Easy MCQ +4 / -1 PYQ · JEE Mains 2025

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

  1. A 5
  2. B 7 Correct answer
  3. C 9
  4. D 4

Solution

<p>$$\begin{aligned} & \text { Mean } \overline{\mathrm{x}}=5.5 \\ & =\sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}=5.5 \times 10=55 \\ & =\sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}^2=371 \\ & \left(\sum \mathrm{x}_{\mathrm{i}}\right)_{\text {new }}=55-(4+5)+(6+8)=60 \\ & \left(\sum \mathrm{x}_{\mathrm{i}}\right)_{\text {new }}=371-\left(4^2+5^2\right)+\left(6^2+8^2\right)=430 \\ & \\ & \text { Variance } \sigma^2=\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{10}-\left(\frac{\sum \mathrm{x}_{\mathrm{i}}}{10}\right)^2 \\ & \\ & \sigma^2=\frac{430}{10}-\left(\frac{60}{10}\right)^2 \\ & \sigma^2=43-36 \\ & \sigma^2=7 \end{aligned}$$</p>

About this question

Subject: Mathematics · Chapter: Statistics · Topic: Measures of Central Tendency

This question is part of PrepWiser's free JEE Main question bank. 91 more solved questions on Statistics are available — start with the harder ones if your accuracy is >70%.

Drill 25 more like these. Every day. Free.

PrepWiser turns these solved questions into a daily practice loop. Chapter-wise drills, full mocks, AI doubt chat. No auto-renew.

Start free →