Question

Suppose 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. What is the probability of observing at least two births in a given hour at the hospital?

Answer 1

Answer

mean occurrence= 1.8 ($\mu$)

we have to find probability of observing at least 2 births in a given hour

$P(x \ge 2) = 1 - [P(x = 1)+P(x=0)]$

where

$P(x=1) = (e^{-\mu} * \mu^x)/x!$

where $\mu$ = 1.8 and x =1

$= (e^{-1.8} * 1.8^1)/1!$

$= 0.2975$

And

$P(x=0) = (e^{-\mu} * \mu^x)/x!$

where $\mu$ = 1.8 and x =0

$= (e^{-1.8} * 1.8^0)/0!$

$= 0.1653$

this implies

$P(x \ge 2) = 1 - [P(x = 1)+P(x=0)]$

$= 1-[0.2975+0.1653]$

$= 0.5372$

Probability is 0.5372