Skip to Content
OpenStax Logo
Introductory Statistics

5.3 The Exponential Distribution

Introductory Statistics5.3 The Exponential Distribution
Buy book
  1. Preface
  2. 1 Sampling and Data
    1. Introduction
    2. 1.1 Definitions of Statistics, Probability, and Key Terms
    3. 1.2 Data, Sampling, and Variation in Data and Sampling
    4. 1.3 Frequency, Frequency Tables, and Levels of Measurement
    5. 1.4 Experimental Design and Ethics
    6. 1.5 Data Collection Experiment
    7. 1.6 Sampling Experiment
    8. Key Terms
    9. Chapter Review
    10. Practice
    11. Homework
    12. Bringing It Together: Homework
    13. References
    14. Solutions
  3. 2 Descriptive Statistics
    1. Introduction
    2. 2.1 Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs
    3. 2.2 Histograms, Frequency Polygons, and Time Series Graphs
    4. 2.3 Measures of the Location of the Data
    5. 2.4 Box Plots
    6. 2.5 Measures of the Center of the Data
    7. 2.6 Skewness and the Mean, Median, and Mode
    8. 2.7 Measures of the Spread of the Data
    9. 2.8 Descriptive Statistics
    10. Key Terms
    11. Chapter Review
    12. Formula Review
    13. Practice
    14. Homework
    15. Bringing It Together: Homework
    16. References
    17. Solutions
  4. 3 Probability Topics
    1. Introduction
    2. 3.1 Terminology
    3. 3.2 Independent and Mutually Exclusive Events
    4. 3.3 Two Basic Rules of Probability
    5. 3.4 Contingency Tables
    6. 3.5 Tree and Venn Diagrams
    7. 3.6 Probability Topics
    8. Key Terms
    9. Chapter Review
    10. Formula Review
    11. Practice
    12. Bringing It Together: Practice
    13. Homework
    14. Bringing It Together: Homework
    15. References
    16. Solutions
  5. 4 Discrete Random Variables
    1. Introduction
    2. 4.1 Probability Distribution Function (PDF) for a Discrete Random Variable
    3. 4.2 Mean or Expected Value and Standard Deviation
    4. 4.3 Binomial Distribution
    5. 4.4 Geometric Distribution
    6. 4.5 Hypergeometric Distribution
    7. 4.6 Poisson Distribution
    8. 4.7 Discrete Distribution (Playing Card Experiment)
    9. 4.8 Discrete Distribution (Lucky Dice Experiment)
    10. Key Terms
    11. Chapter Review
    12. Formula Review
    13. Practice
    14. Homework
    15. References
    16. Solutions
  6. 5 Continuous Random Variables
    1. Introduction
    2. 5.1 Continuous Probability Functions
    3. 5.2 The Uniform Distribution
    4. 5.3 The Exponential Distribution
    5. 5.4 Continuous Distribution
    6. Key Terms
    7. Chapter Review
    8. Formula Review
    9. Practice
    10. Homework
    11. References
    12. Solutions
  7. 6 The Normal Distribution
    1. Introduction
    2. 6.1 The Standard Normal Distribution
    3. 6.2 Using the Normal Distribution
    4. 6.3 Normal Distribution (Lap Times)
    5. 6.4 Normal Distribution (Pinkie Length)
    6. Key Terms
    7. Chapter Review
    8. Formula Review
    9. Practice
    10. Homework
    11. References
    12. Solutions
  8. 7 The Central Limit Theorem
    1. Introduction
    2. 7.1 The Central Limit Theorem for Sample Means (Averages)
    3. 7.2 The Central Limit Theorem for Sums
    4. 7.3 Using the Central Limit Theorem
    5. 7.4 Central Limit Theorem (Pocket Change)
    6. 7.5 Central Limit Theorem (Cookie Recipes)
    7. Key Terms
    8. Chapter Review
    9. Formula Review
    10. Practice
    11. Homework
    12. References
    13. Solutions
  9. 8 Confidence Intervals
    1. Introduction
    2. 8.1 A Single Population Mean using the Normal Distribution
    3. 8.2 A Single Population Mean using the Student t Distribution
    4. 8.3 A Population Proportion
    5. 8.4 Confidence Interval (Home Costs)
    6. 8.5 Confidence Interval (Place of Birth)
    7. 8.6 Confidence Interval (Women's Heights)
    8. Key Terms
    9. Chapter Review
    10. Formula Review
    11. Practice
    12. Homework
    13. References
    14. Solutions
  10. 9 Hypothesis Testing with One Sample
    1. Introduction
    2. 9.1 Null and Alternative Hypotheses
    3. 9.2 Outcomes and the Type I and Type II Errors
    4. 9.3 Distribution Needed for Hypothesis Testing
    5. 9.4 Rare Events, the Sample, Decision and Conclusion
    6. 9.5 Additional Information and Full Hypothesis Test Examples
    7. 9.6 Hypothesis Testing of a Single Mean and Single Proportion
    8. Key Terms
    9. Chapter Review
    10. Formula Review
    11. Practice
    12. Homework
    13. References
    14. Solutions
  11. 10 Hypothesis Testing with Two Samples
    1. Introduction
    2. 10.1 Two Population Means with Unknown Standard Deviations
    3. 10.2 Two Population Means with Known Standard Deviations
    4. 10.3 Comparing Two Independent Population Proportions
    5. 10.4 Matched or Paired Samples
    6. 10.5 Hypothesis Testing for Two Means and Two Proportions
    7. Key Terms
    8. Chapter Review
    9. Formula Review
    10. Practice
    11. Homework
    12. Bringing It Together: Homework
    13. References
    14. Solutions
  12. 11 The Chi-Square Distribution
    1. Introduction
    2. 11.1 Facts About the Chi-Square Distribution
    3. 11.2 Goodness-of-Fit Test
    4. 11.3 Test of Independence
    5. 11.4 Test for Homogeneity
    6. 11.5 Comparison of the Chi-Square Tests
    7. 11.6 Test of a Single Variance
    8. 11.7 Lab 1: Chi-Square Goodness-of-Fit
    9. 11.8 Lab 2: Chi-Square Test of Independence
    10. Key Terms
    11. Chapter Review
    12. Formula Review
    13. Practice
    14. Homework
    15. Bringing It Together: Homework
    16. References
    17. Solutions
  13. 12 Linear Regression and Correlation
    1. Introduction
    2. 12.1 Linear Equations
    3. 12.2 Scatter Plots
    4. 12.3 The Regression Equation
    5. 12.4 Testing the Significance of the Correlation Coefficient
    6. 12.5 Prediction
    7. 12.6 Outliers
    8. 12.7 Regression (Distance from School)
    9. 12.8 Regression (Textbook Cost)
    10. 12.9 Regression (Fuel Efficiency)
    11. Key Terms
    12. Chapter Review
    13. Formula Review
    14. Practice
    15. Homework
    16. Bringing It Together: Homework
    17. References
    18. Solutions
  14. 13 F Distribution and One-Way ANOVA
    1. Introduction
    2. 13.1 One-Way ANOVA
    3. 13.2 The F Distribution and the F-Ratio
    4. 13.3 Facts About the F Distribution
    5. 13.4 Test of Two Variances
    6. 13.5 Lab: One-Way ANOVA
    7. Key Terms
    8. Chapter Review
    9. Formula Review
    10. Practice
    11. Homework
    12. References
    13. Solutions
  15. A | Review Exercises (Ch 3-13)
  16. B | Practice Tests (1-4) and Final Exams
  17. C | Data Sets
  18. D | Group and Partner Projects
  19. E | Solution Sheets
  20. F | Mathematical Phrases, Symbols, and Formulas
  21. G | Notes for the TI-83, 83+, 84, 84+ Calculators
  22. H | Tables
  23. Index

The exponential distribution is often concerned with the amount of time until some specific event occurs. For example, the amount of time (beginning now) until an earthquake occurs has an exponential distribution. Other examples include the length, in minutes, of long distance business telephone calls, and the amount of time, in months, a car battery lasts. It can be shown, too, that the value of the change that you have in your pocket or purse approximately follows an exponential distribution.

Values for an exponential random variable occur in the following way. There are fewer large values and more small values. For example, the amount of money customers spend in one trip to the supermarket follows an exponential distribution. There are more people who spend small amounts of money and fewer people who spend large amounts of money.

Exponential distributions are commonly used in calculations of product reliability, or the length of time a product lasts.

Example 5.7

Let X = amount of time (in minutes) a postal clerk spends with his or her customer. The time is known to have an exponential distribution with the average amount of time equal to four minutes.

X is a continuous random variable since time is measured. It is given that μ = 4 minutes. To do any calculations, you must know m, the decay parameter.

m= 1 μ m= 1 μ . Therefore, m= 1 4 =0.25. m= 1 4 =0.25.

The standard deviation, σ, is the same as the mean. μ = σ

The distribution notation is X ~ Exp(m). Therefore, X ~ Exp(0.25).

The probability density function is f(x) = me-mx. The number e = 2.71828182846... It is a number that is used often in mathematics. Scientific calculators have the key "ex." If you enter one for x, the calculator will display the value e.

The curve is:

f(x) = 0.25e–0.25x where x is at least zero and m = 0.25.

For example, f(5) = 0.25e(-0.25)(5) = 0.072. The value 0.072 is the height of the curve when x = 5. In Example 5.8 below, you will learn how to find probabilities using the decay parameter.

The graph is as follows:

Exponential graph with increments of 2 from 0-20 on the x-axis of μ = 4 and increments of 0.05 from 0.05-0.25 on the y-axis of m = 0.25. The curved line begins at the top at point (0, 0.25) and curves down to point (20, 0). The x-axis is equal to a continuous random variable.
Figure 5.22

Notice the graph is a declining curve. When x = 0,

f(x) = 0.25e(−0.25)(0) = (0.25)(1) = 0.25 = m. The maximum value on the y-axis is m.

Try It 5.7

The amount of time spouses shop for anniversary cards can be modeled by an exponential distribution with the average amount of time equal to eight minutes. Write the distribution, state the probability density function, and graph the distribution.

Example 5.8

a. Using the information in Example 5.7, find the probability that a clerk spends four to five minutes with a randomly selected customer.

Solution 5.8

a. Find P(4 < x < 5).
The cumulative distribution function (CDF) gives the area to the left.
P(x < x) = 1 – e–mx
P(x < 5) = 1 – e(−0.25)(5) = 0.7135 and P(x < 4) = 1 – e(–0.25)(4) = 0.6321

Exponential graph with the curved line beginning at point (0, 0.25) and curves down towards point (∞, 0). Two vertical upward lines extend from points 4 and 5 to the curved line. The probability is in the area between points 4 and 5.
Figure 5.23

NOTE

You can do these calculations easily on a calculator.


The probability that a postal clerk spends four to five minutes with a randomly selected customer is P(4 < x < 5) = P(x < 5) – P(x < 4) = 0.7135 − 0.6321 = 0.0814.

Using the TI-83, 83+, 84, 84+ Calculator

On the home screen, enter (1 – e^(–0.25*5))–(1–e^(–0.25*4)) or enter e^(–0.25*4) – e^(–0.25*5).

b. Half of all customers are finished within how long? (Find the 50th percentile)

Solution 5.8

b. Find the 50th percentile.

Exponential graph with the curved line beginning at point (0, 0.25) and curves down towards point (∞, 0). A vertical upward line extends from point k to the curved line. The probability area from 0-k is equal to 0.50.
Figure 5.24

P(x < k) = 0.50, k = 2.8 minutes (calculator or computer)

Half of all customers are finished within 2.8 minutes.

You can also do the calculation as follows:

P(x < k) = 0.50 and P(x < k) = 1 –e–0.25k

Therefore, 0.50 = 1 − e−0.25k and e−0.25k = 1 − 0.50 = 0.5

Take natural logs: ln(e–0.25k) = ln(0.50). So, –0.25k = ln(0.50)

Solve for k: k= ln(0.50) -0.25 =2.8 k= ln(0.50) -0.25 =2.8 minutes. The calculator simplifies the calculation for percentile k. See the following two notes.

Note

A formula for the percentile k is k= ln(1AreaToTheLeft) m k= ln(1AreaToTheLeft) m where ln is the natural log.

Using the TI-83, 83+, 84, 84+ Calculator

On the home screen, enter ln(1 – 0.50)/–0.25. Press the (-) for the negative.

c. Which is larger, the mean or the median?

Solution 5.8

c. From part b, the median or 50th percentile is 2.8 minutes. The theoretical mean is four minutes. The mean is larger.

Try It 5.8

The number of days ahead travelers purchase their airline tickets can be modeled by an exponential distribution with the average amount of time equal to 15 days. Find the probability that a traveler will purchase a ticket fewer than ten days in advance. How many days do half of all travelers wait?

Collaborative Exercise

Have each class member count the change he or she has in his or her pocket or purse. Your instructor will record the amounts in dollars and cents. Construct a histogram of the data taken by the class. Use five intervals. Draw a smooth curve through the bars. The graph should look approximately exponential. Then calculate the mean.

Let X = the amount of money a student in your class has in his or her pocket or purse.

The distribution for X is approximately exponential with mean, μ = _______ and m = _______. The standard deviation, σ = ________.

Draw the appropriate exponential graph. You should label the x– and y–axes, the decay rate, and the mean. Shade the area that represents the probability that one student has less than $.40 in his or her pocket or purse. (Shade P(x < 0.40)).

Example 5.9

On the average, a certain computer part lasts ten years. The length of time the computer part lasts is exponentially distributed.

a. What is the probability that a computer part lasts more than 7 years?

Solution 5.9

a. Let x = the amount of time (in years) a computer part lasts.
μ = 10 so m= 1 μ = 1 10 =0.1 m= 1 μ = 1 10 =0.1
Find P(x > 7). Draw the graph.
P(x > 7) = 1 – P(x < 7).
Since P(X < x) = 1 –e–mx then P(X > x) = 1 –(1 –e–mx) = e-mx
P(x > 7) = e(–0.1)(7) = 0.4966. The probability that a computer part lasts more than seven years is 0.4966.

Using the TI-83, 83+, 84, 84+ Calculator

On the home screen, enter e^(-.1*7).

Exponential graph with the curved line beginning at point (0, 0.1) and curves down towards point (∞, 0). A vertical upward line extends from point 1 to the curved line. The probability area occurs from point 1 to the end of the curve. The x-axis is equal to the amount of time a computer part lasts.
Figure 5.25

b. On the average, how long would five computer parts last if they are used one after another?

Solution 5.9

b. On the average, one computer part lasts ten years. Therefore, five computer parts, if they are used one right after the other would last, on the average, (5)(10) = 50 years.

c. Eighty percent of computer parts last at most how long?

Solution 5.9

c. Find the 80th percentile. Draw the graph. Let k = the 80th percentile.

Exponential graph with the curved line beginning at point (0, 0.1) and curves down towards point (∞, 0). A vertical upward line extends from point k to the curved line. k is the 80th percentile. The probability area from 0-k is equal to 0.80.
Figure 5.26

Solve for k: k= ln(10.80) 0.1 =16.1 k= ln(10.80) 0.1 =16.1 years

Eighty percent of the computer parts last at most 16.1 years.

Using the TI-83, 83+, 84, 84+ Calculator

On the home screen, enter ln(10.80) 0.1 ln(10.80) 0.1

d. What is the probability that a computer part lasts between nine and 11 years?

Solution 5.9

d. Find P(9 < x < 11). Draw the graph.

Exponential graph with the curved line beginning at point (0, 0.1) and curves down towards point (∞, 0). Two vertical upward lines extend from point 9 and 11 to the curved line. The probability area occurs between point 9 and 11.
Figure 5.27

P(9 < x < 11) = P(x < 11) – P(x < 9) = (1 – e(–0.1)(11)) – (1 – e(–0.1)(9)) = 0.6671 – 0.5934 = 0.0737. The probability that a computer part lasts between nine and 11 years is 0.0737.

Using the TI-83, 83+, 84, 84+ Calculator

On the home screen, enter e^(–0.1*9) – e^(–0.1*11).

Try It 5.9

On average, a pair of running shoes can last 18 months if used every day. The length of time running shoes last is exponentially distributed. What is the probability that a pair of running shoes last more than 15 months? On average, how long would six pairs of running shoes last if they are used one after the other? Eighty percent of running shoes last at most how long if used every day?

Example 5.10

Suppose that the length of a phone call, in minutes, is an exponential random variable with decay parameter 1 12 1 12 . If another person arrives at a public telephone just before you, find the probability that you will have to wait more than five minutes. Let X = the length of a phone call, in minutes.

What is m, μ, and σ? The probability that you must wait more than five minutes is _______ .

Solution 5.10
  • m = 1 12 1 12
  • μ = 12
  • σ = 12

P(x > 5) = 0.6592

Try It 5.10

Suppose that the distance, in miles, that people are willing to commute to work is an exponential random variable with a decay parameter 1 20 1 20 . Let X = the distance people are willing to commute in miles. What is m, μ, and σ? What is the probability that a person is willing to commute more than 25 miles?

Example 5.11

The time spent waiting between events is often modeled using the exponential distribution. For example, suppose that an average of 30 customers per hour arrive at a store and the time between arrivals is exponentially distributed.

  1. On average, how many minutes elapse between two successive arrivals?
  2. When the store first opens, how long on average does it take for three customers to arrive?
  3. After a customer arrives, find the probability that it takes less than one minute for the next customer to arrive.
  4. After a customer arrives, find the probability that it takes more than five minutes for the next customer to arrive.
  5. Seventy percent of the customers arrive within how many minutes of the previous customer?
  6. Is an exponential distribution reasonable for this situation?
Solution 5.11
  1. Since we expect 30 customers to arrive per hour (60 minutes), we expect on average one customer to arrive every two minutes on average.
  2. Since one customer arrives every two minutes on average, it will take six minutes on average for three customers to arrive.
  3. Let X = the time between arrivals, in minutes. By part a, μ = 2, so m = 1 2 1 2 = 0.5.
    Therefore, XExp(0.5).
    The cumulative distribution function is P(X < x) = 1 – e(-0.5)(x).
    Therefore P(X < 1) = 1 – e(–0.5)(1) ≈ 0.3935.

    Using the TI-83, 83+, 84, 84+ Calculator

    1 - e^(–0.5) ≈ 0.3935

    Figure 5.28
  4. P(X > 5) = 1 – P(X < 5) = 1 – (1 – e(-0.50)(5)) = e–2.5 ≈ 0.0821.
    Figure 5.29

    Using the TI-83, 83+, 84, 84+ Calculator

    1 – (1 – e^((-0.50)(5))) or e^( – 5*0.5)

  5. We want to solve 0.70 = P(X < x) for x.
    Substituting in the cumulative distribution function gives 0.70 = 1 – e–0.5x, so that e–0.5x = 0.30. Converting this to logarithmic form gives –0.5x = ln(0.30), or x= ln( 0.30 ) 0.5 2.41 x= ln( 0.30 ) 0.5 2.41 minutes.
    Thus, seventy percent of customers arrive within 2.41 minutes of the previous customer.
    You are finding the 70th percentile k so you can use the formula k = ln(1Area_To_The_Left_Of_k) (m) ln(1Area_To_The_Left_Of_k) (m)
    k = ln(10.70) (0.5) 2.41 ln(10.70) (0.5) 2.41 minutes
    Figure 5.30
  6. This model assumes that a single customer arrives at a time, which may not be reasonable since people might shop in groups, leading to several customers arriving at the same time. It also assumes that the flow of customers does not change throughout the day, which is not valid if some times of the day are busier than others.
Try It 5.11

Suppose that on a certain stretch of highway, cars pass at an average rate of five cars per minute. Assume that the duration of time between successive cars follows the exponential distribution.

  1. On average, how many seconds elapse between two successive cars?
  2. After a car passes by, how long on average will it take for another seven cars to pass by?
  3. Find the probability that after a car passes by, the next car will pass within the next 20 seconds.
  4. Find the probability that after a car passes by, the next car will not pass for at least another 15 seconds.

Memorylessness of the Exponential Distribution

In Example 5.7 recall that the amount of time between customers is exponentially distributed with a mean of two minutes (X ~ Exp (0.5)). Suppose that five minutes have elapsed since the last customer arrived. Since an unusually long amount of time has now elapsed, it would seem to be more likely for a customer to arrive within the next minute. With the exponential distribution, this is not the case–the additional time spent waiting for the next customer does not depend on how much time has already elapsed since the last customer. This is referred to as the memoryless property. Specifically, the memoryless property says that

P (X > r + t | X > r) = P (X > t) for all r ≥ 0 and t ≥ 0

For example, if five minutes have elapsed since the last customer arrived, then the probability that more than one minute will elapse before the next customer arrives is computed by using r = 5 and t = 1 in the foregoing equation.

P(X > 5 + 1 | X > 5) = P(X > 1) = e ( 0.5 )( 1 ) e ( 0.5 )( 1 ) ≈ 0.6065.

This is the same probability as that of waiting more than one minute for a customer to arrive after the previous arrival.

The exponential distribution is often used to model the longevity of an electrical or mechanical device. In Example 5.9, the lifetime of a certain computer part has the exponential distribution with a mean of ten years (X ~ Exp(0.1)). The memoryless property says that knowledge of what has occurred in the past has no effect on future probabilities. In this case it means that an old part is not any more likely to break down at any particular time than a brand new part. In other words, the part stays as good as new until it suddenly breaks. For example, if the part has already lasted ten years, then the probability that it lasts another seven years is P(X > 17|X > 10) = P(X > 7) = 0.4966.

Example 5.12

Refer to Example 5.7 where the time a postal clerk spends with his or her customer has an exponential distribution with a mean of four minutes. Suppose a customer has spent four minutes with a postal clerk. What is the probability that he or she will spend at least an additional three minutes with the postal clerk?

The decay parameter of X is m = 1 4 1 4 = 0.25, so XExp(0.25).

The cumulative distribution function is P(X < x) = 1 – e–0.25x.

We want to find P(X > 7|X > 4). The memoryless property says that P(X > 7|X > 4) = P (X > 3), so we just need to find the probability that a customer spends more than three minutes with a postal clerk.

This is P(X > 3) = 1 – P (X < 3) = 1 – (1 – e–0.25⋅3) = e–0.75 ≈ 0.4724.

This graph shows an exponential distribution. The graph slopes downward. It begins at the point (0, 0.25) on the y-axis and approaches the x-axis at the right edge of the graph. The region under the graph to the right of x = 3 is shaded to represent P(x > 3) = 0.4724.
Figure 5.31

Using the TI-83, 83+, 84, 84+ Calculator

1–(1–e^(–0.25*3)) = e^(–0.25*3).

Try It 5.12

Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight years. If a bulb has already lasted 12 years, find the probability that it will last a total of over 19 years.

Relationship between the Poisson and the Exponential Distribution

There is an interesting relationship between the exponential distribution and the Poisson distribution. Suppose that the time that elapses between two successive events follows the exponential distribution with a mean of μ 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 λ = 1/μ. Recall from the chapter on Discrete Random Variables that if X has the Poisson distribution with mean λ, then P(X=k)= λ k e λ k! P(X=k)= λ k e λ k! . Conversely, if the number of events per unit time follows a Poisson distribution, then the amount of time between events follows the exponential distribution. (k! = k*(k–1*)(k–2)*(k–3)*…3*2*1)

Using the TI-83, 83+, 84, 84+ Calculator

Suppose X has the Poisson distribution with mean λ. Compute P(X = k) by entering 2nd, VARS(DISTR), C: poissonpdf(λ, k). To compute P(Xk), enter 2nd, VARS (DISTR), D:poissoncdf(λ, k).

Example 5.13

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 after a call is received, the next call occurs in less than ten seconds.
  3. Find the probability that exactly five calls occur within a minute.
  4. Find the probability that less than five calls occur within a minute.
  5. Find the probability that more than 40 calls occur in an eight-minute period.
Solution 5.13
  1. On average there are four calls occur per minute, so 15 seconds, or 15 60 15 60 = 0.25 minutes occur between successive calls on average.
  2. Let T = time elapsed between calls. From part a, μ = 0.25, so m = 1 0.25 1 0.25 = 4. Thus, TExp(4).
    The cumulative distribution function is P(T < t) = 1 – e–4t.
    The probability that the next call occurs in less than ten seconds (ten seconds = 1/6 minute) is P( T <  1 6 )=1 e (4)( 1 6 ) 0.4866. P( T <  1 6 )=1 e (4)( 1 6 ) 0.4866.
    This graph shows an exponential distribution. The graph slopes downward. It begins at the point (0, 0.8) on the y-axis and approaches the x-axis at the right edge of the graph. The region under the graph to the left of x = 20 is shaded to represent P(x < 1/6) = 0.4866.
    Figure 5.32
  3. Let X = 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.
    Therefore, XPoisson(4), and so P(X = 5) = 4 5 e 4 5! 4 5 e 4 5! ≈ 0.1563. (5! = (5)(4)(3)(2)(1))

    Using the TI-83, 83+, 84, 84+ Calculator

    poissonpdf(4, 5) = 0.1563.

  4. Keep in mind that X must be a whole number, so P(X < 5) = P(X ≤ 4).
    To compute this, we could take P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).
    Using technology, we see that P(X ≤ 4) = 0.6288.

    Using the TI-83, 83+, 84, 84+ Calculator

    poisssoncdf(4, 4) = 0.6288

  5. Let Y = 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, YPoisson(32). Therefore, P(Y > 40) = 1 – P (Y ≤ 40) = 1 – 0.9294 = 0.0707.

    Using the TI-83, 83+, 84, 84+ Calculator

    1 – poissoncdf(32, 40). = 0.0707

Try It 5.13

In a small city, the number of automobile accidents occur with a Poisson distribution at an average of three per week.

  1. Calculate the probability that there are at most 2 accidents occur in any given week.
  2. What is the probability that there is at least two weeks between any 2 accidents?
Citation/Attribution

Want to cite, share, or modify this book? This book is Creative Commons Attribution License 4.0 and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at https://openstax.org/books/introductory-statistics/pages/1-introduction
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at https://openstax.org/books/introductory-statistics/pages/1-introduction
Citation information

© Sep 19, 2013 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License 4.0 license. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.