Final Solution Chap 2 411s50

Chapter 2 Exercise with solution 1.

A study examined waiting times in the X-ray department for a hospital in Jacksonville, Florida. A clerk recorded the number of patients waiting for service at 9:00 A.M. on 20 consecutive days and obtained the following results. Number of waiting

Number of Days Outcome occurred 2 5 6 4 3 Total 20

0 1 2 3 4

a) Define the experiment the clerk conducted. b) List the experimental outcomes. c) Assign probabilities to the experimental outcomes. d) What method did you use?

Answer 1: a)

This experiment is about an X-ray department in which a clerk identifies the number of patients waiting for the services at 9 am in the department.

b) Following are the experimental outcomes at X-ray department: Numbe r of waiting 0 1 2 3 4 c)

Probabilities of experimental outcomes: Number of waiting 0

1

2

3

Number of Days Outcome occurred 2

5

6

4

Probabilities

2 =0.1 20 5 =0.25 20 6 =0.3 20 4 =0.2 20

4

3

Total 20 d) The Relative Frequency method is used in this experiment

2

1

3 =0.15 20

A company that franchises coffee houses conducted taste tests for a new coffee product. The company prepared four blends and randomly chose individuals to do a taste test and state which of the four blends they liked best. Results of the taste test for 100 individuals are given. Blend 1

Blend 1 2

Preference Test Testers’ Test Testers’ Probabilities Preference 20 20 30

3 2 a) Define being many

3

4

35 30

15

30 100=0.3

35

35 100=0.35

b) Prior to 4

20 100=0.2

15

15 100=0.15

c)

the experiment conducted. How times was it repeated? conducting the experiment, it is reasonable to preferences for blends are equal. probabilities you assign to the

assume the four What Total 100 1 would experimental outcomes prior to conducting the taste test? What method did you use? After conducting the taste test, what probabilities would you assign to the experimental outcomes? What method did you use?

Answer 2: a)

This experiment is about identifying the taste preference of people about the 4 new blends prepared by the coffee company. This experiment was repeated 100 times.

b) As there are 4 blends so the probability before taste test would be

1 =0.25 4 For each blend and for this purpose we have used classical method. c)

Probabilities of experimental outcomes after taste test:

In this probability distribution we have used relative frequency method. 2.

A company that manufactures toothpaste is studying five different package designs. Assuming that one design is just as likely to be selected by a consumer as any other design, what selection probability would you assign to each of the package designs? In an actual experiment, 100 consumers were asked to pick the design they preferred. The following data were obtained. Do the data confirm the belief that one design is just as likely to be selected as another? Explain.

Design 1 2 3 4 5

No. of Times Preferred 5 15 30 40 10

Answer 3: No, the data do not confirm the belief because every design has different preference level of the consumer. If we see the design 4 under the head of No. of times preferred that design is more likely to be selected as compare to any other packing design. 4.

Strom Construction made a bid on two contracts. The owner identified the possible outcomes and subjectively assigned the following probabilities. Experimental Obtain Obtain Probability Outcome Contract 1 Contract 2 1 Yes Yes 0.15 2 Yes No 0.15 3 No Yes 0.30 4 No No 0.25 a) Are these valid probability assignments? Why or why not? If not, what would have to be done to make the probability assignments valid?

Answer 4: a)

No, these are not valid probability assignments because the sum of all is not equal to the 1.

b) To make the all experimental outcomes valid we have to reassign the probabilities. 5.

Suppose that a sample space has five equally likely experimental outcomes:

Let

A= { E1 ,E2 }

B={ E 3 , E4 }

E1 ,E 2 , E3 , E4 , E5 .

C={ E2 ,E 3 ,E5 } a) Find

P=( A ) P=( B )

b) Find

P=( A∪B )

c)

Find

d) Find

,

A

c

,

C

c

,

. Are

, and

and

A

and

P( Ac )

A ∪B c∧P ( A∪Bc )

P=( C )

B P ( Cc )

. mutually exclusive. .

.

Answer 5: a)

Following are the probabilities of A, B and C.

2 P ( A )= =0.4 5 2 P ( B ) = =0.4 5 3 P (C )= =0.6 5 b) Probability of

A ∪B

:

P ( A∪B )= P ( A ) +P ( B ) ¿ P ( E 1 , E2 ) +P ( E3 , E4 ) P ( A∪B )=( {E1 ,E 2 , E3 , E4 )

.

Other method:

P ( A∪B )= P ( A ) +P ( B )

P ( A∪B )=P ( 0.4 )+P ( 0.4 ) P ( A∪B )= 0.80

Yes A and B both are mutually exclusive events because they don’t have any common value.

c)

Following are the compliments.

c

i.

A =( E 3 , E 4 ,E 5)

ii.

C =( E1 ,E 4 )

iii.

P ( A c )=1−P ( A )

c

.

.

P ( A c )=1−P ( 0.40 ) P ( A c )=0.60 iv.

P ( Cc )= 1−P (C ) P ( Cc )= 1−P ( 0.60 ) P ( Cc )= 0.40

d) Following are the compliments.

i.

c

A ∪B ={ E1 ,E 2 , E5 } c

¿ A= { E1 , E2 }+B = { E1 ,E2 , E3 ,E 5 } c

A ∪B ={ E1 ,E 2 , E3 , E5 }

P ( A∪Bc ) =P ( A )+P ( Bc )− P ( A∩Bc ) c

P ( A ∪ B ) =P

2 4 2 +P −P 5 5 5

() () ()

c

P ( A∪B ) =0.8 6

In a recent article U.S. News and World Report rated pediatric hospitals and provided data on several characteristics, including daily inpatient volume and nurse-to-patient ratio, for the top 30 hospitals on its list (U.S. News and World Report, June 9, 2008). Suppose we consider a daily inpatient volume of at least 200 to be high and a nurse-to-patient ratio of at least 3.0 to be high. Sixteen hospitals had a daily inpatient volume of at least 200, one third of the hospitals had a nurse-to-patient ratio of at least 3.0, and seven of the hospitals had both a daily inpatient volume of at least 200 and a nurse-to-patient ratio of at least 3.0. a) Find the probability of a hospital having a daily inpatient volume of at least 200, the probability of a hospital having a nurse-to-patient ratio of at least 3.0, and the probability of a hospital having both a daily inpatient volume of at least 200 and a nurse-to patient ratio of at least 3.0. b) What is the probability that a hospital had a daily inpatient volume of at least 200 or a nurse-to-patient ratio of at least 3.0 or both? c) What is the probability that a hospital had neither a daily inpatient volume of at least 200 nor a nurseto-patient ratio of at least 3.0?

Answer 6: Total No. of hospitals = 30 Daily inpatient volume ≥ 200 Nurse-to-patient ratio ≥ 3.0 Let A = Daily inpatient volume B = Nurse-to-patient ratio

A ∪B

i.

= Hospital having both a daily inpatient volume of at least 200 and a nurse-to patient ratio

of at least 3.0. a) Probabilities: Let A is the probability of Daily inpatient volume P (A) =

ii.

16 =0.533 30

Let B is the probability of nurse-to-patient ratio P (B) =

iii.

10 =0.333 30

Probability of hospital having both a daily inpatient volume of at least 200 and a nurse-to patient ratio of at least 3.0.

P(A∩B)=7/30=0.233 b)

P ( A∪B )=P ( A ) +P ( B )− P ( A∩B ) P(A∪B)=P(0.533)+P(0.333)−P(0.233)

P(A∪B)=0.633 c)

c

P(A∪B) =1−P(A∪B)

c

P(A∪B) =1−0.633 c

P(A∪B) =0.367 7.

A pharmaceutical company conducted a study to evaluate the effect of an allergy relief medicine; 250 patients with symptoms that included itchy eyes and a skin rash received the new drug. The results of the study are as follows: 90 of the patients treated experienced eye relief, 135 had their skin rash clear up, and 45 experienced relief of both itchy eyes and the skin rash. What is the probability that a patient who takes the drug will experience relief of at least one of the two symptoms?

Answer 7: Let A = itchy eye B = skin rash

P(A∪B)=P(A)+P(B)−P(A∩B)

¿ P(90 ⁄ 250)+P(135 ⁄ 250)−P(45 ⁄ 250) P(A∪B)=P(0.36)+P(0.54)−P(0.18) P(A∪B)=0.72 8. For two events A and B, P(A) =0.5, P(B) =0.60, and P(A ∩ B) =0.40. a) Find P(A│ B). b) Find P(B│A). c) Are A and B independent? Why or why not?

Answer 8:

P(A)=0.5 P(B)=0.60 P(A∩B)=0.40 a)

P(A/B)=P(A∩B) ⁄ P(B)

¿0.40 ⁄ 0.60=0.66

b)

P(B/A)=P(A∩B) ⁄ P(A)

¿0.40 ⁄ 0.50=0.8 c)

No, A and B are not independent because

P(A ⁄ B)≠ P(A)

P(0.66)≠P(0.5) 9.

A survey of MBA students obtained the following data on “Students’ first reason for application to the school in which they matriculated. Reasons for Applications Enrolmen t Status

School Quality

School Cost or Convenienc e 393

Other

Totals

Full 421 76 890 Time Part 400 593 46 1039 Time Totals 821 986 122 1929 a) Develop a t probability table using these data. b) b. Use the marginal probabilities of school quality, school cost or convenience, and other to comment on the most important reason for choosing a school. c) If a student goes full time, what is the probability that school quality will be the first reason for choosing a school? d) If a student goes part time, what is the probability that school quality will be the first reason for choosing a school? e) Let A be the event that a student is full time and let B be the event that the student lists school quality as the first reason for applying. Are events A and B independent? Justify your answer.

Answer 9: a)

t probability table. Reasons for Applications School Quality Enrolment Status

School Cost or Convenienc e

Other

Totals

Full 0.22 0.20 0.04 0.46 Time Part 0.21 0.31 0.02 0.54 Time Totals 0.43 0.51 0.06 1 b) The first reason of the most of the students to choose the school is school cost or convenience with the probability of 0.51. At second level the first reason of students while choosing a school is school quality with the probability of 0.43.

c)

P(Quality ⁄ (Fulltime)) ¿

d)

P(Quality ⁄ (Parttime)) ¿

10.

0.22 =0.47 0.46

0.21 =0.38 0.54

Hundreds of thousands of drivers dropped their automobile insurance in 2008 as the un-employment rate rose (Wall Street Journal, December 17, 2008). Sample data representative of the national automobile insurance coverage for individuals 18 years of age and older are shown here.

Auto Mobile Insurance

Age

18 to 34 35 and over

Yes 1500 1900

No 340 260

a) Develop a t probability table for these data and use the table to answer the remaining questions. b) What do the marginal probabilities tell you about the age of the U.S. population? c) What is the probability that a randomly selected individual does not have automobile insurance coverage? d) If the individual is between the ages of 18 and 34, what is the probability that the individual does not have automobile insurance coverage? e) If the individual is age 35 or over, what is the probability that the individual does not have automobile insurance coverage? f) If the individual does not have automobile insurance, what is the probability that the individual is in the 18–34 age group? g) What does the probability information tell you about automobile insurance coverage in the United States?

Answer 10: Let

a)

A= 18 to 34 age group B= 35 and over age group Y = has automobile insurance coverage N = Does not have automobile insurance coverage t probability table. Auto Mobile Insurance

Age

18 to 34 35 and over total

Yes

No

total

0.375 0.475 0.850

0.085 0.065 0.150

0.46 0.54 1

b) Above probabilities describes that the population with the age 35 and over who have automobile insurance with the probability of 0.54 and thy dropped the insurance in 2008. The second class of people is of 18 to 34 who have the automobile insurance and the dropped the insurance in 2008.

c)

P(N)

d)

P(N ⁄ A)

¿0.15

P(N/ A)=P(N ∩ A) ⁄ P(A) P(N/ A)=(0.085 ⁄ 0.46)=0.1848 e)

P(N ⁄ B) P(N/B)=P(N ∩B) ⁄ P(B)

P(N/ A)=(0.065 ⁄ 0.54)=0.1204 f)

P(A ⁄ N)

P(A/N)=P(A∩N) ⁄ P(N) P(A/N)=(0.085 ⁄ 0.15)=0.5667 g) The people are less interested in getting the insurance. 11.

During a recent year, speeding was reported in 12.9% of all automobile accidents in the United States (The World Almanac 2010). Assume the probability that speeding is reported in an accident is 0.129, the probability of an accident in which speeding is reported leading to a fatality is 0.196, and the probability of an accident in which speeding is reported that does not lead to a fatality is 0.05. Suppose you learn of an accident involving a fatality. What is the probability that speeding was reported?

Answer 11: Let: S = Speeding is reported

c

Speeding is not reported

S F =Accident results in fatality for vehicle

P(S)=0.129 P(S)c=0.871

P(F ⁄ S)=0.196 c

P(F ⁄ S) =0.05 Events

S S

Prior Probabilities 0.129

Conditional Probabilities 0.196

t Probabilities 0.0253

Posterior Probabilities 0.37

0.871

0.050

0.0436

0.63

c 1

1

P(F)=0.0687

P(S ⁄ F)=0.0687 12.

i.e., if an accident involved a fatality. The probability speeding was reported is 0.37.

The Wayne Manufacturing Company purchases a certain part from suppliers A, B, and C. Supplier A supplies 60% of the parts, B 30%, and C 10%. The quality of parts varies among the suppliers, with A, B, and C parts having 0.25%, 1%, and 2% defective rates, respectively. The parts are used in one of the company’s major products. a) What is the probability that the company’s major product is assembled with a defective part? Use the tabular approach to Bayes’ theorem to solve. b) When a defective part is found, which supplier is the likely source?

Answer 12: Let, D = Defective Part

A1

a)

= Supplier A

A2

= Supplier B

A3

= Supplier C

Probability of product with defective parts.

Prior Probabilities

Ai

Events

Conditional Probabilities

P(Ai )

A1 A2

P(D ⁄ Ai )

0.0025

0.0015

0.30

0.01

0.003

0.02

Posterior Probabilities

P(Ai ∩D)

0.60

0.10

A3

t Probabilities

P(Ai ⁄ D)

0.0015 ⁄ 0.0065=0.23 0.003 ⁄ 0.0065=0.46

0.002

1.00

1

0.002 ⁄ 0.0065=0.31

P(D)=0.0065 So through above calculation we have found the probability of product with defective part is 0.0065. The supplier 13.

A2

= B is the sours with high probability of 0.46.

Bayes’ theorem and conditional probability can be used in medical diagnosis. Prior probabilities of diseases are based on the physician’s assessment of factors such as geographic location, seasonal influence, and occurrence of epidemics. Assume that a patient is believed to have one of two diseases, denoted D1 and D2, with P(D1) _ 0.60 and P(D2) _ 0.40, and that medical research shows a probability associated with each symptom that may accompany the diseases. Suppose that, given diseases D1 and D2, the probabilities that a patient will have symptoms S1, S2, or S3 are as follows

Symptoms

S1

Disease

D1

S2

S3

0.15

0.10

0.15

0.80

0.15

0.03

D2 P

S3 D1

( )

After finding that a certain symptom is present, the medical diagnosis may be aided by finding the revised probabilities that the patient has each particular disease. Compute the posterior probabilities of each disease for the following medical findings.

a. The patient has symptom S1. b. The patient has symptom S2. c. The patient has symptom S3. d. For the patient with symptom S1 in part (a), suppose that symptom S2 also is present. What are the revised probabilities of D1 and D2?

Answer 13: a)

The patient has symptom S1.

Events

Prior Probabilities

P(D1)

D1 D2

Conditional Probabilities

P(S1 /D1)

t Probabilities

Posterior Probabilities

P(D1 ∩S 1)

P(D1 ⁄ S1 )

0.60

0.15

0.090

0.2195

0.40

0.80

0.320

0.7805

1

1

P(S1 )=0.410 P(D1 ⁄ S1 )=0.2195

P(D2 ⁄ S1 )=0.7805 b) The patient has symptom S2. Events

Prior Probabilities

P(D1)

D1 D2

Conditional Probabilities

P(S2 ⁄ D1 )

t Probabilities

Posterior Probabilities

P(D1 ∩S 2)

P(D1 ⁄ S2 )

0.60

0.10

0.060

0.500

0.40

0.15

0.060

0.500

1

1

P(S2 )=0.120

P(D1 ⁄ S2 )=0.50 P(D2 ⁄ S2 )=0.50

c)

The patient has symptom S3. Prior Probabilities

Events

P ( D 1)

D1 D2

Conditional Probabilities

S3 P D1

0.60

0.15

0.40

0.03

( )

t Probabilities

Posterior Probabilities

P ( D1 ∩S3 ) P

D1 S3

( )

0.090

0.8824

0.012

0.1176

1

1

P ( S 3 )=0.102 P P

D1 =0.8824 S3

( ) ( )

D2 =0.1176 S3

d) Prior probabilities from part (a). Prior Probabilities

Events

P ( D 1)

D1 D2

Conditional Probabilities

S2 P D1

0.2195

0.10

0.7805

0.15

( )

t Probabilities

P ( D1 ∩S2 ) P

( )

( )

0.1582

0.1171

0.8418 1

P ( S 2 )=0.1391 D1 = 0.1582 S 1∧S2

D1 S2

0.0220

1

P

Posterior Probabilities

P

D2 = 0.8418 S 1∧S2

( )