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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Detailed Solution

$\begin{array}{l} \frac{x_{1} + x_{2} + \dots + x_{9} + 50}{10} = 20 \\ x_{1} + x_{2} + \dots + x_{9} = 150 \\ 64 = \frac{x_{1}^{2} + x_{2}^{2} + \dots + x_{9}^{2} + 2500}{10} - 400 \\ x_{1}^{2} + x_{2}^{2} + \dots + x_{9}^{2} = 2140 \end{array}$
$New mean = \frac{150 + 40}{10} = 19$
$New σ = \frac{2140 + 1600}{10} - (19)^{2} σ = 13$

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