# Poisson distribution with large mean

The hits in a website can be modeled as a Poisson process with a rate of 5000 hits per day. Let X be the number of hits in 6 days. Give the distribution of X.

So, is the percentile x = 6 and the mean λ = 5000 * 6 = 30000? If so, should I use another distribution (as the normal distribution) to approximate it?

• The poisson has the interesting property that the sum of poisson random variables is also poisson. That should help – Demetri Pananos Oct 17 '19 at 20:16
• Maybe you need the self-study tag. – Michael R. Chernick Oct 17 '19 at 22:50

Changing the Poisson Rate to Match the Domain of a Model

One Day: It seems reasonable to model the number $$X_1$$ of events (hits) in one day as $$X_1 \sim \mathsf{Pois}(\lambda_1 = 5000),$$ so that $$E(X) = Var(X) = 5000.$$

It is not clear what additional numerical problems about this distribution you want to solve. In R, the 25th, 50th (median), and 75th percentiles of this distribution are $$4952, 5000,$$ and $$5048,$$ as follows:

 qpois(c(.25, .5, .75), 5000)
[1] 4952 5000 5048


[If possible, expand the following image in your browser to resolve individual, evenly-spaced bars.]

For example, if I have data for 300 individual days I might have summary statistics as below:

set.seed(1017)
x1 = rpois(300, 5000)
summary(x1)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
4809    4946    4999    5000    5051    5185


Thus, in this one sample of hits for each of 300 days, we see that about 75% of the days (that's 225 days) had 5051 or fewer hits. Specifically in this example, if we sort the 300 days from smallest to largest numbers of hits, days 224 through 226 in the list had 5050, 5050, and 5053 hits, respectively. (Of course, $$5051$$ in the sample of 300 can't be expected to match exactly $$5048$$ of the population, but the numbers are very close.)

sort(x)[224:226]
[1] 5050 5050 5053


Six days: If you want to know about the distribution of hits in six days, you need to adjust the parameter $$\lambda_1$$ per day to $$\lambda_6 = 6(5000) = 30\,000$$ per six days (assuming that the rate remains the same over six days).

This is called changing the Poisson rate to match the domain of the model. Here we have changed the domain from one day to six days.

Thus, $$E(X_6) = Var(X_6) = 30\,000.$$ And the first, second, and third quantiles are as shown below:

qpois(c(.25, .5, .75), 30000)
[1] 29883 30000 30117


You also say that the probability of no more than $$20\,000$$ hits in six days (less than or equal to $$29\,500$$ is very small---about $$0.00192.$$

ppois(29500, 30000)
[1] 0.001920345


If you want to use a normal distribution to approximate this probability, then you can standardize and use printed tables of the standard normal CDF (or somewhat more accurately, with software):

$$P(X_6 \le 29\,500) = P\left(\frac{X - \lambda_6}{\sqrt{\lambda_6}} < \frac{29\,500 - 30\,000}{\sqrt{30\,000}}\right)\\ \approx P(Z < -2.886751) = 0.00195,$$

where $$Z$$ is standard normal.

pnorm((29500-30000)/sqrt(30000))
[1] 0.001946209