Let a random variable $X$ have a Binomial distribution with mean 8 and variance 4 , If $P(X \leq 2)=\frac{k}{2^{16}}$ then $k$ is equal to

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Aicharya · Accepted Answer

$\begin{aligned} & n p=8, n p q=4 \ & \Rightarrow q=\frac{1}{2}, p=\frac{1}{2} 	ext { and } n=16 \end{aligned}$ Now $P(x \leq 2)=n_{C_0} p^0 q^n+n_{C_1} p^1 q^{n-1}+n_{C_2} p^2 q^{n-2}$ $\begin{aligned} & =\left\{{ }^{16} C_0+{ }^{16} C_1+{ }^{16} C_2ight\}\left(\frac{1}{2}ight)^{16} \ & =\frac{137}{2^{16}} \ & \Rightarrow k=137 \end{aligned}$

Let a random variable $X$ have a Binomial distribution with mean 8 and variance 4 , If $P(X \leq…

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