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A builder requires his electrical subcontractor to finish the electrical wiring in a renovated office building within 28 days. The subcontractor feels he is most likely to finish in about 20 days, but because of unforeseen events, he guesses it might take him about 5 days more or less than this. More precisely, suppose the required time has a normal distribution with 20 days and a standard deviation of 5 days. (a) To provide an incentive to finish on time, the builder specifies a $1000 penalty if the electrical work takes longer than 28 days. What is the chance of this? Solution: Let X be the number of days required to finish the job. Given that X~N(u = 20, o = 5). P(X > 28) = P ("> 2-20) = P(Z > 1.6) = 0.0548

A builder requires his electrical subcontractor to finish the electrical wiring in a renovated office building within 28 days. The subcontractor feels he is most likely to finish in about 20 days, but because of unforeseen events, he guesses it might take him about 5 days more or less than this. More precisely, suppose the required time has a normal distribution with 20 days and a standard deviation of 5 days. (a) To provide an incentive to finish on time, the builder specifies a $1000 penalty if the electrical work takes longer than 28 days. What is the chance of this? Solution: Let X be the number of days required to finish the job. Given that X~N(u = 20, o = 5). P(X > 28) = P ("> 2-20) = P(Z > 1.6) = 0.0548

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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I want to ask how do we know when we need to use the continuity correction like the second image and when will we use the normal distribution like the 1st image? Thank you

A builder requires his electrical subcontractor to finish the electrical wiring in a renovated office building within 28 days. The subcontractor feels he is most likely to finish in about 20 days, but because of unforeseen events, he guesses it might take him about 5 days more or less than this. More precisely, suppose the required time has a normal distribution with 20 days and a standard deviation of 5 days. (а) To provide an incentive to finish on time, the builder specifies a $1000 penalty if the electrical work takes longer than 28 days. What is the chance of this? Solution: Let X be the number of days required to finish the job. Given that X~N(u = 20, o = 5). P(X > 28) = P ("> -20) = P(Z > 1.6) = 0.0548 ||
Transcribed Image Text:A builder requires his electrical subcontractor to finish the electrical wiring in a renovated office building within 28 days. The subcontractor feels he is most likely to finish in about 20 days, but because of unforeseen events, he guesses it might take him about 5 days more or less than this. More precisely, suppose the required time has a normal distribution with 20 days and a standard deviation of 5 days. (а) To provide an incentive to finish on time, the builder specifies a $1000 penalty if the electrical work takes longer than 28 days. What is the chance of this? Solution: Let X be the number of days required to finish the job. Given that X~N(u = 20, o = 5). P(X > 28) = P ("> -20) = P(Z > 1.6) = 0.0548 ||
25+Test... X 2018-11-13+Test2... X Lecture+9+Notes... X Lecture+10+Notes... X Lecture+11+Notes... X Lecture+7+Notes... X Lecture+8+Notes... 2019-06-19+T... U. T A factory which produces chips in lots of ten thousand uses the following scheme to check the quality of its product. From each lot of chips produced, a random sample of size 500 chips is taken. If the sample contains 10 or less defectives, the lot is passed. If the sample contains more than 10 defectives, another random sample of size 500 is chosen from the lot. If this sample contains 10 or less defectives, the lot is passed. Otherwise the lot is rejected. If a lot actually contains 5% defectives, find the probability that it will pass. Solution: Given that the percent of defectives in the lot is p=0.05. In a random sample of 500, the probability of getting 10 or less defectives is 500 500 500 (0.0s)" (0.95)*90 + (0.0s) (0.95)*91 o.os) (0.95)00 ...+ 10 9 which can be approximated using the normal distribution. H = np = 500(0.05)= 25, o² = npq = 500(0.05)(0.95) = 23.75 . X - u 10.5 – 25 Hence P(X <10.5)= P = P(Z <-2.975)= 0.0014 V23.75 The lot will pass if a random sample of 500 contains 10 or less defectives, or (1) (2) the sample in (1) contains more than 10 defectives, and another random sample of 500 contains 10 or less defectives P(lot will pass)= 0.0014 + (1-0.0014)(0.0014) = 0.002798 1 Previous Next Dashboard Calendar Тo Do Notifications Inbox 000
Transcribed Image Text:25+Test... X 2018-11-13+Test2... X Lecture+9+Notes... X Lecture+10+Notes... X Lecture+11+Notes... X Lecture+7+Notes... X Lecture+8+Notes... 2019-06-19+T... U. T A factory which produces chips in lots of ten thousand uses the following scheme to check the quality of its product. From each lot of chips produced, a random sample of size 500 chips is taken. If the sample contains 10 or less defectives, the lot is passed. If the sample contains more than 10 defectives, another random sample of size 500 is chosen from the lot. If this sample contains 10 or less defectives, the lot is passed. Otherwise the lot is rejected. If a lot actually contains 5% defectives, find the probability that it will pass. Solution: Given that the percent of defectives in the lot is p=0.05. In a random sample of 500, the probability of getting 10 or less defectives is 500 500 500 (0.0s)" (0.95)*90 + (0.0s) (0.95)*91 o.os) (0.95)00 ...+ 10 9 which can be approximated using the normal distribution. H = np = 500(0.05)= 25, o² = npq = 500(0.05)(0.95) = 23.75 . X - u 10.5 – 25 Hence P(X <10.5)= P = P(Z <-2.975)= 0.0014 V23.75 The lot will pass if a random sample of 500 contains 10 or less defectives, or (1) (2) the sample in (1) contains more than 10 defectives, and another random sample of 500 contains 10 or less defectives P(lot will pass)= 0.0014 + (1-0.0014)(0.0014) = 0.002798 1 Previous Next Dashboard Calendar Тo Do Notifications Inbox 000
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