Question

If the mean and variance of a binomial variate X are 2 and 1 respectively, find P (X > 1).

 

Answer 1

\[\text{ Mean = 2 , Variance } = 1\]
\[ \therefore q = \frac{\text{ Variance} }{\text{ Mean } } = \frac{1}{2}\]
\[\text{ and p } = 1 - \frac{1}{2} = \frac{1}{2}\]
\[n = \frac{\text{ Mean} }{p} = \frac{2}{\frac{1}{2}} = 4\]
\[\text{ The binomial distribution is given by } \]
\[P(X = r) = ^ {4}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{4 - r} \]
\[ \therefore P(X = 0) = ^{4}{}{C}_0 \left( \frac{1}{2} \right)^0 \left( \frac{1}{2} \right)^{4 - 0} , r = 0, 1, 2, 3, 4\]
\[ = \left( \frac{1}{2} \right)^4 \]
\[P(X > 1) = 1 - P(X = 0) \]
\[ = 1 - \left( \frac{1}{2} \right)^4 \]
\[ = \frac{15}{16}\]