Ch 4 - 16E

Probability and Statistics for Engineering and the Sciences | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole, Raymond H. Myers

Problem 16E Chapter 4

Probability and Statistics for Engineering and the Sciences | 9th Edition

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Probability and Statistics for Engineering and the Sciences | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole, Raymond H. Myers

Probability and Statistics for Engineering and the Sciences | 9th Edition

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Problem 16E

16E

Step-by-Step Solution:

Answer : Step 1 : Given "A Model of Pedestrians' Waiting Times for Street Crossings at Signalized Intersections" suggested that under some circumstances the distribution of waiting time X could be modeled with the following pdf: Now we have to draw a graph f (x; , 80) and given values of = 4, 1, and .5. Put = 4. x 1 We are using excel to find the (1 ) . Here we know that 41 = 80 (1 80 ) . 3 = 0.05(0) . = 0.05 Then we calculated the table is given. x x 1 x x 1 (1 ) (1 ) 0 0.05 40 0.00625 1 0.04814834 41 0.005792871 2 0.046342969 42 0.005358594 3 0.044583301 43 0.004946582 4 0.04286875 44 0.00455625 5 0.04119873 45 0.004187012 6 0.039572656 46 0.003838281 7 0.037989941 47 0.003509473 8 0.03645 48 0.0032 9 0.034952246 49 0.002909277 10 0.033496094 50 0.002636719 11 0.032080957 51 0.002381738 12 0.03070625 52 0.00214375 13 0.029371387 53 0.001922168 14 0.028075781 54 0.001716406 15 0.026818848 55 0.001525879 16 0.0256 56 0.00135 17 0.024418652 57 0.001188184 18 0.023274219 58 0.001039844 19 0.022166113 59 0.000904395 20 0.02109375 60 0.00078125 21 0.020056543 61 0.000669824 22 0.019053906 62 0.000569531 23 0.018085254 63 0.000479785 24 0.01715 64 0.0004 25 0.016247559 65 0.00032959 26 0.015377344 66 0.000267969 27 0.01453877 67 0.000214551 28 0.01373125 68 0.00016875 29 0.012954199 69 0.00012998 30 0.012207031 70 9.76563E-05 31 0.01148916 71 7.11914E-05 32 0.0108 72 0.00005 33 0.010138965 73 3.34961E-05 34 0.009505469 74 2.10938E-05 35 0.008898926 75 1.2207E-05 36 0.00831875 76 6.25E-06 37 0.007764355 77 2.63672E-06 38 0.007235156 78 7.8125E-07 39 0.006730566 79 9.76562E-08 Then f (x; 4, 80) the graph is given below. If you look at the graph of the function (above and to the right). So the function is a decreasing function of X. Put = 1. 1 We are using excel to find th (1 ) . 11 = 80 (1 80) . 0 = 0.0125(1) . = 0.0125 Then we calculated the table is given. x (1 x)1 x (1 x)1 0 0.0125 40 0.0015625 1 0.012037085 41 0.001448218 2 0.011585742 42 0.001339648 3 0.011145825 43 0.001236646 4 0.010717188 44 0.001139063 5 0.010299683 45 0.001046753 6 0.009893164 46 0.00095957 7 0.009497485 47 0.000877368 8 0.0091125 48 0.0008 9 0.008738062 49 0.000727319 10 0.008374023 50 0.00065918 11 0.008020239 51 0.000595435 12 0.007676563 52 0.000535938 13 0.007342847 53 0.000480542 14 0.007018945 54 0.000429102 15 0.006704712 55 0.00038147 16 0.0064 56 0.0003375 17 0.006104663 57 0.000297046 18 0.005818555 58 0.000259961 19 0.005541528 59 0.000226099 20 0.005273438 60 0.000195313 21 0.005014136 61 0.000167456 22 0.004763477 62 0.000142383 23 0.004521313 63 0.000119946 24 0.0042875 64 1E-04 25 0.00406189 65 8.23975E-05 26 0.003844336 66 6.69922E-05 27 0.003634692 67 5.36377E-05 28 0.003432813 68 4.21875E-05 29 0.00323855 69 3.24951E-05 30 0.003051758 70 2.44141E-05 31 0.00287229 71 1.77979E-05 32 0.0027 72 0.0000125 33 0.002534741 73 8.37402E-06 34 0.002376367 74 5.27344E-06 35 0.002224731 75 3.05176E-06 36 0.002079688 76 1.5625E-06 37 0.001941089 77 6.5918E-07 38 0.001808789 78 1.95313E-07 39 0.001682642 79 2.44141E-08 Then f (x; 1, 80) the graph is given below. If you look at the graph of the function (above and to the right). So the function is a decreasing function of X. Put =0.5. x 1 We are using excel to find the ( 1 ) . 0.5 0 41 = 80 ( 1 80) . 0.51 = 0.00625(1) = 0.00625 Then we calculated the table is given. x (1 x)1 x (1 x)1 0 0.00625 40 0.00078125 1 0.006018542 41 0.000724109 2 0.005792871 42 0.000669824 3 0.005572913 43 0.000618323 4 0.005358594 44 0.000569531 5 0.005149841 45 0.000523376 6 0.004946582 46 0.000479785 7 0.004748743 47 0.000438684 8 0.00455625 48 0.0004 9 0.004369031 49 0.00036366 10 0.004187012 50 0.00032959 11 0.00401012 51 0.000297717 12 0.003838281 52 0.000267969 13 0.003671423 53 0.000240271 14 0.003509473 54 0.000214551 15 0.003352356 55 0.000190735 16 0.0032 56 0.00016875 17 0.003052332 57 0.000148523 18 0.002909277 58 0.00012998 19 0.002770764 59 0.000113049 20 0.002636719 60 9.76563E-05 21 0.002507068 61 8.3728E-05 22 0.002381738 62 7.11914E-05 23 0.002260657 63 5.99731E-05 24 0.00214375 64 0.00005 25 0.002030945 65 4.11987E-05 26 0.001922168 66 3.34961E-05 27 0.001817346 67 2.68188E-05 28 0.001716406 68 2.10938E-05 29 0.001619275 69 1.62476E-05 30 0.001525879 70 1.2207E-05 31 0.001436145 71 8.89893E-06 32 0.00135 72 0.00000625 33 0.001267371 73 4.18701E-06 34 0.001188184 74 2.63672E-06 35 0.001112366 75 1.52588E-06 36 0.001039844 76 7.8125E-07 37 0.000970544 77 3.2959E-07 38 0.000904395 78 9.76563E-08 39 0.000841321 79 1.2207E-08 Then f (x; 0.5, 80) the graph is given below. If you look at the graph of the function (above and to the right). So the function is a decreasing function of X. Step 2 : b). We have to find the cumulative distribution function of X. Differentiate with respect to x. F (x) = (1 x) 1 X 2 1 ( 1)(x) F Xx) = 1 d x 1 F (X) = dx (1 ) 1 d x 1 F (X) = dx(1 )= 1 x 1 F (X) = (1 ) Therefore the cumulative distribution function of X is F (x) = (1 x )1 . X

Step 3 of 4

Chapter 4, Problem 16E is Solved
Step 4 of 4

Textbook: Probability and Statistics for Engineering and the Sciences
Edition: 9
Author: Ronald E. Walpole, Raymond H. Myers
ISBN: 9780321629111

The answer to ā€œ16Eā€ is broken down into a number of easy to follow steps, and 1 words. Probability and Statistics for Engineering and the Sciences was written by and is associated to the ISBN: 9780321629111. The full step-by-step solution to problem: 16E from chapter: 4 was answered by , our top Statistics solution expert on 05/06/17, 06:21PM. Since the solution to 16E from 4 chapter was answered, more than 240 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Probability and Statistics for Engineering and the Sciences, edition: 9. This full solution covers the following key subjects: . This expansive textbook survival guide covers 18 chapters, and 1582 solutions.

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