JEE Main 2023 — Statistics Question with Solution

Given,

$9={\mathrm{x}}_1<{\mathrm{x}}_2<\dots <{\mathrm{x}}_7$ be in an A.P. with common difference $d$

And the standard deviation of $x_1,x_2\dots ,x_7$ is 4 and the mean is $\overline{x}$,

Now solving, $9=x_1<x_2<\dots \dots <x_7$ which is an A.P, we get,

$9, 9+d, 9+2d,\dots \dots \dots .9+6d$

Now subtracting $9$ from the series we get,

$0, d, 2d,.......6d$

So, mean will be ${\overline{x}}_{\mathrm{new}}=\frac{21d}{7}=3d$

Now using the formula of variance we get,

${\sigma }^2=\sum _{i=1}^n{\left(x_i\right)}^2-{\left(\overline{x}\right)}^2$

$\Rightarrow 16=\frac{1}{7}\left(0^2+1^2+\dots \dots ..+6^2\right)d^2-9d^2$

$\Rightarrow 16=\frac{1}{7}\left(\frac{6\times 7\times 13}{6}\right)d^2-9d^2$

$\Rightarrow 16=4d^2$

$\Rightarrow d^2=4$

$\Rightarrow d=2$

So, mean $\overline{x}=\frac{9+ 9+d+9+2d\dots \dots \dots .9+6d}{7}=15$

Hence, $\overline{x}+x_6=15+9+5d=15+9+10=34$