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CME 263H1: Quiz 05 – April 6th, 2015 Time: 10:30 pm to 11:15 am (45 minutes) Location: MP202 Total Marks: 45 Student Number:___________________________ Name:_____________________________________ Marks Distribution: Question Number Q1 Q2 Q3 Total marks of Part a 3 4 3 Total marks of Part b 3 4 Total marks of Part c 3 Total marks of Part d 3 Total marks Total marks of Part a Total marks of Part b Total marks of Part c Total marks of Part d Total marks 12 8 3 Total 23 Marks: Question Number Q1 Q2 Q3 Total Q1) Each front tire on a particular type of vehicle is supposed to be filled to a pressure of 26 psi. Suppose the actual air pressure in each tire is a random variable—X for the right tire and Y for the left tire, with joint pdf a. b. c. d. What is the value of K? What is the probability that both tires are underfilled? Determine the (marginal) distribution of air pressure in the right tire alone. Are X and Y independent random variables? ANS: a. b. c. d. is obtained by substituting y for x in (d); clearly not independent. are Q2) Let represent the times necessary to perform three successive repair tasks at a certain service facility. Suppose they are independent normal random variables with expected values respectively. a. If Calculate b. What is Using the given in part (a), calculate ANS: a. b. !" ! = 2.581 Q3) A study of the ability of individuals to walk in a straight line reported that accompanying data on cadence (strides per seconds) for a sample of n – 20 randomly selected healthy men: .95 .92 .81 .96 .93 .92 .95 1.00 .93 .78 .86 1.06 1.05 1.06 .92 .96 .85 .85 .81 .92 A normal probability plot gives substantial support to the assumption that the population distribution of cadence is approximately normal. A descriptive summary of the data is: Variable Cadence a. Mean 0.9255 Median 0.9300 StDev 0.0809 Calculate a 95% confidence interval for a population mean cadence. ANS: a. N 20 A 95% C.I.: SEMean 0.0181 EQUATIONS: