The mean and variance of 10 observations were | vedclass

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(C) Given $n = 10$,incorrect mean $\bar{x} = 15$,and incorrect variance $\sigma^2 = 15$.Incorrect sum of observations: $\sum x_i = n 	imes \bar{x} =

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$2$

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(C) Given $n = 10$,incorrect mean $\bar{x} = 15$,and incorrect variance $\sigma^2 = 15$.Incorrect sum of observations: $\sum x_i = n 	imes \bar{x} = 10 	imes 15 = 150$.Correct sum of observations: $\sum x_{i, 	ext{correct}} = 150 - 25 + 15 = 140$.Correct mean: $\bar{x}_{	ext{correct}} = \frac{140}{10} = 14$.Using the variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$,we have $15 = \frac{\sum x_i^2}{10} - 15^2$.$\frac{\sum x_i^2}{10} = 15 + 225 = 240 \implies \sum x_i^2 = 2400$.Correct sum of squares: $\sum x_{i, 	ext{correct}}^2 = 2400 - 25^2 + 15^2 = 2400 - 625 + 225 = 2000$.Correct variance: $\sigma_{	ext{correct}}^2 = \frac{\sum x_{i, 	ext{correct}}^2}{n} - (\bar{x}_{	ext{correct}})^2 = \frac{2000}{10} - 14^2 = 200 - 196 = 4$.Correct standard deviation: $\sigma_{	ext{correct}} = \sqrt{4} = 2$.

\text{Particulars}	\text{Firm } $A$ \text{ and Firm } $B$
\text{No. of wage earners}	$A: 586, B: 648$
\text{Mean of monthly wages}	$A: Rs. 5253, B: Rs. 5253$
\text{Variance of wages}	$A: 100, B: 121$

The mean and variance of $10$ observations were calculated as $15$ and $15$ respectively by a student who took by mistake $25$ instead of $15$ for one observation. Then,the correct standard deviation is $.....$

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