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 :
Solution
$\bar{x}=10 ~\&~ \sigma^{2}=4$, No. of students $=N$ (let)
<br/><br/>
$$
\therefore \quad \frac{\sum x_{i}}{N}=10 ~\&~ \frac{\sum x_{i}^{2}}{N}-(10)^{2}=4
$$
<br/><br/>
Now if one of $x_{i}$ is changed from 8 to 12 we have
<br/><br/>
New mean $\frac{\sum x_{i}+4}{N}=10+\frac{4}{N}=10.2$
<br/><br/>
$\Rightarrow N=20$
<br/><br/>
and $\sigma_{\text {new }}^{2}=\frac{\sum x_{i}^{2}-(8)^{2}+(12)^{2}}{20}-(10 \cdot 2)^{2}$
<br/><br/>
$$
\begin{aligned}
& =\frac{\sum x_{i}^{2}}{20}+\frac{144-64}{20}-(10 \cdot 2)^{2} \\\\
& =104+4-(10 \cdot 2)^{2} \\\\
& =108-104.04=3.96
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Statistics · Topic: Measures of Central Tendency
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