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Question

# If the mean and variance of a random variable X with a binomial distribution are 4 and 2 respectively, find P (X = 1).

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Solution

## $\mathrm{Given},\mathit{}np=4\mathrm{and}npq=2\phantom{\rule{0ex}{0ex}}\therefore p=1-\frac{\mathrm{Variance}}{\mathrm{Mean}}\phantom{\rule{0ex}{0ex}}=1-\frac{2}{4}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{and}q=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\text{and}n\mathit{}=\frac{np}{p}\phantom{\rule{0ex}{0ex}}=\frac{4}{\frac{1}{2}}\phantom{\rule{0ex}{0ex}}=8\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{the}\mathrm{binomial}\mathrm{distribution}\mathrm{is}\mathrm{given}\mathrm{by}\phantom{\rule{0ex}{0ex}}P\left(X=r\right)={}^{8}C_{r}{\left(\frac{1}{2}\right)}^{r}{\left(\frac{1}{2}\right)}^{8-r},r=0,1,2,3.......8\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore P\left(X=1\right)=8{\left(\frac{1}{2}\right)}^{8}\phantom{\rule{0ex}{0ex}}=\frac{1}{32}$

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