The random variable in case of Poisson distribution is of the type; the number of arrivals of customers per unit of time or the number of defects per unit length of cloth, etc. Alternatively, it is possible to define a random variable, in the context of Poisson Process, as the length of time between the arrivals of two consecutive customers or the length of cloth between two consecutive defects, etc. The probability distribution of such a random variable is termed as Exponential Distribution.
Since the length of time or distance is a continuous random variable, therefore exponential distribution is a continuous probability distribution.
Probability Density Function
The probability density function of a continuous random variable is a function which can be integrated to obtain the probability that the random variable takes a value in a given interval. Let t be a random variable which denotes the length of time or distance between the occurrence of two consecutive events or the occurrence of the first event and m be the average number of times the event occurs per unit of time or length.
Further, let A be the event that the time of occurrence between two consecutive events or the occurrence of the first event is less than or equal to t and f(t) and F(t) denote the probability density function and the distribution (or cumulative density) function of t respectively.
We can write P(A) + P(Ā) =1 or F (t) + P(Ā) = 1. Note that, by definition,F (t) = P(A) Further, P(Ā) is the probability that the length of time between the occurrence of two consecutive events or the occurrence of first event is greater than t. This is also equal to the probability that no event occurs in the time interval t. Since the mean number of occurrence of events in time t is mt, we have , by Poisson distribution
Example : A telephone operator attends on an average 150 telephone calls per hour. Assuming that the distribution of time between consecutive calls follows an exponential distribution, find the probability that (i) the time between two consecutive calls is less than 2 minutes, (ii) the next call will be received only after 3 minutes.
Solution: Here m = the average number of calls per minute = 150/60= 2.5.
Example : The average number of accidents in an industry during a year is estimated to be 5. If the distribution of time between two consecutive accidents is known to be exponential, find the probability that there will be no accidents during the next two months
Here m denotes the average number of accidents per month = 5/12
&there4 P(t>2)= 1- F(2) = e-(5/12)*2 = e-0.833 = 0.4347
Example: The distribution of life, in hours, of a bulb is known to be exponential with mean life of 600 hours. What is the probability that (i) it will not last more than 500 hours, (ii) it will last more than 700 hours?
Since the random variable denote hours, therefore m=1/600
(i) P(t ≤500) = F(500)
= 1- e(-1/600)*500 = 1-e -0.833
(ii) P(t > 700) = 1 - F(700)= e(700/600)= e(-1.1667)= 0.3114