Question

If the mean and variance of a random variable X with a binomial distribution are 4 and 2 respectively, find P (X = 1).

 

Answer 1

\[\text{ Given, np = 4 and npq } = 2\]
\[ \therefore p = 1 - \frac{\text{ Variance } }{\text{ Mean}}\]
\[ = 1 - \frac{2}{4}\]
\[ = \frac{1}{2}\]
\[\text{ and } q = \frac{1}{2} \]
\[\text{ and  }  n = \frac{np}{p}\]
\[ = \frac{4}{\frac{1}{2}}\]
\[ = 8\]
\[\text{ Hence, the binomial distribution is given by } \]
\[P\left( X = r \right) = ^ {8}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{8 - r} , r = 0, 1, 2, 3 . . . . . . . 8\]
\[ \therefore P(X = 1) = 8 \left( \frac{1}{2} \right)^8 \]
\[ = \frac{1}{32}\]