22 Lack of Memory Property Let X denote the time between detections of a particle with a Geiger
Example 4-22 Lack of Memory Property Let X denote the time between detections of a particle with a Geiger
counter and assume that X has an exponential distribution with E X () =. 1 4 minutes. The probabil-
ity that we detect a particle within 30 seconds of starting the counter is
0514 PX< −. ( .
05 . minute ) = F () . =− 1 e =. 0 30
In this calculation, all units are converted to minutes. Now, suppose that we turn on the Geiger counter and wait three minutes without detecting a particle. What is the probability that a particle is detected in the next 30 seconds?
Chapter 4Continuous Random Variables and Probability Distributions
Because we have already been waiting for three minutes, we feel that a detection is “due.’’ That is, the probability of
a detection in the next 30 seconds should be higher than 0.3. However, for an exponential distribution, this is not true.
The requested probability can be expressed as the conditional probability that P X < ( 35 . u X> 3 ) . From the defi nition
of conditional probability,
PX< ( 35 . u X> 3 ) = P (
where
P ( 3 1 F 3 =− − = e 314 ( . () =. 0 117 PX> 3 ) Therefore, PX< ( 35 . u X> 3 ) =. 0 035 0 117 . =. 0 30 Practical Interpretation: After waiting for three minutes without a detection, the probability of a detection in the next 30 seconds is the same as the probability of a detection in the 30 seconds immediately after starting the counter. The fact that we have waited three minutes without a detection does not change the probability of a detection in the next 30 seconds. Example 4-22 illustrates the lack of memory property of an exponential random variable, and a general statement of the property follows. In fact, the exponential distribution is the only continuous distribution with this property. Lack of Memory
Property
For an exponential random variable X,
PX Figure 4-24 graphically illustrates the lack of memory property. The area of region A divided by the total area under the probability density function A ( +++= B C D ) 1 equals P X < t ( 2 ) . The area of region C divided by the area C + equals P X < t t X > t D ( 1 + 2 u 1 ) . The lack of memory property implies that the proportion of the total area that is in A equals the propor- tion of the area in C and D that is in C. The mathematical verifi cation of the lack of memory property is left as a Mind-Expanding exercise. The lack of memory property is not so surprising when we consider the development of a Poisson process. In that development, we assumed that an interval could be partitioned into small intervals that were independent. These subintervals are similar to independent Bernoulli trials that comprise a binomial experiment; knowledge of previous results does not affect the probabilities of events in future subintervals. An exponential random variable is the continu- ous analog of a geometric random variable, and it shares a similar lack of memory property. f (x) FIGURE 4-24 Lack of memory property A of an exponential B
C D
distribution.
t 2 t 1 t 1 + t 2 x
Section 4-8Exponential Distribution
The exponential distribution is often used in reliability studies as the model for the time until failure of a device. For example, the lifetime of a semiconductor chip might be modeled as an exponential random variable with a mean of 40,000 hours. The lack of memory property of the exponential distribution implies that the device does not wear out. That is, regardless of how long the device has been operating, the probability of a failure in the next 1000 hours is the same as the probability of a failure in the first 1000 hours of operation. The lifetime L of a device with failures caused by random shocks might be appropriately modeled as an exponential random variable.
However, the lifetime L of a device that suffers slow mechanical wear, such as bearing wear, is better modeled by a distribution such that P L < t ( +Δ tL>t u ) increases with t. Distri- butions such as the Weibull distribution are often used in practice to model the failure time of this type of device. The Weibull distribution is presented in a later section.
Exercises
FOR SECTION 4-8
Problem available in WileyPLUS at instructor’s discretion.
Tutoring problem available in WileyPLUS at instructor’s discretion.
4-112.
Suppose that X has an exponential distribution
(c) What is the probability that the first call arrives within 5
with λ = 2. Determine the following:
and 10 minutes after opening?
(a) PX ( ≤ ) 0 (b) PX ( ≥ 2 )
(d) Determine the length of an interval of time such that the
(c) PX ( ≤ ) 1 (d) P ( probability of at least one call in the interval is 0.90. 1 2 ) (e) Find the value of x such that P X < x ( ) =.. 0 05
4-118.
The life of automobile voltage regulators has
4-113.
Suppose that X has an exponential distribution
an exponential distribution with a mean life of six years. You
with mean equal to 10. Determine the following:
purchase a six-year-old automobile, with a working voltage
(a) P X >10 ( ) (b) P X > 20 ( ) (c) P X < 30
( regulator and plan to own it for six years. )
(d) Find the value of x such that P X < x ( =.. 0 95
) (a) What is the probability that the voltage regulator fails dur-
ing your ownership?
4-114.
Suppose that X has an exponential distribution
(b) If your regulator fails after you own the automobile three years
with a mean of 10. Determine the following:
(a) PX<5 ( ) (b) PX< ( 15 | X> 10 )
and it is replaced, what is the mean time until the next failure?
4-119.
Suppose that the time to failure (in hours) of fans
(c) Compare the results in parts (a) and (b) and comment on
in a personal computer can be modeled by an exponential dis-
the role of the lack of memory property.
tribution with λ=. 0 0003 .
4-115.
Suppose that the counts recorded by a Geiger
(a) What proportion of the fans will last at least 10,000 hours?
counter follow a Poisson process with an average of two counts
(b) What proportion of the fans will last at most 7000 hours?
per minute.
4-120.
The time between the arrival of electronic mes-
(a) What is the probability that there are no counts in a 30-sec-
sages at your computer is exponentially distributed with a mean
ond interval?
of two hours.
(b) What is the probability that the first count occurs in less
(a) What is the probability that you do not receive a message
than 10 seconds?
during a two-hour period?
(c) What is the probability that the first count occurs between
(b) If you have not had a message in the last four hours, what
one and two minutes after start-up?
is the probability that you do not receive a message in the
4-116.
Suppose that the log-ons to a computer network
next two hours?
follow a Poisson process with an average of three counts per
(c) What is the expected time between your fifth and sixth
minute.
messages?
(a) What is the mean time between counts?
4-121.
The time between arrivals of taxis at a busy inter-
(b) What is the standard deviation of the time between counts?
section is exponentially distributed with a mean of 10 minutes.
(c) Determine x such that the probability that at least one count
(a) What is the probability that you wait longer than one hour
occurs before time x minutes is 0.95.
for a taxi?
4-117.
The time between calls to a plumbing supply busi-
(b) Suppose that you have already been waiting for one hour
ness is exponentially distributed with a mean time between
for a taxi. What is the probability that one arrives within the
calls of 15 minutes.
next 10 minutes?
(a) What is the probability that there are no calls within a
(c) Determine x such that the probability that you wait more
30-minute interval?
than x minutes is 0.10.
(b) What is the probability that at least one call arrives within
(d) Determine x such that the probability that you wait less
a 10-minute interval?
than x minutes is 0.90.
Chapter 4Continuous Random Variables and Probability Distributions
(e) Determine x such that the probability that you wait less
4-127. The time between calls to a corporate office is expo-
than x minutes is 0.50.
nentially distributed with a mean of 10 minutes.
4-122. The number of stork sightings on a route in South
(a) What is the probability that there are more than three calls
Carolina follows a Poisson process with a mean of 2.3 per year.
in one-half hour?
(a) What is the mean time between sightings?
(b) What is the probability that there are no calls within one-
(b) What is the probability that there are no sightings within
half hour?
three months (0.25 years)?
(c) Determine x such that the probability that there are no calls
(c) What is the probability that the time until the first sighting
within x hours is 0.01.
exceeds six months?
(d) What is the probability that there are no calls within a two-
(d) What is the probability of no sighting within three years?
hour interval?
4-123. According to results from the analysis of chocolate bars
(e) If four nonoverlapping one-half-hour intervals are selected,
in Chapter 3, the mean number of insect fragments was 14.4
what is the probability that none of these intervals contains
in 225 grams. Assume that the number of fragments follows a
any call?
Poisson distribution.
(f) Explain the relationship between the results in part (a) and (b).
(a) What is the mean number of grams of chocolate until a
4-128.
Assume that the flaws along a magnetic tape fol-
fragment is detected?
low a Poisson distribution with a mean of 0.2 flaw per meter.
(b) What is the probability that there are no fragments in a
Let
X denote the distance between two successive flaws.
28.35-gram (one-ounce) chocolate bar?
(a) What is the mean of X ?
(c) Suppose you consume seven one-ounce (28.35-gram) bars
(b) What is the probability that there are no flaws in 10 con-
this week. What is the probability of no insect fragments?
secutive meters of tape?
4-124.
The distance between major cracks in a highway
(c) Does your answer to part (b) change if the 10 meters are
follows an exponential distribution with a mean of five miles.
not consecutive?
(a) What is the probability that there are no major cracks in a
(d) How many meters of tape need to be inspected so that the
10-mile stretch of the highway?
probability that at least one flaw is found is 90?
(b) What is the probability that there are two major cracks in a
(e) What is the probability that the first time the distance
10-mile stretch of the highway?
between two flaws exceeds eight meters is at the fifth flaw?
(c) What is the standard deviation of the distance between
major cracks?
(f) What is the mean number of flaws before a distance between
(d) What is the probability that the first major crack occurs
two flaws exceeds eight meters?
between 12 and 15 miles of the start of inspection?
4-129. If the random variable
X has an exponential distribu-
(e) What is the probability that there are no major cracks in
tion with mean θ, determine the following:
(a) PX> ( θ ) (b) PX>2 ( θ ) (c) PX>3 ( θ )
two separate five-mile stretches of the highway?
(f) Given that there are no cracks in the first five miles
(d) How do the results depend on θ?
inspected, what is the probability that there are no major
4-130.
Derive the formula for the mean and variance of
cracks in the next 10 miles inspected?
an exponential random variable.
4-125.
The lifetime of a mechanical assembly in a vibra-
4-131. Web crawlers need to estimate the frequency of changes
tion test is exponentially distributed with a mean of 400 hours.
to Web sites to maintain a current index for Web searches.
(a) What is the probability that an assembly on test fails in less
Assume that the changes to a Web site follow a Poisson process
than 100 hours?
with a mean of 3.5 days.
(b) What is the probability that an assembly operates for more
(a) What is the probability that the next change occurs in less
than 500 hours before failure?
than 2.0 days?
(c) If an assembly has been on test for 400 hours without a fail-
(b) What is the probability that the time until the next change
ure, what is the probability of a failure in the next 100 hours?
is greater 7.0 days?
(d) If 10 assemblies are tested, what is the probability that
(c) What is the time of the next change that is exceeded with
at least one fails in less than 100 hours? Assume that the
probability 90?
assemblies fail independently.
(d) What is the probability that the next change occurs in less than
(e) If 10 assemblies are tested, what is the probability that all
10.0 days, given that it has not yet occurred after 3.0 days?
have failed by 800 hours? Assume that the assemblies fail
4-132. The length of stay at a specific emergency department
independently.
in a hospital in Phoenix, Arizona had a mean of 4.6 hours.
4-126.
The time between arrivals of small aircraft at a county
Assume that the length of stay is exponentially distributed.
airport is exponentially distributed with a mean of one hour.
(a) What is the standard deviation of the length of stay?
(a) What is the probability that more than three aircraft arrive
(b) What is the probability of a length of stay of more than 10
within an hour?
hours?
(b) If 30 separate one-hour intervals are chosen, what is the
(c) What length of stay is exceeded by 25 of the visits?
probability that no interval contains more than three arrivals?
4-133. An article in Journal of National Cancer Institute
(c) Determine the length of an interval of time (in hours) such that
[“Breast Cancer Screening Policies in Developing Countries:
the probability that no arrivals occur during the interval is 0.10.
A Cost-Effectiveness Analysis for India” (2008, Vol.100(18),
Section 4-9Erlang and Gamma Distributions
pp. 1290–1300)] presented a screening analysis model of breast
of healthcare workers to reduce the hazard rate of infl uenza virus
cancer based on data from India. In this analysis, the time that
infection for patients in regular hospital departments. In this anal-
a breast cancer case stays in a preclinical state is modeled to be
ysis, each patient’s length of stay in the department is taken as
exponentially distributed with a mean depending on the state.
exponentially distributed with a mean of 7.0 days.
For example, the time that a cancer case stays in the state of T1C
(a) What is the probability that a patient stays in hospital for
(tumor size of 11–20 mm) is exponentially distributed with a
less than 5.5 days?
mean of 1.48 years.
(b) What is the probability that a patient stays in hospital for
(a) What is the probability that a breast cancer case in India
more than 10.0 days if the patient has currently stayed for
stays in the state of T1C for more than 2.0 years?
7.0 days?
(b) What is the proportion of breast cancer cases in India that
(c) Determine the mean length of stay such that the probability
spend at least 1.0 year in the state of T1C?
is 0.9 that a patient stays in the hospital less than 6.0 days.
(c) Assume that a person in India is diagnosed to be in the state
4-136. An article in Ad Hoc Networks [“Underwater Acous-
of T1C. What is the probability that the patient is in the
tic Sensor Networks: Target Size Detection and Performance
same state six months later?
Analysis” (2009, Vol.7(4), pp. 803–808)] discussed an under-
4-134. Requests for service in a queuing model follow a Pois-
water acoustic sensor network to monitor a given area in an
son distribution with a mean of fi ve per unit time.
ocean. The network does not use cables and does not interfere
(a) What is the probability that the time until the fi rst request is
with shipping activities. The arrival of clusters of signals gen-
less than 4 minutes?
erated by the same pulse is taken as a Poisson arrival process
(b) What is the probability that the time between the second
with a mean of λ per unit time. Suppose that for a specifi c
and third requests is greater than 7.5 time units?
underwater acoustic sensor network, this Poisson process has a
(c) Determine the mean rate of requests such that the probabil-
rate of 2.5 arrivals per unit time.
ity is 0.9 that there are no requests in 0.5 time units.
(a) What is the mean time between 2.0 consecutive arrivals?
(d) If the service times are independent and exponentially distrib-
(b) What is the probability that there are no arrivals within 0.3
uted with a mean of 0.4 time units, what can you conclude
time units?
about the long-term response of this system to requests?
(c) What is the probability that the time until the fi rst arrival
4-135. An article in Vaccine [“Modeling the Effects of Infl uenza
exceeds 1.0 unit of time?
Vaccination of Health Care Workers in Hospital Departments”
(d) Determine the mean arrival rate such that the probability is
(2009, Vol.27(44), pp. 6261–6267)] considered the immunization
0.9 that there are no arrivals in 0.3 time units.
4-9 Erlang and Gamma Distributions
An exponential random variable describes the length until the fi rst count is obtained in a Pois- son process. A generalization of the exponential distribution is the length until r events occur in a Poisson process. Consider Example 4-23.