5.6 Poisson di - university of calgary - base content - v2015  (Page 2/18)

Leah's answering machine receives about six telephone calls between 8 a.m. and 10 a.m. What is the probability that

1. Leah receives one call in the next 15 minutes?
2. Leah receives more than one call in the next 15 minutes?

Let X = the number of calls Leah receives in 15 minutes. (The interval of interest is 15 minutes or $\frac{1}{4}$ hour.)

x = 0, 1, 2, 3, ...

If Leah receives, on the average, six telephone calls in two hours, and there are eight 15 minute intervals in two hours, then Leah receives

$\left(\frac{1}{8}\right)$ (6) = 0.75 calls in 15 minutes, on average. So, μ = 0.75 for this problem.

X ~ P (0.75)

1. P ( X = 1 ) = $(e^{-\mathrm{0.75}})\frac{{\mathrm{0.75}}^1}{1!}$ = 0.3543
2. P ( X >1) = 1- P ( X ≤ 1) = $1-((e^{-\mathrm{0.75}})\frac{{\mathrm{0.75}}^0}{0!}()+(e^{-\mathrm{0.75}})\frac{{\mathrm{0.75}}^1}{1!}())()$ = 1 - 0.8266 = 0.1734

The probability that Leah receives more than one telephone call in the next 15 minutes is about 0.1734:
P ( x >1) = 1 − poissoncdf(0.75, 1).

The graph of X ~ P (0.75) is:

The y -axis contains the probability of x where X = the number of calls in 15 minutes.

According to Baydin, an email management company, an email user gets, on average, 147 emails per day. Let X = the number of emails an email user receives per day. The discrete random variable X takes on the values x = 0, 1, 2 …. The random variable X has a Poisson distribution: X ~ P (147). The mean is 147 emails.