Question

If X follows a binomial distribution with mean 4 and variance 2, find P (X ≥ 5).

Answer 1

Here, mean (np) = 4
variance (npq ) =2

$$\begin{gathered} \therefore q=\frac{1}{2}\mathrm{and}p=\frac{1}{2} \\ n=\frac{\mathrm{Mean}}{p} \\ =4\times 2 \\ =8 \end{gathered}$$
$$\begin{gathered} \mathrm{Hence},\mathrm{the}\mathrm{distribution}\mathrm{is}\mathrm{given}\mathrm{by} \\ P(X=r)={}^{8}C_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{8-r},r=0,1,2......8 \\ P(X\ge 5)=P(X=5)+P(X=6)+P(\mathrm{X}=7)+P(X=8) \\ ={\left(\frac{1}{2}\right)}^8\left[{}^{8}C_5+{}^{8}C_6+{}^{8}C_7+{}^{8}C_8\right] \\ =\frac{(56+28+8+1)}{2^8} \\ =\frac{93}{256} \end{gathered}$$