modeled as a Poisson process. Typically, failures are not caused by components wearing out but by more random failures of the large number of semiconductor circuits in the units. Assume that the units that fail are immediately repaired, and assume that the mean number of failures per hour is 0.0001. Let X denote the time until four failures occur in a system. Determine the probability that X exceeds 40,000 hours.
Let the random variable N denote the number of failures in 40,000 hours of operation. The time until four failures occur exceeds 40,000 hours if and only if the number of failures in 40,000 hours is three or less. Therefore,
P X > ( 40 000 , ) = P N ( ≤ 3 )
The assumption that the failures follow a Poisson process implies that N has a Poisson distribution with E N ( ) = 40 000 0 0001 , ( . ) = 4 failures per 40 000 , hours
Therefore,
P X > , P N e
k
k k
40 000 3 4
0 433
4 0
( ) = ( ≤ ) =
3 −! = .
∑
=Example 4-23
Because P X > x ( ) = − 1 F x , ( ) the probability density function of X equals the negative of the derivative of the right-hand side of the previous equation. After extensive algebraic simplifica- tion, the probability density function of X can be shown to equal
f x x e
r
r r x
( ) =
λ( ) − !
−1 −λ
1 for x > 0 and r = 1, 2, . . . .
This probability density function defines an Erlang random variable. Clearly, an Erlang random variable with r = 1 is an exponential random variable.
It is convenient to generalize the Erlang distribution to allow r to assume any non-negative value. Then the Erlang and some other common distributions become special cases of this generalized distribution. To accomplish this step, the factorial function ( ) r − 1 ! is generalized to apply to any non-negative value of r, but the generalized function should still equal ( ) r − 1 ! when r is a positive integer.
The gamma function is
Γ ( ) r =
∞∫ x e
r−1 −xdx, r >
0
0
for (4-18)
Gamma Function
It can be shown that the integral in the definition of Γ ( ) r is finite. Furthermore, by using inte- gration by parts, it can be shown that
Γ ( ) r = − ( r 1 ) Γ − ( r 1 )
This result is left as an exercise. Therefore, if r is a positive integer (as in the Erlang distribution), Γ ( ) r = − ( r 1 ) !
Also, Γ ( ) 1 = ! = 0 1 and it can be shown that Γ ( ) 1 2 / = π
1 2/. The gamma function can be inter- preted as a generalization to noninteger values of r of the term that is used in the Erlang prob- ability density function. Now the Erlang distribution can be generalized.
The random variable X with probability density function
f x x e
r , x >
r r x
( ) =
λΓ ( )
−1 −λ
0
for (4-19)
is a gamma random variable with parameters
λ> 0 and r > 0 . If r is an integer, X has an Erlang distribution.
Gamma Distribution
The parameters λ and r are often called the scale and shape parameters, respectively. How- ever, one should check the definitions used in software packages. For example, some sta- tistical software defines the scale parameter as 1 / λ . Sketches of the gamma distribution for several values of λ and r are shown in Fig. 4-25. Many different shapes can be generated from changes to the parameters. Also, the change of variable u =
λx and the definition of the gamma function can be used to show that the probability density function integrates to 1.
For the special case when r is an integer and the value of r is not large, Equation (4-17) can be applied to calculate probabilities for a gamma random variable. However, in general, the integral of the gamma probability density function is difficult to evaluate so computer software is used to determine probabilities.
Recall that for an exponential distribution with parameter λ , the mean and variance are 1 / λ
and 1 / λ
2, respectively. An Erlang random variable is the time until the rth event in a Poisson
process and the time between events are independent. Therefore, it is plausible that the mean
Section 4-9/Erlang and Gamma Distributions 141
and variance of a gamma random variable multiply the exponential results by r. This motivates the following conclusions. Repeated integration by parts can be used to derive these, but the details are lengthy and omitted.
0 0.0 0.2 0.4 0.6 0.8 1.0
2 4 6 8 10 12
x
1 8.3 7.5
1 2 3.75
l r
f (x)
FIGURE 4-25 Gamma probability density functions for selected values of
λ
and r.If X is a gamma random variable with parameters λ and r, μ = E X ( ) = r /
λand σ =
2V X ( ) = r /
λ2Mean and Variance
The time to prepare a micro-array slide for high-throughput genomics is a Poisson process with a mean of two hours per slide. What is the probability that 10 slides require more than 25 hours to prepare?
Let X denote the time to prepare 10 slides. Because of the assumption of a Poisson process, X has a gamma distribu- tion with
λ= 1 2 / ,r = 10 , and the requested probability is P X > ( 25 ) . The probability can be obtained from software that provides cumulative Poisson probabilities or gamma probabilities. For the cumulative Poisson probabilities, we use the method in Example 4-23 to obtain
P X > e k
k
k
25
12 512 5
0
( ) =
9( ) .
!
− .
∑
=In software we set the mean = 12.5 and the input = 9 to obtain P X > ( 25 ) = . 0 2014 .
As a check, we use the gamma cumulative probability function in Minitab. Set the shape parameter to 10, the scale parameter to 0.5, and the input to 25. The probability computed is P X ( ≤ 25 ) = . 0 7986, and when this is subtracted from one we match with the previous result that P X > ( 25 ) = . 0 2014 .
What are the mean and standard deviation of the time to prepare 10 slides? The mean time is E X ( ) = r /
λ= 10 0 5 / . = 20
The variance of time is
V X ( ) = r /
λ2= 10 0 5 / . =
240 so that the standard deviation is 40
1/2= 6.32 hours .
The slides will be completed by what length of time with probability equal to 0.95? The question asks for x such that P X ( ≤ x ) = . 0 95
Example 4-24
Furthermore, the chi-squared distribution is a special case of the gamma distribution in which
λ= 1 2 / and r equals one of the values 1/2, 1, 3/2, 2, . . . . This distribution is used extensively in interval estimation and tests of hypotheses that are discussed in subsequent chapters. The chi-squared distribution is discussed in Chapter 7.
where X is gamma with
λ= . 0 5 and r = . 10 In software, we use the gamma inverse cumulative probability function and set the shape parameter to 10, the scale parameter to 0.5, and the probability to 0.95. The solution is
P X ( ≤ . 31 41 ) = . 0 95
Practical Interpretation: Based on this result, a schedule that allows 31.41 hours to prepare 10 slides should be met 95% of the time.
4-137. Use the properties of the gamma function to evalu- ate the following:
(a) Γ
( )
6 (b) Γ( )
5 2/ (c) Γ( )
9 2/4-138. Given the probability density function f x
( )
= .0 013x e2 − .0 01x/Γ
( )
3, determine the mean and variance of the distribution.4-139. Calls to a telephone system follow a Poisson dis- tribution with a mean of fi ve calls per minute.
(a) What is the name applied to the distribution and parameter values of the time until the 10th call?
(b) What is the mean time until the 10th call?
(c) What is the mean time between the 9th and 10th calls?
(d) What is the probability that exactly four calls occur within one minute?
(e) If 10 separate one-minute intervals are chosen, what is the probability that all intervals contain more than two calls?
4-140. Raw materials are studied for contamination. Sup- pose that the number of particles of contamination per pound of material is a Poisson random variable with a mean of 0.01 particle per pound.
(a) What is the expected number of pounds of material required to obtain 15 particles of contamination?
(b) What is the standard deviation of the pounds of materials required to obtain 15 particles of contamination?
4-141. The time between failures of a laser in a cytogenics machine is exponentially distributed with a mean of 25,000 hours.
(a) What is the expected time until the second failure?
(b) What is the probability that the time until the third failure exceeds 50,000 hours?
4-142. In a data communication system, several messages that arrive at a node are bundled into a packet before they are transmitted over the network. Assume that the messages arrive at the node according to a Poisson process with τ =30 mes- sages per minute. Five messages are used to form a packet.
(a) What is the mean time until a packet is formed, that is, until fi ve messages have arrived at the node?
(b) What is the standard deviation of the time until a packet is formed?
(c) What is the probability that a packet is formed in less than 10 seconds?
(d) What is the probability that a packet is formed in less than fi ve seconds?
4-143. Errors caused by contamination on optical disks occur at the rate of one error every 105 bits. Assume that the errors follow a Poisson distribution.
(a) What is the mean number of bits until fi ve errors occur?
(b) What is the standard deviation of the number of bits until fi ve errors occur?
(c) The error-correcting code might be ineffective if there are three or more errors within 105 bits. What is the probability of this event?
4-144. Calls to the help line of a large computer distribu- tor follow a Poisson distribution with a mean of 20 calls per minute. Determine the following:
(a) Mean time until the one-hundredth call (b) Mean time between call numbers 50 and 80
(c) Probability that three or more calls occur within 15 seconds 4-145. The time between arrivals of customers at an auto- matic teller machine is an exponential random variable with a mean of fi ve minutes.
(a) What is the probability that more than three customers arrive in 10 minutes?
(b) What is the probability that the time until the fi fth customer arrives is less than 15 minutes?
4-146. Use integration by parts to show that Γ
( )
r = −( )
r 1Γ −
( )
r 1.4-147. Show that the gamma density function f x, ,r
(
λ)
inte-grates to 1.
4-148. Use the result for the gamma distribution to determine the mean and variance of a chi-square distribution with r=7 2.
4-149. Patients arrive at a hospital emergency department according to a Poisson process with a mean of 6.5 per hour.
FOR SECTION 4-9
EXERCISES
Problem available in WileyPLUS at instructor’s discretion.
Tutoring problem available in WileyPLUS at instructor’s discretion.
Section 4-10/Weibull Distribution 143
(a) What is the mean time until the 10th arrival?
(b) What is the probability that more than 20 minutes is required for the third arrival?
4-150. The total service time of a multistep manufacturing operation has a gamma distribution with mean 18 minutes and standard deviation 6.
(a) Determine the parameters λ and r of the distribution.
(b) Assume that each step has the same distribution for service time. What distribution for each step and how many steps produce this gamma distribution of total service time?
4-151. An article in Sensors and Actuators A: Physical [“Characterization and Simulation of Avalanche PhotoDiodes for Next-Generation Colliders” (2011, Vol.172(1), pp.181–
188)] considered an avalanche photodiode (APD) to detect charged particles in a photo. The number of arrivals in each detection window was modeled with a Poisson distribution with a mean depending on the intensity of beam. For one beam
intensity, the number of electrons arriving at an APD follows a Poisson distribution with a mean of 1.74 particles per detection window of 200 nanoseconds.
(a) What is the mean and variance of the time for 100 arrivals?
(b) What is the probability that the time until the fifth particle arrives is greater than 1.0 nanosecond?
4-152. An article in Mathematical Biosciences [“Influence of Delayed Viral Production on Viral Dynamics in HIV-1 Infected Patients” (1998, Vol.152(2), pp. 143–163)] considered the time delay between the initial infection by immunodeficiency virus type 1 (HIV-1) and the formation of productively infected cells.
In the simulation model, the time delay is approximated by a gamma distribution with parameters r=4 and 1/ = .λ 0 25 days.
Determine the following:
(a) Mean and variance of time delay
(b) Probability that a time delay is more than half a day (c) Probability that a time delay is between one-half and one day
The random variable X with probability density function
f x x x
, x >
( ) = β δ ⎛ ⎝⎜ δ ⎞ ⎠⎟ − ⎡ ⎛ ⎝⎜ δ ⎞ ⎠⎟
⎣ ⎢
⎢
⎤
⎦ ⎥
⎥
β−1 β