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A random variable X follows binomial distribution with mean $\alpha$ and variance $\beta$. Then
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Verified Answer
The correct answer is:
$0 < \beta < \alpha$
Mean, $\mathrm{np}=\alpha$; and variance, $\mathrm{npq}=\beta$ where $\mathrm{n}$ $=$ number of trials and $\mathrm{p}+\mathrm{q}=1$
So, $\frac{\mathrm{npq}}{\mathrm{np}}=\frac{\beta}{\alpha} \Rightarrow \mathrm{q}=\frac{\beta}{\alpha} \Rightarrow(1-\mathrm{p})=\frac{\beta}{\alpha}$
$\because 0 < \mathrm{p} < 1$
$\therefore-1 < -\mathrm{p} < 0 \Rightarrow 0 < 1-\mathrm{p} < 1$
$\Rightarrow 0 < \frac{\beta}{\alpha} < 1 \Rightarrow 0 < \beta < \alpha$
So, $\frac{\mathrm{npq}}{\mathrm{np}}=\frac{\beta}{\alpha} \Rightarrow \mathrm{q}=\frac{\beta}{\alpha} \Rightarrow(1-\mathrm{p})=\frac{\beta}{\alpha}$
$\because 0 < \mathrm{p} < 1$
$\therefore-1 < -\mathrm{p} < 0 \Rightarrow 0 < 1-\mathrm{p} < 1$
$\Rightarrow 0 < \frac{\beta}{\alpha} < 1 \Rightarrow 0 < \beta < \alpha$
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