4.6 Poisson distribution  (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 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)

Find P ( x >1). P ( x >1) = 0.1734 (calculator or computer)

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.