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.

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

The PDF

If \(X\) has an exponential distribution with parameter \(\lambda\) then we write \[X \sim \text{Exp}(\lambda)\] and the probability density function for \(X\) is \[f(x) = \lambda \cdot e^{-\lambda x} \text{ for } x \ge 0 \text{ and } 0 \text{ elsewhere }\] \(\lambda\) is the rate parameter where the mean of \(X\) is \(\frac{1}{\lambda}\).

It can be shown that \(\lambda\) is the same parameter used for the Poisson distribution.

The CDF

\[F(x_0) = P(X \le x_0) = \int_{0}^{x_0} f(x)dx = 1-e^{-\lambda x_0}\] In R, \(P(X \le x) = \text{pexp}(x,\lambda)\)

Example 1: The amount of time (in minutes) a postal clerk spends with his or her customer is known to have an exponential distribution with the average amount of time equal to four minutes.

(a). Find the probability that a clerk spends

  • less than ten minutes with a randomly selected customer
  • more than two minutes with a randomly selected customer
  • between four and five minutes with a randomly selected customer

(b). Find the median time the clerk spends with a randomly selected customer

Let \(X\) be the amount of time (in minutes) a postal clerk spends with a customer. The mean of \(X\) is given as four minutes so \(\lambda = \frac{1}{4} = 0.25\) and the density is \(f(x) = 0.25 \cdot e^{-0.25x}\)


(a).

  • \(P(X < 10) = 1 - e^{-0.25 \cdot 10} = 0.92 = \text{pexp}(10,0.25)\)
  • \(P(X > 2) = 1 - P(X < 2) = 1 - (1 - e^{-0.25 \cdot 2}) = 0.61 = 1 - \text{pexp}(2,0.25)\)
  • \(P(4 < X < 5) = P(X<5)-P(X<4) = \text{pexp}(5,0.25) - \text{pexp}(4,0.25) = 0.08\)

(b). To find the median, we can use the qexp command to find the \(50^{th}\) percentile

> qexp(0.5,0.25)

[1] 2.772589