Answer:
0.5372
Explanation:
Given that the number of births that occur in a hospital can be assumed to have a Poisson distribution with parameter = the average birth rate of 1.8 births per hour.
Let X be the no of births in the hospital per hour
X is Poisson
with mean = 1.8
the probability of observing at least two births in a given hour at the hospital
=
![P(X\geq 2)\\= 1-F(1)\\= 1-0.4628\\= 0.5372](https://img.qammunity.org/2021/formulas/mathematics/college/wx8cws85q1tdye7zqjngyrcdr7i2xo50m7.png)
the probability of observing at least two births in a given hour at the hospital = 0.5372