Verified Solution • 2025–26 Session moderate

QJEEM23Ma6eb4: JEE Main 2023 Mathematics - Class 11 JEEM23Ma6eb4 Solution: The mean and variance of the marks obtained by the students in…

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 :

Multiple Choice Options

  • A 3.92
  • B 4.08
  • C 3.96
  • D 4.04

Step-by-Step Verified Solution

xˉ=10 & σ2=4\bar{x}=10 ~\&~ \sigma^{2}=4 , No. of students =N=N (let)

∴∑xiN=10 & ∑xi2N−(10)2=4\therefore \quad \frac{\sum x_{i}}{N}=10 ~\&~ \frac{\sum x_{i}^{2}}{N}-(10)^{2}=4

Now if one of xix_{i} is changed from 8 to 12 we have

New mean ∑xi+4N=10+4N=10.2\frac{\sum x_{i}+4}{N}=10+\frac{4}{N}=10.2

⇒N=20\Rightarrow N=20

and σnew 2=∑xi2−(8)2+(12)220−(10⋅2)2\sigma_{\text {new }}^{2}=\frac{\sum x_{i}^{2}-(8)^{2}+(12)^{2}}{20}-(10 \cdot 2)^{2}

amp;=∑xi220+144−6420−(10⋅2)2amp;=104+4−(10⋅2)2amp;=108−104.04=3.96\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}

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