$\begin{aligned}
& \frac{\sum x_i}{6}=2 \text { and } \frac{\sum x_i^2}{N}-\mu^2=23 \
& \alpha+\beta=10 \
& \alpha^2+\beta^2=52
\end{aligned}solvingweget\alpha=4, \beta=6\frac{\sum\left|x_i-\bar{x}\right|}{6}=\frac{5+2+5+8+2+4}{6}=\frac{13}{3}$
Let α,β∈R. Let the mean and the variance of 6 observations −3,4,7,−6,α,β be 2 and 23 , respectively. The mean deviation about the mean of these 6 observations is :
Held on 4 Apr 2024 · Verified 6 Jul 2026.
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