Question

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

• A. $1$
• B. $17$
• C. $121$
• D. $137$

Answer 1

The correct option is D $137$

Given that binomial distribution with mean $np=8$ and variance $npq=4$
therefore, $q=\frac{1}{2}\Rightarrow n=16,p=\frac{1}{2},$
Hence $P(X\le 2)$$=P(X=0)+P(X=1)+P(X=2)$
$={}^{16}{C}_0{\left(\frac{1}{2}\right)}^{16}+{}^{16}{C}_1{\left(\frac{1}{2}\right)}^1{\left(\frac{1}{2}\right)}^{15}+{}^{16}{C}_2{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^{14}$
$=\frac{1+16+120}{2^{16}}=\frac{137}{2^{16}}=\frac{k}{2^{16}}$