Let the mean of the data x 1 3 5 7 9 Frequency (f) 4 24 28 8 be 5.

Question

Let the mean of the data

$x$
 1
 3
 5
 7
 9

Frequency ($f$)
 4
 24
 28
 $\alpha $
 8

be 5. If $m$ and ${\sigma }^{2}$ are respectively the mean deviation about the mean and the variance of the data, then $\frac{3\alpha }{m+{\sigma }^{2}}$ is equal to \_\_\_\_\_\_\_\_\_\_

Answer: 8

Accepted Answer

$\begin{gathered} 5=\vec{x}=\frac{\sum {x}_{i}{f}_{i}}{\sum {f}_{i}}=\frac{4+72+140+7\alpha +72}{64+\alpha } \ \Rightarrow 320+5\alpha =288+7\alpha \Rightarrow 2\alpha =32\Rightarrow \alpha =16 \end{gathered}$

$\begin{gathered} \sum {f}_{i} =80 \ \ =\frac{\sum {f}_{i}|{x}_{i}-5|}{\sum {f}_{i}} \ \ =\frac{4+4+24	imes 2+0+16	imes 2+8	imes 4}{80} \ \ =\frac{8}{5} $

$ {\sigma }^{2} =\frac{\sum {f}_{i}|{x}_{i}-5|}{\sum {f}_{i}} \ \ =\frac{4+16+24	imes 4+0+16	imes 4+8	imes 16}{80}=\frac{22}{5} \end{gathered}$

So, $\frac{3\alpha }{m+{\sigma }^{2}}=\frac{3	imes 16}{\frac{8}{5}+\frac{22}{5}}=8$

Solutions · 1