Question

The mean and standard deviation of a distribution of weights of a group of $20$ boys are $40 \,kg$ and $5 \,kg$ respectively.
If two boys of weights $43 \,kg$ and $37\, kg$ are excluded from this group, then the variance of the distribution of weights of the remaining group of boys is

• A. 26.18
• B. 5.27
• C. 26.78
• D. 5.17

Answer 1

Answer: (C)

Here, $n=20$
Mean $=40$
Mean $=\frac{\text { Sum of observation }}{n}$
$\Rightarrow \,40=\frac{\text { Sum of weights }}{20}$
Sum of weights $=20 \times 40=800$
Now, standard deviation, $\sigma=5$
$\because$ variance $=\sigma^{2}=5^{2}=25$
Now, two boys are excluded of weight $43 \,kg$ and $73 \,kg$
New mean $=\frac{\text { Remaining weight }}{20-2}=\frac{800-43-73}{18}$
$\bar{X}_{\text {new }}=$ new mean $=38$
Now, variance $\sigma^{2}=25$
$ \frac{\Sigma x^{2}}{n}-(\bar{X})^{2}=25$
$ \Rightarrow \, \frac{\Sigma x^{2}}{20}-40^{2}=25 $
$ \Rightarrow \, \frac{\Sigma x^{2}}{20}=1600+25$
$ \Rightarrow \, \Sigma x^{2}=1625 \times 20 $
$ \Rightarrow \, \Sigma x^{2}=32500 $
When two boys are removed
$\Sigma x_{\text {new }}^{2}=32500-(43)^{2}-(73)^{2}=25322$.
So, new variance
$\sigma_{\text {new }}^{2}=\frac{\sum x_{\text {new }}^{2}}{18}-\left(\bar{X}_{\text {new }}\right)^{2}=\frac{25322}{18}-(38)^{2}$
$\sigma_{\text {new }}^{2}=26.78$