The mean and standard deviation of 100 observ | vedclass

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Question

(D) Given $n = 100$,incorrect mean $\bar{x} = 40$,incorrect standard deviation $s = 5.1$.Incorrect sum of observations $\sum x_i = 100 	imes 40 = 400

Vedclass · Accepted Answer

$449$

Vedclass · Answer

(D) Given $n = 100$,incorrect mean $\bar{x} = 40$,incorrect standard deviation $s = 5.1$.Incorrect sum of observations $\sum x_i = 100 	imes 40 = 4000$.Correct sum of observations $\sum x_i' = 4000 - 50 + 40 = 3990$.Correct mean $\mu = \frac{3990}{100} = 39.9$.Incorrect variance $s^2 = (5.1)^2 = 26.01$.Using $s^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$,we get $26.01 = \frac{\sum x_i^2}{100} - 40^2$.$\sum x_i^2 = 100(26.01 + 1600) = 100(1626.01) = 162601$.Correct sum of squares $\sum x_i'^2 = 162601 - 50^2 + 40^2 = 162601 - 2500 + 1600 = 161701$.Correct variance $\sigma^2 = \frac{161701}{100} - (39.9)^2 = 1617.01 - 1592.01 = 25$.Correct standard deviation $\sigma = \sqrt{25} = 5$.Therefore,$10(\mu + \sigma) = 10(39.9 + 5) = 10(44.9) = 449$.

The mean and standard deviation of $100$ observations are $40$ and $5.1$,respectively. By mistake,one observation is taken as $50$ instead of $40$. If the correct mean and the correct standard deviation are $\mu$ and $\sigma$ respectively,then $10(\mu+\sigma)$ is equal to

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