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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$.....$
$4$
$6$
$2$
$8$
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 $=\frac{140}{10}=14=\bar{x}_{\text {nev }}$
$\sum_{i=1}^{9} \frac{x_{i}^{2}+25^{2}-15^{2}}{10}=15$
$\sum_{i=1}^{9} x_{i}^{2}+625=2400$
$\sum_{i=1}^{9} x_{i}^{2}=1775$
$\sum_{i=1}^{9} x_{i}^{2}+15^{2}=2000=\left(\sum x_{i}^{2}\right)_{\text {acnaal }}$
$6_{\text {actual }}^{2}=\frac{\left(\sum x_{i}^{2}\right)_{\text {actual }}-\left(\bar{x}_{\text {new }}\right)^{2}}{10}$
$=\frac{2000}{10}-14^{2}$
$=200-196=4$
$(\text { S.D })_{\text {attul }}=6=2$
Similar Questions
Let $\mu$ be the mean and $\sigma$ be the standard deviation of the distribution
$X_i$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ |
$f_i$ | $k+2$ | $2k$ | $K^{2}-1$ | $K^{2}-1$ | $K^{2}-1$ | $k-3$ |
where $\sum f_i=62$. if $[x]$ denotes the greatest integer $\leq x$, then $\left[\mu^2+\sigma^2\right]$ is equal $………$.