engineering statistics and probability chapter 3

Chapter 3 Random Variables and Probability Distributions

Exercises 3.1 Classify the following random variables as discrete or continuous:

X: the number of automobile accidents per year in Virginia.

3.7 The total number of hours, measured in units of 100 hours, that a family runs a vacuum cleaner over a period of one year is a continuous random variable X that has the density function

Y: the length of time to play 18 holes of golf.

x,

the amount of milk produced yearly by a particular

1, 1 < x < 2, elsewhere.

N: the number of eggs laid each month by a hen. Find the probability that over a period of one year, a P: the number of building permits issued each family runs their vacuum cleaner month in a certain city.

the weight of grain produced per acre. 3.2 An overseas shipment of 5 foreign automobiles contains 2 that have slight paint blemishes. If an agency receives 3 of these automobiles at random, list the elements of the sample space S using the letters B and for blemished and respectively; then to each sample point assign a value x of the random variable X representing the number of automobiles purchased by the agency with paint blemishes. 3.3 Let W be a random variable giving the number of heads minus the number of tails in three tosses of a coin. List the elements of the sample space S for the three tosses of the coin and to each sample point assign of W. a value 3.4 A coin is flipped until 3 heads in succession occur. List only those elements of the sample space that require 6 or less tosses. Is this a discrete sample space? Explain. 3.5 Determine the value c so that each of the following functions can serve as a probability distribution of the discrete random variable X: (a) f(x) = c(x2 + 4), for a: =

(b)f(x)

3.6 The shelf life, in days, for bottles of a certain prescribed medicine is a random variable having the density function

elsewhere. Find the probability that a bottle of this medicine will have a shell life of (a) at least 200 days; (b) anywhere from 80 to 120 days.

(a) less than 120 hours; (b) between 50 and 100 hours,

3.8 Find the probability distribution of the random variable W in Exercise 3.3. assuming that the coin is biased so that a head is twice as likely to occur as a tail. The proportion of people who respond to a certain mail-order solicitation is a continuous random variable X that has the function 0,

0
(a) Show that P(0 < X < 1) = (b) Find the probability that more than 1/4 but fewer than 1/2 of the people contacted will respond to this type of solicitation. Find a formula for the probability distribution the random variable X representing the outcome when a single die is rolled once. 3.11 A shipment of 7 television sets contains 2 defective sets. A hotel makes a random purchase of; 3 of the sets. If x is the number of defective sets purchased by the hotel, find the probability distribution of X. Express the results graphically as a probability histogram. firm offers its customers munici3.12 An pal bonds that mature after varying numbers of years. Given that the cumulative distribution function of T, the number of years to maturity for a randomly selected bond, is,

t< 1, F(t) =

1 < t < 3, 3< t <5, 5 <

t>7,

t < 7,

Exercises find (a)P(T = (b) P(T > 3); (c) P(1.4 < T

89 3.21

Consider the density <

X < 1,

elsewhere.

< 6).

3.13 The probability distribution of A, the number of imperfections per 10 meters of a synthetic fabric in continuous rolls of uniform width, is given by 3 4 0 0.41 0.16 0.05 0.01 Construct the cumulative distribution function of X. 3.14 The waiting time, in hours, between successive speeders spotted by a radar unit is a continuous randistribution function dom variable with

x < 0, x > 0. Find the probability of waiting less than 12 between successive speeders (a) using the cumulative distribution function of (b) using the probability density function of

X;

3.15 Find the cumulative distribution function of the random variable X representing the number of defectives in Exercise 3.11. Then using F(x), find (a) P(X = 1); (b) P(0 < X < 2). Construct a graph of the cumulative distribution function of Exercise 3.15. 3.17 A continuous random variable X that can assume values between x = 1 and x 3 has a density function given by f(x) = 1/2. (a) Show that the area under the curve is equal to (b) Find P(2 < X < 2.5). (c) Find P(X < 1.6). 3.18 A continuous random variable X that can assume values between x = 2 and x = 5 has a density = 2(1 + Find function given by (a) P(X < (b) P(3 < X < 4). 3.19 For the density function of Exercise 3.17, find F(x). Use it to evaluate P ( 2 < X < 2.5). 3.20 For the density function of Exercise 3.18, find F(x), and use it to evaluate P ( 3 < X < 4).

(a) Evaluate k. (b) F(x) and use it to evaluate P(0.3 <

X < 0.6).

3.22 Three cards are drawn in succession from a deck without replacement. Find the probability distribution for the number of spades. 3.23 Find the cumulative distribution function of the find random variable W in Exercise 3.8. Using (a) P(W > 0); (b) P ( - l < W < 3 ) . Find the probability distribution for the number of jazz CDs when 4 CDs are selected at random from a collection consisting of 5 jazz CDs, 2 classical CDs, and 3 rock CDs. Express your results by means of a formula. 3.25 From a box containing 4 dimes and 2 nickels, 3 coins are selected at random without replacement. Find the probability distribution for the total T of the 3 coins. Express the probability distribution graphically as a probability histogram. 3.26 From a box containing 4 black balls and 2 green balls, 3 balls are drawn in succession, each ball being replaced in the box before the next draw is made. Find for the number of green the probability balls. 3.27 The time to failure in hours of an important piece of electronic equipment used in a manufactured DVD player the density function

/(

2000

•a:/2000),

x > 0, x<0.

(a) Find F(x). (b) Determine the probability that the component (and thus the DVD player) lasts more than 1000 hours before the component needs to be replaced. (c) Determine the probability that the component fails before 2000 hours. 3.28 A cereal manufacturer is aware that the weight of the product in the box varies slightly from box to box. In fact, considerable historical data has allowed the determination of the density function that describes the probability structure for the (in

Chapter 3

90

ounces). In fact, letting X be the random variable in ounces, the density function can be described as 23.75 < x < 26.25, elsewhere. (a) Verify that this is a valid density (b) Determine the probability that the weight is smaller than 24 ounces. (c) The company desires that the weight exceeding 26 ounces is an extremely rare occurrence. What is the probability that this "rare occurrence" does actually occur?

Random

Variables and Probability

(a) Critics would certainly consider the product a bargain if it is unlikely to require a major repair before the sixth year. Comment on this by determining P(Y > 6). (b) What is the probability that a major repair occurs in the first year? 3.32 The proportion of the budgets for a certain type of industrial company that is allotted to environmental and pollution control is coming under scrutiny. A data collection project determines that the distribution of these proportions is given by

elsewhere. fuel is the 3.29 An important factor in solid particle size distribution. Significant problems occur if the particle sizes are too large. From production data in the past, it has been determined that the particle size (in micrometers) distribution is characterized by > 1, elsewhere. e)

/(

(a) Verify that this is a valid density function. (b) Evaluate F(x). (c) What is the probability that a random particle from the manufactured fuel exceeds 4 micrometers? 3.30 Measurements of scientific systems always subject to variation, some more than others. There are many structures for measurement error and statisticians spend a great deal of time modeling these errors. Suppose the measurement error X of a certain physical quantity is decided by the density function /(•

x) { 0.fc(3

x2),

- 1 < < 1, elsewhere.

f(x) a valid density func(a) Determine k that tion. (b) Find the probability that a random error in measurement is less than 1/2. (c) For this particular measurement, is undesirable if the magnitude of the error (i.e., |a:|), exceeds 0.8. What is the probability that this occurs? 3.31 Based on extensive testing, it is determined by the manufacturer of a washing machine that the time Y (in years) before a major repair is required is characterized by the probability density function

elsewhere.

(a) Verify that the above is a valid density. (b) What is the probability that a company chosen at random expends less than 10% of its budget on environmental and pollution controls? (c) What is the probability that a company selected at random spends more than 50% on environmental and pollution control? 3.33 Suppose a special type of small data processing firm is so specialized that some have difficulty making a profit in their first year of operation. The pdf that characterizes the proportion Y that make a profit is given by 0 < y < 1, elsewhere. (a) What is the value of k that renders the above a valid density function? probability that at most 50% of the firms (b) Find make a profit in the first year. (c) Find the probability at least 80% of the firms make a profit in the first year. are produced from an auto3.34 Magnetron mated assembly line. A sampling plan is used periodically to assess quality on the lengths of the tubes. This measurement is subject to uncertainty. It is thought that the probability that a random tube meets length specification is 0.99. A sampling plan is used in which the lengths of 5 random tubes are measured. (a) Show that the probability function of the numis given ber out of 5 that meet length by the following discrete probability function

for

y = 0.1,2,3,4,5.

3.4 Joint Probability Distributions

91

(b) Suppose random selections made off the line above and 3 are outside specifications. Use either to or refute the conjecture that the probability is 0.99 that a single tube meets cations 3.35 Suppose it is known from large amounts of historical data that X, the of cars that arrive at a specific intersection during a 20 second time period, is characterized by the following discrete probability function x = 0,1.2,.... /(x) = (a) Find the probability that in a specific time period, more than 8 cars arrive at the inter-

3.4

section. that only 2 cars , ,

. ., assignment, the equipment working, the density function of the observed outcome, X is

)

_

2(1 — x), 0 <

x < 1,

(a) Calculate < 1/3). (b) What is the probability that X will exceed 0.5? (c) Given that X > 0.5, what is the probability that X will be less than 0.75?

Joint Probability Distributions Our study of random variables and their probability distributions in the preceding sections is restricted to one-dimensional sample spaces, in that we recorded outcomes of an experiment as values assumed by a single random variable. There will be situations, however, where we may find it desirable to record the simultaneous outcomes of several random variables. For example, we might measure the amount of precipitate P and volume V of gas released from a controlled chemical experiment, giving rise to a two-dimensional sample space consisting of the outor we might be interested in the hardness H and tensile strength T comes (p, of cold-drawn copper resulting in the outcomes (h, t). In a study to determine the likelihood of success in college, based on high school data, we might use a threedimensional sample space and record for each individual his or her aptitude test score, high school rank in class, and average at the end of the freshman year in college. If X and Y are two discrete random variables, the probability distribution for their simultaneous occurrence can be represented by a function with values f(x,y) for any pair of values (x, y) within the range of the random variables X and Y. It is customary to refer to this function as the probability distribution of X and Y. Hence, in the discrete case,

f(x,y) = P(X

x,Y

that is, the values f(x, y) give the probability that outcomes x and y occur at the same time. For example, if a television set is to be serviced and X represents the age to the nearest year of the set and Y represents the number of defective tubes in the set, then is the probability that the television set is 5 years old and needs 3 new tubes.

Exercises

101 type. Indeed many of these are reflected in exercises in b o t h Chapters 2 a n d 3. repeated observations are binary in n a t u r e (e.g., "defective or When not," "survive or not," "allergic or not") with observations 0 or 1, the distribution covering this situation is called the b i n o m i a l d i s t r i b u t i o n and the probability function is known and will be demonstrated in its generality in C h a p t e r 5. Ex3.34 in Section 3.3 and Review Exercise 3.82 are examples and there are others t h a t the reader should recognize. T h e scenario of a continuous distribution in "time to failure" as in Review Exercise 3.71 or Exercise 3.27 on page 89 often suggests distribution type called the e x p o n e n t i a l d i s t r i b u t i o n . These t y p e s of illustrations are merely two of many so-called s t a n d a r d distributions t h a t are used extensively in real world problems because t h e scientific scenario t h a t gives rise to each of t h e m is recognizable and occurs often in practice. Chapters 5 and 6 cover with some underlying theory concerning their use. many of these A second of transition to material in future chapters deals with t h e notion of p o p u l a t i o n p a r a m e t e r s or d i s t r i b u t i o n a l p a r a m e t e r s . Recall in C h a p t e r 1 we discussed the need to use d a t a to provide information a b o u t these parameters. We went to lengths in discussing notion of a m e a n and v a r i a n c e and provided a vision for the concepts in the context of a population. Indeed the population mean and variance are easily found from the probability function for the discrete case or probability density function for the continuous case. These parameters a n d their importance in the solution of many types of real world problems will provide much of the material in Chapters 8 t h r o u g h 17.

Exercises 3.37 Determine the values of c so that the following functions represent joint probability distributions of the random variables A" and Y: (a) f(x, y) — cxy, for x = 1, 2, 3; y = 1, (b) f(x, = c\x - y\, for = y 3.38 If the joint probability distribution of is given by

f(x, y)

,..

o, 1,

X and Y

y = (1, 1, 2,

find (a) P(X

by {(x,y)

+

3.40 A privately owned liquor store operates both a drive-in facility and a walk-in facility. On a randomly selected day, let and respectively, be the proportions of the time that the drive-in and walk-in facilities density function are in use, and suppose that the of these random variables is

+ 2y),

1, 0 < y < 1, elsewhere.

<2,Y = 1); (b) P(X > 2,Y < 1); (c) P(X > Y); (d) P(X + Y = 4).

(a) Find the marginal density of X. (b) Find the marginal density of Y. (c) Find the probability that the drive-in facility is busy less than one-half of the time.

3.39 From a sack of fruit containing 3 oranges, 2 apples, and 3 bananas, a random of 4 pieces of fruit is selected. If X is the number of oranges Y is the number of apples in the sample:, find (a) the joint probability distribution of A' and (b) P[(X, Y) € .4], where A is the region that is given

distributes boxes of choco3.41 A candy lates with a mixture of creams, toffees, and cordials. Suppose the weight of each box is 1 kilogram, but the individual weights of the creams, toffees, and cordials vary from box to box. For a selected box, let X and Y represent the weights of the creams and the toffees, respectively, and suppose the joint

Chapter 3

102 density

of these variables is

Random

Variables and Probability Distributions

so that they range between 0 and 1. Suppose that and Y have the joint density

X

< 1, elsewhere. (a) Find the probability that in a given box the cordials account for more than 1/2 of the weight. (b) Find the marginal density for the weight of the creams. (c) Find the probability that the weight of the toffees in a box is less than 1/8 of a kilogram if it is known that creams constitute 3/4 of the weight. 3.42 Let X and Y denote the lengths of life, in years, of two components in an electronic system. If the joint density function of these variables is

x > 0, y > 0,

1 0, Find

elsewhere.

P(X + Y> 1/2).

3.46 Referring to Exercise 3.38, find (a) the marginal distribution of X; (b) the marginal distribution of 3.47 The amount of kerosene, in thousands of liters, in a tank at the beginning of any day is a random amount Y from which a random amount X is sold during that day. Suppose that the tank is not resupplied during the day so that x < y, and assume that the joint density function of these variables is

elsewhere. findP(0
1

\Y = 2).

elsewhere.

3.43 Let X denote the reaction time, in seconds, to a certain stimulus and Y denote the temperature at which a certain reaction starts to take place. Suppose that two random variables X and Y have the joint density 1, elsewhere. Find (a) P(0 (b) P(X

X <
<1,

and

4 —

— 2 );

(a) Determine if X and Y are independent. (b) Find P ( l / 4 < X < 1/2 | Y = 3/4). 3.48 Referring to Exercise 3.39, find (a) f(y\2) for all values of y; (b) P(Y 0 | X = 2). 3.49 Let X denote the number of times a certain numerical control machine will malfunction: 1, 2, or 3 times on any given day. Let Y denote the number of times a technician is called on an emergency call. Their joint probability distribution is given as

x

3.44 Each rear tire on an experimental airplane is supposed to be filled to a pressure of 40 pound per square inch (psi). Let X denote the actual air pressure for the right tire and Y denote the actual air pressure for the left tire. Suppose that X and Y are random variables with the joint density 30 < < 50; 30 < y < 50, elsewhere. (a) Find k. (b) P(30 < X < 40 and 40 < Y < 50). (c) Find the probability that both tires are underfilled. 3.45 Let X denote the diameter of an armored electric cable and denote the diameter of the ceramic mold that makes the cable. Both X and Y are scaled

1 3 2 0.05 0.05 1 0.1 0.05 0.35 0.1 y 0 0.2 0.1 3 (a) Evaluate the marginal distribution of (b) Evaluate the marginal distribution of (c) Find P(Y = 3 | X = Suppose that X and probability distribution:

f(x,y)

2

Y have the following joint x

4 0.15 0.30 0.15 (a) Find the marginal distribution of A. (b) Find the marginal distribution of Y.

Review

Exercises

103

3.51 Consider an experiment that consists of 2 rolls of a balanced die. If X is the number of 4s and Y is the number of obtained in the 2 rolls of the die, find and (a) the joint probability distribution of (b) P[(X, Y) € where A is the region {(x,y) \ 2x + < 3.52 Let X denote the: number of heads and Y the of heads minus the number of tails when 3 coins are tossed. Find the joint probability distribution of X and 3.53 Three cards are drawn without from the 12 face cards (jacks, queens, and kings) of an ordinary deck of 52 playing cards. Let X be the number of kings selected and Y the number of jacks. Find (a) the joint probability distribution of X and Y; (b) P[(X,Y) e where A is the region given by

{(x,y)

\x + y>2}.

3.54 A coin is tossed twice. Z denote the number of heads on the first toss and W the total number of heads on the 2 tosses. If the coin is unbalanced and a head has a 40% chance of occurring, find (a) the probability distribution of W and Z; (b) the marginal distribution of (c) the marginal distribution of Z; the probability that at least 1 head occurs.

3.57 Determine whether the two random variables of Exercise 3.50 are dependent or independent.

find P ( l <

0 < x < 1, 0 < elsewhere.

y)

(a) Show that X and (b) Find P(X > 0.3 | 3.59 Let X, sity function

z) (a) Find (b)

y< 1

are not independent.

Y = 0.5).

Y, and Z have the joint probability den-

0,'

1; elsewhere.

z <2,

0<

k.

3.61 Determine whether the two random variables of Exercise 3.44 are or 3.62 The joint probability density function of the ranvariables X, and Z is 1; 0< 0 elsewhere.

z < 3,

Find (a) the joint marginal density function of (b) the marginal density of Y;

elsewhere,

Y<3 | X =


x,


3.55 Given the joint density function

f(x, y)

function of the random vari-

3.58 The joint ables X and Y is

(c)

Y and Z:

KZ<2);

= i, 2 = 2).

Review Exercises 3.63 A tobacco company produces blends of tobacco with each blend containing various proportions of Turkish, domestic, and other tobaccos. The proportions of Turkish and domestic in a blend are random variables with joint density function (X = Turkish and Y = domestic) 0 < < 1: elsewhere.

x+

< 1,

(a) Find the probability that in a given box the Turkish tobacco accounts for over half the blend. (b) Find the marginal density function for the proportion of the domestic tobacco. (c) Find the probability that the proportion of Turkish tobacco is less than 1/8 if it is known that the blend contains 3/4 domestic tobacco. 3.64 An insurance company offers its policyholders a

Chapter 3 Random. Variables and Probability Distributions

104

number of different premium payment options. For a randomly selected policyholder, let X be the number of months between successive payments. The cumulative distribution function of X is

F(x) = {

0, 0.4, 0.6, 0.8, 1.0,

if 1 < x < 3, if 3 < x < 5, if 5 < x < 7, if x > 7.

(a) What is the probability mass function of (b) Compute P(4 < X < 7).

X?

x + y,

3.66 A service facility operates with two service lines. On a randomly selected day, let X be the proportion of time that the first line is in use whereas Y is the proportion of time that the second line is in use. Suppose that the joint probability density function for (A, V) is + elsewhere.

(a) Compute the probability that neither line is busy more than half the time. (b) Find the probability that the first line is busy more than 75% of time. 3.67 Let the number of phone calls received by a switchboard during a 5-minute interval be a random variable X with probability function for

x=

(a) Determine the probability that X equals 0, 1, 2, 3, 4, 5, and 6. function for these val(b) Graph the probability ues x. (c) Determine the cumulative distribution function for these values of X.

X and Y with

0 < x,y < 1, elsewhere.

X and

3.69 An industrial process manufactures items that can be classified as either defective or not defective. The probability that an item is defective is 0.1. An experiment is conducted in which 5 items are drawn randomly from the process. Let the random variable X be the number of defectives in this sample of 5. What is the probability mass function of X? 3.70 Consider the following joint probability function of the random variables X and Y: 1 < x < 3, elsewhere.

f{x,y)

0,

(a) Give the marginal density functions for both random variables. (b) What is the probability that both components will exceed 2 hours?

fix

fix, y)

(a) Find the marginal distributions of (b) >0.5).

3.65 Two electronic components of a missile system work in harmony for the success of the total system. Let X and Y denote the life in hours of the two components. The joint density of X and Y is

-{

3.68 Consider the random variables joint density function

<

y < 2,

(a) Find the marginal density functions of (b) Are X and Y independent? (c) Find P(X > 2).

X and Y.

3.71 The life span in hours of an electrical component is a random variable with cumulative distribution function

F(x)

{;-

x > 0, eleswhere.

(a) Determine its probability density function. (b) Determine the probability that the life span of such 70 hours. a component will Pairs of pants are being produced by a particular outlet facility. The pants are by a group of 10 workers. The workers inspect pairs of pants taken Each inspector is randomly from the production assigned a number from 1 through 10. A buyer selects a pair of pants for purchase. Let the random variable X be the inspector number. (a) Give a reasonable probability mass function for X. (b) Plot the cumulative distribution function for X.

3.73 The shelf life of a product is a random variable that is related to consumer acceptance. It turns out that the shelf life Y in days of a certain type of bakery product has a density function

fiv)

0 < y < oc,

elsewhere.

Review

Exercises

105

What fraction of the day would you

of this stocked toto be sellable 3 days from now?

Passenger congestion is a service problem in airports. Trains are installed within the airport to reduce the With the use of the train, the time X that it takes in to travel from the main terminal to a particular concourse has density function

is the probability that component propor< 0.2 and > 0.5? tions produce the results distribution (d) Give the 3.78 Consider the situation of Review Exercise 3.77. But suppose the joint distribution of the two proportions is given by 0 <

0< < elsewhere.

fix)

the pdf above is a valid density function. (a) Show (b) Find the probability that the time it takes a passenger to travel from the main terminal to the concourse will not exceed 7 minutes. in the batch of final product of a 3.75 chemical process often reflect a serious problem. From considerable plant data gathered, it is known that the proportion Y of impurities in a batch has a density function given by 0 < y < 1, elsewhere. (a) Verify that the above is a valid density function. (b) A batch is not sellable and then notacceptable if the percentage of impurities exceeds 60%. With the current quality of the process, what is the percentage of batches that are not acceptable? 3.76 The time Z in minutes between calls to an electrical supply system has the probability density function 0 < z < oo, elsewhere. (a) What is the probability that there are no calls within a 20-minute time interval? (b) What is the probability that the first call comes within 10 minutes of opening? A chemical system that results from a chemical reaction has two important components among others in a blend. The joint distribution describing the proportion and of these two components is given by ,,

(c)

/2,

(0,

0 < < elsewhere.

(a) Give the marginal distribution of (b) Give the marginal distribution of

< 1,

x

X2

<

<

1,

here.

of the pro(a) Give the marginal distribution portion and verify that it is valid density function. (b) What is the probability that proportion is less is 0.7? than 0.5 given that 3.79 Consider the random variables X and Y that represent the number of vehicles that arrive at 2 sepperiod. arate street corners during a certain These street corners arc fairly close together so it is important that traffic engineers deal with them jointly if necessary. The joint distribution of X and Y is known to be

-

• for x = and y = (a) Are the two random variables X and Y independent? Explain why or why not. (b) What is the probability that during the time period in question less than 4 vehicles arrive at the two street corners? The behavior of series of components play a huge role in scientific and engineering reliability problems. The reliability of the entire system is certainly no better than the weakest component in the series. In a series system, the components operate independently of each other. In a particular system containing three components the probability of meeting specification for components 1, 2, and 3, respectively, are 0.99, and 0.92. What is the probability that the entire system works? 3.81 Another type of system that is employed engineering work is a group of parallel components or a parallel system. In this more conservative approach, the probability that the system operates is larger than the probability that any component operates. The system fails only when systems fail. Consider a situation in which there are 4 independent components in a parallel system with probability of operation given by Component 1: 0.95; Component 3: 0.90;

Component 2: 0.94: Component 4:

What is the probability that the system does not fail?

106

Chapter S Random Variables and Probability Distributions

3.82 Consider a system of components in which does have a redundancy built in such that it does not five components, each of which fail if 3 out of the 5 components are operational. What an operational probability of 0.92. system is the probability that the total system is operational?

3.5

Potential Misconceptions and Hazards; Relationship to Material in Other Chapters In future chapters it will become apparent that probability distributions represent the: structure through which probabilities that are computed aid in the evaluation and understanding of a process. For example, in Review Exercise 3.67, the probability distribution that quantifies the probability of a heavy load during certain time periods can be very useful in planning for any changes in the system. Review Exercise 3.71 describes a scenario in which the life span of an electronic: component is studied. Knowledge of the probability structure for the component will contribute significantly toward an understanding of the reliability of a large system of which the component is a part. In addition, an understanding of the general nature of probability distributions will enhance the understanding of the concept of a P-value which was introduced briefly in Chapter 1 and will play a major role: beginning in Chapter 10 and extending throughout the balance of the text. Chapters 4, 5, and 6 depend heavily on the material in this chapter. In Chapter p a r a m e t e r s in probability distributions. 4 we discuss the meaning of These important parameters quantify notions of central t e n d e n c y variability in a system. In fact, knowledge of these quantities themselves, quite apart from the complete distribution, can provide into the nature of the system. Chapters 5 and 6 will deal with engineering, biological, or general scientific scenaridentify special types of For example, the structure of the ios probability function in Review Exercise 3.67 will easily be identified under certain assumptions discussed in Chapter 5. The same holds for the scenario of Review Exercise 3.71. This is a special type of t i m e to failure problem for which the probability density function will be Chapter 6. As far as potential hazards with the use of material in this chapter, the ing" to the reader is not to read more into the material than is evident. The general nature of the probability distribution for a specific scientific: phenomenon is not obvious from what is learned in this chapter. The purpose of this chapter is to learn how to manipulate a probability distribution, not to learn how to identify to a specific type. Chapters 5 and 6 go a long way toward identification general nature of the scientific system.

Chapter 3 Random Variables and Probability Distributions 3.1 Discrete; continuous; continuous; discrete; discrete; continuous. 3.2 A table of sample space and assigned values of the random variable is shown next. Sample Space NNN NNB NBN BNN NBB BNB BBN BBB

x 0 1 1 1 2 2 2 3

3.3 A table of sample space and assigned values of the random variable is shown next. Sample Space HHH HHT HT H T HH HT T T HT TTH TTT

w 3 1 1 1 −1 −1 −1 −3

3.4 S = {HHH, T HHH, HT HHH, T T HHH, T T T HHH, HT T HHH, T HT HHH, HHT HHH, . . . }; The sample space is discrete containing as many elements as there are positive integers. 29

30


3.5 (a) c = 1/30 since 1 =

3 P

c(x2 + 4) = 30c.

x=0

(b) c = 1/10 since

2 X 2 3 2 3 2 3 2 3 1= c =c + + = 10c. x 3−x 0 3 1 2 2 1 x=0 3.6 (a) P (X > 200) =

R∞

20000 200 (x+100)3

(b) P (80 < X < 200) = 3.7 (a) P (X < 1.2) =

R1

R 120

R 1.2

1 x2 2 0

(2 − x) dx = 1 R1 2 (b) P (0.5 < X < 1) = 0.5 x dx = x2 = 0.375. 0

x dx +

= 19 . 120 10000 = dx = − (x+100) 2

dx = −

20000 (x+100)3

80

∞ 10000 (x+100)2 200

1

0.5

80

1000 9801

+ 2x −

x2 2

= 0.1020.

1.2 = 0.68. 1

3.8 Referring to the sample space in Exercise 3.3 and making use of the fact that P (H) = 2/3 and P (T ) = 1/3, we have P (W = −3) = P (T T T ) = (1/3)3 = 1/27; P (W = −1) = P (HT T ) + P (T HT ) + P (T T H) = 3(2/3)(1/3)2 = 2/9; P (W = 1) = P (HHT ) + P (HT H) + P (T HH) = 3(2/3)2(1/3) = 2/9; P (W = 3) = P (HHH) = (2/3)3 = 8/27; The probability distribution for W is then w −3 −1 1 3 P (W = w) 1/27 2/9 2/9 8/27 3.9 (a) P (0 < X < 1) =

R1

2(x+2) 0 5

dx =

R 1/2

2(x+2) 5

(b) P (1/4 < X < 1/2) =

1/4

1

(x+2)2 5

= 1. 2 1/2 dx = (x+2) = 19/80. 5 0

1/4

3.10 The die can land in 6 different ways each with probability 1/6. Therefore, f (x) = 16 , for x = 1, 2, . . . , 6. 5 3.11 We can select x defective sets from 2, and 3 − x good sets from 5 in x2 3−x ways. A 7 random selection of 3 from 7 sets can be made in 3 ways. Therefore, 5 2 f (x) =

In tabular form

x

3−x 7 3

x f (x)

,

x = 0, 1, 2.

0 1 2 2/7 4/7 1/7

31

Solutions for Exercises in Chapter 3

The following is a probability histogram: 4/7

f(x)

3/7

2/7

1/7

1

2

3

x

3.12 (a) P (T = 5) = F (5) − F (4) = 3/4 − 1/2 = 1/4. (b) P (T > 3) = 1 − F (3) = 1 − 1/2 = 1/2. (c) P (1.4 < T < 6) = F (6) − F (1.4) = 3/4 − 1/4 = 1/2. 3.13 The c.d.f. of X  0,      0.41,    0.78, F (x) =  0.94,      0.99,    1,

is for for for for for for

x < 0, 0 ≤ x < 1, 1 ≤ x < 2, 2 ≤ x < 3, 3 ≤ x < 4, x ≥ 4.

3.14 (a) P (X < 0.2) = F (0.2) = 1 − e−1.6 = 0.7981;

(b) f (x) = F ′ (x) = 8e−8x . Therefore, P (X < 0.2) = 8 0.7981.

3.15 The c.d.f. of X  0,    2/7, F (x) =  6/7,    1,

R 0.2 0

is for for for for

0.2

e−8x dx = −e−8x |0

x < 0, 0 ≤ x < 1, 1 ≤ x < 2, x ≥ 2.

(a) P (X = 1) = P (X ≤ 1) − P (X ≤ 0) = 6/7 − 2/7 = 4/7;

(b) P (0 < X ≤ 2) = P (X ≤ 2) − P (X ≤ 0) = 1 − 2/7 = 5/7.

=

32


3.16 A graph of the c.d.f. is shown next. 1 6/7

F(x)

5/7 4/7 3/7 2/7 1/7 0

1

2

x

3 (1/2) dx = x2 1 = 1. 2.5 R 2.5 (b) P (2 < X < 2.5) 2 (1/2) dx = x2 2 = 14 . 1.6 R 1.6 (c) P (X ≤ 1.6) = 1 (1/2) dx = x2 1 = 0.3.

3.17 (a) Area =

R3 1

3.18 (a) P (X < 4) =

R4

2(1+x) 27 2

(b) P (3 ≤ X < 4) = Rx

R4 3

dx =

2(1+x) 27

4

(1+x)2 27

= 16/27. 4 (1+x)2 dx = 27 = 1/3.

3.19 F (x) = 1 (1/2) dt = x−1 , 2 P (2 < X < 2.5) = F (2.5) − F (2) = 3.20 F (x) =

2 27

Rx

(1 + t) dt = 2

2 27

t+

P (3 ≤ X < 4) = F (4) − F (3) = R1√

1.5 2

2

−

3

1 2

= 14 .

x t2 = (x+4)(x−2) , 2 27 2 (8)(2) − (7)(1) = 13 . 27 27

1 x dx = 2k x3/2 0 = 2k . Therefore, k = 32 . 3 3 x Rx√ (b) F (x) = 32 0 t dt = t3/2 0 = x3/2 . P (0.3 < X < 0.6) = F (0.6) − F (0.3) = (0.6)3/2 − (0.3)3/2 = 0.3004.

3.21 (a) 1 = k

0

3.22 Denote by X the number of spades int he three draws. Let S and N stand for a spade and not a spade, respectively. Then P (X = 0) = P (NNN) = (39/52)(38/51)(37/50) = 703/1700, P (X = 1) = P (SNN) + P (NSN) + P (NNS) = 3(13/52)(39/51)(38/50) = 741/1700, P (X = 3) = P (SSS) = (13/52)(12/51)(11/50) = 11/850, and P (X = 2) = 1 − 703/1700 − 741/1700 − 11/850 = 117/850. The probability mass function for X is then x f (x)

0 1 2 3 703/1700 741/1700 117/850 11/850

33


3.23 The c.d.f. of X is   0,      1/27, F (x) = 7/27,    19/27,    1,

for for for for for

w < −3, − 3 ≤ w < −1, − 1 ≤ w < 1, 1 ≤ w < 3, w ≥ 3,

(a) P (W > 0 = 1 − P (W ≤ 0) = 1 − 7/27 = 20/27.

(b) P (−1 ≤ W < 3) = F (2) − F (−3) = 19/27 − 1/27 = 2/3. 3.24 There are 10 ways of selecting any 4 CDs from 10. We can select x jazz CDs from 5 4 5 ways. Hence and 4 − x from the remaining CDs in x5 4−x f (x) =

5 x

5 4−x 10 4

,

x = 0, 1, 2, 3, 4.

3.25 Let T be the total value of the three coins. Let D and N stand for a dime and nickel, respectively. Since we are selecting without replacement, the sample space containing elements for which t = 20, 25, and 30 cents corresponding to the selecting of 2 nickels (2)(4) and 1 dime, 1 nickel and 2 dimes, and 3 dimes. Therefore, P (T = 20) = 2 6 1 = 15 , (3 ) (21)(42) P (T = 25) = 6 = 35 , (3) (43) P (T = 30) = 6 = 15 , (3) and the probability distribution in tabular form is t

20 25 30 P (T = t) 1/5 3/5 1/5 As a probability histogram 3/5

f(x)

2/5

1/5

20

25 x

30

34


3.26 Denote by X the number of green balls in the three draws. Let G and B stand for the colors of green and black, respectively. Simple Event BBB GBB BGB BBG BGG GBG GGB GGG

P (X = x) (2/3)3 = 8/27 (1/3)(2/3)2 = 4/27 (1/3)(2/3)2 = 4/27 (1/3)(2/3)2 = 4/27 (1/3)2 (2/3) = 2/27 (1/3)2 (2/3) = 2/27 (1/3)2 (2/3) = 2/27 (1/3)3 = 1/27

x 0 1 1 1 2 2 2 3

The probability mass function for X is then x P (X = x)

0 1 2 3 8/27 4/9 2/9 1/27

Rx 1 3.27 (a) For x ≥ 0, F (x) = 0 2000 exp(−t/2000) dt = − exp(−t/2000)|x0 = 1 − exp(−x/2000). So ( 0, x < 0, F (x) = 1 − exp(−x/2000), x ≥ 0. (b) P (X > 1000) = 1 − F (1000) = 1 − [1 − exp(−1000/2000)] = 0.6065.

(c) P (X < 2000) = F (2000) = 1 − exp(−2000/2000) = 0.6321. 26.25 R 26.25 3.28 (a) f (x) ≥ 0 and 23.75 52 dx = 25 t 23.75 = 2.5 = 1. 2.5 R 24 (b) P (X < 24) = 23.75 25 dx = 25 (24 − 23.75) = 0.1. R 26.25 (c) P (X > 26) = 26 25 dx = 25 (26.25 − 26) = 0.1. It is not extremely rare. ∞ R∞ −3 3.29 (a) f (x) ≥ 0 and 1 3x−4 dx = −3 x 3 = 1. So, this is a density function. 1 R x −4 (b) For x ≥ 1, F (x) = 1 3t dt = 1 − x−3 . So, ( 0, x < 1, F (x) = −3 1 − x , x ≥ 1.

(c) P (X > 4) = 1 − F (4) = 4−3 = 0.0156. 1 R1 x3 2 3.30 (a) 1 = k −1 (3 − x ) dx = k 3x − 3 = −1

16 k. 3

So, k =

3 . 16

35


Rx x 3 (b) For −1 ≤ x < 1, F (x) = 16 (3 − t2 ) dt = 3t − 31 t3 −1 = −1 1 1 1 3 9 99 So, P X < 21 = 21 − 16 . − 16 2 = 128 2

1 2

+

9 x 16

−

x3 . 16

(c) P (|X| < 0.8) = P (X < −0.8) + P (X > 0.8) = F(−0.8) + 1 − F (0.8) 9 1 9 1 1 3 0.8 − 16 0.83 = 0.164. = 1 + 2 − 16 0.8 + 16 0.8 − 12 + 16 Ry 3.31 (a) For y ≥ 0, F (y) = 14 0 e−t/4 dy = 1 − ey/4 . So, P (Y > 6) = e−6/4 = 0.2231. This probability certainly cannot be considered as “unlikely.” (b) P (Y ≤ 1) = 1 − e−1/4 = 0.2212, which is not so small either. R1 1 3.32 (a) f (y) ≥ 0 and 0 5(1 − y)4 dy = − (1 − y)5 |0 = 1. So, this is a density function. 0.1

(b) P (Y < 0.1) = − (1 − y)5|0 = 1 − (1 − 0.1)5 = 0.4095. (c) P (Y > 0.5) = (1 − 0.5)5 = 0.03125.

3.33 (a) Using integral by parts and setting 1 = k

R1 0

y 4 (1 − y)3 dy, we obtain k = 280.

(b) For 0 ≤ y < 1, F (y) = 56y 5 (1 − Y )3 + 28y 6(1 − y)2 + 8y 7(1 − y) + y 8. So, P (Y ≤ 0.5) = 0.3633. (c) Using the cdf in (b), P (Y > 0.8) = 0.0563.

3.34 (a) The event Y = y means that among 5 selected, exactly y tubes meet the specification (M) and 5 − y (M ′ ) does not. The probability for one combination of such a situation is (0.99)y (1 − 0.99)5−y if we assume independence among the 5! permutations of getting y Ms and 5 − y M ′ s, the tubes. Since there are y!(5−y)! probability of this event (Y = y) would be what it is specified in the problem. (b) Three out of 5 is outside of specification means that Y = 2. P (Y = 2) = 9.8×10−6 which is extremely small. So, the conjecture is false. 0 8 P x 1 8 3.35 (a) P (X > 8) = 1 − P (X ≤ 8) = e−6 6x! = e−6 60! + 61! + · · · + 68! = 0.1528. x=0

(b) P (X = 2) =

2 e−6 62!

= 0.0446. Rx x 3.36 For 0 < x < 1, F (x) = 2 0 (1 − t) dt = − (1 − t)2 |0 = 1 − (1 − x)2 . (a) P (X ≤ 1/3) = 1 − (1 − 1/3)2 = 5/9.

(b) P (X > 0.5) = (1 − 1/2)2 = 1/4. (c) P (X < 0.75 | X ≥ 0.5) = 3.37 (a)

3 P 3 P

f (x, y) = c

x=0 y=0

(b)

PP x

y

f (x, y) = c

3 P 3 P

P (0.5≤X<0.75) P (X≥0.5)

=

(1−0.5)2 −(1−0.75)2 (1−0.5)2

= 34 .

xy = 36c = 1. Hence c = 1/36.

x=0 y=0

PP x

y

|x − y| = 15c = 1. Hence c = 1/15.

3.38 The joint probability distribution of (X, Y ) is

36


x f (x, y) 0 1 2 3 0 0 1/30 2/30 3/30 y 1 1/30 2/30 3/30 4/30 2 2/30 3/30 4/30 5/30 (a) P (X ≤ 2, Y = 1) = f (0, 1) + f (1, 1) + f (2, 1) = 1/30 + 2/30 + 3/30 = 1/5. (b) P (X > 2, Y ≤ 1) = f (3, 0) + f (3, 1) = 3/30 + 4/30 = 7/30. (c) P (X > Y ) = f (1, 0) + f (2, 0) + f (3, 0) + f (2, 1) + f (3, 1) + f (3, 2) = 1/30 + 2/30 + 3/30 + 3/30 + 4/30 + 5/30 = 3/5. (d) P (X + Y = 4) = f (2, 2) + f (3, 1) = 4/30 + 4/30 = 4/15. 3.39 (a) We can x oranges from 3, y apples from 2, and 4 − x − y bananas from 3 select 3 in x3 y2 4−x−y ways. A random selection of 4 pieces of fruit can be made in 84 ways. Therefore, 3 3 2 f (x, y) =

x

y

4−x−y ,

8 4

x = 0, 1, 2, 3;

y = 0, 1, 2;

1 ≤ x + y ≤ 4.

(b) P [(X, Y ) ∈ A] = P (X + Y ≤ 2) = f (1, 0) + f (2, 0) + f (0, 1) + f (1, 1) + f (0, 2) = 3/70 + 9/70 + 2/70 + 18/70 + 3/70 = 1/2. R1 3.40 (a) g(x) = 23 0 (x + 2y) dy = 32 (x + 1), for 0 ≤ x ≤ 1. R1 (b) h(y) = 23 0 (x + 2y) dy = 13 (1 + 4y), for 0 ≤ y ≤ 1. R 1/2 5 (c) P (X < 1/2) = 23 0 (x + 1) dx = 12 .

R 1/2 R 1/2−y R 1/2 3.41 (a) P (X + Y ≤ 1/2) = 0 24xy dx dy = 12 0 0 R 1−x 2 (b) g(x) = 0 24xy dy = 12x(1 − x) , for 0 ≤ x < 1. (c) f (y|x) =

24xy 12x(1−x)2

=

1 2

−y

2

y dy =

1 . 16

2y , (1−x)2

for 0 ≤ y ≤ 1 − x. R 1/8 Therefore, P (Y < 1/8 | X = 3/4) = 32 0 y dy = 1/4. R∞ 3.42 Since h(y) = e−y 0 e−x dx = e−y , for y > 0, then f (x|y) = f (x, y)/h(y) = e−x , for R1 x > 0. So, P (0 < X < 1 | Y = 2) = 0 e−x dx = 0.6321.

R 1/2 R 1/2 R 1/2 3.43 (a) P (0 ≤ X ≤ 1/2, 1/4 ≤ Y ≤ 1/2) = 0 4xy dy dx = 3/8 x dx = 3/64. 1/4 0 R1Ry R1 3 (b) P (X < Y ) = 0 0 4xy dx dy = 2 0 y dy = 1/2. R R 50 R 50 2 R 50 2 50 2 2 3.44 (a) 1 = k 30 30 (x + y ) dx dy = k(50 − 30) 30 x dx + 30 y dy = 392k · 104 . 3 So, k =

3 392

· 10−4 .

37


R 40 R 50 3 (b) P (30 ≤ X ≤ 40, 40 ≤ Y ≤ 50) = 392 · 10−4 30 40 (x2 + y 2) dy dx R 50 2 R 40 2 3 3 503 −403 −3 −3 403 −303 = 392 · 10 ( 30 x dx + 40 y dy) = 392 · 10 + 3 = 3 R 40 R 40 3 (c) P (30 ≤ X ≤ 40, 30 ≤ Y ≤ 40) = 392 · 10−4 30 30 (x2 + y 2) dx dy R 40 3 3 3 3 37 = 2 392 · 10−4(40 − 30) 30 x2 dx = 196 · 10−3 40 −30 = 196 . 3

49 . 196

R 1/4 R 1/2−x 1 dy dx 3.45 P (X + Y > 1/2) = 1 − P (X + Y < 1/2) = 1 − 0 x 1 y 1/4 R 1/4 1 1 =1− 0 ln 2 − x − ln x dx = 1 + 2 − x ln 2 − x − x ln x 0 = 1 + 41 ln 14 = 0.6534. 3.46 (a) From the column totals of Exercise 3.38, we have

x 0 1 2 3 g(x) 1/10 1/5 3/10 2/5 (b) From the row totals of Exercise 3.38, we have y 0 1 2 h(y) 1/5 1/3 7/15 R1 3.47 (a) g(x) = 2 Rx dy = 2(1 − x) for 0 < x < 1; y h(y) = 2 0 dx = 2y, for 0 < y < 1. Since f (x, y) 6= g(x)h(y), X and Y are not independent. (b) f (x|y) = f (x, y)/h(y) = 1/y, for 0 < x < y. R 1/2 Therefore, P (1/4 < X < 1/2 | Y = 3/4) = 43 1/4 dx = 13 . 3.48 (a) g(2) =

2 P

f (2, y) = f (2, 0) + f (2, 1) + f (2, 2) = 9/70 + 18/70 + 3/70 = 3/7. So,

y=0

f (y|2) = f (2, y)/g(2) = (7/3)f (2, y). f (0|2) = (7/3)f (2, 0) = (7/3)(9/70) = 3/10, f (1|2) = 3/5 and f (2|2) = 1/10. In tabular form, y 0 1 2 f (y|2) 3/10 3/5 1/10 (b) P (Y = 0 | X = 2) = f (0|2) = 3/10. 3.49 (a) (b)

x g(x)

1 2 3 0.10 0.35 0.55

y 1 2 3 h(y) 0.20 0.50 0.30

(c) P (Y = 3 | X = 2) =

0.2 0.05+0.10+0.20

= 0.5714.

38


x 3.50

y

f (x, y) 1 3 5 g(x)

2 0.10 0.20 0.10 0.40

4 h(y) 0.15 0.25 0.30 0.50 0.15 0.25 0.60

(a)

x g(x)

(b)

y 1 3 5 h(y) 0.25 0.50 0.25

2 4 0.40 0.60

3.51 (a) Let X be the number of 4’s and Y be the number of 5’s. The sample space consists of 36 elements each with probability 1/36 of the form (m, n) where m is the outcome of the first roll of the die and n is the value obtained on the second roll. The joint probability distribution f (x, y) is defined for x = 0, 1, 2 and y = 0, 1, 2 with 0 ≤ x + y ≤ 2. To find f (0, 1), for example, consider the event A of obtaining zero 4’s and one 5 in the 2 rolls. Then A = {(1, 5), (2, 5), (3, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 6)}, so f (0, 1) = 8/36 = 2/9. In a like manner we find f (0, 0) = 16/36 = 4/9, f (0, 2) = 1/36, f (1, 0) = 2/9, f (2, 0) = 1/36, and f (1, 1) = 1/18. (b) P [(X, Y ) ∈ A] = P (2X + Y < 3) = f (0, 0) + f (0, 1) + f (0, 2) + f (1, 0) = 4/9 + 1/9 + 1/36 + 2/9 = 11/12. 3.52 A tabular form of the experiment can be established as, Sample Space HHH HHT HT H T HH HT T T HT TTH TTT

x 3 2 2 2 1 1 1 0

y 3 1 1 1 −1 −1 −1 −3

So, the joint probability distribution is,

y

x f (x, y) 0 1 2 3 −3 1/8 −1 3/8 1 3/8 1/8 3

39


3.53 (a) If (x, y) represents the selection of x kings and y jacks in 3 draws, we must have x = 0, 1, 2, 3; y = 0, 1, 2, 3; and 0 ≤ x + y ≤ 3. Therefore, (1, 2) represents the selection of 1 king and 2 jacks which will occur with probability f (1, 2) =

4 1

4

2 =

12 3

6 . 55

Proceeding in a similar fashion for the other possibilities, we arrive at the following joint probability distribution: x

y

f (x, y) 0 1 2 3

0 1 2 3 1/55 6/55 6/55 1/55 6/55 16/55 6/55 6/55 6/55 1/55

(b) P [(X, Y ) ∈ A] = P (X + Y ≥ 2) = 1 − P (X + Y < 2) = 1 − 1/55 − 6/55 − 6/55 = 42/55. 3.54 (a) P (H) = 0.4, P (T ) = 0.6, and S = {HH, HT, T H, T T }. Let (W, Z) represent a typical outcome of the experiment. The particular outcome (1, 0) indicating a total of 1 head and no heads on the first toss corresponds to the event T H. Therefore, f (1, 0) = P (W = 1, Z = 0) = P (T H) = P (T )P (H) = (0.6)(0.4) = 0.24. Similar calculations for the outcomes (0, 0), (1, 1), and (2, 1) lead to the following joint probability distribution:

z

w f (w, z) 0 1 2 0 0.36 0.24 1 0.24 0.16

(b) Summing the columns, the marginal distribution of W is 0 1 2 w g(w) 0.36 0.48 0.16 (c) Summing the rows, the marginal distribution of Z is z h(z)

0 1 0.60 0.40

(d) P (W ≥ 1) = f (1, 0) + f (1, 1) + f (2, 1) = 0.24 + 0.24 + 0.16 = 0.64.

40


R4 3.55 g(x) = 81 2 (6 − x − y) dy = 3−x , for 0 < x < 2. 4 f (x,y) 6−x−y So, f (y|x) = g(x) = 2(3−x) , for 2 < y < 4, R3 and P (1 < Y < 3 | X = 1) = 14 2 (5 − y) dy = 58 .

3.56 Since f (1, 1) 6= g(1)h(1), the variables are not independent.

3.57 X and Y are independent since f (x, y) = g(x)h(y) for all (x, y). R 1−y (x,y) 2x 3.58 (a) h(y) = 6 0 x dx = 3(1 − y)2, for 0 < y < 1. Since f (x|y) = fh(y) = (1−y) 2 , for 0 < x < 1 − y, involves the variable y, X and Y are not independent. R 0.5 (b) P (X > 0.3 | Y = 0.5) = 8 0.3 x dx = 0.64. R1R1R2 R1R1 R1 3.59 (a) 1 = k 0 0 0 xy 2 z dx dy dz = 2k 0 0 y 2z dy dz = 2k z dz = k3 . So, k = 3. 3 0 R 1/4 R 1 R 2 2 R 1/4 R 1 2 (b) P X < 14 , Y > 12 , 1 < Z < 2 = 3 0 xy z dx dy dz = 92 0 y z dy dz 1/2 1 1/2 R 1/4 21 21 = 16 z dz = 512 . 0 R1 R1 3.60 g(x) = 4 0 xy dy = 2x, for 0 < x < 1; h(y) = 4 0 xy dx = 2y, for 0 < y < 1. Since f (x, y) = g(x)h(y) for all (x, y), X and Y are independent. R 50 3 50 3.61 g(x) = k 30 (x2 + y 2 ) dy = k x2 y + y3 = k 20x2 + 98,000 , and 3 30 h(y) = k 20y 2 + 98,000 . 3 Since f (x, y) 6= g(x)h(y), X and Y are not independent. R1 3.62 (a) g(y, z) = 94 0 xyz 2 dx = 92 yz 2 , for 0 < y < 1 and 0 < z < 3. R3 (b) h(y) = 29 0 yz 2 dz = 2y, for 0 < y < 1. R 2 R 1 R 1/2 7 (c) P 14 < X < 12 , Y > 31 , Z < 2 = 94 1 1/3 1/4 xyz 2 dx dy dz = 162 . (x,y,z) (d) Since f (x|y, z) = fg(y,z) = 2x, for 0 < x < 1, P 0 < X < 21 | Y = 14 , Z = 2 = R 1/2 2 0 x dx = 14 . R 1−x 3.63 g(x) = 24 0 xy dy = 12x(1 − x)2 , for 0 < x < 1. R1 R1 5 (a) P (X ≥ 0.5) = 12 0.5 x(1 − x)2 dx = 0.5 (12x − 24x2 + 12x3 ) dx = 16 = 0.3125. R 1−y (b) h(y) = 24 0 xy dx = 12y(1 − y)2 , for 0 < y < 1. (c) f (x|y) =

So, P X 3.64 (a)

x f (x)

f (x,y) 24xy = 12y(1−y) 2 h(y) < 18 | Y = 34 =

1 3 5 0.4 0.2 0.2

2x = (1−y) 2 , for 0 < x < 1 − y. R 1/8 R 1/8 2x dx = 32 = 0.25. 0 1/16 0

7 0.2

(b) P (4 < X ≤ 7) = P (X ≤ 7) − P (X ≤ 4) = F (7) − F (4) = 1 − 0.6 = 0.4.

41


∞ R∞ R ∞ −y(1+x) 1 1 3.65 (a) g(x) = 0 ye−y(1+x) dy = − 1+x ye−y(1+x) 0 + 1+x e dy 0 1 −y(1+x) ∞ = − (1+x)2 e 0 1 = (1+x) , for x > 0. 2 R∞ ∞ h(y) = ye−y 0 e−yx dx = −e−y e−yx |0 = e−y , for y > 0. R∞R∞ R∞ R∞ ∞ (b) P (X ≥ 2, Y ≥ 2) = 2 2 ye−y(1+x) dx dy = − 2 e−y e−yx |2 dy = 2 e−3y dy ∞ = − 13 e−3y 2 = 3e16 . 1 ,Y 2

1 2

3 2

R 1/2 R 1/2

2

2

3.66 (a) P X ≤ ≤ = 0 (x + y ) dxdy = 0 R 1/2 1 1 x2 + 12 dx = 16 . = 43 0 R 1 53 (b) P X ≥ 34 = 32 3/4 x2 + 13 dx = 128 .

3.67 (a)

x f (x)

3 2

R 1/2 0

2

x y+

y3 3

1/2 dx 0

0 1 2 3 4 5 6 0.1353 0.2707 0.2707 0.1804 0.0902 0.0361 0.0120

(b) A histogram is shown next. 0.3

f(x)

0.2

0.1

0.0

1

2

3

4

5

6

7

x

x F (x)

0 1 2 3 4 5 6 0.1353 0.4060 0.6767 0.8571 0.9473 0.9834 0.9954 R1 3.68 (a) g(x) = 0 (x + y) dy = x + 12 , for 0 < x < 1, and h(y) = y + 21 for 0 < y < 1. 1 R 1 x2 R1 R1 (b) P (X > 0.5, Y > 0.5) = 0.5 0.5 (x + y) dx dy = 0.5 2 + xy dy 0.5 R1 = 0.5 12 + y − 18 + y2 dy = 38 . 3.69 f (x) = x5 (0.1)x (1 − 0.1)5−x , for x = 0, 1, 2, 3, 4, 5. 2 R 2 3x−y 3xy−y 2 /2 3.70 (a) g(x) = 1 dy = = x3 − 16 , for 1 < x < 3, and 9 9 1 R3 4 2 h(y) = 1 3x−y dx = − y, for 1 < y < 2. 9 3 9 (c)

(b) No, since g(x)h(y) 6= f (x, y). 2 3 R3 (c) P (X > 2) = 2 x3 − 16 dx = x6 − x6 = 23 . 2

42


3.71 (a) f (x) =

d F (x) dx

=

1 −x/50 e , 50

for x > 0.

(b) P (X > 70) = 1 − P (X ≤ 70) = 1 − F (70) = 1 − (1 − e−70/50 ) = 0.2466. 3.72 (a) f (x) =

1 , 10

for x = 1, 2, . . . , 10.

(b) A c.d.f. plot is shown next. 1.0 0.9 0.8 0.7

F(x)

0.6 0.5 0.4 0.3 0.2 0.1 0 1

2

3

5

4

6

7

8

9

10

x

R∞

3.73 P (X ≥ 3) =

1 2

3.77 (a) g(x1 ) =

R1

e−y/2 = e−3/2 = 0.2231. R 10 1 3.74 (a) f (x) ≥ 0 and 0 10 dx = 1. This is a continuous uniform distribution. R 7 1 dx = 0.7. (b) P (X ≤ 7) = 10 0 R1 R1 10 1 3.75 (a) f (y) ≥ 0 and 0 f (y) dy = 10 0 (1 − y)9 dy = − 10 (1 − y) = 1. 10 0 R1 1 (b) P (Y > 0.6) = 0.6 f (y) dy = − (1 − y)10 |0.6 = (1 − 0.6)10 = 0.0001. R ∞ −z/10 1 −z/10 ∞ 3.76 (a) P (Z > 20) = 10 e dz = − e = e−20/10 = 0.1353. 20 20 10 (b) P (Z ≤ 10) = − e−z/10 0 = 1 − e−10/10 = 0.6321. 3

2 dx2 = 2(1 − x1 ), for 0 < x1 < 1. x1 R x2 (b) h(x2 ) = 0 2 dx1 = 2x2 , for 0 < x2 < 1. R 1 R 0.2 (c) P (X1 < 0.2, X2 > 0, 5) = 0.5 0 2 dx1 dx2 = 2(1 − 0.5)(0.2 − 0) = 0.2.

(d) fX1 |X2 (x1 |x2 ) =

f (x1 ,x2 ) h(x2 )

=

2 2x2

=

1 , x2

for 0 < x1 < x2 .

Rx 3.78 (a) fX1 (x1 ) = 0 1 6x2 dx2 = 3x21 , for 0 < x1 < 1. Apparently, fX1 (x1 ) ≥ 0 and R1 R1 f (x1 ) dx1 = 0 3x21 dx1 = 1. So, fX1 (x1 ) is a density function. 0 X1 f (x1 ,x2 ) fX1 (x1 )

= 2 xx22 , for 0 < x2 < x1 . R1 0.5 . So, P (X2 < 0.5 | X1 = 0.7) = 0.72 2 0 x2 dx2 = 25 49

(b) fX2 |X1 (x2 |x1 ) =

=

6x2 3x21

43


3.79 (a) g(x) =

9 (16)4y

∞ P

x=0

1 4x

=

9 1 (16)4y 1−1/4

= 34 · 41x , for x = 0, 1, 2, . . . ; similarly, h(y) = 43 · 41y ,

for y = 0, 1, 2, . . . . Since f (x, y) = g(x)h(y), X and Y are independent. (b) P (X + Y < 4) = f (0, 0) + f (0, 1) + f (0, 2) + f (0, 3) + f (1, 0) + f (1, 1) + f (1, 2) + 9 1 1 1 1 1 1 1 1 1 f (2, 0) + f (2, 1) + f (3, 0) = 16 1 + 4 + 42 + 43 + 4 + 42 + 43 + r2 + 43 + 43 = 9 1 + 24 + 432 + 443 = 63 . 16 64

3.80 P (the system works) = P (all components work) = (0.95)(0.99)(0.92) = 0.86526.

3.81 P (the system does not fail) = P (at least one of the components works) = 1 − P (all components fail) = 1 − (1 − 0.95)(1 − 0.94)(1 − 0.90)(1 − 0.97) = 0.999991. 3.82 Denote by X the number of components (out of 5) work. Then, P is operational) = P (X ≥ 3) = P (X = 3) + P (X = 4) + P (X = (the system 5 5 3 2 5) = 3 (0.92) (1 − 0.92) + 4 (0.92)4 (1 − 0.92) + 55 (0.92)5 = 0.9955.

engineering statistics and probability chapter 3

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