Let the mean and variance of the frequency distribution

x :	${\mathrm{x}}_1=2$	${\mathrm{x}}_2=6$	${\mathrm{x}}_3=8$	${\mathrm{x}}_4=9$
$\text{⨍ :}$	4	4	$\mathrm{\alpha }$	$\mathrm{\beta }$

Be 6 and 6.8 respectively. If ${\mathrm{x}}_3$ is changed from 8 to 7, then the mean for the new data will be:

$\begin{array}{cccc}\mathrm{x} & f& \left|{\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right| & f_{\mathrm{i}}\mathit{ }{\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^2\\ {\mathrm{x}}_1=2& 4& 4& 64\\ {\mathrm{x}}_2=6 & 4& 0& 0\\ {\mathrm{x}}_3=8& \mathrm{\alpha }& 2& 4 \mathrm{\alpha }\\ {\mathrm{x}}_4=9& \mathrm{\beta }& 3& 9 \mathrm{\beta }\end{array}$

Mean $=\frac{\sum f_{\mathrm{i}} {\mathrm{x}}_{\mathrm{i}}}{\sum f_{\mathrm{i}}}=6$

$\begin{array}{rcl} & & \frac{8+24+8\mathrm{\alpha }+9\mathrm{\beta }}{\sum {\mathrm{f}}_{\mathrm{i}}}\\ \frac{8+24+8\mathrm{\alpha }+9\mathrm{\beta }}{8+\mathrm{\alpha }+\mathrm{\beta }}& =& 6\\ 32+8\mathrm{\alpha }+9\mathrm{\beta }& =& 48+6\mathrm{\alpha }+6\mathrm{\beta }\\ 2\mathrm{\alpha }+3\mathrm{\beta }& =& 16 -\left(1\right)\end{array}$

$\begin{array}{rcl}{\mathrm{\sigma }}^2& =& \frac{\sum {\mathrm{f}}_{\mathrm{i}} \left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}{\sum {\mathrm{f}}_{\mathrm{i}}}\\ 6.8& =& \frac{64+4\mathrm{\alpha }+9\mathrm{\beta }}{8+\mathrm{\alpha }+\mathrm{\beta }}\\ \frac{34}{5}& =& \frac{64+4\mathrm{\alpha }+9\mathrm{\beta }}{8+\mathrm{\alpha }+\mathrm{\beta }}\\ 272+34\mathrm{\alpha }+34\mathrm{\beta }& =& 320+20\mathrm{\alpha }+45\mathrm{\beta }\\ 14\mathrm{\alpha }-11\mathrm{\beta }& =& 48 -\left(2\right)\end{array}$

from (1) and (2)

$\mathrm{\alpha }=5, \mathrm{\beta }=2$

$$\begin{gathered} {\mathrm{x}}_{\mathrm{i}} 2 6 7 9 \\ {f}_{\mathrm{i}} 4 4 5 2 \end{gathered}$$

after changing ${\mathrm{x}}_3=8 \mathrm{to} 7$

New Mean $=\frac{\sum f_{\mathrm{i}} {\mathrm{x}}_{\mathrm{i}}}{\sum f_{\mathrm{i}}}$

$\begin{array}{rcl}\overline{\mathrm{X}}& =& \frac{8+24+35+18}{4+4+5+2}\\ & =& \frac{85}{15}\\ & =& \frac{17}{3}\end{array}$