Let the mean of the data be 5.

Question

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

Answer: 8

Accepted Answer

From the recovered working: $5 = \bar x = \dfrac{4+72+140+7\alpha+72}{64+\alpha} \implies 320+5\alpha = 288+7\alpha \implies \alpha=16$

With $\sum f_i = 64+16=80$:

$$\text{M.D.}(\bar x) = \frac{4\times4+24\times2+28\times0+16\times2+8\times4}{80} = \frac{8}{5}$$

$$\text{Variance} = \frac{4\times16+24\times4+0+16\times4+8\times16}{80} = \frac{352}{80}$$

$$\frac{3\alpha}{m+\sigma^2} = \frac{3\times16}{128/80+352/80} = \frac{48}{480/80}=\frac{48}{6}=8$$

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