
as given by Equation (6.40), we have
Comparing this expression with Equation (7.59), we can establish the result
that the interarrival time between Poisson arrivals has an exponential distribu-
tion; the parameter in the distribution of T is the mean arrival rate associated
with Poisson arrivals.
Example 7.6. Problem: referring to Example 6.11 (page 177), determine the
probability that the headway (spacing measured in time) between arriving
vehicles is at least 2 minutes. Also, compute the mean headway.
Answer: in Example 6.11, the parameter was estimated to be 4.16 vehicles
per minute. Hence, if T is the headway in minutes, we have
The mean headway is
Since interarrival times for Poisson arrivals are independent, the time required
for a total of n Poisson arrivals is a sum of n independent and exponentially
distributed random variables. Let T
j
, j 1, 2, . . . , n, be the interarrival time
between the (j 1)th and jth arrivals. The time required for a total of n arrivals,
denoted by X
n
,is
where T
j
,j 1,2,...,n, are independent and exponentially distributed with the
same parameter . In Example 4.16 (page 105), we showed that X
n
has a
gamma distribution with 2 when n 2. The same procedure immediately
shows that, for general n, X
n
is gamma-distributed with n. Thus, as stated,
the gamma distribution is appropriate for describing the time required for a
total of Poisson arrivals.
Example 7.7. Problem: ferries depart for trips across a river as soon as nine
vehicles are driven aboard. It is observed that vehicles arrive independently at
an average rate of 6 per hour. Determine the probability that the time between
trips will be less than 1 hour.
Answer: from our earlier discussion, the time between trips follows a gamma
distribution with 9 and 6. Hence, let X be the time between trips in
216
Fundamentals of Probability and Statistics for Engineers
P[X(0, t) 0] e
t
F
T
t
1 e
t
; for t 0;
0; elsewhere:
(
7:62
PT 2
Z
1
2
f
T
tdt 1 F
T
2e
24:16
0:00024:
m
T
1
1
4:16
minutes 0:24 minutes:
X
n
T
1
T
2
T
n
; 7:63