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Question

Mean $(\bar{x})=10$

$ \Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}=10 $

$ \Sigma \mathrm{x}_{\mathrm{i}}=10 	imes 20=200$

If $8$ is

Vedclass · Accepted Answer

$\sqrt{3.96}$

Vedclass · Answer

Mean $(\bar{x})=10$

$ \Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}=10 $

$ \Sigma \mathrm{x}_{\mathrm{i}}=10 	imes 20=200$

If $8$ is replaced by $12$ , then $\Sigma x_1=200-8+12=204$

$	herefore$ Correct mean $(\overline{\mathrm{x}})=\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}$

$=\frac{204}{20}=10.2$

$ \because$ Standard deviation $=2$

$ 	herefore$ Variance $=( S.D.)^2=2^2=4 $

$ \Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}^2}{20}-\left(\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}ight)^2=4 $

$ \Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}^2}{20}-(10)^2=4 $

$ \Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}^2}{20}=104 $

$ \Rightarrow \Sigma \mathrm{x}_{\mathrm{i}}^2=2080$

Now, replaced $'8'$ observations by $'12'$

$	ext { Then, } \Sigma \mathrm{x}_{\mathrm{i}}^2=2080-8^2+12^2=2160$

$	herefore$ Variance of removing observations

$ \Rightarrow \frac{\Sigma x_i^2}{20}-\left(\frac{\Sigma x_i}{20}ight)^2 $

$ \Rightarrow \frac{2160}{20}-(10.2)^2 $

$ \Rightarrow 108-104.04 $

$ \Rightarrow 3.96$

Correct standard deviation

$=\sqrt{3.96}$

$X_i$	$0$	$1$	$2$	$3$	$4$	$5$
$f_i$	$k+2$	$2k$	$K^{2}-1$	$K^{2}-1$	$K^{2}-1$	$k-3$

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