Stat Spring 2012 Exam 1. Your Name:

Transcription

1 Stat Spring 2012 Exam 1 Your Name: Your Section (circle one): Grant (7:30) Mike C. (7:30) Mike C. (8:30) Jeremy (9:30) Jeremy (10:30) Yen-Ning (11:30) Yen-Ning (12:30) Chris (1:30) Juan (2:30) Juan (3:30) Chris (4:30) Mike L. (4:30) Instructions: Show your work on ALL questions. Unsupported work will NOT receive full credit. Decimal answers should be exact, or to exactly 4 decimal places. (Examples: if it is.25 use.25, if it is say then use.5789.) You are responsible for upholding the Honor Code of Purdue University. This includes protecting your work from other students. Please write legibly. If a grader cannot read your writing, NO credit will be given. You are allowed the following aids: a one-page 8.5" x 11" handwritten (in your handwriting) cheat sheet, a scientific calculator, and pencils or pens. Instructors will not interpret questions for you. If you do have questions, wait until you have looked over the whole exam so that you can ask all of your questions at one time. You must show your student ID (upon request), turn in your cheat sheet and sign the class roster when you turn in your exam to your instructor. Turn off your cell phone before the exam begins. Question Points Possible Points Earned Cheat Sheet 1 Total 100

2 1. Tyler polls 169 Indiana men in order to find a friend to play sports with. 88 men like basketball, 69 like baseball, and 51 like golf. However, 26 of them do not like any of the 3 sports. 33 like baseball and basketball, 29 like basketball and golf, and 15 like baseball and golf. (3 points each) a) Find the number of Indiana men polled that like all 3 sports. Inclusion-Exclusion Principle : 143 = x; so x = 12. After completing this part of the problem, drawing a Venn Diagram could be very useful. b) Find the probability that a man polled likes baseball knowing that they like basketball. P(baseball basketball) = N(baseball and basketball) / N(basketball) = 33 / 88 =.375. c) Find the probability that a man polled likes only golf. P(only golf) = 19 /169 = d) Find the probability that a man polled likes at most one of the above sports. P(at most one) = P(none) + P(only golf) + P(only baseball) + P(only basketball) = ( ) / 169 = e) Find the probability that a man polled does not like basketball nor does he like baseball. P(not basketball and not baseball) = P(none) + P(only golf) = (26+19)/169 =.2663

3 2. Corey is deciding whether or not to go out on a date on Friday night. He will go out with a probability of.92. If he goes out, he will be happy with a probability of.65. If he does not go out, he will only be happy with a probability of.13. a) Find the probability that Cory is happy on Friday night. (4 points) P(happy) = P(date)*P(happy date) + P(stay in)*p(happy stay in) =.92* *.13 = b) Knowing that Cory is happy, what is the probability that he went out on a date? (4 points) P(date happy) = P(date)*P(happy date) / P(happy) =.92*.65 /.6084 = c) What is the probability that Cory neither went out on a date nor is happy? (3 points) There was a difference in opinion on the English interpretation. So both of the following were accepted: P(neither date nor happy) = P(not happy and not on a date) = P(stay in)*p(not happy stay in) =.08*.87 =.0696 P(neither date nor happy) = P(not happy or not on a date) = P(not happy) + P(not on a date) - P(not happy and not on a date) = =.402 d) Are the events going out on a date and being happy independent for Cory? Prove your answer two different ways. (4 points) No, they are dependent. Way 1: P(date happy) =.9829 which is not =.92 = P(date). Also, P(date and happy) =.92*.65 =.598 which is not = 5597 =.92*.6084 = P(date)*P(happy).\

4 3. Zach has a class of 40 students, of which 13 are A or B students. He is picking a student, with replacement, to give a presentation. Zach will stop picking the first time he gets an A or B student. Let T represent the total number of trials (picks). a) Name the distribution and the parameter(s) of T. (2 points) T ~ Geo(p=13/40 =.325) b) Find the probability that T is 4. (2 points) P(T=4) =.325*.675^3 =.1000 c) Find the probability that T is at least 5. (3 points) P(T >= 5) = P(T > 4) =.675^4 = d) Find the probability that T is less than 12, if it is known to be more than 5. (4 points) P(T < 12 T > 5) = P(T < 7) by memoryless property = P(T <=6) = 1- P(T > 6) = ^6 =.9054 e) Suppose you repeat the set-up in this problem 5 times. (Zach can pick the same person 5 times although it is unlikely). Name the distribution and parameter(s) for R, the total number of trials to pick 5 A or B students. Then find the probability that R is 17. (5 points) R ~ NB(r=5, p =.325). P(R = 17) = "16 choose 4" *.325^5 *.675^12 =.0590.

5 4. Elby has a bag of Starburst. It has 3 red, 3 yellow, 2 orange, and 1 pink candies left. a) If she was to put them in a line, what is the probability that both orange are somewhere before the pink candy? (Hint: They do not have to be next to each other.) (2 points) Think of only the 2 orange and the 1 pink candy. Considering that the orange are "identical", then there are only 3 possible arrangements of these candies. They are 00P, OPO, and POO. So, are probability is 1/3 =.3333 b) What is the probability that neither the 2 orange are together nor the 3 red are together. (4 points) Let O be the event the oranges are together and let R be the event the reds are together. We want P(Ocomplement and R-complement). Using De-Morgan's law this is the same as P( (O union R)- complement) or 1 - P(O union R). P(O union R) = P(O) + P(R) - P(O and R) = = So our answer is Each of the above can be found using the MC. The bottom, or N(omega) would be ,3, 2, While the 3 numerators would be: 1120, 420 and 120 3,3,1,1 1,3, 2,1 1,3,1,1 respectively. P(O-complement or R-complement) = P((O and R)-complement) = 1 - P(O and R) = 1-120/5040 = c) Suppose she numbers them 1 through 9. If Elby randomly selects a proper subset of numbers (she can pick as few as 1 or as many as 8) how many different subsets are possible? (2 points) 2^9-2 = 510. Think of the BCR. You have 9 different actions to be taken, i.e. what you are going to do with each #. You have 2 choices per action, which are include or not include. Therefore, you have 2^9 total possibilities. However, since you cannot have a subset of size 0 or 9, we must discount by 2 options. d) Using part c, what is the probability that her chosen subset is just 1 number, i.e. 1 piece of candy (ooh, piece of candy!)? (2 points) 9 / 510 = There are 9 different candies, so there are 9 different subsets of size 1.

6 5. Chris is collecting the quarters featuring different U.S. states on the back. Suppose now he has a jar with 40 quarters, 5 of which are Wisconsin quarters, 6 are Indiana quarters. One day he randomly picks 4 quarters without replacement from the jar. Let X be the number of Wisconsin quarters among the 4 selected ones. a) Name the distribution and the parameter(s) for X. (2 points) X ~ Hyp(N=40, n=4, p=5/40 =.125) b) What is the probability that he picks at least two Wisconsin quarters? (3 points) P(X >=2) = P(X=2, 3, or 4) = 1 - P(X is 0 or 1) = 1 - P(X=0) - P(X=1) = = c) What's the probability he picks 6 Wisconsin quarters? (2 points) P(X = 6) = 0 since X cannot be more than 4 (since he is picking 4 coins). d) What is the probability that he picks 2 Wisconsin quarters and 2 Indiana quarters? (3 points) P(X=2 and I = 2) = "5 choose 2" * "6 choose 2" / "40 choose 4" = e) Six months later, he has 120 quarters in the jar and 12 of them are Wisconsin quarters. Again he randomly picks 4 quarters without replacement. Is an approximation appropriate? Why or why not? Find the probability that at least one of the coins selected is a Wisconsin quarter, but if the approximation is appropriate, use the approximate probability instead. (5 points) Y~ Hyp(N=120, n=4, p = 12/120 =.1) Yes an approximation is appropriate since N = 120 > 80 = 20*4 = 20*n. Approximation Y* ~ Bin(n=4, p =.1). Want P(Y* >= 1) = 1 - P(Y* = 0) = 1 -.9^4 =.3439.

7 6. Assume a page in the book will be considered "defective" if there are more than 3 typos. On average, in one page there is 1 typo. The typos are independent from page to page. Let X denote the number of typos in one page. a) State the distribution of X and its parameter(s). (2 points) X ~ Poi(lambda = 1) b) What is the probability that a randomly picked page is defective? (3 points) P(defective) = P(X >3) = 1 - P(X is 0, 1, 2, or 3) = 1 - ( ) = = c) What is the variability of the number of typos in a 300-page book? (2 points) T ~ Poi(lambda = 300). Var T = lambda = 300. d) Now assume we have a 500-page book, and let Y denote the number of defective pages. State the distribution of Y. Also, state whether the approximated distribution exists, why or why not? (4 points) Y ~ bin(n=500, p = answer from part b =.0190.) An approximation is not appropriate since p >.01 and we do not know the Normal distribution yet. e) How many defective pages can we expect from a 500-page book? (2 points) EY = np = 500 *.0190 = 9.5 ( if you do not round p first).

8 7. The number of bad checks received per day by a store and the respective probabilities are shown below. x P(X=x) a) What is the probability that this store will receive more than 3 bad checks in a day? (2 points) P(X > 3) = P(X is 4 or 5) = =.25 b) Find the expected number of bad checks per day and it's variance. (6 points) EX = 0* *.1 + 2*.2 + 3*.4 + 4* *.1 = EX [ ] = 0* *.1 + 4*.2 + 9* * *.1 = 9.4 Var(X) = *2.8 = c) Define another random variable, Z, as Z=2X-3, find the variance of Z. (2 points) Var(Z) = 4*Var(X) = 4*1.56 = 6.24 d) Suppose this is your father's store and you will be in charge tomorrow. Your father told you that if there is less than 2 bad checks tomorrow, he will give you a $150 bonus. Whereas, if there is at least 2 bad checks, you should pay him a $30 penalty. Should you agree with this deal? Why or why not? (5 points) New variable: Z = payout. Possible values of Z are +150 and -30. Probabilities for Z are.15 and.85 respectively since you get the $150 if X < 2 and P(X < 2) = =.15. E[Z] =.15 * *(-30) = -3. Since you expect to lose money on this deal, you would not agree to it.