# Customers arrive randomly at a super market at an average rate of 3.4 per minute. Assuming that the number of arrivals follows a Poisson distribution, find the probability that one or more...

Customers arrive randomly at a super market at an average rate of 3.4 per minute. Assuming that the number of arrivals follows a Poisson distribution, find the probability that one or more customers will arrive in any given 30 second interval.

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### 2 Answers

The probability mass function (pmf) of the Poisson distribution is given by

`P(X=x) = (e^(-lambdaT)(lambdaT)^x)/(x!)` , `x = 0,1,2, ...`

The parameter `lambda` is the rate of events per time interval length `T`.

Here, we are interested in the number of events in any 30 second interval. Since the expected number of arrivals in a minute is 3.4, the expected number in an interval of length `T=0.5` min is `lambda = 3.4/2 = 1.7`.

Given that the number of arrivals in any 30 second interval is assumed to b Poisson(`lambda`=0.7), the probability that one or more customers arrive in an interval of that length is

`P(X >= 1) = 1- P(X<1) = 1- P(X=0)`

Since, using the pmf given above,

`P(X=0) = (e^(-1.7) (1.7^0))/(0!) = e^(-1.7) = 0.183`

**the probability of one or more customers arriving in any given 30 second interval is 1 - 0.183 = 0.817 to 3dp.**

A Poisson distribution is defined by the following equation:

`P(X=x)=e^-lambdalambda^x/(x!)` ,

where `lambda` is the constant average rate per interval, and x is the interval number. In this case the interval is 30 seconds, therefore we need to convert 3.4/min to the average number of customers per 30 seconds:

`lambda=3.4/2=1.7`

We wish to know the value of P when `X>=1` :

`P(X>=1)=1-P(X<1)=1-P(X=0)`

`P(X=0)=e^-1.7(1.6^0/(0!))=0.183`

`P(X>=1)=1-0.183=0.817`

Therefore the probability that one or more customers will arrive at the supermarket in any 30 second period is 0.817.

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