The number of visitors in a museum during a fixed time interval follows a Poisson distribution. Assume that the mean rate of visitors is 2 per minute. Find the probability that there will be exactly 20 visitors in the next 5 minutes. What I did was first to calculate the mean rate for every 5 minutes which gives me the following: mean rate for every 5 minutes = (2/min) x 5 min = 10
$$P(X=20)= e^{-10}\frac{10^{20}}{20!} $$ I am not 100% certain if my solution is correct, so can someone help me?
Your answer is correct. The standard formula gives us that the probability of $k$ events in an interval in which there are $\lambda$ events on average is $$ P(X=k)= e^{-\lambda} \frac{\lambda^k}{k!}.$$ In your case, since you have an average of $2$ visitors per minute, we can conclude that you will have $2\cdot5=10$ visitors on average in any $5$ minute interval. So, using $\lambda=10$ and $k=20$, we have that the probability of exactly $20$ visitors in a $5$ minute period is $$ P(X=20) = e^{-10} \frac{10^{20}}{20!} \approx 0.0018660... .$$