# 4.4 Poisson distribution  (Page 2/18)

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The formula for computing probabilities that are from a Poisson process is:

$P\left(x\right)=\frac{{\mu }^{x}{e}^{-\mu }}{x!}$

where P(X) is the probability of X successes, μ is the expected number of successes based upon historical data, e is the natural logarithm approximately equal to 2.718, and X is the number of successes per unit, usually per unit of time.

In order to use the Poisson distribution, certain assumptions must hold. These are: the probability of a success, μ, is unchanged within the interval, there cannot be simultaneous successes within the interval, and finally, that the probability of a success among intervals is independent, the same assumption of the binomial distribution.

In a way, the Poisson distribution can be thought of as a clever way to convert a continuous random variable, usually time, into a discrete random variable by breaking up time into discrete independent intervals. This way of thinking about the Poisson helps us understand why it can be used to estimate the probability for the discrete random variable from the binomial distribution. The Poisson is asking for the probability of a number of successes during a period of time while the binomial is asking for the probability of a certain number of successes for a given number of trials.

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

Probability that Leah receives more than one telephone call in the next 15 minutes is about 0.1734.

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 a survey a university professor gets, on average, 7 emails per day. Let X = the number of emails a professor 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 (7). The mean is 7 emails.

1. What is the probability that an email user receives exactly 2 emails per day?
2. What is the probability that an email user receives at most 2 emails per day?
3. What is the standard deviation?
1. $P\left(x=2\right)=\frac{{\mu }^{x}{e}^{\mathrm{-\mu }}}{\mathrm{x!}}=\frac{{7}^{2}{e}^{-7}}{2!}=0.022$
2. $P\left(x\le 2\right)=\frac{{7}^{0}{e}^{-7}}{0!}+\frac{{7}^{1}{e}^{-7}}{1!}+\frac{{7}^{2}{e}^{-7}}{2!}=0.029$
3. Standard Deviation = $\sigma =\sqrt{\mu }=\sqrt{7}\approx 2.65$

Text message users receive or send an average of 41.5 text messages per day.

1. How many text messages does a text message user receive or send per hour?
2. What is the probability that a text message user receives or sends two messages per hour?
3. What is the probability that a text message user receives or sends more than two messages per hour?
1. Let X = the number of texts that a user sends or receives in one hour. The average number of texts received per hour is $\frac{41.5}{24}$ ≈ 1.7292.
2. $P\left(x=2\right)=\frac{{\mu }^{x}{e}^{\mathrm{-\mu }}}{\mathrm{x!}}=\frac{{1.729}^{2}{e}^{-1.729}}{2!}=0.265$
3. $P\left(x>2\right)=1-P\left(x\le 2\right)=1-\left[\frac{{7}^{0}{e}^{-7}}{0!}+\frac{{7}^{1}{e}^{-7}}{1!}+\frac{{7}^{2}{e}^{-7}}{2!}\right]=0.250$

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