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The mean and standard deviation of a binomial distribution are 12 and 2 respectively. What is the number of trials?
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The correct answer is:
18
Mean, $x=n p=12$ ... (i)
Standard deviation
$\Rightarrow \sqrt{\mathrm{npq}}=2$
$\Rightarrow n p q=4$ ... (ii)
$\frac{(\mathrm{ii})}{(\mathrm{i})} \Rightarrow \frac{\not \mathrm{h} \mathrm{pq}}{\not \mathrm{h} \not \mathrm{p}}=\frac{4}{12} \Rightarrow \mathrm{q}=\frac{1}{3}$
$\mathrm{p}=1-\frac{1}{3}=\frac{2}{3}$
Now, (i) $\Rightarrow \mathrm{np}=12$
$\Rightarrow n\left(\frac{2}{3}\right)=12 \Rightarrow 2 n=36 \Rightarrow n=18$
Standard deviation
$\Rightarrow \sqrt{\mathrm{npq}}=2$
$\Rightarrow n p q=4$ ... (ii)
$\frac{(\mathrm{ii})}{(\mathrm{i})} \Rightarrow \frac{\not \mathrm{h} \mathrm{pq}}{\not \mathrm{h} \not \mathrm{p}}=\frac{4}{12} \Rightarrow \mathrm{q}=\frac{1}{3}$
$\mathrm{p}=1-\frac{1}{3}=\frac{2}{3}$
Now, (i) $\Rightarrow \mathrm{np}=12$
$\Rightarrow n\left(\frac{2}{3}\right)=12 \Rightarrow 2 n=36 \Rightarrow n=18$
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