13.Statistics
medium

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$.....$

A

$4$

B

$6$

C

$2$

D

$8$

(JEE MAIN-2022)

Solution

$n =10, \bar{x}=\frac{\sum x_{i}}{10}=15$

$6^{2}=\frac{\sum x_{i}^{2}}{10}-(\bar{x})^{2}=15$

$\sum_{i=1}^{10} x_{i}=150$

$\sum_{i=1}^{9} x_{i}+25=150$

$\sum_{i=1}^{9} x_{i}=125$

$\sum_{i=1}^{9} x_{i}+15=140$

Actual mean $=\frac{140}{10}=14=\bar{x}_{\text {nev }}$

$\sum_{i=1}^{9} \frac{x_{i}^{2}+25^{2}-15^{2}}{10}=15$

$\sum_{i=1}^{9} x_{i}^{2}+625=2400$

$\sum_{i=1}^{9} x_{i}^{2}=1775$

$\sum_{i=1}^{9} x_{i}^{2}+15^{2}=2000=\left(\sum x_{i}^{2}\right)_{\text {acnaal }}$

$6_{\text {actual }}^{2}=\frac{\left(\sum x_{i}^{2}\right)_{\text {actual }}-\left(\bar{x}_{\text {new }}\right)^{2}}{10}$

$=\frac{2000}{10}-14^{2}$

$=200-196=4$

$(\text { S.D })_{\text {attul }}=6=2$

Standard 11
Mathematics

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