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

Let number of trials be n, probability of success be p and that of failure be q.
Given, mean $\left(np\right)=8$, variance $\left(npq\right)=4$

$\Rightarrow q=\frac{4}{8}=\frac{1}{2}\therefore p=1-\frac{1}{2}=\frac{1}{2}\left[\because p+q=1\right]$ So, $n=16$
Now, $P\left(X\le 2\right)=\frac{{ }^{16}{C}_0+{}^{16}{C}_1+{}^{16}{C}_2}{2^{16}}=\frac{1+16+120}{2^{16}}=\frac{137}{2^{16}}=\frac{k}{2^{16}}$

$\therefore k=137$