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The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on it was observed that one observation was recorded as 50 instead of 40 then correct variance is____

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answer is 13.

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Detailed Solution

$\bar{x} = \frac{\sum x_{i}}{n} \Rightarrow 20 = \frac{\sum x_{i}}{10} \Rightarrow \sum x_{i} = 200$
$σ = 8 = \sqrt{\frac{1}{10} \sum {x_{i}}^{2} - {(20)}^{2}} \Rightarrow \sum {x_{i}}^{2} = 4640$
$N e w \sum x_{i} = 190$
$\sum {x_{i}}^{2} = 4640 + 40^{2} - 50^{2} = 3740$
New variance $= \sum \frac{1}{10} \times 3740 - {(19)}^{2} = 13$

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