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The mean and standard deviation of $40$ observations are $30$ and $5$ respectively. It was noticed that two of these observations $12$ and $10$ were wrongly recorded. If $\sigma$ is the standard deviation of the data after omitting the two wrong observations from the data, then $38 \sigma^{2}$ is equal to$.........$
$238$
$239$
$240$
$241$
Solution
Wrong mean $=\mu_{1}=30$
Wrong $S.D$ $=\sigma_{1}=5$
$\frac{\sum x _{ i }}{40}=30$
$\sum x _{ i }=1200$
$\sigma_{1}^{2}=25$
$\frac{\sum x _{ i }^{2}}{40}-30^{2}=25$
$\sum x _{ i }^{2}=925 \times 40=37000$
New sum $=\sum x _{ i }^{\prime}=1200-10-12=1178$
New mean $=\mu_{1}^{\prime}=\frac{1178}{38}=31$
New $\sum x _{ i }^{2}=37000-(10)^{2}-(12)^{2}=36756$
New $S.D$, $\sigma_{1}^{\prime}=\sqrt{\frac{36756}{38}-(31)^{2}}=\sigma$
$36756-(31)^{2} \times 38=38 \sigma^{2}$
$38 \sigma^{2}=238$