# Is it about normalizing Poisson distribution or normal to Poisson Distribution?

Previous research has shown a relationship between the number of emergency room admissions and the level of pollution on a given day. A small local hospital finds the number of daily admissions to the emergency ward on days without high pollution follows a Poisson distribution with a mean of 2.0 admissions per day, and on days with high pollution follows a Poisson distribution with a mean of 4.0 admissions per day. Suppose that each admitted person to the emergency ward stays there for exactly one day and is then discharged so the bed occupied is free again. The hospital is planning a new emergency room facility.

(a) Suppose the hospital wants enough beds in the emergency ward so that on normal pollution days it will have enough beds to cope with the demand with at least 95% probability. What is the smallest number of beds it should have to satisfy this criterion? On normal pollution day the mean are 2 beds.Probability of availability of bed must be 0.95 then what should I do

(b) Suppose the hospital wants enough beds in the emergency ward so that on high pollution days it will have enough beds to cope with the demand with at least 95% probability. What is the smallest number of beds it should have to satisfy this criterion? Same case here

(c) What is the minimum number of beds required so that the demand for beds can be met on a randomly selected day of the year with probability of at least 95%, assuming a year contains 345 normal pollution days and 20 high pollution days?   No idea about this question

• For parts a and b you just have to consult a set of tables of cumulative Poisson – David Quinn Sep 18 '16 at 16:32
• can you explain I am beginner in statistics – Sindhu Bharathi Sep 18 '16 at 16:45

For (a), suppose $X \sim Pois(\lambda = 2).$ You seek $x$ such that $P(X \le x) = .95.$ If you have appropriate tables available, you can just look up the answer as suggested by @DavidQuinn. If not, you can use software. Here is how it could be done using R statistical software.

 x = 1:100
cdf = ppois(x, 2)
min(x[cdf>= .95])
## 5

x       cdf
1 0.4060058
2 0.6766764
3 0.8571235
4 0.9473470
5 0.9834364
6 0.9954662


As it turns out, this problem requires so little computation that it could be done on a calculator that that has an exponential function.

$$P(X \le 5) = e^{-2}[2^0/0! + 2^1/1! + 2^2/2! + 2^3/3! + 2^4/4! + 2^5/5!].$$

It might be argued (as in @Ian's comment) that you could try approximating $Pois(\lambda = 2)$ by $Normal(\mu = 2, \sigma = \sqrt{2}).$ But I think it is dangerous to approximate Poisson by normal when $\lambda$ is so small. Here is a figure that compares $Pois(2)$ (vertical bars) with the approximating normal density function. The fit is better in the right tail than near 2. Added later: A figure comparing the Poisson and normal CDFs. Part (b) is similar.

• Note that to a first approximation you could have estimated Pois(2) $\sim$ N(2,2), so that the desired quantile would be approximately $2+\sqrt{2} \Phi^{-1}(0.95)$. $\Phi^{-1}(0.95)$ is a familiar number in elementary statistics, approximately $1.65$. So the quantile should be around $2+(1.4 \cdot 1.7) \approx 4.3$, and indeed it is. – Ian Sep 18 '16 at 23:09
• @Ian: Can't argue whether it works in this case, because obviously it does. But I don't like to encourage use of normal approximation to Poisson for such small $\lambda,$ especially in an time when exact computations are readily available. – BruceET Sep 18 '16 at 23:48
• Sure. I would not do this to actually answer the question, but rather to get an idea of how much calculation is likely to be involved in more precisely answering the question. Knowing that the quantile from the normal approximation is $4.3$, even with $\lambda=2$ I would be quite confident in saying that the desired quantile is somewhere in $[3,7]$. – Ian Sep 18 '16 at 23:50
• That's fair enough. – BruceET Sep 18 '16 at 23:58
• Note that it's usually better to compare CDF-to-CDF rather than PMF-to-PDF. Comparing CDF-to-CDF the worst deviation is right at $2$ where it is about $0.18$, which is pretty bad...but it catches back up before $x=3$, so it isn't as bad as it sounds. – Ian Sep 19 '16 at 2:30