JEE Main 2023 — Mathematics Probability & Statistics

Given,

$9={x}_{1}<{x}_{2}<\ldots <{x}_{7}$9=x1​<x2​<…<x7​ be in an A.P. with common difference $d$d

And the standard deviation of ${x}_{1},{x}_{2}\ldots ,{x}_{7}$x1​,x2​…,x7​ is 4 and the mean is $\bar{x}$xˉ,

Now solving, $9={x}_{1}<{x}_{2}<\ldots \ldots <{x}_{7}$9=x1​<x2​<……<x7​ which is an A.P, we get,

$$9,9+d,9+2d,\ldots \ldots \ldots .9+6d$$9,9+d,9+2d,……….9+6d

Now subtracting $9$9 from the series we get,

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

So, mean will be ${\bar{x}}_{\mathrm{new}}=\frac{21d}{7}=3d$xˉnew​=721d​=3d

Now using the formula of variance we get,

$${\sigma }^{2}=\sum _{i=1}^{n}{({x}_{i})}^{2}-{(\bar{x})}^{2}$$σ2=i=1∑n​(xi​)2−(xˉ)2

$$\Rightarrow 16=\frac{1}{7}({0}^{2}+{1}^{2}+\ldots \ldots ..+{6}^{2}){d}^{2}-9{d}^{2}$$⇒16=71​(02+12+……..+62)d2−9d2

$$\Rightarrow 16=\frac{1}{7}(\frac{6\times 7\times 13}{6}){d}^{2}-9{d}^{2}$$⇒16=71​(66×7×13​)d2−9d2

$$\Rightarrow 16=4{d}^{2}$$⇒16=4d2

$$\Rightarrow {d}^{2}=4$$⇒d2=4

$$\Rightarrow d=2$$⇒d=2

So, mean $\bar{x}=\frac{9+9+d+9+2d\ldots \ldots \ldots .9+6d}{7}=15$xˉ=79+9+d+9+2d……….9+6d​=15

Hence, $\bar{x}+{x}_{6}=15+9+5d=15+9+10=34$xˉ+x6​=15+9+5d=15+9+10=34