The mean and standard deviation of 20 observa | vedclass

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(C) Given: $n = 20$,$\bar{x} = 10$,$S.D. = 2$.$\Sigma x_i = n 	imes \bar{x} = 20 	imes 10 = 200$.Corrected sum $\Sigma x_i = 200 - 8 + 12 = 204$.Cor

Vedclass · Accepted Answer

$\sqrt{3.96}$

Vedclass · Answer

(C) Given: $n = 20$,$\bar{x} = 10$,$S.D. = 2$.$\Sigma x_i = n 	imes \bar{x} = 20 	imes 10 = 200$.Corrected sum $\Sigma x_i = 200 - 8 + 12 = 204$.Corrected mean $\bar{x}' = \frac{204}{20} = 10.2$.Variance $= (S.D.)^2 = 2^2 = 4$.Since Variance $= \frac{\Sigma x_i^2}{n} - (\bar{x})^2$,we have $4 = \frac{\Sigma x_i^2}{20} - 10^2$.$\frac{\Sigma x_i^2}{20} = 104 \Rightarrow \Sigma x_i^2 = 2080$.Corrected $\Sigma x_i^2 = 2080 - 8^2 + 12^2 = 2080 - 64 + 144 = 2160$.Corrected Variance $= \frac{\Sigma x_i^2}{n} - (\bar{x}')^2 = \frac{2160}{20} - (10.2)^2$.$= 108 - 104.04 = 3.96$.Corrected standard deviation $= \sqrt{3.96}$.

The mean and standard deviation of $20$ observations are found to be $10$ and $2$,respectively. It was later discovered that one observation was mistakenly recorded as $8$ instead of $12$. The correct standard deviation is:

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