I've been revisiting Poisson distributions and I'm creating a question from scratch.
My main point is: when the time interval changes (in the case below from 5 to 8 hours), is it simply a matter of calculating another mean (proportionally)? I'll write, below the question, my answers for it.
Let's say that my daughter usually cries at night once every 5 hours. That said:
a) What is the probability that she cries once in that time frame?
Let's consider now an 8-hour block of sleep time.
b) If she wakes me up zero or once, I'll be fine the next day. What is the probability that I'm fine the next day?
c) If she wakes me up five times or more, I'll skip work that day. What is the probability that that happens?
My answers for the exercises are (not sure these are correct):
a) $\displaystyle P(X=1) = \frac{1^1 \cdot e^{-1}}{1!} \approx 0.368$
b) We need now to find another mean for the 8-hour block, which will be $1.6$ baby cries per hour
$\displaystyle P(X \leq 1) = P(0) + P(1) \approx 0.525$
c) Using the mean found in the exercise above,
$\displaystyle 1 - P(X \leq 4) \approx 0.0237$
Is my reasoning correct?
Yes, that is correct. In general, the probability distribution of $n$ cries in time $t$ is given by
$$P(N(t)=n) = \frac{(\lambda t)^n}{n!} e^{-\lambda t}$$
where $\lambda$ is the average number of cries per unit time. In your case, you chose $\lambda=1$ and your unit time is $5$ hours.
The reasoning behind this is that you are observing a Poisson point process. The Wikipedia page contains all useful interpretations of this process.