Probability and Statistics in Engineering I - Final Exam Questions | IE 23000, Exams of Probability and Statistics

Download Probability and Statistics in Engineering I - Final Exam Questions | IE 23000 and more Exams Probability and Statistics in PDF only on Docsity! IE 230 Seat # ________ Name ___ < KEY > ___ Closed book and notes. 120 minutes. Cover page, five pages of exam. No calculator. No need to simplify answers. (2 points) I agree to complete a course evaluation. ____________________________________________________________ ...sign here... Score ___________________________ Final Exam, Fall 2008 (Dec 16) Schmeiser IE 230 — Probability & Statistics in Engineering I Name ___ < KEY > ___ Closed book and notes. 120 minutes. Throughout this exam, consider an experiment that chooses a random sample X 1, X 2, . . . , Xn from a population. Suppose that the observations are independent and identically distributed with cdf F , mean E(X ) = µ, variance V(X ) = σ2 and median x 0.5 = F −1(0.5). Let Xd denote the sample mean, S 2 = [Σi =1 n (Xi − Xd) 2] / (n − 1) denote the sample variance, and X (i ) the i th order statistic. 1. True or false. (2 points each) (a) T ← F Every statistic is a random variable. (b) T F ← Every random variable is a statistic. (c) T F ← When comparing two point estimators, the point estimator with the larger mean squared error is better. (d) T F ← In statistical inference, data from a population are used to to infer a conclusion about the random sample. (e) T ← F Maximum-likelihood estimation is an example of statistical inference. (f) T F ← The sample standard deviation is a linear combination of the sample observations. (g) T F ← The standard error of a sample mean decreases with one over the sample size n . (h) T ← F The sample median of a sample of size n = 11 is x (6). (i) T F ← A stem-and-leave plot is a type of empirical cdf. (j) T F ← Bayes’s rule assumes independence between events A and B . 2. (3 points each) Fill in the blanks with course jargon (a word or phrase). (Here Θ̂ is a point estimator of θ.) (a) A procedure that leads to a single outcome in the sample space is called a(n) ___ < experiment > ___. (b) For an observed sample x 1, x 2, . . . , xn , the pdf or pmf f X 1, X 2, . . . , Xn (x 1, x 2, . . . , xn ; θ) is called the ___ < likelihood > ___ of the sample for the given value of θ. (c) The expected squared difference between a point estimator and the unknown population characteristic is called the ___ < mean squared error > ___. (d) The expected product of the differences of two random variables and their means is called their ___ < covariance > ___. Final Exam, Fall 2008 (Dec 16) Page 1 of 5 Schmeiser IE 230 — Probability & Statistics in Engineering I Name ___ < KEY > ___ 7. The uniform family of distributions has pdf f X (x ) = 1 / (b − a ) for a ≤ x ≤ b and zero elsewhere. The associated mean is (a + b ) / 2 and variance is (b − a )2 / 12. Suppose that a = 0 is known. Three data points, 4.3, 5.9, and 10.3, are obtained by independent sampling from the population defined by f X , where b has an unknown value. (a) (6 points) For b = 20, write the likelihood function L (4.3,5.9,10.3;b ) = f X 1,X 2,X 3(4.3,5.9,10.3). ____________________________________________________________ f X 1,X 2,X 3(4.3,5.9,10.3) = f X (4.3) f X (5.9) f X (10.3) = [1 / (b − a )] 3 = (1 / 20)3 ← ____________________________________________________________ (b) (6 points) Estimate the value of b using the method of moments. (Begin by setting the sample mean equal to the population mean.) ____________________________________________________________ The sample mean is xd = (4.3 + 5.9 + 10.3) / 3 = 20.5 / 3 Setting the sample mean to the population mean yields xd = 1 / (0 + b̂ ) / 2. Solving for b̂ yields b̂ = 2xd ← ____________________________________________________________ Final Exam, Fall 2008 (Dec 16) Page 4 of 5 Schmeiser IE 230 — Probability & Statistics in Engineering I Name ___ < KEY > ___ 8. (From Montgomery and Runger, fourth edition, 7– 53)A manufacturer of semiconductor devices takes a random sample of 100 chips and tests them, classifying each chip as defective or nondefective. Let Xi = 0 if the i th chip is nondefective and Xi = 1 if the chip is defective. Let Xd = (X 1 + X 2 + . . . + X 100) / 100. Let p denote the true unknown fraction of defective chips in the entire population. (a) (5 points) In words, describe the meaning of Xd in the context of this problem. ____________________________________________________________ The sample mean is the fraction of the sample (of 100 chips) that is defective. ← ____________________________________________________________ (b) (5 points) Suggest a point estimator for p . ____________________________________________________________ p̂ = Xd ← Comment: This is the method of moments, setting the population mean equal to the sample mean. ____________________________________________________________ (c) (5 points) Determine the expected value of X 1. ____________________________________________________________ E(X 1) = (0)(1 − p ) + (1)(p ) = p ← ____________________________________________________________ (d) (2 points) T F ← The sampling distribution of Xd is approximately exponential. Final Exam, Fall 2008 (Dec 16) Page 5 of 5 Schmeiser