# Probability of a event that happens N times in a range of time given others

I'm running an application for making shipments to any place around the globe. I have a set of rules which are like:

A customer makes a shipment...

• Once a week (ie, 50 shipments/year)
• At least once a month (ie, 15 shipments/year)
• At least once a trimester (ie, 6 shipments/year)
• At least once a semester (ie, 4 shipments/year)
• At least once a year (ie, 2 shipments/year)
• Occasionally (ie, 1 shipment/year)

Those are the different profiles of our customers... But I'd like to infer from those rules,

• what is the chance that a customer makes a shipment once a month (ie, 12 shipments/year)?
• What is the chance that a customer makes more than one shipment per month?

Assume I have the data, that one can find in the appendix, permiting that to calculate both the mean and standard desviation.

My approach

I thought about using the Normal Distribution, but turned out that, when doing P(X=12) was null (and that obviously, does not make any sense at all).

I have been exploring on the Internet, and I have found the Poisson Dist, which is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant rate and independently of the time since the last event. The Poisson distribution can also be used for the number of events in other specified intervals such as distance, area or volume (Wikipedia)

$$P(k\ events\ in\ interval\ t)=e^{-\lambda}\frac{\lambda^k}{k!}$$

where:

• $$\lambda$$ is the average number of events per interval (How should I calculate the mean of my data? $$\lambda=\frac{\sum_{1}^{i} X_i*N_i}{TOTAL}$$)?.
• $$k$$ number of events (in my case 12).

Is this last approach the correct one? does it make sense? Should I use different approaches based on the question?

Any help will be much appreciated.

APPENDIX

Mock data of our customers:

[Xi] Profile (shipments/year)        [Ni] Customers
1                                    261
2                                    473
4                                    139
6                                    419
15                                   79
50                                   24
0                                    6
TOTAL: 1401


## 1 Answer

In order to assume that a Poisson distribution is a reasonable approximation, we need the mean and the variance to be close (preferably equal), together with the other assumptions.

Using a notation close to what you used, the total number of customers $$\mathcal{N}$$ is: $$\mathcal{N} = \sum_{i=1}^{n} N_i$$

The mean $$\overline{X}$$ of your data can be computed by: $$\overline{X} = \frac{\sum_{i=1}^{n} N_i X_i}{\mathcal{N}}$$ where $$n$$ is the number of shipment categories.

And the variance $$s^2$$ of your data can be computed by: $$s^2 = \frac{\sum_{i=1}^{n} N_i X_i^2 - \mathcal{N} \overline{X}^2}{\mathcal{N} - 1}$$

In the data you presented, you have $$\mathcal{N} = 1401$$, $$\overline{X} = 4.755$$ and $$s^2 = 46.826$$, so I would say that the Poisson distribution may not have a good fit to your data.

However, from such a large number of customers, you might be able to make some reasonable estimates:

• The probability of finding a customer that makes a shipment in category $$i$$ is, approximately: $$P_i = \frac{N_{i}}{\mathcal{N}}$$ For the "once a month" category, we have a probability near 5%: $$P_i = \frac{79}{1401} = 0.05639$$

• And the probability of a customer making a shipment at least as often as in category $$i$$ is: $$P_{\geq i} = \frac{\sum_{j \geq i} N_{j}}{\mathcal{N}}$$ For the at least "once a month" shipment, we have a probability near 7%: $$P_i = \frac{79+24}{1401} = 0.07352$$