91Ó°ÊÓ

A customer service center receives about ten emails every half -hours. To determined the probability that the customer service receives more than four emails in the next six minutes.

Let X the number of email receives in the next six minutes. 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 and the probability mass function of poisson distribution of X is written as

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

According to the question, a customer service center receives about ten emails every half-hour. Therefore in minute the customer service center receives $\frac{10}{30}=0.333$emails.

Therefore in six minutes average number of email receives is

$0.333×6=2$

So average of Poisson distribution is$\mathrm{μ}=2$

Therefore the required probability that the customer service center receives more than four emails in the next six minutes is determined as:

$\begin{array}{r}P(x>4)=1−P(x≤4)\\ =1−\left[P\right(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4\left)\right]\\ =0.0526\end{array}$

Step in TI-83 + calculator as follows:

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

- Arrow down to poissoncdf. Press ENTER

- Enter localid="1649535297699" $(2.0,4)$

- The result shows

$P(x>4)=0.0526$

Therefore the required probability that the customer service center receives more than four emails in the next six minutes is $0.0526$.