The mean and standard deviation of the marks | vedclass

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(B) Given $n = 10$,$\bar{x} = 50$,and $\sigma = 12$.Sum of marks $\sum x_i = n 	imes \bar{x} = 10 	imes 50 = 500$.Correct sum of marks $\sum x_{i, \

Vedclass · Accepted Answer

$269$

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(B) Given $n = 10$,$\bar{x} = 50$,and $\sigma = 12$.Sum of marks $\sum x_i = n 	imes \bar{x} = 10 	imes 50 = 500$.Correct sum of marks $\sum x_{i, 	ext{correct}} = 500 - 45 - 50 + 20 + 25 = 450$.Correct mean $\bar{x}_{	ext{correct}} = \frac{450}{10} = 45$.Variance $\sigma^2 = 144$,so $\frac{\sum x_i^2}{n} - (\bar{x})^2 = 144$.$\sum x_i^2 = 10 	imes (144 + 50^2) = 10 	imes (144 + 2500) = 26440$.Correct sum of squares $\sum x_{i, 	ext{correct}}^2 = 26440 - 45^2 - 50^2 + 20^2 + 25^2 = 26440 - 2025 - 2500 + 400 + 625 = 22940$.Correct variance $\sigma_{	ext{correct}}^2 = \frac{\sum x_{i, 	ext{correct}}^2}{n} - (\bar{x}_{	ext{correct}})^2 = \frac{22940}{10} - (45)^2 = 2294 - 2025 = 269$.

The mean and standard deviation of the marks of $10$ students were found to be $50$ and $12$ respectively. Later,it was observed that two marks $20$ and $25$ were wrongly read as $45$ and $50$ respectively. Then the correct variance is $............$.

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