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