16 Shaft Diameter The diameter of a shaft in an optical storage drive is normally distributed
Example 4-16 Shaft Diameter The diameter of a shaft in an optical storage drive is normally distributed
with mean 0.2508 inch and standard deviation 0.0005 inch. The specifi cations on the shaft are
0.2500 0.0015 ± inch. What proportion of shafts conforms to specifi cations?
Let X denote the shaft diameter in inches. The requested probability is shown in Fig. 4-18 and
( ⎞
= P ( −. 46 << Z 14 . ) = PZ ( < . ) − ( −. 46 )
= . 0 91924 − 0 0000 == 0 91924 .
Most of the nonconforming shafts are too large because the process mean is located very near to the upper specifi ca- tion limit. If the process is centered so that the process mean is equal to the target value of 0.2500,
( −.
<< 0 2515 ) = P
= P ( − 3 << Z ) = PZ ( < ) − (
=. 0 99865 −. 0 00135 =. 0 9973
Practical Interpretation: By recentering the process, the yield is increased to approximately 99.73.
FIGURE
Distribution of N
4-17
Determining the value of x to
Standardized distribution of N 0.45
meet a specifi ed probability.
FIGURE 4-18
for Example 4-16.
Chapter 4Continuous Random Variables and Probability Distributions
EXERCISES
FOR SECTION 4-6
Problem available in WileyPLUS at instructor’s discretion.
Tutoring problem available in WileyPLUS at instructor’s discretion.
4-63.
Use Appendix Table III to determine the following
4-72.
The time until recharge for a battery in a laptop com-
probabilities for the standard normal random variable Z:
puter under common conditions is normally distributed with a
(a) PZ ( ,1 32 . ) (b) PZ ( ,30 . )
mean of 260 minutes and a standard deviation of 50 minutes.
(c) P Z ( .1 45 . ) (d) PZ ( .− . 2 15 )
(a) What is the probability that a battery lasts more than four
(e) P ( − . 2 34 ,, Z 1 76 )
hours?
4-64. Use Appendix Table III to determine the following
(b) What are the quartiles (the 25 and 75 values) of battery life?
probabilities for the standard normal random variable Z :
(c) What value of life in minutes is exceeded with 95 probability?
(a) P ( − 1 , , (b) P Z ) ( − 2 ,, Z )
4-73. An article in Knee Surgery Sports Traumatol Arthrosc
(c) P ( − 3 , , (d) P Z , 3 Z ) ( )
(e) P ( 0 ,, Z 1 )
[“Effect of Provider Volume on Resource Utilization for Surgi-
cal Procedures” (2005, Vol. 13, pp. 273–279)] showed a mean
4-65.
Assume that Z has a standard normal distribution.
time of 129 minutes and a standard deviation of 14 minutes for
Use Appendix Table III to determine the value for z that solves
anterior cruciate ligament (ACL) reconstruction surgery at high-
each of the following:
volume hospitals (with more than 300 such surgeries per year).
(a) PZ ( , z ) =09 . (b) P Z ( , z ) =05 .
(a) What is the probability that your ACL surgery at a high-
(c) PZ ( . z ) =01 . (d) P Z ( . z ) =09 .
volume hospital requires a time more than two standard (e) P ( − . 1 24 ,, Z z ) = 08 deviations above the mean?
4-66.
Assume that Z has a standard normal distribution.
(b) What is the probability that your ACL surgery at a high-
Use Appendix Table III to determine the value for z that solves
volume hospital is completed in less than 100 minutes?
each of the following:
(c) The probability of a completed ACL surgery at a high-vol-
(a) P ( − z ,, Z z ) = . 0 95 (b) P ( − z ,, Z z ) = . 0 99
ume hospital is equal to 95 at what time?
(c) P ( − z ,, Z z ) = . (d) 0 68 P ( − z ,, Z z ) = . 0 9973
(d) If your surgery requires 199 minutes, what do you conclude
4-67.
Assume that X is normally distributed with a mean
about the volume of such surgeries at your hospital? Explain.
of 10 and a standard deviation of 2. Determine the following:
4-74. Cholesterol is a fatty substance that is an important part
(a) PZ ( ,13 ) (b) PZ ( .9 )
(c) P ( 6 ,, X 14 ) (d) P ( 2 ,, X 4 ) (e) P ( − 2 ,, X 8 )
of the outer lining (membrane) of cells in the body of animals.
Its normal range for an adult is 120–240 mgdl. The Food and
4-68.
Assume that X is normally distributed with a mean
Nutrition Institute of the Philippines found that the total cho-
of 10 and a standard deviation of 2. Determine the value for x
lesterol level for Filipino adults has a mean of 159.2 mgdl and
that solves each of the following:
84.1 of adults have a cholesterol level less than 200 mgdl
(a) PX ( . x ) =05 .
(b) P X ( . x ) = 0 95 . (http:www.fnri.dost.gov.ph). Suppose that the total choles-
(c) Px ( , ,10 X ) = . (d) P x X 0 ( − , 10 x ) = . 0 95
terol level is normally distributed.
(e) P ( − x , X 10 x ) = . 0 99
(a) Determine the standard deviation of this distribution.
4-69.
Assume that X is normally distributed with a mean
(b) What are the quartiles (the 25 and 75 percentiles) of
of 5 and a standard deviation of 4. Determine the following:
this distribution?
(a) PX ( ,11 ) (b) PX ( .0 ) (c) P ( 3 ,, X 7 )
(c) What is the value of the cholesterol level that exceeds 90
(d) P ( − 2 , , (e) P X 9 ) ( 2 ,, X 8 )
of the population?
4-70. Assume that X is normally distributed with a mean of
(d) An adult is at moderate risk if cholesterol level is more
5 and a standard deviation of 4. Determine the value for x that
than one but less than two standard deviations above the
solves each of the following:
mean. What percentage of the population is at moderate
(a) PX ( . x ) =05 .
(b) P X ( . x ) = 0 95 .
risk according to this criterion?
(c) Px ( ,,9 X ) = . (d) 02 P ( 3 ,, X x ) =. 0 95
(e) An adult whose cholesterol level is more than two standard
(e) P ( − x , X 5 x ) = . 0 99
deviations above the mean is thought to be at high risk.
4-71.
The compressive strength of samples of cement can
What percentage of the population is at high risk?
be modeled by a normal distribution with a mean of 6000 kilo-
(f) An adult whose cholesterol level is less than one standard
grams per square centimeter and a standard deviation of 100
deviations below the mean is thought to be at low risk.
kilograms per square centimeter.
What percentage of the population is at low risk?
(a) What is the probability that a sample’s strength is less than
4-75.
The line width for semiconductor manufacturing is
6250 Kgcm 2
assumed to be normally distributed with a mean of 0.5 microm-
(b) What is the probability that a sample’s strength is between
eter and a standard deviation of 0.05 micrometer.
5800 and 5900 Kgcm 2 ?
(a) What is the probability that a line width is greater than 0.62
(c) What strength is exceeded by 95 of the samples?
micrometer?
Section 4-6Normal Distribution
(b) What is the probability that a line width is between 0.47
(b) What thickness is exceeded by 95 of the samples?
and 0.63 micrometer?
(c) If the specifications require that the thickness is between
(c) The line width of 90 of samples is below what value?
1.39 cm and 1.43 cm, what proportion of the samples meets
4-76.
The fill volume of an automated filling machine used
specifications?
for filling cans of carbonated beverage is normally distributed
4-82.
The demand for water use in Phoenix
with a mean of 12.4 fluid ounces and a standard deviation of 0.1
in 2003 hit a high of about 442 million gallons per day on June
fluid ounce.
27 (http:phoenix.govWATERwtrfacts.html). Water use in the
(a) What is the probability that a fill volume is less than 12
summer is normally distributed with a mean of 310 million gal-
fluid ounces?
lons per day and a standard deviation of 45 million gallons per
(b) If all cans less than 12.1 or more than 12.6 ounces are
day. City reservoirs have a combined storage capacity of nearly
scrapped, what proportion of cans is scrapped?
350 million gallons.
(c) Determine specifications that are symmetric about the
(a) What is the probability that a day requires more water than
mean that include 99 of all cans.
is stored in city reservoirs?
4-77. In the previous exercise, suppose that the mean of the
(b) What reservoir capacity is needed so that the probability
filling operation can be adjusted easily, but the standard devia-
that it is exceeded is 1?
tion remains at 0.1 fluid ounce.
(c) What amount of water use is exceeded with 95 probability?
(a) At what value should the mean be set so that 99.9 of all
(d) Water is provided to approximately 1.4 million people.
cans exceed 12 fluid ounces?
What is the mean daily consumption per person at which
(b) At what value should the mean be set so that 99.9 of all
the probability that the demand exceeds the current reser-
cans exceed 12 fluid ounces if the standard deviation can be
voir capacity is 1? Assume that the standard deviation of
reduced to 0.05 fluid ounce?
demand remains the same.
4-78.
4-83. The life of a semiconductor laser at a constant power is
A driver’s reaction time to visual stimulus is nor-
mally distributed with a mean of 0.4 seconds and a standard
normally distributed with a mean of 7000 hours and a standard
deviation of 0.05 seconds.
deviation of 600 hours. (a) What is the probability that a laser fails before 5000 hours?
(a) What is the probability that a reaction requires more than
(b) What is the life in hours that 95 of the lasers exceed?
0.5 seconds?
(c) If three lasers are used in a product and they are assumed to
(b) What is the probability that a reaction requires between 0.4
fail independently, what is the probability that all three are
and 0.5 seconds?
still operating after 7000 hours?
(c) What reaction time is exceeded 90 of the time?
4-84. The diameter of the dot produced by a printer is nor-
4-79. The speed of a file transfer from a server on campus to a
mally distributed with a mean diameter of 0.002 inch and a
personal computer at a student’s home on a weekday evening is
standard deviation of 0.0004 inch.
normally distributed with a mean of 60 kilobits per second and
a standard deviation of four kilobits per second.
(a) What is the probability that the diameter of a dot exceeds
(a) What is the probability that the file will transfer at a speed
of 70 kilobits per second or more?
(b) What is the probability that a diameter is between 0.0014
and 0.0026?
(b) What is the probability that the file will transfer at a speed
of less than 58 kilobits per second?
(c) What standard deviation of diameters is needed so that the
(c) If the file is one megabyte, what is the average time it will
probability in part (b) is 0.995?
take to transfer the file? (Assume eight bits per byte.)
4-85. The weight of a sophisticated running shoe is normally
4-80. In 2002, the average height of a woman aged 20–74 years
distributed with a mean of 12 ounces and a standard deviation
was 64 inches with an increase of approximately 1 inch from 1960
of 0.5 ounce.
(http:usgovinfo.about.comodhealthcare). Suppose the height
(a) What is the probability that a shoe weighs more than
of a woman is normally distributed with a standard deviation of
13 ounces?
two inches.
(b) What must the standard deviation of weight be in order for
(a) What is the probability that a randomly selected woman in
the company to state that 99.9 of its shoes weighs less
this population is between 58 inches and 70 inches?
than 13 ounces?
(b) What are the quartiles of this distribution?
(c) If the standard deviation remains at 0.5 ounce, what must
(c) Determine the height that is symmetric about the mean that
the mean weight be for the company to state that 99.9 of
includes 90 of this population.
its shoes weighs less than 13 ounces?
(d) What is the probability that five women selected at random
4-86. Measurement error that is normally distributed with a
from this population all exceed 68 inches?
mean of 0 and a standard deviation of 0.5 gram is added to the
4-81. In an accelerator center, an experiment needs a 1.41-cm-
true weight of a sample. Then the measurement is rounded to
thick aluminum cylinder (http:puhep1.princeton.edumumu
the nearest gram. Suppose that the true weight of a sample is
targetSolenoid_Coil.pdf). Suppose that the thickness of a cyl-
165.5 grams.
inder has a normal distribution with a mean of 1.41 cm and a
(a) What is the probability that the rounded result is 167 grams?
standard deviation of 0.01 cm.
(b) What is the probability that the rounded result is 167 grams
(a) What is the probability that a thickness is greater than 1.42 cm?
or more?
Chapter 4Continuous Random Variables and Probability Distributions
4-87. Assume that a random variable is normally distributed
4-91.
A signal in a communication channel is detected when
with a mean of 24 and a standard deviation of 2. Consider an
the voltage is higher than 1.5 volts in absolute value. Assume
interval of length one unit that starts at the value a so that the
that the voltage is normally distributed with a mean of 0. What
interval is[ , + 1] aa . For what value of a is the probability of
is the standard deviation of voltage such that the probability of
the interval greatest? Does the standard deviation affect that
a false signal is 0.005?
choice of interval?
4-92. An article in Microelectronics Reliability [“Advanced
4-88.
A study by Bechtel et al., 2009, described in the Archives
Electronic Prognostics through System Telemetry and Pattern
of Environmental Occupational Health considered polycy-
Recognition Methods” (2007, Vol.47(12), pp. 1865–1873)] pre-
clic aromatic hydrocarbons and immune system function in beef
sented an example of electronic prognosis. The objective was to
cattle. Some cattle were near major oil- and gas-producing areas
detect faults to decrease the system downtime and the number of
of western Canada. The mean monthly exposure to PM1.0 (par-
unplanned repairs in high-reliability systems. Previous measure-
ticulate matter that is 3 <μ 1 m in diameter) was approximately
ments of the power supply indicated that the signal is normally
μ 7.1 gm with standard deviation 1.5. Assume that the monthly
distributed with a mean of 1.5 V and a standard deviation of 0.02 V.
exposure is normally distributed.
(a) Suppose that lower and upper limits of the predetermined (a) What is the probability of a monthly exposure greater than 3 specifications are 1.45 V and 1.55 V, respectively. What is
μ 9 gm ?
the probability that a signal is within these specifications? (b) What is the probability of a monthly exposure between 3 3 (b) What is the signal value that is exceeded with 95 probability?
and 8 gm μ ?
(c) What is the probability that a signal value exceeds the
(c) What is the monthly exposure level that is exceeded with
mean by two or more standard deviations?
probability 0.05?
4-93. An article in International Journal of Electrical Power (d) What value of mean monthly exposure is needed so that 3 Energy Systems [“Stochastic Optimal Load Flow Using a
the probability of a monthly exposure more than 9 gm μ is
Combined Quasi–Newton and Conjugate Gradient Technique”
(1989, Vol.11(2), pp. 85–93)] considered the problem of opti-
4-89. An article in Atmospheric Chemistry and Physics “Rela-
mal power flow in electric power systems and included the
tionship Between Particulate Matter and Childhood Asthma—
effects of uncertain variables in the problem formulation. The
Basis of a Future Warning System for Central Phoenix” (2012,
method treats the system power demand as a normal random
Vol. 12, pp. 2479–2490)] reported the use of PM10 (particulate
variable with 0 mean and unit variance.
matter < μ 10 m diameter) air quality data measured hourly from
(a) What is the power demand value exceeded with 95
sensors in Phoenix, Arizona. The 24-hour (daily) mean PM10 3 probability?
for a centrally located sensor was 50.9 gm μ with a standard
(b) What is the probability that the power demand is positive?
deviation of 25.0. Assume that the daily mean of PM10 is nor-
(c) What is the probability that the power demand is more than
mally distributed.
– 1 and less than 1? (a) What is the probability of a daily mean of PM10 greater 3 4-94. An article in the Journal of Cardiovascular Magnetic
than 100 gm μ ?
Resonance [“Right Ventricular Ejection Fraction Is Better (b) What is the probability of a daily mean of PM10 less than 3 Reflected by Transverse Rather Than Longitudinal Wall Motion
μ 25 gm ?
in Pulmonary Hypertension” (2010, Vol.12(35)] discussed a
(c) What daily mean of PM10 value is exceeded with prob-
study of the regional right ventricle transverse wall motion in
ability 5?
patients with pulmonary hypertension (PH). The right ventricle
4-90. The length of stay at a specific emergency department
ejection fraction (EF) was approximately normally distributed
in Phoenix, Arizona, in 2009 had a mean of 4.6 hours with
with a mean and a standard deviation of 36 and 12, respec-
a standard deviation of 2.9. Assume that the length of stay is
tively, for PH subjects, and with mean and standard deviation
normally distributed.
of 56 and 8, respectively, for control subjects.
(a) What is the probability of a length of stay greater than 10 hours?
(a) What is the EF for PH subjects exceeded with 5 probability?
(b) What length of stay is exceeded by 25 of the visits?
(b) What is the probability that the EF of a control subject is
(c) From the normally distributed model, what is the probabil-
less than the value in part (a)?
ity of a length of stay less than 0 hours? Comment on the
(c) Comment on how well the control and PH subjects can be
normally distributed assumption in this example.
distinguished by EF measurements.
4-7 Normal Approximation to the Binomial and
Poisson Distributions
We began our section on the normal distribution with the central limit theorem and the normal distribution as an approximation to a random variable with a large number of trials. Consequently, it should not be surprising to learn that the normal distribution can be used
Section 4-7Normal Approximation to the Binomial and Poisson Distributions
to approximate binomial probabilities for cases in which n is large. The following exam- ple illustrates that for many physical systems, the binomial model is appropriate with an extremely large value for n. In these cases, it is diffi cult to calculate probabilities by using the binomial distribution. Fortunately, the normal approximation is most effective in these cases. An illustration is provided in Fig. 4-19. The area of each bar equals the binomial probability of x. Notice that the area of bars can be approximated by areas under the normal probability density function.
From Fig. 4-19, it can be seen that a probability such as P ( 3 ≤≤ X 7 ) is better approxi-
mated by the area under the normal curve from 2.5 to 7.5. Consequently, a modifi ed interval is used to better compensate for the difference between the continuous normal
distribution and the discrete binomial distribution. This modifi cation is called a continu-
ity correction .