to be 10 and 4, respectively. On rechecking, it was found that an observation 9 was incorrect and the correct observation was 11. Then the correct variance is:

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a

4.02

b

4.01

c

3.99

d

3.98

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{\sum x_{i}}{20} = 10 \dots \dots \dots \\ \frac{\sum x_{i}^{2}}{20} - 100 = 4 \dots \dots (i i) \\ \sum x_{i}^{2} = 104 \times 20 = 2080 \\ Actual mean = \frac{200 - 9 + 11}{20} = \frac{202}{20} \end{array}$

$\begin{array}{l} Actual Variance = \frac{2080 - 81 + 121}{20} - {(\frac{202}{20})}^{2} \\ = \frac{2120}{20} - (10.1)^{2} = 106 - 102.01 = 3.99 \end{array}$

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