Download Chapter 5 Continuous Random Variables (2) (連續隨機變數)

Chapter 5
Continuous Random Variables (2)
(連續隨機變數)
5.4 Exponential Distribution (指數分佈)
We now turn to a continuous distribution that is related to the discrete Poisson distribution. We
will see the relationship between them at a later time. Let us first define what is an exponential
distribution with mean μ .
Definition 5.12： Let μ be any positive real number. A continuous random variable X
with probability density function f(x) = 0, for x < 0,
f(x)
=
1
μ
e−x / μ ,
and
for
x > 0,
is said to have an exponential distribution with mean μ . We denote X ~ Exp( μ ).
then f(x) = e − x for x > 0, and X is said to have a unit exponential distribution.
If μ = 1,
Exercise 5.13：Verify that the above function f(x) is a probability density function. Sketch
the p.d.f. and c.d.f. of an exponential distribution with mean μ .
As with any continuous probability distribution, the area under the curve corresponding to
some interval provides the probability that the random variable takes on value in that interval.
In order to compute the exponential probabilities, we make use of the following formulas:
(a)
P( X ≤ c )
=
1 − e−c/μ .
(b)
P( X ≥ c )
=
e−c/μ .
(c)
P( c ≤ X ≤ d )
=
e−c/μ − e− d /μ .
Definition 5.14：The mean and variance of the exponential distribution are
E(X)
= μ
and
Exercise 5.15： With the definition E(X) =
density function of the random variable X.
Var(X)
∫
=
μ 2.
∞
−∞
x f(x) dx, where f(x) is the probability
Show that
E(X) = μ if X is exponential.
Exercise 5.16： Suppose X has an exponential distribution with mean equal to 10.
Determine the following.
(a)
P( X > 10 )
(b)
P( X < 20 )
(c)
P( 15 < X < 30 )
(d)
Find the value of a such that
P( X < a ) = 0.95.
Exercise 5.17： Suppose that every three months, on average, an earthquake occurs in
Indonesia. What is the probability that the next earthquake occurs after three and before seven
months?
Exercise 5.18： Suppose that a system contains a certain type of component whose time in
years to failure is given by the random variable X, distributed exponentially with mean time 5
to failure. If 5 of these components are installed in different systems, what is the probability
that at least 2 are still functioning at the end of 8 years?
Relationship between the Poisson and Exponential Distributions
Since the Poisson distribution as a discrete probability distribution that is often useful when
dealing with the number of occurrences of an event over a specified interval of time or space.
Recall that the Poisson probability density function is given by
e
=
P( X = k )
−λ
λk
k!
,
where
k = 0, 1, 2, …. ,
where
λ = expected value (or mean number) of occurrences in an interval.
The continuous exponential probability distribution is related to the discrete Poisson
distribution such that if the Poisson distribution provides an appropriate description of the
number of occurrences per interval, then the exponential distribution provides a description of
the length of the interval between occurrences.
Example 5.19： To provide an example that illustrates this relationship, suppose that the
number of cars that arrive at a gas station during 1 hour is described by a Poisson probability
distribution with a mean of λ = 10 (cars per hour). Thus the Poisson probability density
function that provides the probability of k arrivals per hour is
P( X = k )
e
=
−λ
λk
k!
=
e
−10
10 k
.
k!
Since the average number of arrivals is 10 cars per hour, the average time between cars
arriving is given by
μ
=
1
λ
1hour
10cars
=
=
0.1 hour / car.
Thus the corresponding exponential distribution that describes the time between the arrival of
cars has a mean of μ = 0.1 (hour per car); the appropriate exponential probability density
function (pdf) is given by
f(x)
=
1
μ
e−x / μ
=
1 − x / 0.1
e
0.1
and the cumulative distribution function (cdf) is given by
=
10 e −10 x ,
P( X ≤ a )
=
1 − e −10 a .
Exercise 5.20： At an intersection, there are two accidents per day, on average. What is the
probability that after the next accident there will be no accidents at all for the next two days?
Exercise 5.21： The manager for HK-Line, a company that sells tickets to concerts, has
determined that the time between people arriving at the box office on a typical day is
exponentially distributed with an arrival rate of 12 per hour. It takes approximately 4 minutes
to process a ticket request. Thus, if customers arrive in the intervals that are shorter than 4
minutes, they will have to wait. Assuming that a customer has just arrived and the ticket agent
is starting to serve that customer, what is the probability that the next customer who arrives
will have to wait in line?
Exercise 5.22： The number of customers arriving at a teller’s window at a bank follows the
Poisson distribution with a mean rate of 0.75 customer per minute. If the time between
arrivals is less than or equal to three minutes, then the teller can provide banking services
without irritating customers with annoying waiting times.
(a)
Find the mean and standard deviation of X, namely the time between customer
arrivals at the teller’s window.
(b)
Find the proportion of customers for whom the teller provides service without an
annoying delay.