HW5_SOHAIB_KHURAM_10042023

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Virginia Commonwealth University *

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500

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Industrial Engineering

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Dec 6, 2023

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docx

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BIOS 500 Homework 5 Name: Sohaib Khuram Although you are encouraged to work together in groups, collaboration does not mean copying a classmate’s response to a question. What you turn in must represent your understanding of the concepts learned. 1. Rule of Total Probability and Bayes Rule . Do Exercise 4.162 (p. 195). P(D~T) = .60(.40)/((.6*.4) + (.80*.32) + (.30*.28)) = .4138 or 41.38% probability that if someone favors increased spending to combat terrorism that they are a Democrat 2. Probability distribution . Do Exercise 5.7 (pp. 217-218) a-e. (a) 2 through 8 (b) X = 7 (c) 2/96 = 1/48 or 2.08 percent chance that there would be a crew size of 4 (d) P(X=2) = 4/96 = .0417 i. P(X=3) = 1/96 = .0104 ii. P(X=4) = 2/96 = .0208 iii. P(X=5) = 36/96 = .375 iv. P(X=6) = 18/96 = .1875 v. P(X=7) = 33/96 = .3438 vi. P(X=8) = 2/96 = .0208 (e) 1 0 0.2 0.4 0.6 0.8 1 1.2 1
3. Binomial distribution . Do Exercise 5.63 (p. 238) a-d. (a) N=8, p=.4, calculations made using excel i. Exactly 3: P(X=3) = (8 3) .4 3 (.6) 8-3 = 8!/(3!*(8-3)!)*.4 3 *.6 5 = . 2787 ii. At least 3: P(X>=3) = P(X=3) + P(X=4) + … + P(X=8) = .2787 + . 2322 + .1239 + .0413 + .0079 + .0007 = .6846 iii. At most 3: P(X<=3) =P(X=1) + P(X=2) + P(X=3) = . 0896+.2090+.2787 = .5773 (b) P(2 <= X <=4) = P(X=2) + P(X=3) + P(X=4) = .72 (c) Mean = np = 8*.40 = 3.2; on average 3.2 out of 8 traffic fatalities involve an intoxicated or alcohol-impaired driver or nonoccupant (d) std dev=sqrt(np(1-p)) = sqrt(3.2*(1-.4)) = sqrt(1.92) = 1.386 4. Poisson distribution . Do Exercise 5.88 (p. 247) a-e. (a) P(X = k) = e -lambda * Lambda k /k! i. P(X=1) = .3106 (b) P(X<=2) = P(X=0) + P(X=1) + P(X=2) = .7572 (c) P(X>=2) = 1 – P(X < 2) = 1 - .4932 = .5068 (d) Mean = ux = lambda = 1.7 (e) variance = lambda so std dev = sqrt(lambda) = 1.3038 5. Normal probability plot . Do Exercise 6.120 (p. 283). Plot is not roughly linear so it is not normally distributed 6. Normal probabilities and z -scores . Do 6.60a (p. 268) and 6.68 (p. 269). (a) .9875 - .1894 = .7981 (b) -2.33 7. Normal probabilities and quantiles . Read Exercise 6.32 (p. 261). (a) Draw the normal curve using SAS. Attach the graph to your homework. 2
(b) Do Exercise 6.32 d & e. i. Z = (x – u)/σ = (150-206)/44.7 = -1.25; (250-206)/44.7; area under the standard normal curve is between -1.25 and .98 ii. Z = (220 – 206) / 44.7 = .31 (c) Do Exercise 6.94 b & d (p. 276). Note: The 4th decile is the 40th percentile. i. Z = (220 – 206)/44.7 = .31 and from table the percent is .6217 or 62.17% ii. Z score for 40% is -.25; x = u +z σ = 206 - .25(44.7) = 194.83 (d) At least how high are the top 20% serum cholesterol levels among adult females? 243.62 (e) Below what value are the lowest 20% of serum cholesterol levels among adult females? 168.38 8. The prevalence of prostate cancer in the U.S. in 2013 was 0.9%. Two methods are commonly used to screen for prostate cancer: PSA (a blood test), and digital rectal exam (DRE). Researchers found that the PSA test had a sensitivity of 67%. In other words, two-thirds of all the cases who truly have prostate cancer were detected. One- third of the cases of prostate cancer go undiagnosed. On the other hand, when the PSA test indicated no disease, in almost all cases there was no disease. The specificity was 97%. DRE has a low sensitivity of 50%, and the specificity of 94% is also lower than PSA. 3
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