The Exponential Distributions


Recall the context of the Poisson random variable: An event has a small probability of occurring, a large number of independent trials take place, and the average number of occurrences µ over a period of time is known. In this setting, an exponential distribution measures the time needed for the first occurrence.

The average time needed for the first occurrence is given by Q = 1 / µ . This parameter Q defines the exponential random variable which is denoted by X ~exp(Q). The probability density function (pdf) is given by

f(x) = (1 / Q)* e ^-(x / Q) , for x >= 0.

If X ~ exp(Q), then the graph of the distribution function is a typical exponential decay function for x >= 0.

A formula for the cumulative distribution is also known. If X ~ exp(Q), then for x >= 0,

P(X <= x) = 1 - e ^ -(x / Q),

and for 0 <= j <= k, we have

P(j<= X <= k) = e ^ -(j / Q) - e ^ -(k / Q).

Using the EXPONEN Program

The EXPONEN program can be used to compute probabilities such as P(X <=k), P(X >= k), or P(j <= X <= k). To execute the program, first enter 1, 2, or 3 to designate the type of probability you wish to compute. Next, enter the lower and upper bounds j and k (or just the single bound k for a tail probability). The program next asks if you want to see a shaded graph of the pdf. If so, enter 1. If not, then enter 0. After the graph, we receive a display of the desired probability.

Example. On weeknight shifts between 6 pm and 10 pm, there is an average of 5.2 calls to 911 for medical emergencies. Let X measure the time needed for the first call on such a shift. Find the probability that the first call arrives (a) between 6:15 and 6:45 (b) before 6:30. Also find the median time needed for the first call.

Solution. First, we must determine the correct average of this exponential distribution. If we consider the time interval to be 240 minutes, then, on average, there is a call every 240 / 5.2 (or 46.15) minutes. Then X ~ exp(240 / 5.2) measures the time in minutes after 6:00 pm until the first call.

Call up the EXPONEN program and enter 3 for MIDDLE PROB. Next, enter 240/5.2 for AVERAGE, enter 15 for LOWER BOUND, and enter 45 for UPPER BOUND. We see that P(15 <= X <= 45) = 0.345335.

For part (b), reexecute the program but enter 1 for LEFT PROB. When prompted, enter 30 for BOUND. We see that P(0 <= X <= 30) = 0.478. Thus, there is nearly a 48% chance that the first call will come within the first half hour.

To find the median, we must find x such that P(X <=x) = 0.5. That is, we must solve the equation 1 - e ^ -(x/Q) = 0.5. Solving for x, we obtain x = -Q ln(0.5) = -240 / 5.2 * ln(0.5) = 31.99. So around half of the time, the first call will come within the first 32 minutes.

Other Features:

1. If you entered 1 to see a graph, then the partially shaded graph of the pdf initially appears, but then is replaced by the display of the computed probability. To see the graph again, press GRAPH. Then to see the probability value again, press CLEAR. On the TI-86, press CLEAR twice, or just press EXIT to remove the graph. To exit the graph on the TI-89, press HOME. Then press F5 to see the program output again.

2. If you initially enter either 1 or 2 to compute a tail probability, then both the left-tail and right-tail probabilities will be displayed in either case. However only the desired region will be shaded in the graph.


1. Customers arrive at a certain store at an average of 15 per hour. What is the probability that the manager must wait at least 5 minutes for the first customer?

2. The exponential distribution is often used in probability to model (remaining) lifetimes of mechanical objects for which the average lifetime is known and for which the probability distribution is assumed to decay exponentially.

Suppose that after the first 6 hours, the average remaining lifetime of batteries for a portable compact disc player is 8 hours. Find the probability that a set of batteries lasts between 12 and 16 hours.


1. Here the average waiting time is 60 / 15 = 4 minutes. Thus, X ~ exp(4) and we want P(X >= 5). So call up the EXPONEN program, enter 2 for RIGHT PROB, enter 4 for AVERAGE, and enter 5 for BOUND. We obtain a right-tail value of 0.2865. So around 28.65% of the time, the store must wait at least 5 minutes for the first customer.

2. Here the remaining lifetime can be assumed to be X ~ exp(8). For the total lifetime to be from 12 to 16, then the remaining lifetime (after the first 6 hours) is from 6 to 10. We execute the EXPONEN program with 8 for AVERAGE, 6 for LOWER BOUND, and 10 for UPPER BOUND. We find that P(6 <= X <= 10) = 0.18586.

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