Question

The mean number of births per minute in a country in a recent year was about nine. Find the probability that the

number of births in any given minute is (a) exactly six, (b) at least six, and (c) more than six.

Answer 1

Let $X=$ the number of births : $X\sim Po(\lambda).$

Given $\lambda=9.$

(a)

$$P(X=6)=\dfrac{e^{-9}(9)^6}{6!}=0.09109$$

(b)

$$P(X\ge6)=1-P(X=0)-P(X=1)$$

$$-P(X=2)-P(X=3)$$

$$-P(X=4)-P(X=5)$$

$$=1-\dfrac{e^{-9}(9)^0}{0!}-\dfrac{e^{-9}(9)^1}{1!}$$

$$-\dfrac{e^{-9}(9)^2}{2!}-\dfrac{e^{-9}(9)^3}{3!}$$

$$-\dfrac{e^{-9}(9)^4}{4!}-\dfrac{e^{-9}(9)^5}{5!}=0.88431$$

(c)

$$P(X>6)=P(X\ge 6)-P(X=6)$$

$$=0.79322$$