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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.
The exponential distribution is widely used in the field of reliability. Reliability deals with the amount 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.
$m=\frac{1}{\mu}$ . 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 ^{–0.25 x } where x is at least zero and m = 0.25.
For example, f (5) = 0.25 e ^{−(0.25)(5)} = 0.072. The postal clerk spends five minutes with the customers.
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.
X ~ Exp (0.125); f ( x ) = 0.125e ^{–0.125 x } ;
a. Using the information in [link] , find the probability that a clerk spends four to five minutes with a randomly selected customer.
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
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.
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:
P ( x < k ) = 0.50 and P ( x < k ) = 1 – e ^{–0.25 k }
Therefore, 0.50 = 1 − e ^{−0.25 k } and e ^{−0.25 k } = 1 − 0.50 = 0.5
Take natural logs: ln ( e ^{–0.25 k } ) = ln (0.50). So, –0.25 k = ln (0.50)
Solve for k : $k=\frac{ln(0.50)}{-0.25}=2.8$ minutes. The calculator simplifies the calculation for percentile k . See the following two notes.
A formula for the percentile k is $k=\frac{ln(1-AreaToTheLeft)}{-m}$ where ln is the natural log.
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.
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