JEE MAIN - Mathematics (2025 - 4th April Evening Shift - No. 3)
Let the mean and the standard deviation of the observation $2,3,3,4,5,7, a, b$ be 4 and $\sqrt{2}$ respectively. Then the mean deviation about the mode of these observations is :
$\frac{1}{2}$
$\frac{3}{4}$
1
2
Explanation
$$\begin{aligned} &\begin{aligned} & \frac{2+3+3+4+5+7+a+b}{8}=4 \\ & \Rightarrow a+b=8 \\ & (\sqrt{2})^2=\frac{2^2+3^2+3^2+4^2+5^2+7^2+a^2+b^2}{8}-16 \\ & 112+a^2+b^2=18 \times 8 \\ & \Rightarrow a^2+b^2=32 \\ & \Rightarrow a=b=4 \end{aligned}\\ &\text { Now numbers be }\\ &2,3,3,4,4,4,5,7 \end{aligned}$$
Mode $=4$
Mean deviation about mode :
$$\begin{aligned} & \frac{|2-4|+|3-4|+|3-4|+0+0+0+|5-4|+|4-7|}{8} \\ = & \frac{2+1+1+1+3}{8}=\frac{8}{8}=1 \end{aligned}$$
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