The mean and standard deviation of 50 observa | vedclass

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Question

(C) Let $n = 50$. The incorrect mean $\bar{x} = 15$ and standard deviation $\sigma = 2$.Sum of incorrect observations $\sum x_i = 50 	imes 15 = 750$.

Vedclass · Accepted Answer

$43$

Vedclass · Answer

(C) Let $n = 50$. The incorrect mean $\bar{x} = 15$ and standard deviation $\sigma = 2$.Sum of incorrect observations $\sum x_i = 50 	imes 15 = 750$.Let the incorrect observation be $x_1$ and the correct observation be $x_1'$.Given $x_1 + x_1' = 70$ and the correct mean $\bar{x}' = 16$.Sum of correct observations $\sum x_i' = 50 	imes 16 = 800$.Thus,$x_1' - x_1 = \sum x_i' - \sum x_i = 800 - 750 = 50$.Solving $x_1' + x_1 = 70$ and $x_1' - x_1 = 50$,we get $x_1' = 60$ and $x_1 = 10$.Incorrect variance $\sigma^2 = 4 = \frac{\sum x_i^2}{n} - \bar{x}^2$ $\Rightarrow 4 = \frac{\sum x_i^2}{50} - 225$ $\Rightarrow \sum x_i^2 = 50 	imes 229 = 11450$.Sum of squares of remaining $49$ observations: $\sum_{i=2}^{50} x_i^2 = 11450 - x_1^2 = 11450 - 100 = 11350$.Correct sum of squares $\sum x_i'^2 = 11350 + (x_1')^2 = 11350 + 3600 = 14950$.Correct variance $\sigma'^2 = \frac{\sum x_i'^2}{n} - (\bar{x}')^2 = \frac{14950}{50} - 16^2 = 299 - 256 = 43$.

The mean and standard deviation of $50$ observations are $15$ and $2$ respectively. It was found that one incorrect observation was taken such that the sum of the correct and incorrect observations is $70$. If the correct mean is $16$,then the correct variance is equal to

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