Verified Solution • 2025–26 Session moderate

QJEEM25Ma4e85: JEE Main 2025 Mathematics - Class 11 JEEM25Ma4e85 Solution: For a statistical data of 10 values, a student obtained the…

For a statistical data x1,x2,…,x10\mathrm{x}_1, \mathrm{x}_2, \ldots, \mathrm{x}_{10} of 10 values, a student obtained the mean as 5.5 and ∑i=110xi2=371\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

Multiple Choice Options

  • A 5
  • B 7
  • C 9
  • D 4

Step-by-Step Verified Solution

amp; Mean x‾=5.5amp;=∑i=110xi=5.5×10=55amp;=∑i=110xi2=371amp;(∑xi)new =55−(4+5)+(6+8)=60amp;(∑xi)new =371−(42+52)+(62+82)=430amp;amp; Variance σ2=∑xi210−(∑xi10)2amp;amp;σ2=43010−(6010)2amp;σ2=43−36amp;σ2=7\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}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring