Question
Let a random variable $X$ have a binomial distribution with mean $8$ and variance $4$. If $P\left( {X \le 2} \right) = \frac{k}{{{2^{16}}}}$, then $k$ is equal to
$\mathrm{npq}=4$
$\mathrm{q}=\frac{1}{2} \Rightarrow \mathrm{p}=\frac{1}{2}$
$n=16$
$p(x=r)=^{16} C_{r}\left(\frac{1}{2}\right)^{16}$
$p(x \leq 2)=\frac{^{16} C_{0}+^{16} C_{1}+^{16} C_{2}}{2^{16}}$
$=\frac{137}{2^{16}}$
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