Question

Let X be a binomially distributed variate with mean 10 and variance 5. Then p(x>10) is

• A. $\frac{1}{2^{20}}{\sum }_{k=11}^{20}$${}^{20}$$C_k$
• B. $\frac{1}{2^{20}}{\sum }_{k=1}^{11}$${}^{20}{C}_k$
• C. $\frac{1}{2^{20}}{\sum }_{k=11}^{20}$${}^{10}{C}_k$
• D. ${\sum }_{k=11}^{20}$ ${}^{20}{C}_k{\left(\frac{2}{3}\right)}^{30-k}$

Answer 1

The correct option is A $\frac{1}{2^{20}}{\sum }_{k=11}^{20}$${}^{20}$$C_k$

Mean $=\mu =np=10$ and Variance = ${\sigma }^2=np(1-p)=5$
So, $10\times (1-p)=5=>p=\frac{1}{2}$
$n=\frac{10}{p}=10\times 2=20$
p(x=k)= ${}^n{C}_k{p}^k{(1-p)}^{n-k}$

p(x>10) = ${}^{20}{C}_{11}{\frac{1}{2}}^{11}{(1-\frac{1}{2})}^{20-11}{+}^{20}{C}_{12}{\frac{1}{2}}^{12}{(1-\frac{1}{2})}^{20-12}+.....{+}^{20}{C}_{20}{\frac{1}{2}}^{20}{(1-\frac{1}{2})}^{20-20}$

${=}^{20}{C}_{11}\left(\frac{1}{2}{)}^{20}{+}^{20}{C}_{12}\right(\frac{1}{2}{)}^{20}{+}^{20}{C}_{13}(\frac{1}{2}{)}^{20}+....{+}^{20}{C}_{20}(\frac{1}{2}{)}^{20}$

$={\frac{1}{2}}^{20}{\sum }_{k=11}^{20}{}^{20}{C}_k$