STA 5325, University of Florida, 2009

Page 1 of 13

Ramin Shamshiri STA 5325, HW #1 Due 05/18/09

STA 5325, Homework #1

Due May 18, 2009

Ramin Shamshiri

UFID # 90213353 Note: Problem numbers are according to the 6

th text edition. Selected problem are highlighted.

Assignment for Monday, May 11th

, 2009

1.4. Acquired immunodeficiency syndrome (AIDS) has become one of the most devastating diseases in modern

society. The number of cases of AIDS (in thousands) reported in 25 major cities in the United States during 1992 are as follows:

38.3 6.2 3.7 2.6 2.1

14.6 5.6 3.7 2.3 2 11.9 5.5 3.4 2.2 2

6.6 4.6 3.1 2.2 1.9 6.3 4.5 2.7 2.1 1.8

a. Construct a relative frequency histogram to describe these data. b. What portion of these cities reported more than 10,000 cases of aids in 1992

c. If one of the cities is selected at random from the 25 for which preceding data were taken, what is the

probability that it will have reported fewer than 3000 cases of AIDS in 1992?

Solution a:

Solution b: 3 out of 25 cities which is 12%

Solution c: Total number of cities less than 3000 cases = 11, so probability = 11/25=0.44 or 44%

Page 2 of 13


2.2. Suppose that A and B are two events. Write expressions involving unions, intersections, and complements

that describe the following:

a. Both events occur.

b. At least one occurs. c. Neither occurs

d. Exactly one occurs.

Solution a: Both events occur means A AND B, which is intersection, A∩B

Solution b: At least one event occurs means either A OR B, which is union, A∪B

Solution c: Neither occurs means NOT A OR B, which is 𝐴 ∪ 𝐵

Solution d: Exactly one occurs means A AND NOT B , or B AND NOT A, which is A∩𝐵 or B∩𝐴 2.12. A survey classified a large number of adults according to whether they were diagnosed as needing eyeglasses to correct their reading vision and whether they use eyeglasses when reading. The proportions falling

into the four resulting categories are given in the table.

Uses Eyeglasses for Reading

Needs Glasses Yes No

Yes 0.44 0.14

No 0.02 0.40

If a single adult is selected from the large group, find the probabilities of the events defined below:

a. The adult needs glasses.

b. The adult needs glasses but does not use them. c. The adult uses glasses whether the glasses are needed or not.

Solution a: P(needs glasses)= 0.44+0.14=0.58 Solution b: P (adult needs glasses BUT not use them)= 0.14

Solution c: P(use glasses)= 0.44+0.02=0.46

2.15. Hydraulic landing assemblies coming from an aircraft rework facility are each inspected for defects.

Historical records indicate that 8% have defects in shaft only, 6% have defects in bushings only, and 2% have

defects in both shafts and bushings. One of the hydraulic assemblies is selected randomly. What is the probability that the assembly has

a. A bushing defect? b. A shaft or bushing defect?

c. Exactly one of the two types of defects?

d. Neither type of defect?

Solution a: P(bushing defect)=0.06+0.02=0.08

Solution b: P(shaft or bushing)=0.08+0.06+0.02=0.16

Solution c: P(exactly one types of defects)=0.08+0.06=0.14 (both type of defect should not be included) Solution d: P(Neither type)=1-P(defect)=1-0.16=0.84

Page 3 of 13


2.16. Suppose two balanced coins are tossed and the upper faces are observed.

a. List the sample points for this experiment.

b. Assign a reasonable probability to each sample point. ( Are the sample points equally likely?) c. Let A denote the event that exactly one head is observed and B the even that at least one head is

observed. List the sampling points in both A and B.

d. From your answer to (c) find P(A), P(B), P(A∩B), P(A∪B), and P(A ∪B)

H=Head, B=Back Solution a: Set of all samples points are: S={HB,BH,HH,BB}

Solution b: if the coins are normal, we can assign equal probability to each sample point which is 0.25.

Solution c: A={HB,BH } and B={HB,BH,BB}.

Solution d:

P(A)=2/4

P(B)=3/4

P(A∩B)=1/2

P(A∪B)=3/4

P(A ∪B)=1

Page 4 of 13


Assignment for Tuesday, May 12th

, 2009

2.21. Two additional jurors are needed to complete a jury for a criminal trial. There are six prospective jurors,

two women and four men. Two jurors are randomly selected from the six available.

a. Define the experiment and describe one sample point. Assume that you need describe only the two

jurors chosen and not the order in which they were selected.

b. List the sample space associated with this experiment. c. What is the probability that both of the jurors selected are women?

Solution a: The experiment is to randomly select two jurors out of a group of 6 jurors that includes two women and 4 men.

Solution b: Let w1 and w2 denotes women and m1, m2 and… denote men.

S={(w1,m1),(w1,m2),(w1,m3),(w1,m4),(w2,m1),(w2,m2),(w2,m3),(w2,m4),(m1,m2),(m1,m3),(m1,m4),

(m2,m3),( m2,m4),( m3,m4),( w1,w2)}

Solution c:

Probability that the first juror is a woman = 2/6

Probability that the second juror is a woman=1/5 Probability that both jurors selected are woman = (2/6)*( 1/5)=1/15

2.27. An airline has six flights from New York to California and seven flights from California to Hawaii per

day. If the flights are to be made on separate days, how many different flight arrangements can the airline offer

from New York to Hawaii?

Solution:

Using m.n rule, there would be 6*7=42 possible ways to fly from NY to CA.

2.33. How many different seven-digit telephone numbers can be formed if the first digit cannot be zero?

Solution:

1 2 3 4 5 6 7

9 10 10 10 10 10 10

There would be 9*106 possible phone numbers with seven digits.

Page 5 of 13


2.44. A group of three undergraduate and five graduate students are available to fill certain student government

posts. If four students are to be randomly selected from this group, find the probability that exactly two undergraduates will be among the four chosen.

Solution:

Combination of 4 out of 8,

𝑛

𝑟 =

8

4 =

8!

4! 4!= 70

Number of ways that exactly two undergrad students can be selected:

3

2 =

3!

2!= 3

5

2 =

5!

2! 3!= 10

P(exactly two undergrad students):

3 × 10

70=

3

7

2.50. A balanced die is tossed six times and the number on the uppermost face is recorded each time. What is the probability that the numbers recorded are 1,2,3,4,5 and 6 in any order?

Solution: 1 2 3 4 5 6

1 5/6 4/6 3/6 2/6 1/6

Probability that the numbers recorded are 1,2,3,4,5 or 6 in any order is

(1)(5/6)(4/6)(3/6)(2/6)(1/6)=5!

65 =120

7776= 0.01543209

Page 6 of 13


Assignment for Wednesday, May 13th

, 2009

2.57. If two events, A and B, are such that P(A)=0.5, P(B)=0.3 and P(A∩B)-0.1, find the following:

a. P(A׀B) b. P(B׀A)

c. P(A׀A∪B)

d. P(A׀A∩B)

e. P(A∩B׀A∪B)

Solution a:

𝑃 𝐴׀𝐵 =𝑃 𝐴 ∩ 𝐵

𝑃 𝐵 =

0.1

0.3=

1

3

Solution b:

𝑃 𝐵׀𝐴 =𝑃 𝐵 ∩ 𝐴

𝑃 𝐴 =

0.1

0.5=

1

5

Solution c:

𝑃 A׀A ∪ B =𝑃 𝐴 ∩ A ∪ B

𝑃 A ∪ B

=𝑃 𝐴 ∩ 𝐴 ∪ 𝐴 ∩ 𝐵

𝑃 𝐴 + 𝑃 𝐵 − 𝑃 𝐴 ∩ 𝐵

=𝑃 𝐴 ∪ 𝐴 ∩ 𝐵

𝑃 𝐴 + 𝑃 𝐵 − 𝑃 𝐴 ∩ 𝐵

=𝑃 𝐴 + 𝑃 𝐴 ∩ 𝐵 − 𝑃(𝐴 ∩ (𝐴 ∩ 𝐵))

𝑃 𝐴 + 𝑃 𝐵 − 𝑃(𝐴 ∩ 𝐵)

=𝑃 𝐴

𝑃 𝐴 + 𝑃 𝐵 − 𝑃(𝐴 ∩ 𝐵)=

0.5

0.7=

5

7

Solution d:

𝑃 A׀A ∩ B =𝑃 𝐴 ∩ A ∩ B

𝑃 A ∩ B

=𝑃(A ∩ B)

𝑃(A ∩ B)= 1

Solution e:

𝑃 A ∩ B׀A ∪ B =𝑃 A ∩ B ∩ (A ∪ B)

𝑃 A ∪ B

=𝑃[( A ∩ B ∩ A ) ∪ ( A ∩ B ∩ B)]

𝑃 𝐴 + 𝑃 𝐵 − 𝑃 A ∩ B

=𝑃[(A ∩ B) ∪ (A ∩ B)]

𝑃 𝐴 + 𝑃 𝐵 − 𝑃 A ∩ B

=𝑃[(A ∩ B)

𝑃 𝐴 + 𝑃 𝐵 − 𝑃 A ∩ B =

0.1

0.7=

1

7

Page 7 of 13


2.60. A survey of consumers in a particular community shows that 10% were dissatisfied with plumbing jobs done in their home. Half the complaints dealt with plumber A, who does 40% of the plumbing jobs in the town.

a. Find the probability that a consumer will obtain an unsatisfactory plumbing job, given that the plumber was A.

b. Find the probability that a consumer will obtain a satisfactory plumbing job, given that the plumber was

A.

Solution a:

P(A)= Probability that plumber A does the job is 40%

P(B)=Probability that consumers are dissatisfied is 10%

Half the complains dealt with plumber A, means probability that plumber A does the job given the probability

that consumers are dissatisfied, So P(A׀B)=50%

Finding the probability that a consumer will obtain an unsatisfactory plumbing job, given that the plumber was

A, is

𝑃 𝐵׀𝐴 =𝑃 𝐵 ∩ 𝐴

𝑃 𝐴 =

𝑃 P(A׀B) .𝑃(𝐵)

0.4=

0.5(0.1)

0.4=

0.5

4= 0.125

Solution b:

𝑃 𝐵 ׀𝐴 = 1 − 𝑃 𝐵׀𝐴 = 1 − 0.125 = 0.875

2.65. If P(A)>0, P(B)>0, and P(A)<P(A׀B), show that P(B)<P(B׀A).

Solution:

If P(A)<P(A׀B) then we have: 𝑃 𝐴 <𝑃 𝐴∩𝐵

𝑃 𝐵

Or 𝑃 𝐵 <𝑃 𝐴∩𝐵

𝑃 𝐴

This is equivalent as: 𝑃 𝐵 <𝑃 𝐵∩𝐴

𝑃 𝐴 , or 𝑃 𝐵 < P(B׀A)

Page 8 of 13


2.69. In s game, a participant is given three attempts to hit a ball. On each try, she either scores a hit, H or miss,

M. The game requires that the player must alternate which hand she uses in successive attempts. That is, if she makes her first attempt with her right hand, she must use her left hand for the second attempt and her right hand

for the third. Her chance of scoring a hit with her right hand is 0.7 and with her left hand is 0.4. Assume that the

results of successive attempts are independent and that she wins the game if she scores at least two hits in a row. If she makes her first attempt with her right hand, what is the probability that she wins the game?

Solution:

RIGHT LEFT RIGHT

H H M (0.7)(0.4)(0.3)=0.084

M H H (0.3)(0.4)(0.7)=0.084 H H H (0.7)(0.4)(0.7)=0.196

0.084+0.084+0.196=0.364

2.73. Consider the following portion of an electric circuit with three relays. Current will flow from point a to

point b if there is at least one closed path when the relays are activated. The relays may malfunction and not

close when activated. Suppose that the relays act independently of one another and close properly when activated, with a probability of 0.9

a. What is the probability that current will flow when the relays are activated? b. Given that current flowed when the relays were activated, what is the probability that relay 1

functioned?

Solution a:

Probability of closing properly for each relay is 90%, therefore probability of not closing is 1-0.9=0.1

So the probability that current will flow when relays are activated is 1-P(not closing 3 relays) which is:

1-(0.1*0.1*0.1)=1-(0.001)=0.999

Solution b:

Probability of current flowed=0.999 Probability of relay 1 functioned if current flowed is:

𝑃 𝐴׀𝐵 =𝑃 𝐴 ∩ 𝐵

0.999=

0.9

0.999= 0.9009

Page 9 of 13


Assignment for Thursday, May 14th

, 2009

2.71. Two events A and B are such that P(A)=0.2, P(B)=0.3 and 𝑃 𝐴 ∪ 𝐵 = 0.4. Find the following:

a. 𝑃 𝐴 ∩ 𝐵 b. 𝑃 𝐴 ∪ 𝐵 c. 𝑃 𝐴 ∩ 𝐵

d. 𝑃 𝐴 ׀𝐵

Solution a: 𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 − 𝑃 𝐴 ∩ 𝐵 𝑃 𝐴 ∩ 𝐵 = 0.2 + 0.3 − 0.4 = 0.1

Solution b: 𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 − 𝑃 𝐴 ∩ 𝐵 = 1 − 𝑃 𝐴 + 1 − 𝑃 𝐵 − (1 − 𝑃 𝐴 ∪ 𝐵 )

= 0.8 + 0.7 − 0.6 = 0.9

Solution c: 𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐴 ∪ 𝐵 = 1 − 𝑃 𝐴 ∪ 𝐵 = 1 − 0.4 = 0.6

Solution d:

𝑃 𝐴 ׀𝐵 =𝑃(𝐴 ∩ 𝐵)

𝑃(𝐵)=

𝑃 𝐵 − 𝑃(𝐴 ∩ 𝐵)

𝑃(𝐵)=

0.3 − 0.1

0.3=

2

3

2.87. An advertising agency notices that approximately 1 in 50 potential buyers of a product sees a given magazine ad, and 1 in 5 sees a corresponding ad on television. One in 100 sees both. One in 3 actually

purchases the product after seeing the ad, 1 in 10 without seeing it. What is the probability that a randomly

selected potential customer will purchase the product?

Solution:

P(magazine ad)=P(A)=1/50

P(TV add)=P(B)=1/5

P(A∩B)=1/100

P(ad)=P(A∪B)=1/50+1/5-1/100=21/100

P(NOT ad)=P(NOT A AND NOT B)=P(𝐴 ∩ 𝐵 )=1- P(A∪B)=79/100

P(buy after ad)=P(buy׀ ad)=1/3=𝑃(buy ∩𝑎𝑑 )

𝑃(𝑎𝑑 )

So, 𝑃 buy ∩ 𝑎𝑑 =1

3.

21

100=

7

100

P(buy׀ NOT ad)=1/10=𝑃(buy ∩𝑁𝑂𝑇 𝑎𝑑 )

𝑃(𝑁𝑂𝑇 𝑎𝑑 )

So, 𝑃 buy ∩𝑁𝑂𝑇 𝑎𝑑 =79

100.

1

10=

79

1000

P(buy)= 𝑃 buy ∩ 𝑎𝑑 + 𝑃 buy ∩𝑁𝑂𝑇 𝑎𝑑 =7/100+79/1000=0.149

Page 10 of 13


2.88. Three radar sets, operating independently, are set to detect any aircraft flying through a certain area. Each set has a probability of 0.02 of failing to detect a plane in its area.

a. If an aircraft enters the area, what is the probability that it goes undetected? b. If an aircraft enters the area, what is the probability that it is detected by all three radar sets?

Solution a: (0.02)( 0.02)( 0.02)=8*10-6

Solution b: (0.98)( 0.98)( 0.98)=0.941192

2.90. A lie detector will shows a positive reading (indicate a lie) 10% of the time when a person is telling the

truth and 95% of the time when the person is lying. Suppose two people are suspected in a one-person crime

and (for certain) one is guilty and will lie. Assume further that the lie detector operates independently for the truthful person and the liar.

a. What is the probability that the detector shows a positive reading for both suspects?

b. What is the probability that the detector shows a positive reading for the guilty suspect and a negative reading for the innocent suspect?

c. What is the probability that the detector is completely wrong-that is, that it gives a positive reading for

the innocent suspect and a negative reading for the guilty? d. What is the probability that it gives a positive reading for either or both of the two suspects?

Solution:

From the problem, 10% of times that the person is really saying the truth, the lie detector shows lie which is

Positive. So 90% of times that a person is really saying truth, the lie detector gives Negative.

Also, 95% of the time that a person is saying lying, the lie detector shows lie which is positive and 5% of times that a person is lying, the lie detector shows Negative.

For each person, we have four cases: 1- He is saying the truth and lie detector shows it as a lie (Positive) : 10%

2- He is saying the truth and lie detector shows it as a truth (Negtive): 90%

3- He is lying and lie detector shows it as a lie (Positive) :95%

4- He is lying and lie detector shows it as a truth (Negative): 5%

Bases on this, we can answer the questions:

Solution a: (0.1)(0.95)=0.095

Solution b: (0.9)(0.95)=0.855

Solution c: (0.1)(0.05)=0.005 Solution d: (0.1)+(0.95)-(0.1)(0.95)=0.955

Page 11 of 13


2.104. A study of Georgia residents suggests that those who worked in shipyards during World War II were subjected to a significantly higher risk of lung cancer. It was found that approximately 22% of those persons

who had lung cancer worked at some prior time in a shipyard. In contrast, only 14% of those who had no lung

cancer worked at some prior time in a shipyard. Suppose that the proportion of all Georgians living during World War II who have or will have contracted lung cancer is 0.04%. Find the percentage of Georgians living

during the same period who will contract (or have contracted) lung cancer, given that they have at some prior

time worked in shipyard.

Solution:

Pr(Cancer)=P(C)=0.0004

P(𝐶 )=0.9996 Pr(work given cancer)=P(B׀C)=0.22

Pr(work given NOT cancer)=P(B׀𝐶 )=0.14 What is the probability of Cancer given work, P(C ׀B)=?

𝑃 C׀B =𝑃 𝐶 ∩ 𝐵

𝑃 𝐵 =

𝑃 𝐶 .𝑃 B׀C

𝑃 𝐶 ∩ 𝐵 + 𝐶 ∩ 𝐵

=𝑃 𝐶 .𝑃 B׀C

𝑃 𝐶 . B׀C + 𝐶 . B׀C =

0.0004 0.22

0.0004 0.22 + 0.9996 0.14 = 0.0006

2.111. A student answers a multiple-choice examination question that offers four possible answers. Suppose

that the probability that the student knows the answer to the question is 0.8 and the probability that the student will guess is 0.2. Assume that if the student guesses, the probability of selecting the correct answer is 0.25. If

the student correctly answers a question, what is the probability that the student really knew the correct answer?

Solution:

Pr(guess)=P(A)=0.2

P(know)=P(B)=0.8 P(correct)=P(C)

P(C׀A)=0.25

P(C׀B)=1

P(C)= (0.8)(1)+(0.2)(0.25)=0.85

P(B׀C)=P(NOT guess ׀ C)

=P(𝐴 ׀C)

=𝑃 𝐴 ∩ 𝐶

𝑃 𝐶 =

𝑃 𝐶 ׀𝐴 𝑃 𝐴

0.85=

1 0.8

0.85= 0.9412

Page 12 of 13


Assignment for Friday, May 15th

, 2009

3.1. When the health department tested private wells in a county for two impurities commonly found in drinking

water, it found that 20% of the wells had neither impurity, 40% had impurity A, and 50% had impurity B.

(Obviously, some had both impurities.) If a well is randomly chosen from those in the county, find the probability distribution for Y, the number of impurities found in the well.

Solution:

𝑃 𝐴 ∩ 𝐵 = 0.2

𝑃 𝐴 = 0.4

𝑃 𝐵 = 0.5

𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐴 ∪ 𝐵 => 0.2 = 1 − 𝑃 𝐴 ∪ 𝐵 Or 𝑃 𝐴 ∪ 𝐵 = 0.8

𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 − 𝑃 𝐴 ∪ 𝐵 = 0.5 + 0.4 − 0.8 = 0.1 P(Y=0)=0.2

P(Y=2)=0.1

P(Y=1)=0.7, (since P(Y) should sum to 1 over all Ys)

3.4. Consider a system of water flowing through valves from A to B. Valves 1,2, and 3 operate independently, and each correctly opens on signal with probability 0.8. Find the probability distribution for Y, the number of

open paths from A to B after the signal is given. (Note that Y can take on the values 0,1, and 2.)

Solution:

P(each valve open)=0.8

P(all valve open correctly)==> there would be two paths from A to B, or P(Y=2)= (0.8)(0.8)(0.8)=0.512 P(valve 1 does not open AND either valve 2 OR 3 does not open)

=P(Not openV1∩ (𝑁𝑜𝑡 𝑜𝑝𝑒𝑛𝑉2 ∪𝑁𝑜𝑡 𝑂𝑝𝑒𝑛𝑉3))

=P(Not open V1)*P(𝑁𝑜𝑡 𝑂𝑝𝑒𝑛 𝑉2 ∪ 𝑁𝑜𝑡 𝑂𝑝𝑒𝑛 𝑉3)=0.2*(0.2+0.2-0.04)=0.072 So, P(Y=0)=0.072

Consequently P(Y=1)=1-(0.512+0.072)=0.416

3.10. Let Y be a random variable with p(y) given in the accompanying table. Find E(Y), E(1/Y), E(Y2-1), and

V(Y).

y 1 2 3 4

P(y) 0.4 0.3 0.2 0.1

Solution:

𝐸 𝑌 = 𝑦𝑝 𝑦 = 0.4 + 0.6 + 0.6 + 0.4 = 2

𝐸 1

𝑌 =

1

𝑦𝑝 𝑦 = 0.4 + 0.15 + 0.2/3 + 0.1/4 = 0.6417

𝐸 Y2 − 1 = 𝐸 𝑌2 − 1 = 0.4 + 4 × 0.3 + 9 × 0.2 + 16 × 0.1 − 1 = 4

𝑉 𝑌 = 𝐸 𝑌2 − 𝐸 𝑌 2

= 5 − 22 = 1

Page 13 of 13


3.21. A potential customer for an $85,000 fire insurance policy possesses a home in an area that according to

experience, may sustain a total loss in a given year with probability of 0.001 and 50% loss with probability 0.01. Ignoring all other partial losses, what premium should the insurance company charge for a year policy in

order to break even on all $85,000 policies in this area?

Solution:

P(total loss)=0.001

P(50% loss)=0.01

P(0% loss)=1-(0.001-0.01)=0.989

𝐸 𝑌 = 𝑦𝑝 𝑦 = 85000 0.001 + 0.5 ∗ 85000 0.01 + 0 ∗ 85000 0.989 = 510

3.23. Let Y be a discrete random variable with mean 𝜇 and variance 𝜎2. If a and b are constants, use Theorem

3.3 through 3.6 to prove that:

a) E(aY+b)=aE(Y)+b=a 𝜇+b

b) V(aY+b)=a2V(Y)=a

2𝜎2

Solution a:

Let g1(Y)=aY and g2(Y)=b

E(aY+b)=E[g1(Y)+ g2(Y)]=E(g1(Y))+E(g2(Y))=E(aY)+E(b)=aE(Y)+E(b)= a 𝜇+b

Solution b:

V(aY+b)=E[Ay+b-( a 𝜇+b)]2=E[aY-a 𝜇+b-b]

2=E[a(Y- 𝜇)]2

=E[a2(Y- 𝜇)

2]

=a2E(Y- 𝜇)

2

=a2V(Y)

=a2𝜎2

Page 1 of 12



Due June 01, 2009

Ramin Shamshiri



Assignment for Tuesday, May 26th

, 2009

4.5. Suppose that Y posses the density function

𝑓 𝑦 = 𝑐𝑦, 0 ≤ 𝑦 ≤ 20, 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒.

a. Find the value of c that makes f(y) a probability density function.

b. Find F(y)

c. Graph f(y) and F(y)

d. Use F(y) to find 𝑃(1 ≤ 𝑌 ≤ 2).

e. Use f(y) and geometry to find 𝑃(1 ≤ 𝑌 ≤ 2).

Solution a: According to the properties of density function,

𝑓 𝑦 ≥ 0

𝑓(𝑦)∞

−∞𝑑𝑦 = 1

Therefore:

𝐹 ∞ = 𝑓(𝑦)∞

−∞

𝑑𝑦 = 𝑐𝑦2

0

𝑑𝑦 = 𝑐𝑦2

2

0

2

= 1 => 𝑐 =1

2

𝑓 𝑦 =

𝑦

2, 0 ≤ 𝑦 ≤ 2

0, 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒.

Solution b:

According to definition 4.3,

𝑓 𝑦 =𝑑𝐹(𝑦)

𝑑𝑦= 𝐹′(𝑦)

So, F(y) can be written as:

𝐹 𝑦 = 𝑓(𝑡)𝑦

−∞

𝑑𝑡 = 𝑡

2

𝑦

−∞

𝑑𝑡 =𝑦2

4

Solution c:

Plots are shown in Figures 4.5a and 4.5b.

Fig 4.5.a Fig 4.5.b

Page 2 of 12


Solution d:

𝑃 1 ≤ 𝑌 ≤ 2 = 𝐹(𝑦) 12 = 𝑦

2

4 1

2

= 1 −1

4=

𝟑

𝟒

Solution e:

According to the figure below, the area corresponding to (1 ≤ 𝑌 ≤ 2) can be calculated by calculating the area

corresponding to (0 ≤ 𝑌 ≤ 2) minus (0 ≤ 𝑌 ≤ 1).

Area under 0 ≤ 𝑌 ≤ 2 = 2 1

2= 1

Area under 0 ≤ 𝑌 ≤ 1 = 1 0.5

2=

1

4

1 −1

4=

𝟑

𝟒

Page 3 of 12


4.7. A supplier of kerosene has a 150-gallon tank that is filled at the begging of each week. His weekly demand

shows a relative frequently behavior that increases steadily up to 100 gallons and then levels off between 100 and 150 gallons. If Y denotes weekly demand in hundreds of gallons, the relative frequency of demand can be

modeled by

𝑓 𝑦 = 𝑦, 0 ≤ 𝑦 ≤ 1

1, 0 < 𝑦 ≤ 1.50, 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒.

a. Find F(y)

b. Find 𝑃(1 ≤ 𝑌 ≤ 0.5).

c. Find 𝑃(0.5 ≤ 𝑌 ≤ 1.2).

Solution a:

𝐹 𝑦 = 𝑓 𝑡 𝑦

−∞

𝑑𝑡

For (𝑦 < 0), 𝐹 𝑦 = 𝑃(𝑌 ≤ 𝑦) = 𝟎

For (0 ≤ 𝑦 ≤ 1), 𝐹 𝑦 = 𝑡𝑦

0𝑑𝑡 = 𝑡

2

2

0

𝑦

=𝒚𝟐

𝟐

For 1 ≤ 𝑦 ≤ 1.5 , 𝐹 𝑦 = 𝑓 𝑡 1

0𝑑𝑡 + 𝑓 𝑡

𝑦

1𝑑𝑡

= 𝑡1

0

𝑑𝑡 + 1𝑦

1

𝑑𝑡 =1

2+ 𝑦 − 1 = 𝒚 −

𝟏

𝟐

For 𝑦 ≥ 1.5 , 𝐹 𝑦 = 1

𝐹 𝑦 =

0, 𝑦 < 0

𝒚𝟐

𝟐, 0 ≤ 𝑦 ≤ 1

𝒚 −𝟏

𝟐, 1 ≤ 𝑦 ≤ 1.5

1, 𝑦 > 1.5

Solution b:

𝑃 1 ≤ 𝑌 ≤ 0.5 = 𝐹 0.5 =1

8

Solution c:

𝑃 0.5 ≤ 𝑌 ≤ 1.2 = 𝐹 1.2 − 𝐹 0.5 = 1.2 − 0.5 − 1

8 = 0.575

Page 4 of 12


4.14. If, as in Exercise 4.10, Y has density function

𝑓 𝑥 = 1/2 2 − 𝑦 , 0 ≤ 𝑦 ≤ 20, 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒.

find the mean and variance of Y.

Solution:

The mean of Y is the expected value of Y:

𝜇 = 𝐸 𝑌 = 𝑦.∞

−∞

𝑓 𝑦 𝑑𝑦 =1

2 𝑦.

2

0

2 − 𝑦 𝑑𝑦 =𝟐

𝟑

The variance can be calculated as:

𝜎2 = 𝐸 𝑌2 − 𝜇2

𝐸 𝑌2 = 𝑦2.∞

−∞

𝑓 𝑦 𝑑𝑦 =1

2 𝑦2.

2

0

2 − 𝑦 𝑑𝑦 =2

3

𝜎2 = 𝐸 𝑌2 − 𝜇2 =2

3−

2

3

2

=𝟐

𝟗

4.16. If, as in Exercise 4.12, Y has density function

𝑓 𝑥 = 0.2, − 1 ≤ 𝑦 ≤ 0

0.2 + 1.2 𝑦, 0 < 𝑦 ≤ 10, 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒.

find the mean and variance of Y.

Solution:

𝜇 = 𝐸 𝑌 = 𝑦.∞

−∞

𝑓 𝑦 𝑑𝑦 = 𝑦.0

−1

0.2 𝑑𝑦 + 𝑦.1

0

0.2 + 1.2 𝑦 𝑑𝑦 = 𝟎. 𝟒

𝜎2 = 𝐸 𝑌2 − 𝜇2

𝐸 𝑌2 = 𝑦2.0

−1

0.2 𝑑𝑦 + 𝑦2.1

0

0.2 + 1.2 𝑦 𝑑𝑦 = 0.4333

𝜎2 = 𝐸 𝑌2 − 𝜇2 = 0.4333 − 0. 42 = 𝟎. 𝟐𝟕𝟑𝟑

Page 5 of 12


Assignment for Wednesday, May 27th

, 2009

4.36. If a point is randomly located in an interval (a,b) and if Y denotes the location of the point, then Y is

assumed to have a uniform distribution over (a,b). A plant efficiency expert randomly selects a location along a

500-foot assembly line from which to observe the work habits of the workers on the line. What is the probability that the point she selects

a. is whiting 25 feet of the end of the line?

b. is within 25 feet of the beginning of the line? c. is closer to the beginning of the line that to the end of the line?

Solution: According to the definition of uniform probability distribution, a random variable Y is said to have a

continuous uniform probability distribution on the interval (𝜃1, 𝜃2) if and only if the density function of Y is:

𝑓 𝑦 =1

𝜃2−𝜃1 𝜃1<y<𝜃2

Here we have: 𝜃1 = 0 and 𝜃2 = 500, therefore 𝑓 𝑦 =1

500

Solution a: Probability that the selected point is within 25 of the end of the line means P(475<y<500), so we have:

= 𝑓 𝑦 500

475

𝑑𝑦 = 1

500

500

475

𝑑𝑦 = 1 −475

500= 0.05

Solution b:

Probability that the selected point is within 25 of the beginning of the line means P(0<y<25), so we have:

= 𝑓 𝑦 25

0

𝑑𝑦 = 1

500

25

0

𝑑𝑦 =25

500= 0.05

Solution c: P(0<y<250),

= 𝑓 𝑦 250

0

𝑑𝑦 = 1

500

250

0

𝑑𝑦 = 250/500 = 0.5

Page 6 of 12


4.38. Beginning at 12:00 midnight, a computer center is up for 1 hour and then down for 2 hours on a regular

cycle. A person who is unaware of this schedule dials the center at a random time between 12:00 midnight and 5:00 AM. What is the probability that the center is up when the person’s call comes in?

Solution:

Let 12:00 midnight be 𝜃1 = 0, then we will have 𝜃2 = 5, therefore 𝑓 𝑦 =1

5

Beginning at 𝜃1 = 0, the computer is up for 1 hour, so: P(0<Y<1)

The computer is down for 2 hours and then up for another one hour, so P(3<Y<4)

The probability that the center is up when the person call is: P(0<Y<1) + P(3<Y<4)

= 𝑓 𝑦 1

0

𝑑𝑦 + 𝑓 𝑦 4

3

𝑑𝑦

= 1

5

1

0

𝑑𝑦 + 1

5

4

3

𝑑𝑦 =1

5+

4

5−

3

5=

𝟐

𝟓

4.49. A company that manufactures and bottles apple juice uses a machine that automatically fills 16-ounces bottles. There is some variation, however, in the amounts of liquid dispensed into the bottles that are filled. The

amount dispensed has been observed to be approximately normally distributed with mean 16 ounces and

standard deviation 1 ounce. What portion of bottles will have more than 17 ounces dispenses into them?

Solution:

Mean = 16

Standard deviation=1

Transforming normal random variable Y to a standard normal random variable Z:

𝑧 =𝑦 − 𝜇

𝜎=

17 − 16

1= 1

Using Table 4, page 792, P(Z>1)=0.1587

Page 7 of 12


4.57. Wires manufacture for use in a computer system are specified to have resistance of between 0.12 and 0.14 ohms. The actual measured resistances of the wires produced by company A have a normal probability

distribution with mean 0.13 ohm and standard deviation 0.005 ohm.

a. What is the probability that a randomly selected wire from company A’s production will meet the

specifications?

b. If four of these wires are used in each computer system and all are selected from company A, what is

the probability that all four in a randomly selected system will meet the specifications?

Solution a:

Mean=0.13 SD=0.005

Probability to meet the specification, P(0.12<Z<0.14)

Transforming

𝑧1 =𝑦 − 𝜇

𝜎=

0.12 − 0.13

0.005= −2

𝑧2 =𝑦 − 𝜇

𝜎=

0.14 − 0.13

0.005= 2

Using table 4;

P(-2<Z<2)=2[0.5-P(Z>2)]=2[0.5-0.0228]=0.9544

Solution b:

Probability that all four meet the specifications is (0.9544)4=0.829

4.61. A soft-drink machine can be regulated so that it discharges and average of 𝜇 ounces per cup. If the ounces

of fill are normally distributed with standard deviation 0.3 ounce, give the setting for 𝜇 so that 8-ounce cups will

overflow only 1% of the time.

Solution:

Y is a normal random variable with (𝜇, 𝜎 = 0.3)

From table 4.1, P(Z>z)=0.01, so z should be 2.33.

𝑧 =𝑦 − 𝜇

𝜎=

8 − 𝜇

0.3= 2.33

Therefore 𝝁 = 𝟕. 𝟑𝟎𝟏

Page 8 of 12


Assignment for Thursday, May 28th

, 2009

4.74. One-hour carbon monoxide concentrations in air samples from a large city have an approximately

exponential distribution with mean 3.6 parts per million.

a. Find the probability that the carbon monoxide concentration exceeds 9 parts per million during a

randomly selected 1-hour period.

b. A traffic control strategy reduced the mean to 2.5 parts per million. Now find the probability that the

concentration exceeds 9 parts per million.

Solution

This is an exponential distribution with Mean= 𝛽= 3.6. The density function of random variable Y in

exponential distribution is:

𝑓 𝑦 =

1

𝛽𝑒

−𝑦𝛽 , 0 ≤ 𝑦 < ∞

0, 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒

Solution a:

Probability of exceeding 9 means P(Y>9), therefore we will have:

𝑃 𝑌 > 9 = 1

𝛽𝑒

−𝑦𝛽

∞

9

𝑑𝑦 = 1

3.6𝑒−

𝑦3.6

∞

9

𝑑𝑦 = −𝑒−𝑦

3.6 9

∞

= 0.0821

Solution b:

𝛽= 2.5

𝑃 𝑌 > 9 = 1

𝛽𝑒

−𝑦𝛽

∞

9

𝑑𝑦 = 1

2.5𝑒−

𝑦2.5

∞

9

𝑑𝑦 = −𝑒−𝑦

2.5 9

∞

= 0.0273

4.76. Suppose that a random variable Y has a probability density function given by

𝑓 𝑦 = 𝑘𝑦3𝑒−𝑦2 , 𝑦 > 0


Find the value of k that makes f(y) a density function.

Solution:

From definition of gamma distribution:

A random variable Y is said to have gamma distribution with parameters 𝛼 > 0 and 𝛽 > 0 if and only if the

density function of Y is:

𝑓 𝑦 = 𝑦𝛼−1𝑒−𝑦/𝛽

𝛽𝛼Γ(𝛼) 0 ≤ 𝑦 < ∞


Comparing this with the given probability density function;

𝛼 − 1 = 3 => 𝛼 = 4

𝛽 = 2

Γ 𝑛 = 𝑛 − 1 ! => Γ 4 = 6

𝛽𝛼 = 24 = 16 Therefore:

𝑘 =1

𝛽𝛼Γ(𝛼)=

1

6 × 16=

𝟏

𝟗𝟔

Page 9 of 12


4.78. Consider the plant of exercise 4.77. How much of the bulk product should be stocked so that the plant’s

chance of running out of the product is only 0.05?

Solution:

From problem 4.77, the amount of product used in one day can be modeled by an exponential distribution with

𝛽 = 4.

Plant’s chance of running out of the product is 0.05, which means P(Y>x)=0.05

P Y > 𝑥 = 1

𝛽𝑒

−𝑦𝛽

∞

𝑥

𝑑𝑦 = 1

4𝑒−

𝑦4

∞

𝑥

𝑑𝑦 = −𝑒−𝑦4

𝑥

∞

= 𝑒−𝑥4 = 0.05

𝐿𝑛(𝑒−𝑥4) = 𝐿𝑛(0.05)

−𝑥

4= −2.995

=> 𝒙 = 𝟏𝟏. 𝟗𝟖𝟐

4.88. If Y has a probability density function given by

𝑓 𝑦 = 4𝑦2𝑒−2𝑦 , 𝑦 > 0


Obtain E(Y) and V(Y) by inspection.

Solution:

Comparing the given density function with gamma distribution function, 𝑦𝛼−1𝑒−𝑦 /𝛽

𝛽𝛼Γ(𝛼)

𝛼 − 1 = 2 => 𝜶 = 𝟑

𝜷 =𝟏

𝟐

According to Theorem 4.8, the mean and variance of gamma distribution is:

𝜇 = 𝐸 𝑌 = 𝛼𝛽

𝜎2 = 𝑉 𝑌 = 𝛼𝛽2

Therefore:

𝜇 = (3) 1

2 =

𝟑

𝟐

𝑉 𝑌 = 3 . 1

2

2

=𝟑

𝟒

Page 10 of 12


4.101. The proportion of time per day that all checkout counters in a supermarket are busy is a random variable

Y with a density function given by

𝑓 𝑦 = 𝑐𝑦2 1 − 𝑦 4, 0 ≤ 𝑦 ≤ 10, 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒.

a. Find the value of c that makes f(y) a probability density function.

b. Find E(Y).

Solution a:

A random variable Y is said to have a beta probability distribution with parameters 𝛼 > 0 and 𝛽 > 0 if and only if the density function of Y is:

𝑓 𝑦 = 𝑦𝛼−1 1 − 𝑦 𝛽−1

𝐵(𝛼, 𝛽), 0 ≤ 𝑦 ≤ 1


𝐵 𝛼, 𝛽 =Γ α Γ(β)

Γ(α + β)

Comparing the given density function with beta probability distribution function:

𝛼 − 1 = 2 => 𝛼 = 3

𝛽 − 1 = 4 => 𝛽 = 5 and

𝑐 =1

𝐵 𝛼, 𝛽 =

Γ α + β

Γ α Γ β

=> 𝑐 =1

𝐵 3,5 =

Γ(3 + 5)

Γ 3 Γ(5)=

7.6.5

2=

210

2= 𝟏𝟎𝟓

Knowing that Γ 𝑛 = 𝑛 − 1 !

Solution b: According to Theorem 4.11, the mean of beta distribution is:

𝜇 = 𝐸 𝑌 =𝛼

𝛼 + 𝛽=

3

8= 𝟎. 𝟑𝟕𝟓

Page 11 of 12


Assignment for Friday, May 29th

, 2009

4.104. Suppose that the waiting time for the first customer to enter a retail shop after 9:00 AM is a random

variable Y with an exponential density function given by

𝑓 𝑦 = 1

𝜃 𝑒−𝑦/𝜃 , 𝑦 > 0


a. Find the moment-generating function for Y.

b. Use the answer from (a) to find E(Y) and V(Y).

Solution a:

As an special case of gamma-distributed random variables, the gamma density function in which 𝛼 = 1 is

called the exponential density function. A random variable is said to have an exponential distribution with

parameter 𝛽 > 0 if and only if the density function Y is:

𝑓 𝑦 =

1

𝛽𝑒

−𝑦𝛽 , 0 ≤ 𝑦 < ∞


Comparing the given density function with exponential density function: 𝛽 = 𝜃

The moment-generating function for a gamma distributed random variable, according to example 4.13, page

190 of the text book is:

𝑚 𝑡 =1

1 − 𝛽𝑡 𝛼

Replacing 𝛽 = 𝜃 and 𝛼 = 1, we will have:

𝒎 𝒕 =𝟏

𝟏 − 𝜽𝒕

Solution b:

𝐸 𝑌 = 𝜇 = 𝑑𝑚(𝑡)

𝑑𝑡 𝑡=0

= 𝜃

1 − 𝜃𝑡 2 𝑡=0

= 𝜽

𝐸 𝑌2 = 𝑑2𝑚(𝑡)

𝑑𝑡2 𝑡=0

= 2𝜃2

1 − 𝜃𝑡 3 𝑡=0

= 2𝜃2

𝑉 𝑌 = 𝐸 𝑌2 − 𝜇2 = 2𝜃2 − 𝜃2 = 𝜽𝟐

Page 12 of 12


4.107. The moment generating function of a normally distributed random variable, Y, with mean 𝜇 and variance

𝜎2 was shown in Exercise 4.106 to be 𝑚 𝑡 = 𝑒𝜇𝑡 +1

2𝑡2𝜎2

. Use the result in Exercise 4.105 to derive the moment

generating function of = −3𝑌 + 4 . What is the distribution of X? Why?

Solution:

Solving exercise 4.105 for 𝑈 = −3𝑌 + 4, (𝑎 = −3, 𝑏 = 4)

According to the theorem 4.12, the moment generating function of g(Y) is given by:

𝑚 𝑡 = 𝐸 𝑒𝑡𝑔 𝑌 Let 𝑔 𝑦 = 𝑋 = −3𝑌 + 4, then:

𝑚𝑋 𝑡 = 𝐸 𝑒(−3𝑌+4)𝑡 = 𝐸 𝑒−3𝑡𝑌+4𝑡 = 𝑒4𝑡𝐸 𝑒−3𝑡𝑌

=> 𝒎𝑿 𝒕 = 𝒆𝟒𝒕𝒎𝒀(−𝟑𝒕)

Solving exercise 4.106,

According to Example 4.16, if we let 𝑔 𝑦 = 𝑌 − 𝜇, where Y is a normally distributed random variable with

mean 𝜇 and variance 𝜎2, the moment generating function for g(Y) is:

𝑚𝑌−𝜇 𝑡 = 𝑒 𝑡2

2 𝜎2

In the other words,

𝑚𝑌−𝜇 𝑡 = 𝐸 𝑒(𝑌−𝜇)𝑡 = 𝐸 𝑒𝑌𝑡−𝜇𝑡 = 𝑒−𝜇𝑡 𝐸 𝑒𝑡𝑌 = 𝒆−𝜇𝒕𝒎𝒀(𝒕)

Therefore we have:

𝑒 𝑡2

2 𝜎2

= 𝑒−𝜇𝑡 𝑚𝑌(𝑡)

Which leads to:

𝒎𝒀 𝒕 = 𝒆𝝁𝒕. 𝒆 𝒕𝟐

𝟐 𝝈𝟐

Or

𝒎𝒀 𝒕 = 𝒆𝝁𝒕+

𝒕𝟐

𝟐 𝝈𝟐

Now, having 𝑚𝑋 𝑡 = 𝑒4𝑡𝑚𝑌(−3𝑡)

𝑚𝑋 𝑡 = 𝑒4𝑡𝑒𝜇 (−3𝑡)+

(−3𝑡)2

2 𝜎2

=> 𝒎𝑿 𝒕 = 𝒆(𝟒−𝟑𝝁)𝒕𝒆 𝒕𝟐

𝟐 𝟗𝝈𝟐

X has a normal distribution with mean equal to 4 − 3𝜇 and variance of 9𝜎2.

This is because of the uniqueness of moment generating function.

Page 1 of 20



Due June 15, 2009

Ramin Shamshiri



Assignment for Monday, June 1st , 2009

5.4. Given here is the joint probability function associated with data obtained in a study of automobile accidents

in which a child (under age 5) was in the car and at least one fatality occurred. Specifically, the study focused

on whether or not the child survived and what type of seatbelt (if any) he or she used. Define:

𝑌1 = 0, 𝑖𝑓 𝑡𝑕𝑒 𝑐𝑕𝑖𝑙𝑑 𝑠𝑢𝑟𝑣𝑖𝑣𝑒𝑑1, 𝑖𝑓 𝑛𝑜𝑡

and 𝑌1 =

0, 𝑖𝑓 no belt used1, 𝑖𝑓 adult belt used2, 𝑖𝑓 car − seat belt used.

Notice that Y1 is the number of fatalities per child and, since children’s car seats usually utilize two belts, Y2 is

the number of seatbelts in use at the time of accident.

y2

y1

0 1

0 0.38 0.17 0.55

1 0.14 0.02 0.16

2 0.24 0.05 0.29

0.76 0.24 1.00

a. Verify that the preceding probability function satisfies theorem 5.1.

b. Find F(1,2). What is the interpretation of this value?

Solution a:

Theorem 5.1 states that if Y1 and Y2 are discrete random variables with joint probability function p(y1,y2),

then

𝑃(𝑦1, 𝑦2) ≥ 0 for all y1,y2

𝑝 𝑦1, 𝑦2 = 1𝑦1 ,𝑦2, where the sum is over all values (y1,y2) that are assigned nonzero probabilities.

It can be seen from the table values that all of the probabilities are at least 0 and sum of them are 1.

Solution b:

According to definition 5.2, for any random variable Y1 and Y2, the joint distribution function F(y1,y2) is

given by: 𝐹 𝑦1, 𝑦2 = 𝑃(𝑌1 ≤ 𝑦1, 𝑌2 ≤ 𝑦2).

Therefore,

𝐹 1,2 = 𝑃 𝑌1 ≤ 1, 𝑌2 ≤ 2 = 𝑃 0,0 + 𝑃 0,1 + 𝑃 0,2 + 𝑃 1,0 + 𝑃 1,1 + 𝑃(1,2)

= 0.38 + 0.14 + 0.24 + 0.17 + 0.02 + 0.05 = 𝟎. 𝟕𝟔 + 𝟎. 𝟐𝟒 = 𝟏

It means that every child in the experiment either survived or didn’t and use either 0,1 or 2 seatbelts.

Page 2 of 20


5.8. An environment engineer measures the amount (by weight) of particular pollution in air samples of a

certain volume collected over two smokestacks at a coal-operated power plant. One of the stacks is equipped with a cleaning device. Let Y1 denote the amount of pollutant per sample collected above the stack that has no

cleaning device, and let Y2 denote the amount of pollutant per sample collected above the stack that is equipped

with the cleaning device. Suppose that the relative frequency behavior of Y1 and Y2 can be modeled by

𝑓(𝑦1, 𝑦2) = 𝑘, 0 ≤ y1 ≤ 2, 0 ≤ y2 ≤ 1, 2y2 ≤ y1


That is, Y1 and Y2 are uniform distributed over the region inside the triangle bounded by y1=2, y2=0, and

2y2=y1

a. Find the value of k that makes this function a probability density function.

b. Find 𝑃(𝑌1 ≥ 3𝑌2). That is, find the probability that the cleaning device reduces the amount of pollutant by one-their or more.

Solution a:

𝑘2

2𝑦2

1

0

𝑑𝑦1𝑑𝑦2 = 1

𝑘(2 − 2𝑦2)1

0

𝑑𝑦2 = 1

2𝑘𝑦2 − 𝑦22

0

1= 1

2𝑘 − 1 = 1 => 𝒌 = 𝟏

Solution b:

𝑦2 =1

3𝑦1

𝑃 𝑌1 ≥ 3𝑌2 = 1

13𝑦1

0

2

0

𝑑𝑦2𝑑𝑦1 = 1

3𝑦1

2

0

𝑑𝑦1

= 𝑦12

6

0

2

=4

6=

𝟐

𝟑

Page 3 of 20


5.13. The management at a fast-food outlet is interested in the joint behavior of the random variables Y1,

defined as the total time between a customer’s arrival at the store and departure from the service window, and Y2, the time a customer waits in line before reaching the service windows. Because Y1 includes the time a

customer waits in line, we must have 𝑌1 ≥ 𝑌2. The relative frequency distribution of observed values of Y1 and

Y2 can be modeled by the probability density function

𝑓(𝑦1, 𝑦2) = 𝑒−𝑦1 , 0 ≤ y2 ≤ y1 < ∞

0, elsewhere

With time measured in minutes.

a. Find 𝑃(𝑌1 < 2, 𝑌2 > 1). b. Find 𝑃(𝑌1 ≥ 2𝑌2). c. Find 𝑃(𝑌1 − 𝑌2 ≥ 1). (Notice that Y1-Y2 denotes the time spent at the service window)

Solution a:

𝑃 𝑌1 < 2, 𝑌2 > 1 = 𝑒−𝑦1

2

𝑦2

2

1

𝑑𝑦1𝑑𝑦2

Or we could say 𝑃 𝑌1 < 2, 𝑌2 > 1 = 𝑒−𝑦1𝑦1

1

2

1𝑑𝑦2𝑑𝑦1

Both are the same. Solving the integral,

𝑃 𝑌1 < 2, 𝑌2 > 1 = 𝑒−1 − 2𝑒−2

Solution b:

𝑃 𝑌1 ≥ 2𝑌2 = 𝑒−𝑦1

∞

2𝑦2

∞

0

𝑑𝑦1𝑑𝑦2

Or we could say:

𝑃 𝑌1 ≥ 2𝑌2 = 𝑒−𝑦1

∞

0

∞

2𝑦2

𝑑𝑦2𝑑𝑦1

Both are the same. Solving the integral

𝑃 𝑌1 ≥ 2𝑌2 =1

2

Solution c:

𝑃 𝑌1 − 𝑌2 ≥ 1 = 𝑒−𝑦1

∞

𝑦2+1

∞

0

𝑑𝑦1𝑑𝑦2 = 𝑒−1

Page 4 of 20


5.20. In exercise 5.4, you were given the following joint probability function for

𝑌1 = 0, 𝑖𝑓 𝑡𝑕𝑒 𝑐𝑕𝑖𝑙𝑑 𝑠𝑢𝑟𝑣𝑖𝑣𝑒𝑑1, 𝑖𝑓 𝑛𝑜𝑡

and 𝑌1 =

0, 𝑖𝑓 no belt used1, 𝑖𝑓 adult belt used2, 𝑖𝑓 car − seat belt used.

y2

y1

0 1

0 0.38 0.17 0.55

1 0.14 0.02 0.16

2 0.24 0.05 0.29

0.76 0.24 1.00

a. Give the marginal probability function for Y1 and Y2. b. Give the conditional probability function for Y2 given Y1=0.

c. What is the probability that a child survived given that he or she was in a car-seat belt?

Solution a:

𝑃1 𝑦1 = 𝑝(𝑦1, 𝑦2)

𝑦2

P1(Y1=0)=P(0,0)+P(0,1)+P(0,2)=0.38+0.14+0.24=0.76 P1(Y1=1)=P(1,0)+P(1,1)+P(1,2)=0.17+0.02+0.05=0.24

So:

𝑃1(𝑌1) = 0.76, 𝑦1 = 00.24, 𝑦1 = 1

𝑃2 𝑦2 = 𝑝(𝑦1, 𝑦2)

𝑦1

𝑃2(𝑌2) = 0.55, 𝑦0 = 00.16, 𝑦1 = 10.29, 𝑦2 = 2

Solution b:

𝑃 𝑌2 𝑌1 = 0 =𝑃(𝑦1 = 0, 𝑦2)

𝑃1(𝑦1 = 0)=

𝑃 0,0 = 0.38

0.76= 0.5 , 𝑦2 = 0

𝑃 0,1 = 0.14

0.76= 0.184, 𝑦2 = 1

𝑃 0,2 = 0.24

0.76= 0.315, 𝑦2 = 2

Solution c:

𝑃 𝑌1 = 0 𝑌2 = 2 =𝑃(0,2)

𝑃2(2)=

0.24

0.29= 0.827

Page 5 of 20


5.21. In example 5.4 and exercise 5.5 we considered the joint density of Y1, the proportion of the capacity of the

tank that is stocked at the beginning of the week, and Y2, the proportion of the capacity sold during the week, given by

𝑓(𝑦1, 𝑦2) = 3𝑦1, 0 ≤ y2 ≤ y1 < 1

0, elsewhere

a. Find the marginal density function for Y2.

b. For what values of y2 is the conditional density 𝑓( 𝑦1 𝑦2) defined?

c. What is the probability that more than half of a tank is sold given that three-fourths of a tank is stocked?

Solution a:

𝑓2 𝑦2 = 𝑓(𝑦1, 𝑦2)∞

−∞

𝑑𝑦1 = 3𝑦1

1

𝑦2

𝑑𝑦1 = 3𝑦12

2 𝑦2

1

=3

2−

3𝑦22

2

hence:

𝑓2 𝑦2 = 3

2−

3𝑦22

2, 0 ≤ y2 ≤ 1

0, elsewhere

Solution b: Denominator should not be zero,

𝑓 𝑦1 𝑦2 =𝑓(𝑦1, 𝑦2)

𝑓2(𝑦2) ≠ 0

Therefore, 0 ≤ y2 ≤ 1

Solution c:

𝑓1 𝑦1 = 𝑓(𝑦1, 𝑦2)∞

−∞

𝑑𝑦2 = 3𝑦1

𝑦1

0

𝑑𝑦2 = 3𝑦1𝑦2 0𝑦1 = 3𝑦1

2

𝑃 𝑌2 >1

2 𝑌1 =

3

4 = 𝑓 0.5 0.75 =

𝑓(0.75,0.5)

𝑓1(0.75)=

3𝑦1

𝑓1(0.75)=

94

3𝑦12 𝑦1=0.75

=

94

3 34

2 =𝟒

𝟑

Page 6 of 20


Assignment for Tuesday, June 2nd

, 2009

5.42. In exercise 5.4 you were given the following joint probability function for

𝑌1 = 0, 𝑖𝑓 𝑐𝑕𝑖𝑙𝑑 𝑠𝑢𝑟𝑣𝑖𝑣𝑒𝑑1, 𝑖𝑓 𝑛𝑜𝑡

and 𝑌2 =

0, 𝑖𝑓 𝑛𝑜 𝑏𝑒𝑙𝑡 𝑢𝑠𝑒𝑑1, 𝑖𝑓 𝑎𝑑𝑢𝑙𝑡 𝑏𝑒𝑙𝑡 𝑢𝑠𝑒𝑑2, 𝑖𝑓 𝑐𝑎𝑟 − 𝑠𝑒𝑎𝑡 𝑏𝑒𝑙𝑡 𝑢𝑠𝑒𝑑

y2

y1

0 1

0 0.38 0.17 0.55

1 0.14 0.02 0.16 2 0.24 0.05 0.29

0.76 0.24 1.00

Are Y1 and Y2 independent? Why or why not?

Solution: We should check P(y1,y2)=P1(y1).P2(y2)

Lets pick examine P(0,1), from table, we can see that P(0,1)=0.14. P1(0)=0.76 and P2(1)=0.16

Since 0.14≠0.76)(0.16), we conclude that Y1 and Y2 are not independent. So they are dependent.

5.43. In example 5.4 and exercise 5.5 we considered the joint density of Y1, the proportion of the capacity of

the tank that is stocked at the beginning of the week and Y2, the proposition of the capacity sold during the week, given by

𝑓(𝑦1, 𝑦2) = 3𝑦1, 0 ≤ 𝑦2 ≤ 𝑦1 ≤ 1


Show that Y1 and Y2 are dependent.

Solution: Quick check:

We examine the joint density function and see if it satisfies:

a. The region in which the joint density is positive must be a rectangle (possibly infinite) b. The function equation must factor into a product of two functions-one depends only on y1 and the other

depends only on y2. (One or both could be constant)

Since the region in which the joint density is positive is not a rectangle (it is triangle), then part (a) fails and we

conclude that Y1 and Y2 are dependent. (However part (b) holds)

Detailed solution:

𝑓2 𝑦2 = 𝑓(𝑦1, 𝑦2)∞

−∞

𝑑𝑦1 = 3𝑦1

1

𝑦2

𝑑𝑦1 = 3𝑦12

2 𝑦2

1

=3

2−

3𝑦22

2

hence:

𝑓2 𝑦2 = 3

2−

3𝑦22

2, 0 ≤ y2 ≤ 1

0, elsewhere

𝑓1 𝑦1 = 𝑓(𝑦1, 𝑦2)∞

−∞

𝑑𝑦2 = 3𝑦1

𝑦1

0

𝑑𝑦2 = 3𝑦1𝑦2 0𝑦1 = 3𝑦1

2

𝑓1 𝑦1 = 3𝑦1

2, 0 ≤ y1 ≤ 10, elsewhere

𝑓 𝑦1, 𝑦2 = 3𝑦1 ≠ 𝑓1 𝑦1 . 𝑓2 𝑦2

Hence, Y1 and Y2 are dependent.

Page 7 of 20


5.44. In exercise 5.6, we derived the fact that

𝑓(𝑦1, 𝑦2) = 4𝑦1𝑦2, 0 ≤ 𝑦1 ≤ 1, 0 ≤ 𝑦2 ≤ 1


is a valid joint probability density function. Are Y1 and Y2 independent?

Quick check: We examine the joint density function and see if it satisfies:

a. The region in which the joint density is positive must be a rectangle (possibly infinite)

b. The function equation must factor into a product of two functions-one depends only on y1 and the other depends only on y2. (One or both could be constant)

Since the region in which the joint density is positive is a rectangle and since the function equation can be

factored into a product of two functions 4y1 and y2m both criteria hold and we conclude that Y1 and Y2 are

independent. (However part (b) holds)

Detailed solution:

𝑓2 𝑦2 = 𝑓(𝑦1, 𝑦2)∞

−∞

𝑑𝑦1 = 4𝑦1𝑦2

1

0

𝑑𝑦1 = 2𝑦12𝑦2

0

1= 2𝑦2

hence:

𝑓2 𝑦2 = 2𝑦2, 0 ≤ y2 ≤ 1

0, elsewhere

𝑓1 𝑦1 = 𝑓(𝑦1, 𝑦2)∞

−∞

𝑑𝑦2 = 4𝑦1𝑦2

1

0

𝑑𝑦2 = 2𝑦1𝑦22

0

1= 2𝑦1

𝑓1 𝑦1 = 2𝑦1, 0 ≤ y1 ≤ 1

0, elsewhere

𝒇 𝒚𝟏, 𝒚𝟐 = 𝟒𝒚𝟏𝒚𝟐 = 𝒇𝟏 𝒚𝟏 . 𝒇𝟐 𝒚𝟐

Page 8 of 20


5.65. In exercise 5.7, we determined that

𝑓(𝑦1, 𝑦2) = 6(1 − 𝑦2), 0 ≤ 𝑦1 ≤ 𝑦2 ≤ 1


is a valid joint probability density function.

a. Find E(Y1) and E(Y2).

b. Find V(Y1) and V(Y2).

c. Find E(Y1-3 Y2).

Solution a:

𝐸 𝑌1 = 𝑦1

∞

−∞

∞

−∞

𝑓(𝑦1, 𝑦2)𝑑𝑦1𝑑𝑦2

𝑦1

𝑦2

0

1

0

6 − 6𝑦2 𝑑𝑦1𝑑𝑦2 =𝟏

𝟒

𝐸 𝑌2 = 𝑦2

∞

−∞

∞

−∞

𝑓(𝑦1, 𝑦2)𝑑𝑦1𝑑𝑦2

𝑦2

𝑦2

0

1

0

6 − 6𝑦2 𝑑𝑦1𝑑𝑦2 =𝟏

𝟐

Solution b:

𝑉 𝑌1 = 𝐸 𝑌12 − 𝐸(𝑌1) 2

𝐸 𝑌12 = 𝑦1

2𝑦2

0

1

0

6 − 6𝑦2 𝑑𝑦1𝑑𝑦2 =𝟏

𝟏𝟎

𝑉 𝑌1 =1

10−

1

4

2

=3

80

𝑉 𝑌2 = 𝐸 𝑌22 − 𝐸(𝑌2) 2

𝐸 𝑌22 = 𝑦2

2𝑦2

0

1

0

6 − 6𝑦2 𝑑𝑦1𝑑𝑦2 =𝟑

𝟏𝟎

𝑉 𝑌2 =3

10−

1

2

2

=1

20

Solution c:

𝐸 𝑌1 − 3𝑌2 = 𝐸 𝑌1 − 3𝐸 𝑌2 =1

4−

3

2= −

𝟓

𝟔

Page 9 of 20


5.70. In exercise 5.32 we determined that the joint density function for Y1, the weight in tons of a bulk item

stocked by a supplier, and Y2, the weight of the item sold by the supplier, has joint density

𝑓(𝑦1, 𝑦2) =

1

𝑦1, 0 ≤ 𝑦2 ≤ 𝑦1 ≤ 1


In this case, the random variable Y1-Y2 measures the amount of stock remaining at the end of the week, a quantity of great importance to the supplier. Find E(Y1-Y2).

Solution:

𝐸 𝑌1 − 𝑌2 = 𝐸 𝑌1 − 𝐸 𝑌2

𝐸 𝑌1 = 𝑦1

∞

−∞

∞

−∞

𝑓 𝑦1, 𝑦2 𝑑𝑦1𝑑𝑦2

= 11

𝑦2

1

0

𝑑𝑦1𝑑𝑦2 =1

2

𝐸 𝑌2 = 𝑦2

∞

−∞

∞

−∞

𝑓 𝑦1, 𝑦2 𝑑𝑦1𝑑𝑦2

= 𝑦2

𝑦1

1

𝑦2

1

0

𝑑𝑦1𝑑𝑦2 =1

4

𝐸 𝑌1 − 𝑌2 =1

2−

1

4=

𝟏

𝟒

Page 10 of 20


Assignment for Wednesday, June 3rd

, 2009

5.75. In Exercise 5.1 we determined that the joint distribution of Y1, the number of contracts awarded to firm A,

and Y2, the number of contracts awarded to firm B, is given by the entries in the following table

y2

y1

0 1 2

0 1/9 2/9 1/9 1 2/9 2/9 0

2 1/9 0 0

Find Cov(Y1, Y2). Does it surprise you that Cov(Y1, Y2) is negative? Why?

Solution:

Cov(Y1, Y2) = E(Y1Y2)-E(Y1).E(Y2)

𝐸 𝑌1𝑌2 = 𝑦1𝑦2𝑝 𝑦1, 𝑦2 =

𝑦2𝑦1

0 0 1

9 + 1 0

2

9 + 2 0

1

9 + 0 1

2

9 + 1 1

2

9

+ 0 2 1

9 =

2

9

E(Y1)= E(Y2)=2(1/3)=2/3 since Y1 and Y2 are both binomial with n=2 and p=1/3

Cov(Y1, Y2) = 2/9-(2/3)(2/3)= -2/9

The covariance is negative; it means that the Y2 decreases as the value of Y1 increases.

5.78. In Exercise 5.7, we determined that

𝑓 𝑦1, 𝑦2 = 6(1 − 𝑦2), 0 ≤ 𝑦1 ≤ 𝑦2 ≤ 1


is a valid joint probability density function. Find Cov(Y1,Y2). Are Y1 and Y2 independent?

Solution:

From Exercise 5.65, E(Y1)=1/4 and E(Y2)=1/2

𝐸 𝑌1𝑌2 6𝑦1𝑦2 1 − 𝑦2 𝑑𝑦1𝑑𝑦2

𝑦2

0

1

0

= 3 𝑦23 − 𝑦2

4 𝑑𝑦2 =3

4−

3

5=

3

20

1

0

Cov(Y1, Y2) = 3/20-1/8=1/40

Since Cov(Y1, Y2) ≠ 0, Y1 and Y2 are not independent.

Page 11 of 20


5.87. Assume that Y1,Y2 and Y3 are random variables, with

E(Y1)=2 E(Y2)= -1 E(Y3)= 4

V(Y1)=4 V(Y2)=6 V(Y3)= 8

Cov(Y1,Y2)= 1 Cov(Y1,Y3)= -1 Cov(Y2,Y3)= 0

Find E(3Y1+4Y2-6Y3) and V(3Y1+4Y2-6Y3)

Solution:

If U=a1Y1+a2Y2+…+anYn Then

E U = 𝑎𝑖 . 𝐸[𝑌𝑖]

𝑛

𝑖=1

E(3Y1+4Y2-6Y3) = E (3Y1)+E(4Y2)+E(-6Y3)

= 3E (Y1)+4E(Y2)-6E(Y3) =3(2)+4(-1)-6(4)=-22

V U = 𝑎𝑖2. 𝑉[𝑌𝑖]

𝑛

𝑖=1

+ 2 𝑎𝑖𝑎𝑗𝐶𝑜𝑣[𝑌𝑖 , 𝑌𝑗 ]

V(3Y1+4Y2-6Y3)= 9(4)+16(6)+36(8)+2.3.4.1+2.3.6(-1)+0=480

Page 12 of 20


5.92. If Y1 is the total time between a customer’s arrival in the store and departure from the service window and

if Y2 is the time spent in line before reaching the window, the joint density of these variables was given in Exercise 5.13 to be

𝑓 𝑦1, 𝑦2 = 𝑒−𝑦1, 0 ≤ 𝑦2 ≤ 𝑦1 ≤ 1


The random variable Y1-Y2 represents the time spent at the service window. Find E(Y1-Y2) and V(Y1-Y2). Is it

highly likely that a randomly selected customer would spend more than 4 minutes at the service window?

Solution:

From exercise 5.29,

𝑓1 𝑦1 = 𝑦1𝑒−𝑦1 is a gamma distribution with 𝛼 = 2, 𝛽 = 1.

Hence E(Y1)=2(1)=2 and V(Y1)= 𝛼𝛽2 = 2.

𝑓2 𝑦2 = 𝑒−𝑦1

∞

𝑦2

𝑑𝑦1 = −𝑒−𝑦1 𝑦2∞ = 𝑒−𝑦2

Which has a gamma distribution with 𝛼 = 𝛽 = 1. Hence E(Y2)=V(Y2)=1.

𝐸 𝑌1𝑌2 = 𝑦1𝑦2𝑒−𝑦1𝑑𝑦2𝑑𝑦1

𝑦1

0

1

0

= 𝑦1

3

2𝑒−𝑦1𝑑𝑦1 =

Γ 4 4

2= 3

∞

0

Cov(Y1, Y2) = 3-(1)(2)=1

E(Y1-Y2) = 2-1=1

V(Y1-Y2) = 2+1-2(1)=1

Page 13 of 20


Assignment for Monday, June 8th

, 2009

5.99. A learning experiment requires a rat to run a maze (a network of pathways) until it locates one of three

possible exists. Exit 1 presents a reward of food, but exit 2 and 3 do not. (If the rat eventually select exit 1

almost every time, learning may have taken place.) Let Yi denote the number of times exit i is chosen in successive runnings. For the following, assume that the rat chooses an exit at random on each run.

a. Find the probability than n=6 runs result in Y1=3, Y2=1 and Y3=2.

b. For general n, find E(Y1) and V(Y1). c. Find Cov(Y2,Y3) for general n.

d. To check for the rat’s preference between exit 2 and 3, we may look at Y2-Y3. Find E(Y2-Y3) and V(Y2-

Y3) for general n.

Solution:

Y1 denote the number of times that exit 1 is chosen. The probability that exit 1, 2 or 3 is chosen is equal and is 1/3.

Solution a:

𝑃 𝑦1, 𝑦2, 𝑦3 = 𝑃 3,1,2 =6!

3! .1! .2!

1

3

3

. 1

3

1

. 1

3

2

= 0.823

Solution b:

𝑃 𝑌𝑖 = 𝑛𝑃𝑖 => 𝐸 𝑌1 = 𝑛𝑃1 =1

3𝑛

𝑉 𝑌𝑖 = 𝑛𝑝𝑖 . 𝑞𝑖 => 𝑉 𝑃1 = 𝑛 1

3

2

3 =

2

9𝑛

Solution c:

𝐶𝑜𝑣 𝑌𝑠 , 𝑌𝑡 = −𝑛. 𝑝𝑠𝑝𝑡

𝐶𝑜𝑣 𝑌2, 𝑌3 = −𝑛. 1

3

1

3 =

1

9𝑛

Solution d:

𝐸 𝑌2 − 𝑌3 = 𝐸 𝑌2 − 𝐸 𝑌3 = 𝑛𝑃2 − 𝑛𝑃3 =𝑛

3−

𝑛

3= 0

𝑉 𝑌2 − 𝑌3 = 𝑉 𝑌2 + 𝑉 𝑌3 + 2𝐶𝑜𝑣 𝑌2, 𝑌3 =2

9𝑛 +

2

9𝑛 − 2 −

1

9𝑛 =

6

9𝑛

Page 14 of 20


5.103. The National Fire Incident Reporting Service stated that, among residential fires, 73% are in family

homes, 20% are in apartments, and 7% are in other types of dwellings. If four residential fires are independently reported on a single day, what is the probability that two are in family homes, one is in an apartment, and one is

in another type of dwelling?

Solution:

n=4

𝑃 𝑦1, 𝑦2, 𝑦3 = 𝑃 2,1,1 =4!

2! .1! .1! 0.73 2. 0.2 1. 0.07 1 = 0.08953

5. 115. In Exercise 5.35, we considered a quality control plan that calls for randomly selecting three items from

the daily production (assumed large) of a certain machine and observing the number of defectives. The

proportion 𝜌 of defectives produced by the machine varies from day to day and has a uniform distribution on

the interval (0,1).

a. Find the expected number of defectives observed among the three samples items.

b. Find the variance of the number of defectives among the three sampled.

Solution:

Y=(# of defectives)

P= (Proportion of defectives)

n=3

P is uniform on interval (0,1), 𝑓 𝑝 = 1, 0 ≤ 𝑝 ≤ 10, 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒

E[P]=1/2

Solution a:

𝐸 𝑌 = 𝐸[𝐸[ 𝑌 𝑃]], given that n=3, 𝐸[ 𝑌 𝑃]=np=3p

𝐸 𝑌 =E[3P]=3E[P]=3(1/2)=3/2

Solution b:

𝑉 𝑌 = 𝑉 𝐸 𝑌 𝑃 + 𝐸[𝑉[ 𝑌 𝑃]] 𝑉[ 𝑌 𝑃]=np(1-p)=3p(1-p)

𝑉 𝑌 =V[3p]+E[3P(1-P)]=9V[P]+3[E(P)-E(P2)]

E[P]=1/2 and V[P]=1/12 (since P is uniform (0,1)) and E[P2]=V[P]+(E[P])

2=1/12+(1/2)

2=1/3

Therefore,

𝑉 𝑌 = 9 1

12 + 3

1

2−

1

3 = 𝟏. 𝟐𝟓

Page 15 of 20


5.118. Assume that Y denotes the number of bacteria per cubic centimeter in a particular liquid and that Y has a

Poisson distribution with parameter 𝜆. Further assume that 𝜆 varies from location to location and has a gamma

distribution with parameters 𝛼 and 𝛽, where 𝛼 is a positive integer. If we randomly select a location,

a. What is the expected number of bacteria per cubic centimeter? b. What is the standard deviation of the number of bacteria per cubic centimeter?

Solution a:

𝑝 𝑦 𝜆 =𝜆𝑦𝑒−𝜆

𝑦!

𝐸 𝑌 𝜆 = 𝜆 𝜆 has a gamma distribution with parameters 𝛼 and 𝛽. The expected number of bacteria per cubic centimeter is

given by 𝛼𝛽.

Solution b:

𝐸 𝑌 𝜆 = 𝜆 and 𝑉 𝑌 𝜆 = 𝜆.

𝐸 𝜆 = 𝛼𝛽

𝑉 𝜆 = 𝛼𝛽2

𝑉 𝑌 = 𝐸 𝑉 𝑌 𝜆 + 𝑉 𝐸 𝑌 𝜆 = 𝐸 𝜆 + 𝑉 𝜆 = 𝛼𝛽 + 𝛼𝛽2 Therefore:

𝜍 = 𝛼𝛽(1 + 𝛽)

Page 16 of 20


6.7. Suppose that a unit of mineral ore contains a proportion Y1 of metal A and a proportion Y2 of metal B.

Experience has shown that the joint probability density function of Y1 and Y2 is uniform over the region

0 ≤ 𝑦1 ≤ 1, 0 ≤ 𝑦2 ≤ 1, 0 ≤ 𝑦1 + 𝑦2 ≤ 1. Let U=Y1+Y2, the proportion of either metal A or B per unit.

a. Find the probability density function for U. b. Find E(U) by using the answer to (a).

c. Find E(U) by using only the marginal density of Y1 and Y2.

Solution: Y1=(Proportional of metal A)

Y2=(Proportional of metal B)

𝑓 𝑦1, 𝑦2 = 2, 0 ≤ 𝑦1 ≤ 1 , 0 ≤ 𝑦2 ≤ 1 , 0 ≤ 𝑦1 + 𝑦2 ≤ 1 0 , 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒

Solution a:

The range of U is 0 ≤ 𝑈 ≤ 1 , taking u such that 0 ≤ 𝑢 ≤ 1 , we will have:

𝐹𝑈 𝑢 = 𝑃 𝑈 ≤ 𝑢 = 𝑃 𝑌1 + 𝑌2 ≤ 𝑢 = 𝑃 𝑌1 ≤ 𝑢 − 𝑌2 = 𝑃[𝑌2 ≤ 𝑢 − 𝑌1]

= 2𝑑𝑦1𝑑𝑦2

𝑢−𝑌2

0

= 2𝑦1 0𝑢−𝑦2𝑑𝑦2 = 2 𝑢 − 𝑦2 𝑑𝑦2

𝑢

0

𝑢

0

𝑢

0

= 2(𝑢𝑦 −𝑦2

2

2)

0

𝑢

= 2 𝑢2 −𝑢2

2 = 2

𝑢2

2 = 𝒖𝟐

𝑓𝑈 𝑢 =𝑑

𝑑𝑢𝐹𝑈 𝑢 =

𝟐𝒖, 0 ≤ 𝑢 ≤ 1 𝟎, 𝒆𝒍𝒔𝒆𝒘𝒉𝒆𝒓𝒆

Solution b:

𝐹𝑈 = 𝑢. 2𝑢. 𝑑𝑢 = 2𝑢2𝑑𝑢1

0

= 2𝑢3

3

0

1

=𝟐

𝟑

1

0

Page 17 of 20


6.30. A density function sometimes used by engineers to model lengths of life of electronic components is the

Rayleigh density, given by

𝑓 𝑦 = 2𝑦

𝜃 𝑒

−𝑦2

𝜃 , 𝑦 > 0

0 , 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒.

a. If Y has the Rayleigh density, find the probability density function for U=Y

2.

b. Use the result of (a) to find E(Y) and V(Y)

Solution a:

Y has the Rayleigh density and U=Y2

Let y2=h(y)=u, then 𝑦 = 𝑕−1 𝑢 = 𝑢 and

𝑑

𝑑𝑢𝑕−1 𝑢 =

1

2𝑢−

1

2

𝑓𝑈 𝑢 = 𝑓𝑌 𝑕−1 𝑢 𝑑

𝑑𝑢𝑕−1 𝑢 =

2 𝑢

𝜃 𝑒

− 𝑢

2

𝜃 1

2𝑢−

12 =

2𝑢12

𝜃

𝑢−12

2 𝑒

−𝑢𝜃

𝒇𝑼 𝒖 = 𝟏

𝜽𝒆

−𝒖𝜽 , 𝑢 ≥ 0

𝟎 , 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒.

This is an exponential density with mean 𝜃

Solution b:

𝐸 𝑌 = 𝐸 𝑈 = 𝑢.𝑒

−𝑢𝜃

𝜃

∞

0

𝑑𝑢 =Γ

32 𝜃3/2

𝜃 𝑢.

𝑒−𝑢𝜃

Γ 32 𝜃3/2

∞

0

𝑑𝑢 = Γ 3

2 𝜃1/2 =

1

2Γ

1

2 𝜃1/2

Γ 1

2 = 𝜋

∴ 𝑬 𝒀 = 𝝅𝜽

𝟐

𝐸 𝑌2 = 𝐸 𝑈 = 𝜃

𝐸 𝑌 = 𝐸 𝑌2 − 𝐸 𝑌 2 = 𝜃 −𝜋𝜃

4= 𝜽(𝟏 −

𝝅

𝟒)

Page 18 of 20


6.46. Let Y1, Y2, …Yn be independent Poisson random variables with means 𝜆1, 𝜆2, …,𝜆𝑛 , respectively.

a. Find the probability function of 𝑌𝑖𝑛𝑖=1

b. Find the conditional probability function of Y1 given 𝑌𝑖𝑛𝑖=1 = 𝑚

c. Find the conditional probability function of Y1+Y2 given 𝑌𝑖𝑛𝑖=1 = 𝑚

Solution a:

Let 𝑈 = 𝑌𝑖𝑛𝑖=1

Since each Yi is Poisson distributed with mean 𝜆𝑖; we have: 𝑚𝑌1 𝑡 = 𝑒𝜆(𝑒 𝑡−1). By Theorem 6.2:

𝑚𝑉 𝑡 = 𝑚𝑌1 𝑡 . 𝑚𝑌2

𝑡 …𝑚𝑌𝑛 𝑡 = 𝑚𝑌𝑖

𝑡

𝑛

𝑖=1

= 𝑒𝜆1 𝑒 𝑡−1 . 𝑒𝜆2 𝑒 𝑡−1 …𝑒𝜆𝑛 𝑒 𝑡−1

= 𝒆(𝒆𝒕−𝟏) 𝝀𝒊𝒏𝒊=𝟏

This is the moment generating function of a Poisson with mean 𝜆𝑖𝑛𝑖=1 .

Solution b: Let k be an arbitrary value for Y1, then:

𝑃 𝑌1 = 𝑘 𝑌𝑖 = 𝑚

𝑛

𝑖=1

=𝑃 𝑌1 = 𝑘, 𝑌𝑖 = 𝑚𝑛

𝑖=1

𝑃[ 𝑌𝑖 = 𝑚𝑛𝑖=1 ]

=𝑃[𝑌1 = 𝑘, 𝑌𝑖 = 𝑚 − 𝑘𝑛

𝑖=2 ]


=𝑃 𝑌1 = 𝑘 𝑃[ 𝑌𝑖 = 𝑚 − 𝑘𝑛

𝑖=2 ]


by independence

=

𝜆1𝑘𝑒𝜆1

𝑘!. 𝜆𝑖

𝑛𝑖=2 𝑚−𝑘𝑒 𝜆𝑖

𝑛𝑖=2

(𝑚 − 𝑘)!

𝜆𝑖𝑛𝑖=1

𝑚𝑒 𝜆𝑖

𝑛𝑖=1

𝑚!

=𝑚!

𝑘! 𝑚 − 𝑘 !.𝜆1

𝑘 𝜆𝑖𝑛𝑖=2 𝑚−𝑘

𝜆𝑖𝑛𝑖=1

𝑚 = 𝑚

𝑘 .

𝜆1𝑘 . 𝜆𝑖

𝑛𝑖=2 𝑚−𝑘

𝜆𝑖𝑛𝑖=1

𝑘 𝜆𝑖

𝑛𝑖=1

𝑚−𝑘

= 𝑚

𝑘

𝜆1

𝜆𝑖𝑛𝑖=1

𝑘

𝜆𝑖

𝑛𝑖=2

𝜆𝑖𝑛𝑖=1

𝑚−𝑘

This is a binomial probability function with probability of success 𝑃 = 𝜆1

𝜆𝑖𝑛𝑖=1

𝑘

Page 19 of 20


Solution c:

Let W= Y1+Y2 and k be an arbitrary value for W. Then:

𝑃 𝑊 = 𝑘 𝑌𝑖 = 𝑚

𝑛

𝑖=1

=𝑃 𝑊 = 𝑘, 𝑌𝑖 = 𝑚𝑛

𝑖=1


=𝑃 𝑊 = 𝑘 𝑃[ 𝑌𝑖 = 𝑚 − 𝑘𝑛

𝑖=3 ]


by independence

=

𝜆1 + 𝜆2 𝑘𝑒𝜆1+𝜆2

𝑘!. 𝜆𝑖

𝑛𝑖=2 𝑚−𝑘𝑒 𝜆𝑖

𝑛𝑖=3

(𝑚 − 𝑘)!

𝜆𝑖𝑛𝑖=1

𝑚𝑒 𝜆𝑖

𝑛𝑖=1

𝑚!

=𝑚!

𝑘! 𝑚 − 𝑘 !. 𝜆1 + 𝜆2 𝑘 𝜆𝑖

𝑛𝑖=3 𝑚−𝑘

𝜆𝑖𝑛𝑖=1

𝑚

= 𝑚

𝑘 .

𝜆1 + 𝜆2 𝑘 . 𝜆𝑖𝑛𝑖=2 𝑚−𝑘

𝜆𝑖𝑛𝑖=1

𝑘 𝜆𝑖

𝑛𝑖=1

𝑚−𝑘

= 𝑚

𝑘

𝜆1 + 𝜆2

𝜆𝑖𝑛𝑖=1

𝑘

𝜆𝑖

𝑛𝑖=3

𝜆𝑖𝑛𝑖=1

𝑚−𝑘

This is a binomial probability function with probability of success 𝑃 = 𝜆1+𝜆2 𝜆𝑖

𝑛𝑖=1

Page 20 of 20


6.58. Let Y1 and Y2 be independent and uniformly distributed over the interval (0,1).

a. Find the probability density function of U1=min(Y1,Y2)

b. Find E(U1) and V(U1)

Solution:

Given that Y1 and Y2 are independent and each uniformly distributed over the interval (0,1)

𝑓 𝑦 = 1, 0 < 𝑦 < 10, 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒

𝐹 𝑦 = 0, 𝑦 < 0𝑦, 0 ≤ 𝑦 ≤ 11, 𝑦 > 0

Solution a:

U1=min(Y1,Y2)

𝐺1 𝑢 = 𝑃 𝑈1 ≤ 𝑢 = 1 − 𝑃[𝑈 ≤ 𝑢] = 1 − 𝑃[𝑌1 > 𝑢, 𝑌2 > 𝑢] = 1 − 𝑃 𝑌1 > 𝑢 𝑃[𝑌2 > 𝑢] (by independence)

The density function of U1 is:

𝑔1 𝑢 = 2 1 − 𝐹 𝑢 − 𝑓 𝑢 = 𝟐(𝟏 − 𝒖), 0 ≤ 𝑢 ≤ 1

𝟎, 𝒆𝒍𝒔𝒆𝒘𝒉𝒆𝒓𝒆

Solution b:

𝐸[𝑈1] = 𝑢. 2 1 − 𝑢 𝑑𝑢1

0

= 2 𝑢 − 𝑢2 𝑑𝑢 = 2 𝑢2

2−

𝑢3

3

0

1 1

0

= 1 −2

3=

𝟏

𝟑

𝐸[𝑈12] = 𝑢2 . 2 1 − 𝑢 𝑑𝑢

1

0

= 2 𝑢2 − 𝑢3 𝑑𝑢 = 2 𝑢3

3−

𝑢4

4

0

1 1

0

=2

3−

1

2=

𝟏

𝟔

Thus

𝑉[𝑈1] = 𝐸[𝑈12] − 𝐸[𝑈1] 2 =

1

6−

1

3

2

=𝟏

𝟏𝟖

STA 5325, University of Florida, 2009

Documents

STA 5325, University of Florida, 2009