91Ó°ÊÓ

In a large population, the Poisson distribution is used to characterize the distribution of unusual events.

We are given information,

X = the number of births in an hour and The maternity ward at the hospital in Manila is one of the busiest in the world with an average of $60$ births per day.

The average or mean of births in one hour is:

$$\begin{gathered} p=\frac{60}{24} \\ =2.5 \end{gathered}$$

The standard deviation is

$$\begin{gathered} s=\sqrt{p} \\ s=\sqrt{2.5} \\ s=1.5811 \end{gathered}$$

The graph of the probability distribution is:

The probability that the maternity ward will deliver three babies in one hour is:

$$\begin{gathered} P_n\left(X\right)=\frac{{\mathrm{λ}}^n}{n!}{e}^{-\mathrm{λ}} \\ P(X=3)=\frac{2.5^3}{3!}{e}^{-2.5} \\ P(X=3)=0.21376 \end{gathered}$$

The probability that the maternity ward will deliver at most three babies in one hour is:

$$\begin{gathered} P(X≤3)=P(X=0)+P(X=1)+P(X=2)+P(X=3) \\ P(X≤3)=\frac{2.5^0}{0!}{e}^{-2.5}+\frac{2.5^1}{1!}{e}^{-2.5}+\frac{2.5^2}{2!}{e}^{-2.5}+\frac{2.5^3}{3!}{e}^{-2.5} \\ P(X≤3)=0.08208+0.20521+0.25652+0.21376 \\ P(X≤3)=0.756 \end{gathered}$$

The probability that the maternity ward will deliver more than five babies in one hour is:

$$\begin{gathered} P(X>5)=1-P (X≤5) \\ P(X>5)=1-P (X=0)-P(X=1)-P(X=2)-P(X=3)-P(X=4)-P(X=5) \\ P(X>5)=1-0.756-0.1336-0.0668 \\ P(X>5)=0.04203 \end{gathered}$$