The mean and standard deviation of 40 observa | vedclass

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Question

(A) Given,$n = 40$,$\mu = 30$,and $\sigma = 5$.Sum of observations $\sum x_i = 40 	imes 30 = 1200$.Variance $\sigma^2 = 25$,so $\frac{\sum x_i^2}{40}

Vedclass · Accepted Answer

$238$

Vedclass · Answer

(A) Given,$n = 40$,$\mu = 30$,and $\sigma = 5$.Sum of observations $\sum x_i = 40 	imes 30 = 1200$.Variance $\sigma^2 = 25$,so $\frac{\sum x_i^2}{40} - (30)^2 = 25$.$\sum x_i^2 = 40 	imes (900 + 25) = 40 	imes 925 = 37000$.After omitting observations $10$ and $12$,the new number of observations $n' = 38$.New sum $\sum x_i' = 1200 - 10 - 12 = 1178$.New mean $\mu' = \frac{1178}{38} = 31$.New sum of squares $\sum (x_i')^2 = 37000 - 10^2 - 12^2 = 37000 - 100 - 144 = 36756$.New variance $\sigma^2 = \frac{\sum (x_i')^2}{n'} - (\mu')^2 = \frac{36756}{38} - (31)^2$.Multiplying by $38$,we get $38 \sigma^2 = 36756 - 38 	imes 961 = 36756 - 36518 = 238$.

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