JEE MAIN - Mathematics (2023 - 13th April Morning Shift - No. 16)
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
Explanation
$$
\begin{aligned}
& 5=\bar{x}=\frac{\sum x_i f_i}{\sum f_i}=\frac{4+72+140+7 \alpha+72}{64+\alpha} \\\\
& \Rightarrow 320+5 \alpha=288+7 \alpha \Rightarrow 2 \alpha=32 \Rightarrow \alpha=16
\end{aligned}
$$
$$ \begin{aligned} \sum f_i & =80 \\\\ \text { M.D } & =\frac{\sum f_i\left|x_i-5\right|}{\sum f_i} \\\\ & =\frac{4+4+24 \times 2+0+16 \times 2+8 \times 4}{80} \\\\ & =\frac{8}{5} \end{aligned} $$
$$ \begin{aligned} \sigma^2 & =\frac{\sum f_i\left|x_i-5\right|}{\sum f_i} \\\\ & =\frac{4+16+24 \times 4+0+16 \times 4+8 \times 16}{80}=\frac{22}{5} \end{aligned} $$
So, $$ \frac{3 \alpha}{m+\sigma^2}=\frac{3 \times 16}{\frac{8}{5}+\frac{22}{5}}=8 $$
$$ \begin{aligned} \sum f_i & =80 \\\\ \text { M.D } & =\frac{\sum f_i\left|x_i-5\right|}{\sum f_i} \\\\ & =\frac{4+4+24 \times 2+0+16 \times 2+8 \times 4}{80} \\\\ & =\frac{8}{5} \end{aligned} $$
$$ \begin{aligned} \sigma^2 & =\frac{\sum f_i\left|x_i-5\right|}{\sum f_i} \\\\ & =\frac{4+16+24 \times 4+0+16 \times 4+8 \times 16}{80}=\frac{22}{5} \end{aligned} $$
So, $$ \frac{3 \alpha}{m+\sigma^2}=\frac{3 \times 16}{\frac{8}{5}+\frac{22}{5}}=8 $$
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