Chapter 5: Continuous Random Variables

5.5 Using Technology for the Exponential Distribution

Learning Objectives

By the end of this section, the student should be able to:

  • use technology to solve problems involving the Exponential Distribution

This is another section to show technology tools that can be used to help with the statistics concepts taught in this chapter. This section will cover tools to help solve problems involving the Exponential Distribution.

Graphing Calculator

Currently, the TI-83+ and TI-84 do not have built-in exponential probability distribution functions. We will have to rely on the additional technology tools to assist as with these problems.

Relationship between the Poisson and the Exponential Distribution

We saw in 5.3 The Exponential Distribution there is an interesting relationship between the exponential distribution and the Poisson distribution. Since there is no graphing calculator function for the exponential distribution, we could use the poisson distribution from the graphing calculator for certain problems. Here was the set-up:

  • Suppose that the time that elapses between two successive events follows the exponential distribution with a mean of [latex]\mu[/latex] units of time.
  • Also assume that these times are independent, meaning that the time between events is not affected by the times between previous events.

If these assumptions hold, then the number of events per unit time follows a Poisson distribution with mean [latex]\lambda = \frac{1}{\mu}[/latex]. Conversely, if the number of events per unit time follows a Poisson distribution, then the amount of time between events follows the Exponential distribution.

Recall from the 4.6 Poisson Distribution that if [latex]X[/latex] has the Poisson distribution with mean [latex]\lambda[/latex], then we had to do by-hand calculations for [latex]P\left(X=k\right)=\frac{{\lambda }^{k}{e}^{-\lambda }}{k!}[/latex], where [latex](k! = k \cdot (k-1) \cdot (k-2) \cdot (k-3) ... 3 \cdot 2 \cdot 1)[/latex]. Not anymore! We learned in the Using Technology page how to use the graphing calculator for the Poisson Distribution, so we will use that again here.

 

Note

To compute poisson probabilities on a graphing calculator, the syntax is as follows:

  • To calculate [latex]P(X = x)[/latex]: poissonpdf([latex]\lambda[/latex], x).
  • To calculate [latex]P(X \leq x)[/latex]: poissoncdf([latex]\lambda[/latex], x).
  • If you wanted to instead find [latex]P(X>x)[/latex], use 1 - poissoncdf([latex]\lambda[/latex], x).

Remember, the TI calculators use [latex]\lambda[/latex] (lambda) for the mean.

 

Let's see this in action.

Example

At a police station in a large city, calls come in at an average rate of four calls per minute. Assume that the time that elapses from one call to the next has the exponential distribution. Take note that we are concerned only with the rate at which calls come in, and we are ignoring the time spent on the phone. We must also assume that the times spent between calls are independent. This means that a particularly long delay between two calls does not mean that there will be a shorter waiting period for the next call. We may then deduce that the total number of calls received during a time period has the Poisson distribution.

  1. Find the average time between two successive calls.
  2. Find the probability that exactly five calls occur within a minute.
  3. Find the probability that less than five calls occur within a minute.
  4. Find the probability that more than 40 calls occur in an eight-minute period.

Solution

  1. On average there are four calls that occur per minute, so 15 seconds, or [latex]\frac{15}{60}[/latex] = 0.25 minutes, occur between successive calls on average.
  2. Let [latex]X =[/latex] the number of calls per minute. As previously stated, the number of calls per minute has a Poisson distribution, with a mean of four calls per minute. We can use the graphing calculator to find poissonpdf(4, 5) = 0.1563.
  3. Using technology, we find poisssoncdf(4, 4) = 0.6288.
  4. Let [latex]Y =[/latex] the number of calls that occur during an eight-minute period. Since there is an average of four calls per minute, there is an average of (8)(4) = 32 calls during each eight-minute period. Hence, [latex]Y \sim \text{Poisson}(32)[/latex]. Therefore, [latex]P(Y > 40) = 1 – P (Y \le 40)[/latex]. Using the graphing calculator, we have 1 – poissoncdf(32, 40). = 0.0707.

Your Turn!

In a small city, the number of automobile accidents occur with a Poisson distribution at an average of three per week. Calculate the probability that there are at most 2 accidents that occur in any given week.

Solution

Let [latex]X =[/latex] the number of accidents per week, so that [latex]X \sim \text{Poisson}(3)[/latex]. We need to find [latex]P(X \le 2) = \text{poissoncdf}(3, 2) \approx 0.4232[/latex]

 

Additional Technology Tools

In addition to the graphing calculator, there are some additional technology tools that can be used for the concepts covered on this page. Below are links to helpful videos for those tools:

 

Section Practice

The time (in years) after reaching age 60 that it takes an individual to retire is approximately exponentially distributed with a mean of about five years. Suppose we randomly pick one retired individual. We are interested in the time after age 60 to retirement.

  1. Define the random variable. [latex]X= \underline{\hspace{2cm}}[/latex].
  2. Is X continuous or discrete?
  3. [latex]X \sim \underline{\hspace{2cm}}[/latex]
  4. [latex]\mu = \underline{\hspace{2cm}}[/latex]
  5. [latex]\sigma = \underline{\hspace{2cm}}[/latex]
  6. Draw a graph of the probability distribution. Label the axes.
  7. Find the probability that the person retired after age 70.
  8. Do more people retire before age 65 or after age 65?
  9. In a room of 1,000 people over age 80, how many do you expect will NOT have retired yet?
Solution
  1. [latex]X=[/latex] the time (in years) after reaching age 60 that it takes an individual to retire
  2. [latex]X[/latex] is continuous.
  3. [latex]X \sim \text{Exp}\left(\frac{1}{5}\right)[/latex]
  4. five
  5. five
  6. Check student’s solution.
  7. 0.1353
  8. before
  9. 18.3

The cost of all maintenance for a car during its first year is approximately exponentially distributed with a mean of $150.

  1. Define the random variable. [latex]X= \underline{\hspace{2cm}}[/latex].
  2. [latex]X \sim \underline{\hspace{2cm}}[/latex]
  3. [latex]\mu = \underline{\hspace{2cm}}[/latex]
  4. [latex]\sigma = \underline{\hspace{2cm}}[/latex]
  5. Draw a graph of the probability distribution. Label the axes.
  6. Find the probability that a car required over $300 for maintenance during its first year.

According to the American Red Cross, about one out of nine people in the U.S. have Type B blood. Suppose the blood types of people arriving at a blood drive are independent. In this case, the number of Type B blood types that arrive roughly follows the Poisson distribution.

  1. If 100 people arrive, how many on average would be expected to have Type B blood?
  2. What is the probability that over 10 people out of these 100 have type B blood?
  3. What is the probability that more than 20 people arrive before a person with type B blood is found?
Solution
  1. [latex]\frac{100}{9} = 11.11[/latex]
  2. [latex]P(x > 10) = 1- P(X \le 10) = 1 - \text{Poissoncdf}(11.11, 10) \approx 0.5532[/latex].
  3. The number of people with Type B blood encountered roughly follows the Poisson distribution, so the number of people X who arrive between successive Type B arrivals is roughly exponential with mean [latex]\mu = 9[/latex] and [latex]m=\frac{1}{9}[/latex]. The cumulative distribution function of X is [latex]P(X \ lt x)=1-{e}^{-\frac{x}{9}}[/latex]. Thus, [latex]P(X > 20) = 1 - P(X \le 20) = 1-(1-{e}^{-\frac{20}{9}}) \approx 0.1084[/latex].

Note: We could also deduce that each person arriving has a 8/9 chance of not having Type B blood. So the probability that none of the first 20 people arrive have Type B blood is [latex]{\left(\frac{8}{9}\right)}^{20}\approx 0.0948[/latex]. (The geometric distribution is more appropriate than the exponential because the number of people between Type B people is discrete instead of continuous.)

A web site experiences traffic during normal working hours at a rate of 12 visits per hour. Assume that the duration between visits has the exponential distribution.

  1. Find the probability that the duration between two successive visits to the website is more than ten minutes.
  2. The top 25% of durations between visits are at least how long?
  3. Suppose that 20 minutes have passed since the last visit to the web site. What is the probability that the next visit will occur within the next 5 minutes?
  4. Find the probability that less than 7 visits occur within a one-hour period.

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