Question

Let the mean and variance of the frequency distribution
$\begin{array}{ccccc}x:& x_1=2& x_2=6& x_3=8& x_4=9\\ f:& 4& 4& \alpha & \beta \end{array}$
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. $5$
• B. $4$
• C. $\frac{17}{3}$
• D. $\frac{16}{3}$

Answer 1

The correct option is C $\frac{17}{3}$

$\overline{x}=\frac{\sum x_i{f}_i}{\sum f_i}=\frac{8+24+8\alpha +9\beta }{8+\alpha +\beta }=6\Rightarrow 2\alpha +3\beta =16\dots \left(1\right)$
${\sigma }^2=\frac{\sum x_i^2{f}_i}{\sum f_i}-(\overline{x}{)}^2\Rightarrow \frac{16+144+64\alpha +81\beta }{8+\alpha +\beta }-36=6.8\Rightarrow 106\alpha +191\beta =912\dots (2)$
From $\left(1\right)$ and $\left(2\right),$ $\alpha =5$ and $\beta =2$
Now, new mean when $x_3=7$ is
$\frac{8+24+35+18}{15}=\frac{17}{3}$