Question

Let the mean and variance of the frequency distribution

$x:$	$x_1=2$	$x_2=6$	$x_3=8$	$x_4=9$
$f:$	$4$	$4$	$\alpha$	$\beta$

be $6$ and $6.8$ respectively. If $x_3$ is changed from $8$ to $7$ , then the mean for the new data will be:

• A. $4$
• B. $5$
• C. $\frac{17}{3}$
• D. $\frac{16}{3}$

Answer 1

Answer: (C)

Given $32+8\alpha +9\beta =(8+\alpha +\beta )\times 6$
$\Rightarrow 2\alpha +3\beta =16\dots \dots$ (i)
Also, $4\times 16+4\times \alpha +9\beta =(8+\alpha +\beta )\times 6.8$
$\Rightarrow 640+40\alpha +90\beta =544+68\alpha +68\beta$
$\Rightarrow 28\alpha -22\beta =96$
$\Rightarrow 14\alpha -11\beta =48$
from (i) & (ii) $\alpha =5\&\beta =2$
so, new mean $=\frac{32+35+18}{15}$
$=\frac{85}{15}=\frac{17}{3}$