6.
0 0.5 1 1.5 2 2.5 3
0.001
0.01
0.05
0.1
0.25
0.5
0.75
0.9
0.95
0.99
0.999
8.
3 3.5 4 4.5 5 5.5 6
0.001
0.01
0.05
0.1
0.25
0.5
0.75
0.9
0.95
0.99
0.999
Section 4.11
Page 208
1. (a) Let X1, …, X144 be the volumes in the 144 bottles.
2. (a) Let X1, …, X250 denote the thicknesses of the 250 sheets of paper.
Let S=X1+···+X250 be the total thickness of the 250 pages.
3. The mean number of red lights encountered per day is
µX= 0(0.1) + 1(0.3) + 2(0.3) + 3(0.2) + 4(0.1) = 1.9.
4. (a) Let X1, …, X625 be the amounts of tax on the 625 forms.
(b) Let Ybe the number of forms whose tax is greater than $3000.
5. (a) Let X1, …, X60 be the weights of the 60 bags.
(b) Let x70 denote the 70th percentile
(c) Let nbe the necessary sample size.
6. (a) Let X1, …, X200 be the amounts of warpage in the 200 wafers.
Page 210
PROPRIETARY MATERIAL. c
The McGraw-Hill Companies, Inc. All rights reserved. No part of this
Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written
permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by
McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
SECTION 4.11 211
(b) Let x25 denote the 25th percentile.
(c) Let nbe the necessary sample size.
7. (a) Let X1, …, X50 be the times taken by each of the 50 customers. Let S=X1+···+X50 .
(b) Let X1, …, X50 be the times taken by each of the 50 customers. Let S=X1+···+X50.
8. (a) Let X1, …, X50 be the amounts of solution in 50 drums. Then µXi= 30.01 and σXi= 0.1.
(b) Let X1, …, X80 be the amounts of solution in 80 drums. Then µXi= 30.01 and σXi= 0.1.
Page 211
PROPRIETARY MATERIAL. c
The McGraw-Hill Companies, Inc. All rights reserved. No part of this
Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written
permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by
McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
212 CHAPTER 4
(c) Let sbe the required amount of solution. Then P(T < s) = 0.9, so sis the 90th percentile of the
9. Let nbe the required number of measurements. Let Xbe the average of the nmeasurements.
10. Let Xrepresent the number of adults who have a college degree.
11. (a) Let Xrepresent the number of nondefective bearings in a shipment.
Page 212
SECTION 4.11 213
(b) Let Yrepresent the number of shipments out of 300 that are acceptable.
(c) Let pbe the required proportion of defective bearings, and let Xrepresent the number of defective
bearings in a
shipment.
12. (a) Let Xbe the number of O-rings that meet the specification.
Then X∼Bin(1000,0.9), so Xis approximately normal with mean µX= 1000(0.90) = 900 and
(b) Let x60 denote the 60th percentile
(c) From part (a), the probability that fewer than 890 O-rings meet the specification on any given day is
Page 213
214 CHAPTER 4
Y∼Bin(5,0.1335).
13. (a) Let Xbe the number of particles emitted by mass A and let Ybe the number of particles emitted by
mass B in a five-minute time period. Then X∼Poisson(100) and Y∼Poisson(125).
(b) Let Xbe the number of particles emitted by mass A and let Ybe the number of particles emitted by
mass B in a two-minute time period. Then X∼Poisson(40) and Y∼Poisson(50).
Now Xis approximately normal with mean 40 and variance 40, and Yis approximately normal with
14. (a) Let Xbe the number of particles contained in a 2 ml sample. Then X∼Poisson(60),
(b) Let Ybe the number of samples that contain more than 50 particles.
Page 214
SECTION 4.11 215
(c) Let Wbe he number of samples that contain more than 50 particles.
From part (a), the probability that a sample contains more than 50 particles is 0.9015.
15. (a) Let Xbe the number of particles withdrawn in a 5 mL volume.
Then the mean of Xis 50(5) = 250, so X∼Poisson(250), and Xis approximately normal with mean
(b) Since the withdrawn sample contains 5 mL, the average number of particles per mL will be between
48 and 52 if the total number of particles is between 5(48) = 240 and 5(52) = 260.
(c) Let Xbe the number of particles withdrawn in a 10 mL volume.
Page 215
216 CHAPTER 4
(d) Let vbe the required volume. Let Xbe the number of particles withdrawn in a volume of vmL.
Then X∼Poisson(50v), so Xis approximately normal with mean µX= 50vand standard deviation
σX=√50v.
16. (a) If the claim is true, then Xis approximately normal with mean µX= 40 and σX= 5/√100 = 0.5.
The z-score of 36.7 is (36.7−40)/0.5 = −6.6.
17. (a) If the claim is true, then X∼Bin(1000,0.05), so Xis approximately normal with mean µX=
Page 216
SECTION 4.11 217
(e) No. More than 1/3 of the samples of size 1000 will have 53 or more nonconforming tiles if the goal
has been reached.
18. Let X1, …, X100 be the times taken on machine 1 by the 100 parts. Let Y1, …, Y100 be the times taken
on machine 2 by the 100 parts.
Page 217
218 CHAPTER 4
(c) Let T=SX+SYbe the total time used by both machines.
Since SXand SYare independent and approximately normal, Tis approximately normal with mean
19. Let Xbe the number of rivets from vendor A that meet the specification, and let Ybe the number
of rivets from vendor B that meet the specification.
20. (a) λsolves the equation 11,000 = (ln 15.3−ln λ)/0.0001210.
Therefore λ= 4.0425 emissions per minute.
Page 218
SECTION 4.12 219
Section 4.12
1. (a) X∼Bin(100,0.03), Y∼Bin(100,0.05)
(b) Answers will vary.
2. (a) ≈0.84
220 CHAPTER 4
4. (a) ≈36
5. (a) ≈0.25
6. (a–d) Answers will vary.
7. (a–c) Answers will vary.
8. (a,b) Answers will vary.
Page 220
9. (a) Answers will vary.
10. (a) Answers will vary.
11. (a) Answers will vary.
Page 221