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

\[\therefore q = \frac{1}{2}\text{ and } p = \frac{1}{2} \]

\[n = \frac{\text{ Mean } }{p}\]

\[ = 4 \times 2\]

\[ = 8\]

\[\text{ Hence, the distribution is given by } \]

\[P(X = r) = ^{8}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{8 - r} , r = 0, 1, 2 . . . . . . 8\]

\[P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)\]

\[ = \left( \frac{1}{2} \right)^8 \left[ ^{8}{}{C}_5 + ^{8}{}{C}_6 + ^{8}{}{C}_7 + ^{8}{}{C}_8 \right]\]

\[ = \frac{(56 + 28 + 8 + 1)}{2^8}\]

\[ = \frac{93}{256}\]