Question

The mean and standard deviation of a distribution of weights of a group of $20$ boys are $40kg$ and $5kg$ respectively.
If two boys of weights $43kg$ and $37kg$ 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 $43kg$ and $73kg$
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
${\sigma }_{\text{new }}^2=\frac{\sum x_{\text{new }}^2}{18}-{\left({\bar{X}}_{\text{new }}\right)}^2=\frac{25322}{18}-(38{)}^2$
${\sigma }_{\text{new }}^2=26.78$