The mean and variance of 10 observations were | vedclass

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Question

$n =10, \bar{x}=\frac{\sum x_{i}}{10}=15$

$6^{2}=\frac{\sum x_{i}^{2}}{10}-(\bar{x})^{2}=15$

$\sum_{i=1}^{10} x_{i}=150$

$\sum_{i=1}^{9} x_{i

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$n =10, \bar{x}=\frac{\sum x_{i}}{10}=15$

$6^{2}=\frac{\sum x_{i}^{2}}{10}-(\bar{x})^{2}=15$

$\sum_{i=1}^{10} x_{i}=150$

$\sum_{i=1}^{9} x_{i}+25=150$

$\sum_{i=1}^{9} x_{i}=125$

$\sum_{i=1}^{9} x_{i}+15=140$

Actual mean $=\frac{140}{10}=14=\bar{x}_{	ext {nev }}$

$\sum_{i=1}^{9} \frac{x_{i}^{2}+25^{2}-15^{2}}{10}=15$

$\sum_{i=1}^{9} x_{i}^{2}+625=2400$

$\sum_{i=1}^{9} x_{i}^{2}=1775$

$\sum_{i=1}^{9} x_{i}^{2}+15^{2}=2000=\left(\sum x_{i}^{2}ight)_{	ext {acnaal }}$

$6_{	ext {actual }}^{2}=\frac{\left(\sum x_{i}^{2}ight)_{	ext {actual }}-\left(\bar{x}_{	ext {new }}ight)^{2}}{10}$

$=\frac{2000}{10}-14^{2}$

$=200-196=4$

$(	ext { S.D })_{	ext {attul }}=6=2$

	Height	Weight
Mean	$162.6\,cm$	$52.36\,kg$
Variance	$127.69\,c{m^2}$	$23.1361\,k{g^2}$

The mean and variance of $10$ observations were calculated as $15$ and $15$ respectively by a student who took by mistake $25$ instead of $15$ for one observation. Then, the correct standard deviation is$.....$

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