Question

The mean and the standard deviation of a data of $8$ items are $25$ and $5$ respectively. If two items $15$ and $25$ are added to this data, then the variance of the new data is

• A. 29
• B. 24
• C. 26
• D. $\sqrt{29}$

Answer 1

Answer: (A)

We have, $n=8, \bar{x}=25$ and $\sigma=5$
$\therefore \bar{x}=\frac{\Sigma x_{i}}{n}$
$\Rightarrow \Sigma x_{i}=n \bar{x}=8 \times 25=200$
$\Rightarrow $ In corrected $\Sigma x_{i}=200$
and $\sigma=5$
$\Rightarrow \sigma^{2}=25$
$\Rightarrow \frac{1}{n} \Sigma x_{i}^{2}-(\text { mean })^{2}=25$
$\Rightarrow \frac{\Sigma x_{i}^{2}}{8}-625=25$
$\Rightarrow \Sigma x_{i}^{2}=5200$
Corrected $\Sigma x_{i}^{2}=5200+225+625=6050$
and corrected mean $=200+15+25=240$
$ \therefore $ Corrected variance $=\frac{6050}{10}-\left(\frac{240}{10}\right)^{2} $
$=605-(24)^{2}=605-576=29 $