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(B) Given $n = 100$,$\bar{x} = 20$,$\sigma = 3$.Sum of observations $\sum x_i = 100 	imes 20 = 2000$.Sum of squares of observations $\sum x_i^2 = n(\

Vedclass · Accepted Answer

Mean $= 20.1$,Standard Deviation $= 3.02$

Vedclass · Answer

(B) Given $n = 100$,$\bar{x} = 20$,$\sigma = 3$.Sum of observations $\sum x_i = 100 	imes 20 = 2000$.Sum of squares of observations $\sum x_i^2 = n(\sigma^2 + \bar{x}^2) = 100(3^2 + 20^2) = 100(9 + 400) = 40900$.Incorrect observations are $21, 21, 18$. Sum $= 60$. Sum of squares $= 21^2 + 21^2 + 18^2 = 441 + 441 + 324 = 1206$.New $n = 100 - 3 = 97$.New sum $\sum x_i' = 2000 - 60 = 1940$.New mean $\bar{x}' = \frac{1940}{97} = 20$.New sum of squares $\sum x_i'^2 = 40900 - 1206 = 39694$.New variance $\sigma'^2 = \frac{\sum x_i'^2}{n'} - (\bar{x}')^2 = \frac{39694}{97} - 20^2 = 409.216 - 400 = 9.216$.New standard deviation $\sigma' = \sqrt{9.216} \approx 3.036$.

$x_i$	$2$	$4$	$6$	$8$	$10$	$12$	$14$	$16$
$f_i$	$4$	$4$	$\alpha$	$15$	$8$	$\beta$	$4$	$5$

The mean and standard deviation of a group of $100$ observations were found to be $20$ and $3$,respectively. Later on,it was found that three observations were incorrect,which were recorded as $21, 21,$ and $18$. Find the mean and standard deviation if the incorrect observations are omitted.

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