Verified Solution • 2025–26 Session moderate

QJEEM24Maa94f: JEE Main 2024 Mathematics - Class 11 JEEM24Maa94f Solution: Let . Let the mean and the variance of 6 observations be 2 and…

Let α,β∈R\alpha, \beta \in \mathbf{R} . Let the mean and the variance of 6 observations −3,4,7,−6,α,β-3,4,7,-6, \alpha, \beta be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is :

Multiple Choice Options

  • A 163\frac{16}{3}
  • B 113\frac{11}{3}
  • C 143\frac{14}{3}
  • D 133\frac{13}{3}

Step-by-Step Verified Solution

amp; Mean =−3+4+7+(−6)+α+β6=2amp;⇒α+β=10amp; Variance =∑xi2n−(xˉn)2=23amp;⇒∑xi2=27×6amp;⇒9+16+49+36+α2+β2=162amp;⇒α2+β2=52\begin{aligned}& \text { Mean }=\frac{-3+4+7+(-6)+\alpha+\beta}{6}=2 \\& \Rightarrow \alpha+\beta=10 \\& \text { Variance }=\frac{\sum x_i^2}{n}-\left(\frac{\bar{x}}{n}\right)^2=23 \\& \Rightarrow \sum x_i^2=27 \times 6 \\& \Rightarrow 9+16+49+36+\alpha^2+\beta^2=162 \\& \Rightarrow \alpha^2+\beta^2=52\end{aligned}

We get α\alpha and β\beta as 4 and 6

So, mean deviation about mean

amp;=∣−3−2∣+∣4−2∣+∣7−2∣+∣−6−2∣+∣4−2∣+∣6−2∣6amp;=5+2+5+8+2+46amp;=133\begin{aligned}& =\frac{|-3-2|+|4-2|+|7-2|+|-6-2|+|4-2|+|6-2|}{6} \\& =\frac{5+2+5+8+2+4}{6} \\& =\frac{13}{3}\end{aligned}

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