Verified Solution • 2025–26 Session moderate

QJEEM23Ma6020: JEE Main 2023 Mathematics - Class 11 JEEM23Ma6020 Solution: Let the mean and variance of 12 observations be and 4…

Let the mean and variance of 12 observations be 92\frac{9}{2} and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is mn\frac{m}{n} , where m\mathrm{m} and n\mathrm{n} are coprime, then m+n\mathrm{m}+\mathrm{n} is equal to :

Multiple Choice Options

  • A 317
  • B 316
  • C 314
  • D 315

Step-by-Step Verified Solution

amp; Since, Mean =92amp;⇒Σx=92×12=54\begin{aligned}& \text { Since, Mean }=\frac{9}{2} \\\\& \Rightarrow \Sigma x=\frac{9}{2} \times 12=54\end{aligned}

Also, variance =4=4

amp;⇒∑x212=[∑xi12]2=4amp;⇒∑x212=4+814=974amp;⇒∑x2=291\begin{aligned}& \Rightarrow \frac{\sum x^2}{12}=\left[\frac{\sum x_i}{12}\right]^2=4 \\\\& \Rightarrow \frac{\sum x^2}{12}=4+\frac{81}{4}=\frac{97}{4} \\\\& \Rightarrow \sum x^2=291\end{aligned}

amp;∑x′=54−(9+10)+7+14amp;=54−19+21=56amp; and ∑x2=291−(81+100)+49+196amp;=291−181+49+196=355\begin{aligned}& \sum x^{\prime}=54-(9+10)+7+14 \\\\& =54-19+21=56 \\\\& \text { and } \sum x^2=291-(81+100)+49+196 \\\\& =291-181+49+196=355\end{aligned}

amp; So, σnew 2=∑xnew 212−(∑xnew 12)2amp;=35512−(5612)2amp;=4260−3136144=1124144=28136amp;=mnamp;⇒m=281,n=36amp;⇒m+n=281+36=317\begin{aligned}& \text { So, } \sigma_{\text {new }}^2=\frac{\sum x_{\text {new }}^2}{12}-\left(\frac{\sum x_{\text {new }}}{12}\right)^2 \\\\& =\frac{355}{12}-\left(\frac{56}{12}\right)^2 \\\\& =\frac{4260-3136}{144}=\frac{1124}{144}=\frac{281}{36} \\\\& =\frac{m}{n} \\\\& \Rightarrow m=281, n=36 \\\\& \Rightarrow m+n=281+36=317\end{aligned}

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