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.
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. 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 (Figure) below, you will learn how to find probabilities using the decay parameter. The graph is as follows: 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.
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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.
a. Using the information in (Figure), find the probability that a clerk spends four to five minutes with a randomly selected customer.
a. Find P(4 < x < 5).
NOTE You can do these calculations easily on a 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)
b. Find the 50th 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: 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:
Note A formula for the percentile k is
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 50th percentile is 2.8 minutes. The theoretical mean is four minutes. The mean is larger.
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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)).
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?
a. Let x = the amount of time (in years) a computer part lasts. 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.
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 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?
c. Find the 80th percentile. Draw the graph. Let k = the 80th percentile. Solve for k: Eighty percent of the computer parts last at most 16.1 years.
On the home screen, enter
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.
On the home screen, enter e^(–0.1*9) – e^(–0.1*11).
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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?
Suppose that the length of a phone call, in minutes, is an exponential random variable with decay parameter
What is m, μ, and σ? The probability that you must wait more than five minutes is _______ .
P(x > 5) = 0.6592
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Suppose that the distance, in miles, that people are willing to commute to work is an exponential random variable with a decay parameter
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.
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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.
In (Figure) 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) = 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 (Figure), 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.
Refer to (Figure) 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 = = 0.25, so X ∼ Exp(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.
1–(1–e^(–0.25*3)) = e^(–0.25*3).
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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.
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
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(X ≤ k), enter 2nd, VARS (DISTR), D:poissoncdf(λ, k).
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.
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In a small city, the number of automobile accidents occur with a Poisson distribution at an average of three per week.
If X has an exponential distribution with mean μ, then the decay parameter is m = , and we write X ∼ Exp(m) where x ≥ 0 and m > 0 . The probability density function of X is f(x) = me-mx (or equivalently The exponential distribution has the memoryless property, which says that future probabilities do not depend on any past information. Mathematically, it says that P(X > x + k|X > x) = P(X > k). If T represents the waiting time between events, and if T ∼ Exp(λ), then the number of events X per unit time follows the Poisson distribution with mean λ. The probability density function of X is
Exponential: X ~ Exp(m) where m = the decay parameter
Use the following information to answer the next ten exercises. A customer service representative must spend different amounts of time with each customer to resolve various concerns. The amount of time spent with each customer can be modeled by the following distribution: X ~ Exp(0.2)
What type of distribution is this?
Are outcomes equally likely in this distribution? Why or why not?
No, outcomes are not equally likely. In this distribution, more people require a little bit of time, and fewer people require a lot of time, so it is more likely that someone will require less time.
What is m? What does it represent?
What is the standard deviation?
State the probability density function.
Find the 70th percentile.
What is the probability density function?
What is the cumulative distribution function?
Find the 30th percentile.
Which is larger, the mean or the median?
The mean is larger. The mean is Use the following information to answer the next 16 exercises. Carbon-14 is a radioactive element with a half-life of about 5,730 years. Carbon-14 is said to decay exponentially. The decay rate is 0.000121. We start with one gram of carbon-14. We are interested in the time (years) it takes to decay carbon-14.
What is being measured here?
Are the data discrete or continuous?
In words, define the random variable X.
What is the decay rate (m)?
The distribution for X is ______.
Find the amount (percent of one gram) of carbon-14 lasting less than 5,730 years. This means, find P(x < 5,730).
Find the percentage of carbon-14 lasting longer than 10,000 years.
Thirty percent (30%) of carbon-14 will decay within how many years?
Suppose that the length of long distance phone calls, measured in minutes, is known to have an exponential distribution with the average length of a call equal to eight minutes.
Suppose that the useful life of a particular car battery, measured in months, decays with parameter 0.025. We are interested in the life of the battery.
The percent of persons (ages five and older) in each state who speak a language at home other than English is approximately exponentially distributed with a mean of 9.848. Suppose we randomly pick a state.
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.
The cost of all maintenance for a car during its first year is approximately exponentially distributed with a mean of ?150.
The decay rate is:
What is the probability that a phone will fail within two years of the date of purchase?
What is the median lifetime of these phones (in years)?
Let X ~ Exp(0.1).
Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight years.
Let T = the life time of a light bulb. The decay parameter is m = 1/8, and T ∼ Exp(1/8). The cumulative distribution function is
At a 911 call center, calls come in at an average rate of one call every two minutes. Assume that the time that elapses from one call to the next has the exponential distribution.
In major league baseball, a no-hitter is a game in which a pitcher, or pitchers, doesn’t give up any hits throughout the game. No-hitters occur at a rate of about three per season. Assume that the duration of time between no-hitters is exponential.
Let X = the number of no-hitters throughout a season. Since the duration of time between no-hitters is exponential, the number of no-hitters per season is Poisson with mean λ = 3.
NOTE You could let T = duration of time between no-hitters. Since the time is exponential and there are 3 no-hitters per season, then the time between no-hitters is season. For the exponential, µ = .
During the years 1998–2012, a total of 29 earthquakes of magnitude greater than 6.5 have occurred in Papua New Guinea. Assume that the time spent waiting between earthquakes is exponential.
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.
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
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.
At an urgent care facility, patients arrive at an average rate of one patient every seven minutes. Assume that the duration between arrivals is exponentially distributed.
Let T = duration (in minutes) between successive visits. Since patients arrive at a rate of one patient every seven minutes, μ = 7 and the decay constant is m = . The cdf is P(T < t) =
decay parameter The decay parameter describes the rate at which probabilities decay to zero for increasing values of x. It is the value m in the probability density function f(x) = me(-mx) of an exponential random variable. It is also equal to m = , where μ is the mean of the random variable. memoryless property For an exponential random variable X, the memoryless property is the statement that knowledge of what has occurred in the past has no effect on future probabilities. This means that the probability that X exceeds x + k, given that it has exceeded x, is the same as the probability that X would exceed k if we had no knowledge about it. In symbols we say that P(X > x + k|X > x) = P(X > k). Poisson distribution If there is a known average of λ events occurring per unit time, and these events are independent of each other, then the number of events X occurring in one unit of time has the Poisson distribution. The probability of k events occurring in one unit time is equal to |