1 Copyright 015 by Sean Connolly All rights reserved. No part of this publication may be reproduced, distributed, or transmitted in any form or by any means, including photocopying, recording, or other electronic or mechanical methods, without the prior written permission of the publisher, except in the case of brief quotations embodied in critical reviews and certain other noncommercial uses permitted by copyright law. ISBN 978-0-9933047-0-5 Portions of information contained in this documentation are printed with permission of Minitab Inc. All such material remains the exclusive property and copyright of Minitab Inc. All rights reserved. IBM SPSS Statistics software ("SPSS" have kindly granted permission to include screen-prints generated from the running of the software package IBM SPSS Statistics. SPSS Inc. was acquired by IBM in October, 009. IBM, the IBM logo, ibm.com, and SPSS are trademarks or registered trademarks of International Business Machines Corporation, registered in many jurisdictions worldwide. Other product and service names might be trademarks of IBM or other companies. A current list of IBM trademarks is available on the Web at IBM Copyright and trademark information at www.ibm.com/legal/copytrade.shtml. Microsoft Microsoft Excel 013 is used with permission from Microsoft.
Chapter 5 1 (a 0.341345 (b 0.433193 (c 0.4775 (d 0.49379 (e 0.49865 3 (a P( X < 35 35 40 < σ 5... as µ 40 and σ 5 P( Z < 1 0.158655 (b P( 36 X 44 36 40 X µ 44 40 5 σ 5... as µ 40 and σ 5 P( 0.8 Z 0.8 0.57689 (c P( 36 X 44 36 40 X µ 39 40 5 σ 5... as µ 40 and σ 5 P( 0.8 Z 0. 0.08885 5 (a We wish to find two central values, call them a and b, such that 90% or 0.90 of yearly returns exit between them. Start As P( a X b P( X b 0.90 this implies that 15 0.45 as shown below. 45% of yearly returns exist between a and 15 45% of yearly returns exist between 15 and b a 15 b Z Step b µ + zσ general formula to be used in the calculation of b µ 0.15 σ 0.08 z 1.65 as P 0 < Z < 1.65 0.45 ( Step b 15 + 1.65( 8 8. Step Now that we have obtained b 11.6 let s find the value of a. Look at the diagram above! a and b are two central values about 100. Therefore both values are an equal distance from 100. As b 8. is a distance of 13. from 15, a 15 13. 1.8. End Range [1.8, 8.]
3 (b P( X < 0 0 15 < σ 8 P Z < ( 1.88 0.0301... as µ 15 and σ 8 (c The shaded area in the graph below represents the required probability: 0% of the population is beyound b Step P( X b Step 15 0.3 deduced from the graph above In the next step we calculate b. b µ + zσ µ 15 σ 8 0.84 z... as P( Z 0.84 End b 15 + 0.84( 8 1.7 0 < < 0.30 (obtained from the normal table 30% of the population exists between 15 and 1.7. Alternatively we may state that 0% of the population exists above 1.7. 7 (a 15 b X P( X > 37 37 8 > σ 5 P Z ( > 1.80 0.0359 (b P( X 0 0 8 σ 5 P Z 1.60 0.0548 (... as µ 8 and σ 5... as µ 8 and σ 5
4 (c First find ( 50 P X >. P( X > 50 50 8 > σ 5 P Z > 0 ( 4.4... as µ 8 and σ 5 The probability of a person waiting more than 50 days for a reimbursement is approximately 0. Thus If a person is waiting longer than 50 days then it suggests that the accountant never received the submission in the first place. 9 (a P( X > 180 180 180 > σ 10 P( Z > 0 05.... as µ 180 and σ 10 Therefore True. (b P( X > 00 00 180 > σ 10 P Z > ( 0.08... as µ 180 and σ 10 Therefore True, as the question states that about % of men weigh more than 00 pounds (c P( X < 170 170 180 < σ 10 P Z < ( 1 0.1587... as µ 180 and σ 10 Therefore False (d False 11 a Mean np ( µ 100 0.5 5 Standard deviation np( p ( ( σ 1 100 0.5 1 0.5 18.75 4.33
5 b np 100( 0.5 5 and n( p ( to use the approximation. 1 100 1 0.5 75. As both of these values are greater than 5 it s safe c ( 8 ( 8 1 8 + 1 P X P ( 7.5 X 8.5 7.5 5 X µ 8.5 5 4.33 σ 4.33 ( 0.58 Z 0.81 P 0. 071987... as µ 5 and σ 4.33 X µ... as Z σ d ( 5 ( 1 5 + 1 P Y P X P( 1.5 X 5.5 1.5 5 X µ 5.5 5 4.33 σ 4.33... as µ 5 and σ 4.33 P( 0.81 Z 0.1 X µ... as Z σ 0.3388 e ( 0 P( X 0 + 1 ( 0.5 P X 0.5 5 σ 4. 33 P Z ( 1.04 0.14917... as µ 5 and σ 4.33 13 np 60( 0.15 9 and n( p ( 1 60 1 0.15 51. As both of these values are greater than 5 it s safe to use the approximation. Mean µ np 60( 0.15 9 Standard deviation np( p ( ( σ 1 60 0.15 1 0.15 7.65.77
6 ( 8 P( X 8 + 1 ( 8.5 P X 8.5 9 σ.77 P Z ( 0.18 0.48576... as µ 9 and σ.77 15 np 900( 0.0 180 and n( p ( 1 900 1 0.0 70. As both of these values are greater than 5 it s safe to use the approximation. Mean µ np 900( 0.0 180 Standard deviation np( p ( ( ( 170 P( X 170 1 17 a σ 1 900 0.0 1 0.0 144 1 ( 169.5 P X σ P Z 0.81057 ( 0.88 169.5 180 1 np 500( 0.03 15 and n( p (... the third approximation is used... as µ 180 and σ 1 1 500 1 0.03 485. As both of these values are greater than 5 it s safe to use the approximation. µ np 500 0.03 15 Mean ( Standard deviation np( p ( ( ( 1 P( X 1 1 σ 1 500 0.03 1 0.03 14.55 3.81 P( X 0.5 0.5 15 σ 3.81 P( Z 1.44 0.074934... as µ 15 and σ 3.81
7 b ( 10 9 ( 10 1 9 + 1 P Y P X P ( 9.5 X 9.5 9.5 15 X µ 5.5 15 3.81 σ 3.81 P ( 1.44 Z.76 0.9176... as µ 15 and σ 3.81 19 (a ( P X 8 1 e λ( 8 0.1( 8 1 e... λ 1 1 0.10 mean 10 0.550671 b P X > 8 1 P X 8 ( ( ( 1 0.550671 in part (a we obtained P X 8 0.550671 0.44939 c P X 5 λ( 5 1 e ( 0.1( 5 1 e... λ 1 1 0.10 mean 10 0.393469 d P 9 X 1 P X 1 P X 9 ( ( ( 0.1( 1 0.1( 9 1 e 1 e... λ 1 1 0.10 mean 10 0.698806 0.59343 0.105376 1 a P X 9 λ( 9 1 e ( 0.1( 9 1 e... λ 1 1 0.067 mean 15 0.4583 b P X 0 1 P X 19 ( ( 0.067( 19 1 1 e... λ 1 1 0.10 mean 10 1 0.7001 0.7999
8 c P 9 X 18 P X 18 P X 9 ( ( ( 0.067( 18 0.067( 9 1 e 1 e... λ 1 1 0.067 mean 15 0.700608 0.4583 0.47776 3 a P X 100 1 P X 100 ( ( 0.015( 100 1 1 e... λ 1 1 0.015 mean 80 1 0.713495 0.86505 (b P X 50 λ( 50 1 e ( 0.015( 50 1 e... λ 1 1 0.015 mean 80 0.464739 5 (a P X < 1 λ( 1 1 e ( 0.05( 1 1 e... λ 1 1 0.05 mean 0 0.451188 (b P X > 35 1 P X 35 ( ( 0.05( 35 1 1 e... λ 1 1 0.05 mean 0 1 0.866 0.173774 c P 10 < X < 50 P X < 50 P X < 10 ( ( ( 0.05( 50 0.05( 10 1 e 1 e... λ 1 1 0.05 mean 0 0.917915 0.393469 0.54446 7 (a 0.11507 (b 0.841345 (c 0.998074 (d 0.87644 (e 0.3818
9 9 Start From the graph below we observe P( µ X b 0.48 50% of population exists below µ 48% of population exists between µ and b µ b 80 Z Step In the next step we find µ b µ + zσ b 80 provided in the question σ 5 provided in the question.05 P 0 < Z <.05 0.48 (obtained from the normal table z... as ( Step 80 µ +.05( 5 End After some algebra we obtain µ 69.75 We can now proceed to find the probability that a value will assume a value in excess of 74. P( X > 74 74 69.75 > σ 5 P Z > ( 0.85 0.197663... as µ 69.75 and σ 5 31 (a np 50( 0.33 16.5 and n( p ( 1 50 1 0.33 33.5. As both of these values are greater than 5 it s safe to use the approximation. Mean µ np 50( 0.33 16.5 Standard deviation np( p ( ( ( 11 P( X 11 + 1 σ 1 50 0.33 1 0.33 11.055 3.3 P( X 11.5 11.5 16.5 σ 3.3 P( Z 1.51 0.0655... as µ 16.5 and σ 3. 3
10 (b ( 1 P( X 1 1 ( 0.5 P X 0.5 16.5 σ 3.3... as µ 16.5 and σ 3.3 P( Z 1.0 0.11507 c P( 15 Y 5 P( 15 1 X 5 + 1 P( 14.5 X 5.5 14.5 16.5 X µ 5.5 16.5 3.3 σ 3.3 P( 0.60 Z.71 0.738 33 (a np 150( 0.0 30 and n( p (... as µ 16.5 and σ 3.3 1 150 1 0. 10. As both of these values are greater than 5 it s safe to use the approximation. Mean µ np 150( 0.0 30 Standard deviation np( p ( ( σ 1 150 0.0 1 0.0 4 4.90 ( ( 1 + 1 P X P ( 1.5 X.5 1.5 30 X µ.5 30 4.90 σ 4.90 P ( 1.73 Z 1.53 0.01193... as µ 30 and σ 4.90 (b ( P( X 1 ( 1.5 P X ( 1.73 1.5 30 σ 4.9... a s µ 30 and σ 4.9 P Z 0.958185
11 c ( P( X + 1 (.5 P X.5 30 σ 4.9 P Z ( 1.53 0.063008... as µ 30 and σ 4.9 35 a P X < 00 1 e ( λ( 00 0.00( 00 1 e... λ 1 1 0.00 mean 500 0.3968 b P X < 450 λ( 450 1 e ( 0.00( 00 1 e... λ 1 1 0.00 mean 500 0.59343 (c We wish to find the probability that at least parts are defective Start Let s restate the question: find the probability that at least parts are defective i.e. find the probability of obtaining one of the red numbers shown below: 0 1 3 4 5 6 7 8 9 10 Step Let X denote the number of failures. X is a Binomial variable. We require P( X. Step P( X 1 P( X 1 n x As 10 parts are chosen, n 10. In part a we found the probability of a part failing in less than 00 hours is 0.3968 10 0 for X 0, ( 0 0.3698 ( 1 0.3698 10 P X 0 0.01907 0 10 1 for X 1, ( 1 0.3698 ( 1 0.3698 10 P X 1 0.0963 1 We use these calculated values in the next step. P X 1 P X 1 1 P( X 0 + P( X 1 1 ( 0.01907 + 0.093... obtained in previous se tp 0.88863 x X is a binomial variable. The formula to calculate its probability is: P( X x p ( 1 p End ( ( n x
(d Let X denote the number of failures. X is a Binomial variable. We require P( X 5 as we require the probability that all 5 parts fail. n x As 5 parts are chosen, n 5. In part b we found the probability of a part failing in less than 450 hours is 0.59343 5 5 Thus ( 5 0.59343 ( 1 0.59343 5 P X 5 0.07359 5 x X is a binomial variable. The formula to calculate its probability is: P( X x p ( 1 p n x 1