Mean and standard deviation of 40 observations are 30 and 5 respectively. It was noticed that two of

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13.Statistics

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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$.........$

A

$238$

B

$239$

C

$240$

D

$241$

(JEE MAIN-2022)

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 $S.D$, $\sigma_{1}^{\prime}=\sqrt{\frac{36756}{38}-(31)^{2}}=\sigma$

$36756-(31)^{2} \times 38=38 \sigma^{2}$

$38 \sigma^{2}=238$

Standard 11

Mathematics

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