Help with Poisson Distribution asking for 90% chance

Question

Hello StackExchangers,

I had a question on calculating a percent chance that an event occurs under a Poisson distribution. The essential information is as follows. $X$ = # of patients needing a hospital bed tomorrow. $E(X) = 2.8.$ How many beds needed so that there is at least a 90% chance there will not be a shortage of beds.

My attempt:

In class, we have yet to discuss Poisson distributions for a percentage and just learned integer calculations thus far. However, I tried to conceptualize this percentage into the Poisson distribution formula and got the following: $P(X\le k)\ge 0.9.$ However, when attempting to solve for $k$ (the # of beds needed for a 90% chance), I was unable to calculate $k!.$ Any and all knowledge and assistance would be appreciated, thanks!

Answer 1

I'm not sure an algebraic solution is the best approach to this problem.

You have $X \sim \mathsf{Pois}(\lambda = 2.8).$ The PDF of $X$ is illustrated below:

You can use trial and error to see that $P(X \le 5) = 0.9349,$ but $P(X \le 4) = 0.8477,$ so that the answer is $x = 5$ beds.

Computations in R statistical software, where ppois denotes a Poisson CDF are shown below. Some statistical calculators will do such computations. Alternatively, perhaps there are some tables of the Poisson CDF in the back of your text. And just doing some hand computations of $P(X = x)$ for $x = 0, 1, 2, 3, 4, 5$ is not prohibitively difficult. (See note.)

ppois(5, 2.8)
## 0.9348897
ppois(4, 2.8)
## 0.8476761

The 'quantile' function in the inverse CDF. If you have access to software that computes quantile functions, you can get the answer directly, In R, a Poisson quantile function is denoted qpois.

qpois(.9, 2.8)
## 5

Note: $P(X \le 5) = e^{-2.8}\left(1 + 2.8 + \frac{2.8^2}{2!} + \frac{2.8^3}{3!} + \frac{2.8^4}{4!} + \frac{2.8^5}{5!}\right);\,$ a bit messy, but not messy enough to spoil your day.