JEE MAIN - Mathematics (2024 - 4th April Morning Shift - No. 6)
Let $$\alpha, \beta \in \mathbf{R}$$. Let the mean and the variance of 6 observations $$-3,4,7,-6, \alpha, \beta$$ be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is :
$$\frac{16}{3}$$
$$\frac{11}{3}$$
$$\frac{14}{3}$$
$$\frac{13}{3}$$
Explanation
$$\begin{aligned} & \text { Mean }=\frac{-3+4+7+(-6)+\alpha+\beta}{6}=2 \\ & \Rightarrow \alpha+\beta=10 \\ & \text { Variance }=\frac{\sum x_i^2}{n}-\left(\frac{\bar{x}}{n}\right)^2=23 \\ & \Rightarrow \sum x_i^2=27 \times 6 \\ & \Rightarrow 9+16+49+36+\alpha^2+\beta^2=162 \\ & \Rightarrow \alpha^2+\beta^2=52 \end{aligned}$$
We get $$\alpha$$ and $$\beta$$ as 4 and 6
So, mean deviation about mean
$$\begin{aligned} & =\frac{|-3-2|+|4-2|+|7-2|+|-6-2|+|4-2|+|6-2|}{6} \\ & =\frac{5+2+5+8+2+4}{6} \\ & =\frac{13}{3} \end{aligned}$$
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