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13.Statistics
hard
Let the six numbers $a_1, a_2, a_3, a_4, a_5, a_6$ be in $A.P.$ and $a_1+a_3=10$. If the mean of these six numbers is $\frac{19}{2}$ and their variance is $\sigma^2$, then $8 \sigma^2$ is equal to
A
$220$
B
$210$
C
$200$
D
$105$
(JEE MAIN-2023)
Solution
$a_1+a_3=10=a_1+d \Rightarrow 5$
$a_1+a_2+a_3+a_4+a_5+a_6=57$
$\Rightarrow \frac{6}{2}\left[a_1+a_6\right]=57$
$\Rightarrow a_1+a_6=19$
$\Rightarrow 2 a_1+5 d=19 \text { and } a_1+d=5$
$\Rightarrow a_1=2, d=3$
$\text { Numbers }: 2,5,8,11,14,17$
$\text { Variance }=\sigma^2=\text { mean of squares }-\text { square of mean }$ $=\frac{2^2+5^2+8^2+(11)^2+(14)^2+(17)^2}{6}-\left(\frac{19}{2}\right)^2 ~\\ =\frac{699}{6}-\frac{361}{4}=\frac{105}{4}$
$8 \sigma^2=210$
Standard 11
Mathematics
