JEE MAIN - Mathematics (2023 - 25th January Morning Shift - No. 4)
The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12. If the new mean of the marks is 10.2, then their new variance is equal to :
3.92
4.08
3.96
4.04
Explanation
$\bar{x}=10 ~\&~ \sigma^{2}=4$, No. of students $=N$ (let)
$$ \therefore \quad \frac{\sum x_{i}}{N}=10 ~\&~ \frac{\sum x_{i}^{2}}{N}-(10)^{2}=4 $$
Now if one of $x_{i}$ is changed from 8 to 12 we have
New mean $\frac{\sum x_{i}+4}{N}=10+\frac{4}{N}=10.2$
$\Rightarrow N=20$
and $\sigma_{\text {new }}^{2}=\frac{\sum x_{i}^{2}-(8)^{2}+(12)^{2}}{20}-(10 \cdot 2)^{2}$
$$ \begin{aligned} & =\frac{\sum x_{i}^{2}}{20}+\frac{144-64}{20}-(10 \cdot 2)^{2} \\\\ & =104+4-(10 \cdot 2)^{2} \\\\ & =108-104.04=3.96 \end{aligned} $$
$$ \therefore \quad \frac{\sum x_{i}}{N}=10 ~\&~ \frac{\sum x_{i}^{2}}{N}-(10)^{2}=4 $$
Now if one of $x_{i}$ is changed from 8 to 12 we have
New mean $\frac{\sum x_{i}+4}{N}=10+\frac{4}{N}=10.2$
$\Rightarrow N=20$
and $\sigma_{\text {new }}^{2}=\frac{\sum x_{i}^{2}-(8)^{2}+(12)^{2}}{20}-(10 \cdot 2)^{2}$
$$ \begin{aligned} & =\frac{\sum x_{i}^{2}}{20}+\frac{144-64}{20}-(10 \cdot 2)^{2} \\\\ & =104+4-(10 \cdot 2)^{2} \\\\ & =108-104.04=3.96 \end{aligned} $$
Comments (0)
