The change in rate and Poisson distribution

Question

I was given the births in a country follow a Poisson process in which on average number of babies born in $24$ hours is $11.7$.

I figured out this indicates $0.4875$ babies are born per hour.

So how can I find the probability of more than $3$ hours between births?

Is this a change in rate? I'm new to probability and statistics.

The answer is $0.2317$, but how was it calculated ?

Answer 1

As per your information, births occur according to Poisson Process with rate parameter $\lambda=0.4875$.

In a Poisson process, the inter occurrence times follow Exponential distribution. We require the probability of the event: the time elapsed between two successive births is more than 3 hours.

This means that we require probability of having no births in an interval of length 3 hours. From Poisson Process this probability is $e^{-\lambda t }=e^{-0.4875 \times 3}= 0.2316564.$

Answer 2

If you have a Poisson process with rate $\lambda$, then the time $T$ between the events (child births in this case) is an exponential random variable with parameter $1/\lambda$. Letting $T \sim \operatorname{Expo}(1/\lambda)$, $$ P(\text{more than 3 hours between births}) = P(T > 3). $$

Answer 3

$$ \frac{3\text{ hours}}{24 \text{ hours}} = \frac 1 8. $$ Therefore the average number of births in $3$ hours is $$ \frac 1 8 \times 11.7 = 1.4625. $$ The probability that the number of births during that time is $0$ is therefore $$ \frac{1.4624^0 e^{-1.4625}}{0!} = 0.2316564\ldots. $$