The maternity ward at Dr. Jose Fabella Memorial Hospital in Manila in the Philippines is one of the busiest in the world with an average of 60 births per day. Let X = the number of births in an hour.

a. Find the mean and standard deviation of X.

b. Sketch a graph of the probability distribution of X.

c. What is the probability that the maternity ward will deliver three babies in one hour?

d. What is the probability that the maternity ward will deliver at most three babies in one hour?

e. What is the probability that the maternity ward will deliver more than five babies in one hour?

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:

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

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

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

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

$$\begin{gathered} P(X>5)=1-P\hspace{0.17em}(X\le 5) \\ P(X>5)=1-P\hspace{0.17em}(X=0)-P(X=1)-P(X=2)-P(X=3)-P(X=4)-P(X=5) \\ P(X>5)=1-0.756-0.1336-0.0668 \\ P(X>5)=0.04203 \end{gathered}$$