Princeton University Department of Operations Research and Financial Engineering ORF 45 Fundamentals of Engineering Statistics Midterm Exam March 06, 009 0:00am-0:50am PLEASE DO NOT TURN THIS PAGE AND START THE EAM UNTIL YOU ARE TOLD TO DO SO. Instructions: This exam is open book and open notes. Calculators are allowed, but not computers or the use of statistical software packages. Write all your work in the space provided after each question. Use the other side of the page, if necessary. Explain as thoroughly and as clearly as possible all your steps in answering each question. Full or partial credit can only be granted if intermediate steps are clearly indicated. Name: Pledge: I pledge my honor that I have not violated the honor code during this examination. Pledge: Signature: : (0) 3: (0) : (30) 4: (30) Total: (00)
. (0 pts.) A company divides its production of 4-foot by 8-food woodgrained panels into three different quality groups (I, II, and III). Samples of ten sets of 00 panels each were randomly taken from each group and the number of imperfections in each set was counted. The box-plot displays of the imperfection counts in each group sample are given in the following figure (the numbers printed by the box-plots are the corresponding readings in the vertical axis they were added for convenience only). a) What was the median number of imperfections observed for each one of the group samples? I: ; II: 4; III: 0. b) What was the inter quartile range (IQR) for each group? I: ; II: 5; III: 7. c) What were the minimum and maximum observed values in each group? Are there any outliers? If so, what are their values? I: 0, 3; II: 0, 9; III: 4,8; there are NO outliers. d) What can you conclude about the quality of each group, as measured by the number of imperfections found? Comment in terms of both measures: centeredness and variability of the data. Clearly the # of imperfections increases from group I to III, as observed by the increasing median. But the data spread (its variability) also increases from I to III, as shown by the progression of the IQR s.
. (30 pts.) Consider the following probability questions. (Note that correct answers expressed in terms of factorials or combinations are acceptable.) a) Four engineers, A, B, C, and D, have been scheduled for job interviews at 0 am on Friday, Jan 3, at Random Sampling, Inc. The personnel manager has scheduled the four for interview rooms,, 3, and 4, respectively. However, the manager s secretary does not know this, and as a result assigns them to the four rooms in a completely random fashion (what else!). What is the probability that none of the four ends up in the correct room? # of assignments none is in correct room P (none in correct room) = total # of possible room assignments Enumerating the # of assignments in which none of the engineers is in the correct room, we get: BADC, BCDA, BDAC, CADB, CDAB, CDBA, DABC, DCAB, DCBA 9 9 Thus: P (none in correct room) = = = 0.375 4! 4 b) A bowl contains 0 chips, of which 9 are red, 8 are white, and 3 are blue. Six chips are taken at random and without replacement. What is the probability that there is at least white and at least blue chip among the 6 selected? { } { } Let A= at least white and B= at least blue, we want P( A B). C C C ( ) ( ) C C C C ( ) ( ) ( ) PA ( B) = P( A B) = P A B = P A + P B P A B C 0 94 But: P( A ) = P( no white) = = = 0.0384, 6 6 38760 C 7 0 376 P( B ) = P( no blue) = = = 0.393, 6 6 38760 C C 9 0 84 and P( A B ) = P( no white and no blue) = = = 0.0067. 6 6 38760 0 7 0 9 0 Thus: PA ( B) = + = 6 6 6 6 6 6 = 0.0384 0.393 + 0.0067 = 0.6590.
c) A lot of 00 fuses is accepted if at least 9 of the 0 fuses taken at random blow at the correct amperage. If there are exactly 0 defectives among the 00 fuses (that is, 0 that will blow at the wrong amperage), what is the probability of accepting the lot? Let = {# of defectives in 0 randomly chosen fuses} 80 80 0 0 9 Then P( accept lot) = P( = 0) + P( = ) = + = 0.363. 00 00 0 0 80 79 7 7 80 79 7 0 0 Alternate: P( accept lot ) =... +... 00 99 9 9 00 99 9 9 = 0.0956 + 0.67933 = 0.363049.
3. (0 pts.) Box A contains three cubes and two balls, and box A contains two cubes and five balls. A fair die is cast. If the outcome is five or six, one object is taken from A ; otherwise an object is taken from A. a) Compute the probability that a ball (say, B ) is taken. Let A = one object is taken from box A, i=, and i { i} { } { object taken is a ball }. B = We know that PA ( ) = 3 and PA ( ) = 3. We want PB ( ). But PB ( ) = PB ( A) PA ( ) + PB ( A) PA ( ) 5 64 = + = = 0.6095. 5 3 7 3 05 b) Given that a ball is observed, compute the conditional (posterior) probability that it came from A. P( B A) P( A ) 7 P( A B) = = 5 3 = = 0.88. P( B) 64 3 05 c) Compare P( A B ) determined in part (b) with the prior probability PA ( ). Does the result agree with your intuition? Yes. Box A has significantly fewer balls () than box A (5), thus knowing that the picked object is a ball reduces the likelihood that the ball came from A. Namely, the posterior probability of picking an object from A, after knowing that the object is a ball, is smaller than the prior probability of picking an object from A.
4. (30 pts.) A certain market has both an express and a superexpress checkout line. Let denote the number of customers in line at the express checkout at a particular time of the day, and let denote the number of customers in line at the superexpress checkout at the same time. Suppose the joint p.m.f. of and is as given in the accompanying table. x x 0 3 0.08.07.04.00.06.5.05.04.05.04.0.06 3.00.03.04.07 4.00.0.05.06 a) What is P( = ), that is, the probability that the number of customers in the two lines are identical? P( = ) = p(0,0) + p(,) + p(,) + p(3,3) =.08 +.5 +.0 +.07 =.40. b) Let A denote the event that there are at least two more customers in one line than in the other line. Express A in terms of and and calculate the probability of this event. A= {( x, x): x + x} {( x, x): x + x} P( A) = p(,0) + p(3,0) + p(4,0) + p(3,) + p(4,) + p(4,) + p(0,) + p(0,3) + p(,3) =., c) Determine the marginal p.m.f. of. P (0) = P( = 0) = p(0,0) + p(0,) + p(0,) + p(0,3) =.9 P () = P( = ) = p(, 0) + p(,) + p(, ) + p(,3) =.30, etc. x 0 3 4 p ( x ).9.30.5.4.
d) Calculate the expected number of customers in line at the express checkout. E = 0.9 +.30 +.5 + 3.4 + 4. =.7 e) Find the variance of the number of customers in line at the express checkout. ( ) = var( ) E E E = 0.9 +.30 +.5 + 3.4 + 4. = 4.48 var( ) = 4.48.7 =.59 f) By inspection of the probabilities P ( = 4), P ( = 0), and P ( = 4, = 0), are and independent random variables? Explain. p(4,0) = 0, yet p (4) =. > 0 and p (0) =.9 > 0, so px (, x) p ( x) p ( x) for every ( x, x), and the two variables are not independent.