91Ó°ÊÓ

An electronics store expects to have ten returns per day on average. The manager wants to know the probability of the store getting fewer than eight returns on any given day.

Let X be the number that the store getting retunrs on any given day. Therefore the sample size X takes on the values $0,1,2,3,4,5,6,7,8,9,10…$

Here X follows a poisson distribution with the probability mass function of poisson distribution as

$P(X=x)=\frac{e^{-\mathrm{μ}}{\mathrm{μ}}^x}{x!}$

According to the question, the expected number of returns ten per day on average. So average of Poisson distribution is $\mathrm{μ}=10$.

Therefore the probability of the store getting fewer than eight returns on any given day is determined as:

$P(X≤8)=0.33282$

Step in TI-83 + calculator as follows:

- Press 1 -and then press $2^{\text{nd}}$DISTR

- Arrow down to poissoncdf. Press ENTER

- Enter $(2.0,4)$

- The result shows

Therefore the required probability that the store getting fewer than eight returns on any given day is$0.3328.$