JEE MAIN - Mathematics (2025 - 24th January Morning Shift - No. 17)
For a statistical data $\mathrm{x}_1, \mathrm{x}_2, \ldots, \mathrm{x}_{10}$ of 10 values, a student obtained the mean as 5.5 and $\sum_{i=1}^{10} x_i^2=371$. He later found that he had noted two values in the data incorrectly as 4 and 5 , instead of the correct values 6 and 8 , respectively. The variance of the corrected data is
5
7
9
4
Explanation
$$\begin{aligned}
& \text { Mean } \overline{\mathrm{x}}=5.5 \\
& =\sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}=5.5 \times 10=55 \\
& =\sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}^2=371 \\
& \left(\sum \mathrm{x}_{\mathrm{i}}\right)_{\text {new }}=55-(4+5)+(6+8)=60 \\
& \left(\sum \mathrm{x}_{\mathrm{i}}\right)_{\text {new }}=371-\left(4^2+5^2\right)+\left(6^2+8^2\right)=430 \\
& \\
& \text { Variance } \sigma^2=\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{10}-\left(\frac{\sum \mathrm{x}_{\mathrm{i}}}{10}\right)^2 \\
& \\
& \sigma^2=\frac{430}{10}-\left(\frac{60}{10}\right)^2 \\
& \sigma^2=43-36 \\
& \sigma^2=7
\end{aligned}$$
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