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 $32+8 \alpha+9 \beta=(8+\alpha+\beta) \times 6$
$\Rightarrow 2 \alpha+3 \beta=16 \ldots \ldots$ (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 $=\frac{32+35+18}{15}$
$=\frac{85}{15}=\frac{17}{3}$