Question

Let $9=x_1<x_2<\dots <x_7$ be in an A.P. with common difference $d$. If the standard deviation of $x_1,x_2$ $\dots ,x_7$ is 4 and the mean is $\bar{x}$, then $\bar{x}+x_6$ is equal to :

• A. $18\left(1+\frac{1}{\sqrt{3}}\right)$
• B. $2\left(9+\frac{8}{\sqrt{7}}\right)$
• C. 34
• D. 25

Answer 1

Answer: (C)

$9=x_1<x_2<\dots \dots <x_7$
$9,9+d,9+2d,\dots \dots .9+6d$
$0,d,2d,\dots \dots .6d$
${\overline{x}}_{\text{new }}=\frac{21d}{7}=3d$
$16=\frac{1}{7}\left(0^2+1^2+\dots \dots .+6^2\right)d^2-9d^2$
$=\frac{1}{/7}\left(\frac{6\times /7\times 13}{/6}\right)d^2-9d^2$
$16=4d^2$
$d^2=4$
$d=2$
$\bar{x}+x_6=6+9+10+9$