Statistics Midterm 1623 2 Answer

Statistics / Midterm Exam (Duration: 90 minutes)

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Q1. A multiple-choice test has nine questions. For each question, there are four possible answers from which to select. One point is awarded for each correct answer, and points are not subtracted for incorrect answers. The instructor awards a bonus point if the student spells his or her name correctly. A student who has not studied for this test decides to choose at random an answer for each question. (a) Find the expected number of correct answers for the student on these nine questions. (b) (b) Find ind the the stan stand dard ard devi deviat atio ion n of th the numb number er of of corr correc ectt answ answer erss for for the the stud studen entt on these nine questions. (c) The student spells his name correctly. (i) Find the expected total score on the test for this student, (ii) Find the standard deviation of his total score on the test. Answer: a) 2.25 b) 1.30 c) 3.25, 1.30 Solution: X=B(9,0.25) a) E(X) = np = 9(0.25) = 2.25  b) STD(X) = SQRT(np(1-p)) = SQRT(9(0.25)(0.75)) = 1.30 c) Y = X +1 E(Y) = E(X) + 1 = 3.25, STD(Y) = STD(X) = 1.30

Q2. The World Series of baseball is to be played by team A and team B. The first team to win four games wins the series. Suppose that team A is the better team, in the sense that the  probability is .6 that team A will win any specific game. Assume also that the th e result of any game is independent of that of any other. (a) (a) What What is the the prob probab abil ilit ity y tha thatt tea team m A wil willl win win the the seri series es?? (b) What is the the pro probabil bility tha thatt a seventh gam game will will be be nee needed to det determine th the winner? (c) (c) Supp Suppos osee tha that, t, in fact fact,, eeac ach h tea team m wins wins two two of the the firs firstt fou fourr gam games es.. i. What hat is is the the prob probab abil ilit ity y tha thatt te team A wil willl win win the the ser serie ies? s? ii. ii. What hat is the the prob probab abil ilit ity y that that a seven eventh th gam game will will be need needed to dete determ rmin inee the winner? Answer: a) 0.710 .71021 21 b) 0.27 0.2764 648 8 c) 0.648 .648,, 0.4 0.48 8 Solution: Fact: A best-of-seven playoff , also known by the name seven-game series , puts two teams against each other other for as many games games (or sets) sets) as needed needed for one team to win four games (or sets). sets). (source: (source: wikipedia) a) Probability = P(A win the series at 4 th game) + ... + P(A win the series at 7 th game) Write P(k) = P(A win the series at k th game) for short.

P(4) = (0.6) 4 = 0.1296, P(5) = (0.6) 4(0.4)Choose(4,3) = 0.20736, explain: win 3 other games from the first 4 games P(6) = (0.6) 4(0.4)2Choose(5,3) = 0.20736 P(7) = (0.6)4(0.4) 3Choose(6,3) = 0.165888 Thus, Probability = P(4)+...+P(7) = 0.71021  b) Probability = P(A will win the series at 7 th game) + P(B will win the series at 7 th game) = (0.6)4.(0.4)3.Choose(6,3) + (0.4)4.(0.6)3.Choose(6,3) = 0.165888 + 0.110592 = 0.27648

c) (i) There are the following cases: AA* (no 7 th game), ABA, BAA. Thus, the probability is: (0.6)2 + 2(0.6)2(0.4) = 0.648

(ii) There are 4 cases for the last 3 games: ABA, BAA, BAB, ABB. Thus, the probability is: 2(0.6)2(0.4) + 2(0.4)2(0.6) = 0.48

Q3. A hamburger stand sells burgers for $1.45 each. Daily sales have a distribution with mean 530 and standard deviation 69. (a) Find the mean daily total revenues from the sale of hamburgers. (b) Find the standard deviation of total revenues from the sale of hamburgers. (c) Daily costs (in dollars) are given by C = 100 + 0.95X where X is the number  of hamburgers sold. Find the mean and standard deviation of daily profits from sales. Answer: a) $768.5 b) $100.05 c) $165, $34.5 Solution: Revenue R = (1.45).X, where daily sales X = N(530;69 2) a) E(R) = (1.45)530 = 768.5 ($)  b) STD(R) = (1.45)69 = 100.05 ($) c) Profit P = R – C = 1.45X – (100+0.95X) = 0.5X – 100 E(P) = (0.5)530 – 100 = 165 ($) STD(P) = (0.5)69 = 34.5 ($) Q4. A pizza delivery service delivers to a campus dormitory. Delivery times follow a normal distribution with mean 20 minutes and standard deviation 4 minutes. (a) What is the probability that a delivery will take between 15 and 25 minutes? (b) The service does not charge for the pizza if deliver takes more than 30 minutes. What is the  probability of getting a free pizza from a single order? (c) During final exams week, a student plans to order pizza five consecutive evenings. Assume that these delivery times are independent of each other. What is the probability that the student will get at least one free pizza? (d) Find the shortest range of times that includes 40% of all deliveries from this service. (e) For a single delivery, state in which of the following ranges (expressed in minutes) delivery time is most likely to lie: 18-20 19-21 20-22 21-23 (f)For a single delivery, state in which of the following ranges (expressed in minutes) delivery time is least likely to lie: 18-20 19-21 20-22 21-23 Answer: a) 0.7888 b) 0.0062 c) 0.0306 d) 17.9 to 22.1 minutes e) 19-21 f) 2123 Solution: Delivery time X = N(20,4 2) a) P(15≤X≤25) = P(X≤25) – P(X≤15) = 0.8944 - 0.1056 = 0.7888  b) P(X>30) = 0.0062 c) Probability = 1 – (1-0.0062)5 = 0.0306 d) The range should be of the form [20-x,20+x].  Need to find x such that P(20-x≤X≤20+x) = 0.4 It is equivalent to P(-x/4 ≤ Z ≤ x/4) = 0.4 P(0≤Z≤x/4) = 0.2 P(Z≤x/4) = 0.7 From the table of standard normal distribution, we find x/4 = 0.524.

Thus, the interval is [17.9,22.1]. e) These ranges have same length. Thus the range time is most likely to lie should contain the mean 20, that is 19-21. f) Thus the range time is least likely to lie should farthest from the mean, that is 21-23.

Q5. It is estimated that amounts of money spent on gasoline by customers at a gas station follow a normal distribution with standard deviation $2.50. It was also found that 10% of all customers spend more than $15. What percentage of customers spend less than $10? Answer: 23.58% Solution: X=N(µ,2.52) We know P(X>15) = 0.1, and want to find P(X(15-µ)/2.5) = 0.1 P(Z≤(15-µ)/2.5) = 0.9 Hence (15-µ)/2.5 = 1.28, so µ = 11.8. Thus, X=N(11.8,2.52). P(X