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# Continuous Random Variables: The Exponential Distribution

Summary: This module introduces the properties of the exponential distribution, the behavior of probabilities that reflect a large number of small values and a small number of high values.

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 that spend less money and fewer people that spend large amounts of money.

The exponential distribution is widely used in the field of reliability. Reliability deals with the amount of time a product lasts.

## Example 1

Illustrates the exponential distribution: Let XX = amount of time (in minutes) a postal clerk spends with his/her customer. The time is known to have an exponential distribution with the average amount of time equal to 4 minutes.

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

m= 1μ m=1μ. Therefore, m= 14 =0.25 m=14=0.25

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

The distribution notation is X~Exp(m)X~Exp(m) size 12{X "~" ital "Exp" $$m$$ } {}. Therefore, X~Exp(0.25)X~Exp(0.25) size 12{X "~" ital "Exp" $$m$$ } {}.

The probability density function is f(x) = m e -m⋅x f(x)=m e -m⋅x The number ee = 2.71828182846... It is a number that is used often in mathematics. Scientific calculators have the key "exex." If you enter 1 for xx, the calculator will display the value ee.

The curve is:

f ( x ) = 0.25 e − 0.25⋅x f(x)=0.25 e − 0.25⋅x where xx is at least 0 and mm = 0.25.

For example, f ( 5 ) = 0.25 e − 0.25⋅5 = 0.072 f(5)=0.25 e − 0.25⋅5 =0.072

The graph is as follows:

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

f ( x ) = 0.25 e − 0.25⋅0 = 0.25 1 = 0.25 = m f(x)=0.25 e − 0.25⋅0 =0.251=0.25=m

### Example 2

#### Problem 1

Find the probability that a clerk spends four to five minutes with a randomly selected customer.

##### Solution

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

The cumulative distribution function (CDF) gives the area to the left.

P ( x < x ) = 1 - e -m⋅x P ( x x ) =1- e -m⋅x

P ( x < 5 ) = 1 - e -0.25⋅5 = 0.7135 P ( x 5 ) =1- e -0.25⋅5 =0.7135 and P ( x < 4 ) = 1 - e -0.25⋅4 = 0.6321 P ( x 4 ) =1- e -0.25⋅4 =0.6321

###### 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 P(4 x 5)= P(x 5)- P(x 4)=0.71350.6321=0.0814

###### Note:
TI-83+ and TI-84: On the home screen, enter (1-e^(-.25*5))-(1-e^(-.25*4)) or enter e^(-.25*4)-e^(-.25*5).

#### Problem 2

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

##### Solution

Find the 50th percentile.

P(x<k)=0.50 P(x k) 0.50, kk = 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 P(x k) 0.50 and P ( x < k ) = 1 - e -0.25⋅k P ( x k ) = 1 - e -0.25⋅k

Therefore, 0.50 = 1 e −0.25⋅k 0.50=1 e −0.25⋅k and e −0.25⋅k = 1 0.50 = 0.5 e −0.25⋅k =10.50=0.5

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

Solve for kk: k = ln(.50) -0.25 = 2.8 k= ln(.50) -0.25 =2.8 minutes

###### Note:
A formula for the percentile kk is k=LN(1−AreaToTheLeft) −mk=LN(1−AreaToTheLeft) −m where LN is the natural log.
###### Note:
TI-83+ and TI-84: On the home screen, enter LN(1-.50)/-.25. Press the (-) for the negative.

#### Problem 3

Which is larger, the mean or the median?

##### Solution

Is the mean or median larger?

From part b, the median or 50th percentile is 2.8 minutes. The theoretical mean is 4 minutes. The mean is larger.

## Optional Collaborative Classroom Activity

Have each class member count the change he/she has in his/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 5 intervals. Draw a smooth curve through the bars. The graph should look approximately exponential. Then calculate the mean.

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

The distribution for XX is approximately exponential with mean, μμ = _______ and mm = _______. The standard deviation, σσ = ________.

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/her pocket or purse. (Shade P(x<0.40) P(x 0.40)).

### Example 3

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

#### Problem 1

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

##### Solution

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

μ=10μ=10 so m= 1 μ = 1 10 = 0.1 m= 1 μ = 1 10 =0.1

Find P(x>7) P(x 7). Draw a graph.

P(x >7)= 1-P(x< 7)P(x>7)=1-P(x<7).

Since P ( X < x ) = 1 - e -mx P ( X x ) =1- e -mx then P ( X > x ) = 1 - ( 1 - e -m⋅x ) = e -m⋅x P ( X x ) =1-(1- e -m⋅x )= e -m⋅x

P ( x > 7 ) = e -0.1⋅7 = 0.4966 P ( x 7 ) = e -0.1⋅7 =0.4966. The probability that a computer part lasts more than 7 years is 0.4966.

###### Note:
TI-83+ and TI-84: On the home screen, enter e^(-.1*7).

#### Problem 2

On the average, how long would 5 computer parts last if they are used one after another?

##### Solution

On the average, 1 computer part lasts 10 years. Therefore, 5 computer parts, if they are used one right after the other would last, on the average,

(5) (10) = 50 (5)(10)=50 years.

#### Problem 3

Eighty percent of computer parts last at most how long?

##### Solution

Find the 80th percentile. Draw a graph. Let kk = the 80th percentile.

Solve for kk: k = ln(1-.80) -0.1 = 16.1 k= ln(1-.80) -0.1 =16.1 years

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

###### Note:
TI-83+ and TI-84: On the home screen, enter LN(1 - .80)/-.1

#### Problem 4

What is the probability that a computer part lasts between 9 and 11 years?

##### Solution

Find P(9<x<11) P(9 x 11). Draw a 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 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. (calculator or computer)

The probability that a computer part lasts between 9 and 11 years is 0.0737.

###### Note:
TI-83+ and TI-84: On the home screen, enter e^(-.1*9) - e^(-.1*11).

### Example 4

Suppose that the length of a phone call, in minutes, is an exponential random variable with decay parameter = 112112. If another person arrives at a public telephone just before you, find the probability that you will have to wait more than 5 minutes. Let XX = the length of a phone call, in minutes.

#### Problem 1

What is mm, μμ, and σσ? The probability that you must wait more than 5 minutes is _______ .

##### Solution
• mm = 112112
• μμ = 12
• σσ = 12

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

#### Note:

A summary for exponential distribution is available in "Summary of The Uniform and Exponential Probability Distributions".

## Glossary

Exponential Distribution:
A continuous random variable (RV) that appears when we are interested in the intervals of time between some random events, for example, the length of time between emergency arrivals at a hospital. Notation: X~Exp(m)X~Exp(m) size 12{X "~" ital "Exp" $$m$$ } {}. The mean is μ=1mμ=1m size 12{μ= { {1} over {m} } } {} and the standard deviation is σ = 1 m σ= 1 m . The probability density function is f(x)=memx,f(x)=memx, size 12{f $$x$$ = ital "me" rSup { size 8{- ital "mx"} } ," "} {} x 0 x 0 and the cumulative distribution function is P(Xx)=1emxP(Xx)=1emx size 12{P $$X <= x$$ =1-e rSup { size 8{- ital "mx"} } } {}.

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