Let the mean and the standard deviation of the observation $2,3,3,4,5,7$, a, b be 4 and $\sqrt{2}$…
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 :
$1$
$\frac{3}{4}$
$2$
$\frac{1}{2}$
Solution
$\begin{aligned} & \frac{24+a+b}{8}=4 \\ & a+b=8 \\ & 2=\frac{4+1+1+0+1+9+(a-4)^2+(b-4)^2}{8} \\ & 16=48+a^2+b^2-8 a-8 b \\ & a^2+b^2=32 \\ & 32=2 a b \\ & a b=16 \\ & a=4 b=4 \\ & \text { mode }=4 \\ & \text { mean deviation }=\frac{2+1+1+0+1+3+0+0}{8}=1 \\ & \text { option }(1)\end{aligned}$
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