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Let the mean of the data
| $X$ | $1$ | $3$ | $5$ | $7$ | $9$ |
| $(f)$ | $4$ | $24$ | $28$ | $\alpha$ | $8$ |
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 $..........$.
$7$
$6$
$8$
$5$
Solution
$5=\bar{x}=\frac{\sum x_i f_i}{\sum f_i}=\frac{4+72+140+7 \alpha+72}{64+\alpha}$
$\Rightarrow 320+5 \alpha=288+7 \alpha \Rightarrow 2 \alpha=32 \Rightarrow \alpha=16$
$\text { M.. }(\bar{x})=\frac{\sum f _{ i }\left| x _{ i }-\overline{ x }\right|}{\sum f _{ i }} \text { where } \sum f _{ i }=64+16=80$
$\text { M.D. }(\bar{x})=\frac{4 \times 4+24 \times 2+28 \times 0+16 \times 2+8 \times 4}{80}$
$=\frac{8}{5}$
$\text { Variance }=\frac{\sum f _{ i }\left( x _{ i }-\overline{ x }\right)^2}{\sum f _{ i }}$
$=\frac{4 \times 16+24 \times 4+0+16 \times 4+8 \times 16}{80}=\frac{352}{80}$
$\therefore \frac{3 \alpha}{m+\sigma^2}=\frac{3 \times 16}{\frac{128}{80}+\frac{352}{80}}=8$
Similar Questions
Let $\mu$ be the mean and $\sigma$ be the standard deviation of the distribution
| $X_i$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ |
| $f_i$ | $k+2$ | $2k$ | $K^{2}-1$ | $K^{2}-1$ | $K^{2}-1$ | $k-3$ |
where $\sum f_i=62$. if $[x]$ denotes the greatest integer $\leq x$, then $\left[\mu^2+\sigma^2\right]$ is equal $………$.
