Q35 / 75 · JEE Main 2025 (24 Jan Shift 1) · Mathematics
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
- A$9$
- B$5$
- Correct answerC$7$
- D$4$
Solution
$\begin{aligned} & \text { Mean } \overline{\mathrm{x}}=5.5 \\ & =\sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}=5.5 \times 10=55 \\ & =\sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}^2=371 \\ & \left(\sum \mathrm{x}_{\mathrm{i}}\right)_{\text {new }}=55-(4+5)+(6+8)=60 \\ & \\ & \left(\sum \mathrm{x}_{\mathrm{i}}^2\right)_{\text {new }}=371-\left(4^2+5^2\right)+\left(6^2+8^2\right)=430\end{aligned}$
$\begin{aligned} & \text { Variance } \sigma^2=\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{10}-\left(\frac{\sum \mathrm{x}_{\mathrm{i}}}{10}\right)^2 \\ & \sigma^2=\frac{430}{10}-\left(\frac{60}{10}\right)^2 \\ & \sigma^2=43-36 \\ & \sigma^2=7\end{aligned}$
Hence, the correct answer is option: 'C'.