STE3A01 HW Solutions Chapter 2 Part 2 (1)

STE3A01: Part 2 Solutions Chapter 2 Semester 1, 2015

Statistics for Engineers 3A (STE3A01)Part 2 Solutions to Selected Homework Exercises from Chapter 2 of

Probability and Statistics for Engineering and the Sciencesby Jay L. Devore

Section 2.2, p.62

Question 20, p.63

A swing shift is a shift that starts in the afternoon and ends at night.

a. The simple events are:

E1 = {Day shift, Unsafe conditions}; E2 = {Day shift, Unrelated conditions};E3 = {Swing shift, Unsafe conditions}; E4 = {Swing shift, Unrelated conditions};E5 = {Night shift, Unsafe conditions}; E6 = {Night shift, Unrelated conditions}.

b. The probability that the selected accident was attributed to unsafe conditions can be calculated as

P (Unsafe conditions) = 10% + 8% + 5% = 23%.

c. The probability that the selected accident did not occur on the day shift is given by

P (Not day shift) = 1 P (Day shift) = 1 (10% + 35%) = 55%.

Question 22, p.63

Assume that the traffic signals are not synchronised and operate independently of one another.

Let S1 represent the event that the motorist must stop at traffic signal 1. Then, P (S1) = 0.4. Let S2represent the event that the motorist must stop at traffic signal 2. Then, P (S2) = 0.5. We have alsobeen given that P (S1 S2) = 0.6.

a. Probability of stopping at both signals:

P (S1 S2) = P (S1) + P (S2) P (S1 S2) manipulated proposition, p.60

= 0.4 + 0.5 0.6= 0.3

b. Probability of stopping at first signal, but not second one:

P (S1 S2) = P (S1) P (S1 S2)

= 0.4 0.3 from question (a) above

= 0.1

c. Probability of stopping at exactly one signal:

P (S1 S2) + P (S1 S2) = P (S1 S2) P (S1 S2)

= 0.6 0.3= 0.3

University of Johannesburg Page 1 of 12


Section 2.4, p.80

Question 50, p.81

a. Probability of a medium, long-sleeved, print shirt:

P (M Long Pr) = 0.05 read off from Long-sleeved table

b. Probability of a medium, print shirt (which could be long-sleeved or short-sleeved):

P (M Pr) = P (M Pr Long) + P (M Pr Short)= 0.05 + 0.07= 0.12

c. Probability of a short-sleeved shirt:

P (Short) = 0.04 + 0.02 + 0.05 + . . .0.08 = 0.56

Probability of a long-sleeved shirt:

P (Long) = 1 P (Short) department store only sells short-sleeved shirts or long-sleeved shirts

= 1 0.56= 0.44

d. Probability that size of next shirt is a medium:

P (Medium) = P (M Short) + P (M Long)= (0.08 + 0.07 + 0.12) + (0.10 + 0.05 + 0.07)= 0.49

Probability of pattern of next shirt sold is a print:

P (Print) = P (Print Short) + P (Print Long)= (0.02 + 0.07 + 0.07) + (0.02 + 0.05 + 0.02)= 0.25

e. Probability of a medium shirt if it is known that the shirt selected is a short-sleeved plaid shirt:

P (M (Short Pl)) =0.08

0.04 + 0.08 + 0.03= 0.533

f. Probability of a short-sleeved shirt if it is known that the shirt selected is a medium plaid shirt:

P (Short (M Pl)) =0.08

0.08 + 0.10= 0.444

Probability of a long-sleeved shirt if it is known that the shirt selected is a medium plaid shirt:

P (Long (M Pl)) =0.1

0.08 + 0.10= 0.556



Question 53, p.81

Refer to p.32 of the Appendix at the back of the textbook.

Question 55, p.81


Question 60, p.81

Let D represent the event of a light aircraft that disappear while in flight subsequently being discov-ered.The complement of this event, D, is therefore the event of a light aircraft that disappear while inflight subsequently not being discovered.Let L represent the event of a light aircraft having an emergency locator.The complement of this event, L, is therefore the event of a light aircraft not having an emergencylocator.

We have been given

P (D) = 70%, from which we can calculate P (D) = 1 P (D) = 1 0.7 = 0.3.

P (L D) = 60%, from which we can calculate P (L D) = 1 P (L D) = 1 0.6 = 0.4.

P (L D) = 90%, from which we can calculate P (L D) = 1 P (L D) = 1 0.9 = 0.1.

a. We need to find P (D L), which has reversed conditionality when compared to P (L D). Thereforeuse Bayes Theorem to solve:

P (D L) =P (L D)P (D)

P (L D)P (D) + P (L D)P (D)Bayes Theorem, Formula (2.6), p.79

=0.1 0.3

0.1 0.3 + 0.6 0.7= 0.0667

b. We need to find P (D L), which has reversed conditionality when compared to P (L D). Thereforeuse Bayes Theorem to solve:

P (D L) =P (L D)P (D)

P (L D)P (D) + P (L D)P (D)Bayes Theorem, Formula (2.6), p.79

=0.4 0.7

0.4 0.7 + 0.9 0.3= 0.509

Question 62, p.82

Let B represent the event that a purchaser buys the basic video camera model. It has been given thatP (B) = 0.4.LetD represent the event that a purchaser buys the deluxe video camera model. Because this companyonly manufactures two models of video camera (basic and deluxe), we can calculate P (D) by usingthe complement rule: P (D) = 1 P (B) = 1 0.4 = 0.6.



Let W represent the event that a purchaser buys an extended warranty. We have been given theconditional probability P (W B) = 0.3, from which it follows that P (W B) = 1 0.3 = 0.7. We havealso been given that P (W D) = 0.5, from which it follows that P (W D) = 1 0.5 = 0.5.

We need to calculate P (B W ), which has reversed conditionality when compared to P (W B). There-fore use Bayes Theorem to solve:

P (B W ) =P (W B)P (B)

P (W B)P (B) + P (W D)P (D)Bayes Theorem, Formula (2.6), p.79

=0.3 0.4

0.3 0.4 + 0.5 0.6= 0.286

Question 63, p.81


Question 69, p.81


Section 2.5, p.86

Question 71, p.86


Question 75, p.87


Question 77, p.87


Question 78, p.87

Let Ai denote the event that boiler valve i opens on demand, for i = 1, 2, . . . ,5. We have been giventhat P (Ai) = 0.95, so that P (Ai) = 1 P (Ai) = 1 0.95 = 0.05.

P (At least one valve opens) = 1 P (No valves open) first proposition, p.59

= 1 P (A1 A2 A

3 A

4 A

5)

= 1 P (A1) P (A2) . . . P (A

5) valves operate indepently

= 1 0.055

= 0.9999996875 1



P (At least one valve fails to open) = 1 P (All valves open) first proposition, p.59

= 1 P (A1 A2 A3 A4 A5)

= 1 P (A1) P (A2) . . . P (A5) valves operate indepently

= 1 0.955

= 0.226

Question 80, p.87

Let W represent the event that a component works. We have been given that P (W ) = 0.9.Let N represent the event that a component does not work. It follows that P (N) = 1 P (W ) =1 0.9 = 0.1.Also, WNNW will represent the event that components 1 and 4 work, but 2 and 3 not.

The sample space will contain 16 outcomes: there are two possibilities per component (Work, Notwork) and there are 4 components. This gives us 24 = 16 possible outcomes, which have all been listedbelow:

Component Status System Status Component Status System Status

WWWW Works NWNW Works

NWWW Works WNWN Works

WNWW Works NWWN Works

WWNW Works NNNW Does not work

WWWN Works NNWN Does not work

NNWW Works NWNN Works

WNNW Works WNNN Works

WWNN Works NNNN Does not work

Therefore,

P (System works) = 1 P (System does not work) first proposition, p.59

= 1 P (NNNW ) P (NNWN) P (NNNN)= 1 P (N) P (N) P (N) P (W )

P (N) P (N) P (W ) P (N)

P (N) P (N) P (N) P (N) components work independently

= 1 0.053 0.95 0.053 0.95 0.054

= 0.99975625 0.9998

Question 83, p.87


Question 84, p.87

Let E represent the event that a vehicle passes its emissions test. We have been given that P (E) = 0.7.



a. P (All of the next three vehicles inspeted pass) can be solved as follows:

P (EEE) = P (E) P (E) P (E) vehicles pass or fail independently of one another

= 0.7 0.7 0.7= 0.343

b. P (At least one of the next three inspected fails) can be solved by working with the complement:1 - P (All of the next three vehicles inspected pass)

1 P (EEE) = 1 0.343 from question (a) above

= 0.757

c. P (Exactly one of the next three inspected passes) can be solved as listed below:

P (Only vehicle 1 passes) or P (Only vehicle 2 passes) or P (Only vehicle 3 passes)= P (EEE) + P (EEE) + P (EEE)

= [P (E) P (E) P (E)] + [P (E) P (E) P (E)] + [P (E) P (E) P (E)]

vehicles pass or fail independently of one another

= 3 0.7 0.32

= 0.189

d. P (At most one of the next three vehicles inspected passes) is the same as exactly one vehicle passesor no vehicles passes:

P ((Exactly one vehicle passes) or (No vehicles passes))= P (Exactly one vehicle passes) + P (No vehicles passes)

= 0.189 + P (EEE) from question (b) above

= 0.189 + P (E) P (E) P (E) vehicles pass or fail independently of one another

= 0.189 + 0.33

= 0.216

e. P (All three pass at least one passes) is a conditional probability which can be solved as follows:

P (All three pass at least one passes)

=P (All three pass and at least one passes)

P (At least one passes)Formula (2.3), p.74

=P (All three pass)

P (At least one passes)

=P (All three pass)1 P (All three fail)

first proposition, p.59

=0.343

1 0.33from question (a) above

= 0.353

Question 85, p.88




Question 86, p.88

a. The probability of selecting two boards has been illustrated below by using a tree diagram:

BoardOne

BoardTwo

P (A B) = 8 00010 000

7 9999 999

= 0.63998

B A 7 9999 999

P (A B) = 8 00010 000

2 0009 999

= 1 6009 999

B A

2 000

9 999

A 8 00010 000

BoardTwo

P (A B) = 2 00010 000

8 0009 999

= 1 6009 999

B A8 0009 999

P(A B) = 2 00010 000

1 9999 999

= 1 99949 995

B A

1 999

9 999

A

2 000

10000

Check: 1 99949 995

+ 1 6009 999

+ 1 6009 999

+ 0.63998 = 1. -,

The lumber company has already selected 1 green board. There remains, therefore,9 999 boards of which 1 999 are green.

The lumber company has already selected 1 green board. There remains, therefore,9 999 boards of which 8 000 are usable in construction.

The lumber company has already selected 1 board that is usable in construction.There remains, therefore, 9 999 boards of which 2 000 are green.

The lumber company has already selected 1 board that is usable in construction.There remains, therefore, 9 999 boards of which 7 999 are usable in construction.

P (A) =2 000

10 000= 0.2. see tree diagram at

P (B) = P (A B) + P (A B)

=1 999

49 995+1; 600

9 999= 0.2. see tree diagram at and

P (A B) =1999

49995= 0.03998. see tree diagram at

P (A) P (B) = 0.2 0.2 = 0.04 P (A B).

It can therefore be concluded that A and B are not statistically independent.

b. P (A B) = P (A) P (B) = 0.2 0.2 = 0.04. if A and B were independentThe answer in part (a) is 0.000016 smaller than the answer in part (b), which is insignificant.Because of the size of the lot, we can in this case assume independence in order to obtain essentiallythe correct probability.



c. For a lot of size 10 of which 2 boards are green, we have:

P (A) =2

10= 0.2.

P (B A) =1

9.

P (A B) = P (A) P (B A) = 0.2 1

9= 0.0222 0.04.

The critical difference between (c) and (a) is the lot size. For very small sample sizes we cannotassume independence, but for very large samples we might just do it, because the difference inanswer is insignificantly small.

Question 89, p.88


Supplementary Exercises, p.88

Question 90, p.88

a. The order in which the 3 machinists are selected is not important, as all 3 of them will be servingon the night crew. There are therefore C3,20 = 1140 possible night crews that can be selected.

b. Remove the best machinist from the 20 machinists to leave 19 machinists. Now, C3,19 = 969 nightcrews would not have the best machinist.

c. Number of crews with at least one member from the ten best = total number of different crews number of crews with no one from the top ten:

All possible crewsucurly1140

No. of crews with no one from the top 10ucurlyleftudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymoducurlymidudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymoducurlyrightC0,10dcurlyTop 10

C3,10dcurly

Bottom 10

= 1140 1 120udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod

No. of crews with at least one member frrom the top 10

= 1020

d. Probability that the best machinist will not work:

P (Best machinist will not work on a particular night)

=number of crews without best machinist

total number of possible crews

=969

1140= 0.85

Question 91, p.88




Question 94, p.89

View the video solution on Dropbox by scanning the QR code or entering the short URL in the addressbar of your web browser:

Question 94(a): http://qrs.ly/gd4ka8d

Question 94(b): http://qrs.ly/u44ka8p

Question 94(c): http://qrs.ly/fu4ka9b

Question 96, p.89

We have been given the equation

Pd(c) =(c/c)

1 + (c/c). (A)

a. Substitute c with c in Equation (A) above:

Pd(c) =(c/c)

1 + (c/c)

=1

1 + 1

=1

2= 0.5

b. Substitute c with 2c and with 4 in Equation (A) above:

Pd(2c) =(2c/c)4

1 + (2c/c)4

=24

1 + 24

=16

17= 0.941



c. Let D represent the event of a crack being detected. Then,

P (Exactly one of two cracks being detected)= P (D D) + P (D D)

= P (D) P (D) + P (D) P (D) inspections are independent

= 0.5 (1 0.941) + (1 0.5) 0.941= 0.5

d. As , Pd(c) 1.

Question 97, p.89


Question 101, p.90


Question 102, p.90

We have been given thatP (E1) = 0.4; P (L E1) = 0.02.

We need to calculate P (E1 L):

P (E1 L) = P (L E1)P (E1) multiplication rule, p.75

= 0.02 0.4= 0.008

Question 103, p.90


Question 104, p.90

Let R represent the event that a component needs rework. It has been given that

P (A1) = 0.5; P (A2) = 0.3; P (A3) = 0.2; P (R A1) = 0.05; P (R A2) = 0.08; P (R A3) = 0.1.

P (Randomly selected component came from assembly line A1, given it needs rework)= P (A1 R)

=P (R A1)P (A1)

P (R A1)P (A1) + P (R A2)P (A2) + P (R A3)P (A3)Bayes Theorem, Formula (2.6), p.79

=0.05 0.5

0.05 0.5 + 0.08 0.3 + 0.1 0.2= 0.362




=P (R A2)P (A2)


=0.08 0.3

0.05 0.5 + 0.08 0.3 + 0.1 0.2= 0.348


=P (R A3)P (A3)


=0.1 0.2

0.05 0.5 + 0.08 0.3 + 0.1 0.2= 0.290

Check: 0.362 + 0.348 + 0.290 = 1, which gives us 100% of reworked components. -,

Question 107, p.90


Question 109, p.91


Question 110, p.91

Let A represent the event that the flight on this airline from Chicago to New York is full. We havebeen given that P (A) = 0.6, so that P (A) = 1 P (A) = 0.4.Let B represent the event that the flight on this airline from Chicago to Atlanta is full. We have beengiven that P (B) = 0.5, so that P (B) = 1 P (B) = 0.5.Let C represent the event that the flight on this airline from Chicago to Los Angeles is full. We havebeen given that P (C) = 0.4, so that P (C ) = 1 P (C) = 0.6.A, B and C are independent events.

a. The probability that all three flights are full:

P (A B C) = P (A) P (B) P (C) A, B and C are independent events

= 0.5 0.4 0.3= 0.06

The probability that at least one flight is not full can be written as P ((A B C)).

P (A B C) + P ((A B C)) = 1P ((A B C)) = 1 P (A) P (B) P (C)

= 1 0.06 from question (a) above

= 0.94



b. The probability that only the New York flight is full (P (A B C )) can be calculated as follows:

P (A B C ) = P (A) P (A B) P (A C) + P (A B C)

draw your Venn diagram!

= P (A) [P (A) P (B)] [P (A) P (C)] + [P (A) P (B) P (C)]

A, B and C are independent events

= 0.6 [0.6 0.5] [0.6 0.4] + [0.6 0.5 0.4]= 0.18

The probability that exactly one of the three flights is full, can be calculated as follows:

P (Exactly one of the three flights is full)

= P (A B C) P (A B) P (A C) P (B C) + 2P (A B C)= [P (A) + P (B) + P (C) P (A B) P (A C) P (B C) + P (A B C)] P (A B) P (A C) P (B C) + 2P (A B C)= 0.6 + 0.5 + 0.4 0.6 0.5 0.6 0.4 0.5 0.4 + 0.6 0.5 0.4 0.6 0.5 0.6 0.4 0.5 0.4 + 2 0.6 0.5 0.4= 0.38

Or you just could have calculated P (A B C ) and P (A B C) and added the two answersto P (A B C ) calculated above, to give you an answer of 0.38 as well -,.

Question 113, p.91



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STE3A01 HW Solutions Chapter 2 Part 2 (1)