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  • Let the mean of the data <img alt="\mathrm{x}" src="https://entrancecorner.oncodecogs.com/gif

Let the mean of the data

\mathrm{x} 1 3 5 7 9
Frequency (F) 4 24 28 \mathrm{\alpha } 8

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

Option: 1

8


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\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} & \text { M.D. }(\overline{\mathrm{x}})=\frac{\sum \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\sum \mathrm{f}_{\mathrm{i}}} \text { where } \sum \mathrm{f}_{\mathrm{i}}=64+16=80 \\ & \text { M.D. }(\overline{\mathrm{x}})=\frac{4 \times 4+24 \times 2+28 \times 0+16 \times 2+8 \times 4}{80}=\frac{8}{5} \end{aligned}

\begin{aligned} & \text { Variance }=\frac{\sum \mathrm{f}_{\mathrm{i}}\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2}{\sum \mathrm{f}_{\mathrm{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 \end{aligned}

Posted by

Gautam harsolia

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