Closed book and notes. 120 minutes. Cover page, five pages of exam. No calculators.

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1 IE 230 Seat # Closed book and notes. 120 minutes. Cover page, five pages of exam. No calculators. Score Final Exam, Spring 2005 (May 2) Schmeiser

2 Closed book and notes. 120 minutes. Consider an experiment that chooses a random sample X 1, X 2,...,X n from a population. Suppose that the observations are independent and identically distributed with cdf F, mean E(X ) =µ, variance V(X ) =σ 2 and 90th percentile F 1 (0.9) = x 0.9. Let X denote the sample mean and S denote the sample standard deviation. 1. True or false. (2 points each) (a) T F E(X ) =µ. (b) T F E(S 2 ) =σ 2. (c) T F X =µ. (d) T F V(X ) =σ 2 /n. (e) T F ste(s 2 ) = std(s 2 ). (f) T F F (x 0.9 ) = 0.9. (g) T F S =σ. (h) T F For every sample size n, X has a normal distribution. (i) T F The mean squared error of S as a point estimator of σ is MSE(S, σ) = E[(S σ) 2 ]. (j) T F The empirical cdf is obtained by creating a scatter plot of the observed i th order statistic, x (i ), with n/(i + 1) for i = 1, 2,..., n. 2. (2 points each) Fill in the blanks with the name of the appropriate distribution family name. (a) The number of Bernoulli trials until one success: < geometric >. (b) The number of successes in n Bernoulli trials: < binomial >. (c) The number of successes in a sample of size n from a population of size N containing K successes: < hypergeometric >. (d) The time between occurrences from a Poisson process with rate λ: < exponential >. (e) The number of occurrences from a Poisson process with rate λ and interval length t : < Poisson >. Final Exam, Spring 2005 (May 2) Page 1 of 5 Schmeiser

3 3. (Montgomery and Runger, third edition, Problem 7 19). Consider the Poisson pmf f (x ; µ) = e µ µ x /x! for x = 0, 1, 2,..., which has mean and variance µ. The value of µ is unknown. We have a random sample of three observations: 7, 0, 3. (a) (4 points) What is the value of f (0.5; µ)? f (0.5; µ) = P(X = 0.5) = 0 (b) (4 points) Suggest a point estimate of µ. Both the method of moments and the maximum likelihood estimates of the Poisson mean are µˆ = x = ( ) / 3 = 10 / 3 (Other arguments could be made, for example based on estimating the variance.) (c) (4 points) For µ=6, determine the value of the likelihood function n L (µ) =Π i =1 f (x i ;µ). n 3 L (µ) =Π i =1 f (x i ;µ) =Π i =1 e µ µ x i / (x i )! = e 6 [((6 7 ) / 7!) ((6 0 ) / 0!) ((6 3 ) / 3!)]. (d) (5 points) Compute the observed value of the sample variance. n Σi =1 (x i x )2 s 2 = = (7 10 / 3) 2 + (0 10 / 3) 2 + (3 10 / 3) 2 n Final Exam, Spring 2005 (May 2) Page 2 of 5 Schmeiser

4 4. (Montgomery and Runger, third edition, Problem 3 23). The distributor of a machine for cytogenics has developed a new model. The company estimates that when it is introduced into the market, it will be "very successful" with probability 0.6, "moderately successful with probability 0.3, and "not successful" otherwise. The estimated yearly profit associated with "very successful" is $15 million, with "moderately successful" a $5 million profit, and with "not successful" a $500,000 loss. Let X denote the year profit of the new model. (a) (4 points) Write the probability mass function. ( 500,000) = 0.1 (5,00,000) = 0.3 (15,00,000) = 0.6 (x ) = 0.0 elsewhere. (b) (4 points) Determine the value of E(X ). E(X ) = ( 500,000)(0.1) + (5,000,000)(0.3) + (15,000,000)(0.6) dollars (c) (4 points) Determine the cdf value F X (0). F X (0) = P(X 0) = P(X = 500,000) = 0.1 Final Exam, Spring 2005 (May 2) Page 3 of 5 Schmeiser

5 5. (Montgomery and Runger, third edition, Problem 7 37). The compressive strength of concrete is normally distributed with mean µ=2500 psi and standard deviation σ=50 psi. A random sample of n = 9 specimens is collected. (a) (4 points) Determine the value of the mean of the sample mean. E(X ) = E(X ) = 2500 psi (b) (4 points) Determine the value of the standard deviation of the sample mean. 50 V(X ) = V(X ) /n = 2. 9 Therefore, std(x ) = 50 / 3 psi (c) (5 points) Determine the (approximate) value of the probability that the sample mean lies between 2480 and 2520 psi. These two values are 20 / (50 / 3) = 1.2 standard errors from the mean. We memorized that about 68% are within one standard deviation from the mean, so the answer is a bit larger. Maybe 0.77 Final Exam, Spring 2005 (May 2) Page 4 of 5 Schmeiser

6 6. Definitions. (a) (4 points) What is a random variable? A function that assigns a real number to every outcome in the sample space. (b) (4 points) (choose two) A cumulative distribution function of a random variable X is P(X c ) for every real number c when... c =/ x. X is discrete. X is continuous. c = x. (c) (4 points) The distribution of a random variable X is memoryless if... the remaining lifetime always has the same distribution. or P(X >t 1 + t 2 X>t 1 ) = P(X >t 2 ) (d) (4 points) The median of a sample of observations x 1, x 2,...,x 11 is... the sixth order statistic, x (6) or the observation with five others smaller and five others larger. 7. (2 points each) Suppose that X 1 and X 2 are identically distributed with mean µ and standard deviation σ and correlation ρ=0.0. (a) T F E(X 1 X 2 ) = 2µ. (b) T F V(X 1 X 2 ) = 2σ 2. (c) T F std(x 1 X 2 ) = 2σ. Final Exam, Spring 2005 (May 2) Page 5 of 5 Schmeiser