### Introduction

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.

$m=\frac{1}{\mu}$. Therefore, $m=\frac{1}{4}=\mathrm{0.25.}$

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 "*e ^{x}*." If you enter one for

*x*, the calculator will display the value

*e*.

The curve is:

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

For example, *f*(5) = 0.25*e*^{(−0.25)(5)} = 0.072. The probability that the postal clerk spends five minutes with the customers is 0.072.

The graph is as follows:

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

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

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.

a. Find *P*(4 < *x* < 5).

*m*

*x*

*P*

*x*

*e*

*P*

*x*

*e*

*P*

*x*

*x*

*e*

*m*

*x*

*P*

*x*

*e*

*P*

*x*

*e*

### 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 50^{th} percentile.)

b. Find the 50^{th} percentile.

*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:

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, –0.25*k* = *ln*(0.50).

Solve for *k*: $k=\frac{ln(0.50)}{\u20130.25}=2.8$ minutes. The calculator simplifies the calculation for percentile *k*. See the following two notes:

### Note

A formula for the percentile *k* is $k=\frac{ln(\mathrm{}1-\text{Area to the left})}{-m}$ where *ln* is the natural log.

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

#### Collaborative Exercise

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?

c. From Part b, the median or 50^{th} percentile is 2.8 minutes. The theoretical mean is four minutes. The mean is larger.

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 10 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 $0.40 in his or her pocket or purse. (Shade *P*(*x* < 0.40)).

### Example 5.9

On average, a certain computer part lasts 10 years. The length of time the computer part lasts is exponentially distributed.

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

a. Let *x* = the amount of time (in years) a computer part lasts.

Find *P*(*x* > 7). Draw the graph.

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).

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

b. On the average, one computer part lasts 10 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?

c. Find the 80^{th} percentile. Draw the graph. Let *k* = the 80^{th} percentile.

Solve for *k*:

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 $\frac{\mathrm{ln}(1\u20130.80)}{\u20130.1}$.

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

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

*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).

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 $\frac{1}{12}$. 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 _______ .

*m*= $\frac{1}{12}$*μ*= 12*σ*= 12

*P*(*x* > 5) = 0.6592

Suppose that the distance, in miles, that people are willing to commute to work is an exponential random variable with a decay parameter $\frac{1}{20}$. 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.

- On average, how many minutes elapse between two successive arrivals?
- When the store first opens, how long on average does it take for three customers to arrive?
- After a customer arrives, find the probability that it takes less than one minute for the next customer to arrive.
- After a customer arrives, find the probability that it takes more than five minutes for the next customer to arrive.
- Seventy percent of the customers arrive within how many minutes of the previous customer?
- Is an exponential distribution reasonable for this situation?

- Since we expect 30 customers to arrive per hour (60 minutes), we expect on average one customer to arrive every two minutes on average.
- Since one customer arrives every two minutes on average, it will take six minutes on average for three customers to arrive.
- Let
*X*= the time between arrivals, in minutes. By Part a,*μ*= 2, so*m*= $\frac{1}{2}$ = 0.5.*X*~*Exp*(0.5).*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 *P*(*X*> 5) = 1 –*P*(*X*< 5) = 1 – (1 –*e*^{(−0.5)(5)}) =*e*^{–2.5}≈ 0.0821.### Using the TI-83, 83+, 84, 84+ Calculator

$$1\u2013(1\u2013{e}^{\text{(\u2013}0.50)(5)})\text{}\text{or}\text{}{e}^{\text{(\u2013}0.50)(5)}$$- 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 –0.5*x*=*ln*(0.30), or $x=\frac{ln\left(0.30\right)}{\u20130.5}\approx 2.41$ minutes.Thus, 70 percent of customers arrive within 2.41 minutes of the previous customer.

You are finding the 70

^{th}percentile*k*so you can use the formula.

$$k\text{=}\frac{ln(1\u2013\text{Area to the left of}\phantom{\rule{1em}{0ex}}k)}{(\u2013m)}$$$$k\text{=}\frac{ln(1\u20130.70)}{(\u20130.5)}\approx 2.41\text{minutes}$$ - 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.

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.

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