Math Key Homework 3 (Chapter 4)

Math 3339 - Key Homework 3 (Chapter 4) Name: PeopleSoft ID: Instructions: Homework will NOT be accepted through email or in person. Homework must be submitted through CourseWare BEFORE the deadline. Print out this file and complete the problems. Use blue or black ink or a dark pencil. Write your solutions in the space provided. You must show all work for full credit. Submit this assignment at http://www.casa.uh.edu under Assignments" and choose HW3. Total points: 15. 1. *For each random variable defined here, describe the set of possible values for the variable, and state whether the variable is discrete. a. X = the number of unbroken eggs in a randomly chosen standard egg carton. b. Y = the number of students on a class list for a particular course who are absent on the first day of classes. c. U = the number of times a duffer has to swing at a golf ball before hitting it. d. X = the length of a randomly selected rattlesnake. e. Z = the amount of royalties earned from the sale of a first edition of 10,000 textbooks. f. Y= the ph of a randomly chosen soil sample. g. X = the tension (psi) at which a randomly selected tennis racket has been strung. h. X = the total number of coin tosses required for three individuals to obtain a match (HHH or TTT). a. X = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} discrete. b. Y = {0, 1, 2, 3, } discrete. c. U 0 continuous d. X > 0 continuous e. Z = {$0.01, $0.02, $0.03, } discrete. f. Y 0 continuous g. X 0 continuous h. X = {0, 1, 2, 3, } discrete

2. *Airlines sometimes overbook flights. Suppose that for a plane with 50 seats, 55 passengers have tickets. Define the random variable Y as the number of ticketed passengers who actually show up for the flight. The probability mass function f(y) appears in the accompanying table. Y 45 46 47 48 49 50 51 52 53 54 55 f(y) 0.05 0.10 0.12 0.14 0.25 0.17 0.06 0.05 0.03 0.02 0.01 a. What is the probability that the fight will accommodate all ticketed passengers who will show up? b. What is the probability that not all ticketed passengers who show up can be accommodated? c. If you are the first person on the standby list (which means you will be the first one to get on the plane if there are any seats available after all ticketed passengers have been accommodated), what is the probability that you will be able to take the flight? What is this probability if you are the third person on the list? a. Accommodate all who will show up. Let X = number of people that show up. If X 50, P(X 50) = 0.05+0.1+0.12+0.14 +0.25 +0.17 = 0.83. If all 55 people show up, the probability that the plane can accommodate them is ZERO. b. P(Y 51) = 0.17 is the probability that a ticketed passenger will not have a seat. c. Being the first person on the standby list and gets in that means that only 49 passengers showed up: P(Y 49) = 0.66 Probability the third person that means that 47 P(Y 47) = 0.27

3. Section 4.3, Problem 1. A fair coin is tossed until either a head occurs or 6 tails in a row have occurred. Let X denote the number of tosses. Find the frequency function, mean, and variance of X. This means that X = {1, 2, 3, 4, 5, 6} If X = 1 H; P(X = 1) = P(H) = 0.5 If X = 2 TH: P(X = 2) = P(TH) = 0.5*0.5 = 0.25 If X = 3 TTH; P(X = 3) = P(TTH) = 0.5*0.5*0.5 = 0.125 If X = 4 TTTH; P(X = 4) = P(TTTH) = 0.5*0.5*0.5*0.5 = 0.0625 If X = 5 TTTTH; P(X = 5) = P(TTTTH) = 0.5*0.5*0.5*0.5*0.5 = 0.03125 If X = 6 TTTTTH or TTTTT; P(X = 6) = P(TTTTTH or TTTTTT) = (0.5)^6 + 0.5^6 = 0.03125 The frequency function is the following table X 1 2 3 4 5 6 P(X) = f(x) 0.5 0.25 0.125 0.0625 0.03125 0.03125 E(X) = 1*0.5 + 2*0.25 + 3*0.125 + 4 *0.0625 +5*0.03125 + 6*0.03125 = 1.96875 E(X 2 ) = 1*.5 + 4*.25 + 9*0.125 + 16*.0625 + 25*0.03125 + 36*0.03125 = 5.53125 V(X) = 5.53125 1.96875 2 = 1.655273

4. Section 4.3, Problem 2. Verify Chebyshev s inequality for k = 2 and k = 3 when X is the total number of spots on two rolls of a fair 6-sided die. Given from example 4.7 in textbook E(X) = 7, SD(X) = 2.415 For k = 2; P( X 7 > 2*2.415) ¼ X 7 > 4.83 X 7 < - 4.83 or X 7 > 4.83 X < 2.17 or X > 11.83 P(X < 2.17 or X > 11.83) = P(X = 2 or X = 12) = 1/36+1/36 = 0.0556 which is less than ¼ = 0.25 thus the Chebyshev s inequality is verified for k = 2 For k = 3; P( X 7 > 3*2.415) 1/9 X 7 > 7.245 X 7 < - 7.245 or X 7 > 7.245 X < -0.245 or X > 14.245 P(X < -0.245 or X > 14.245) = 0 since there are no possible values less than -0.245 or greater than 14.245 which is less than 1/9 thus the Chebyshev s inequality is verified for k = 3.

5. *An appliance dealer sells three different models of upright freezers having 13.5, 15.9, and 19.1 cubic feet of storage space, respectively. Let X = the amount of storage space purchased by the next customer to buy a freezer. Suppose that X has pmf (f(x)), X 13.5 15.9 19.1 f(x) 0.2 0.5 0.3 a. Compute E(X), E(X 2 ) and V(X). b. If the price of a freezer having capacity X cubic feet is 25X 8.5, what is the expected price paid by the next customer to buy a freezer? c. What is the variance of the price 25X 8.5 paid by the next customer? d. Suppose that although the rated capacity of a freezer is X, the actual capacity is h(x) = X 0.01X 2. What is the expected actual capacity of the freezer purchased by the next customer? a. E(X) = 13.5*.2 + 15.9*.5 +19.1*.3 = 16.38 E(X 2 ) = 13.5 2 *0.2 + 15.9 2 *0.5 + 19.1 2 *0.3 = 272.298 V(X) = 272.298 16.38 2 = 3.9936 b. E(25X 8.5) = 25*E(X) 8.5 = 25*16.38 8.5 = 401 c. V(25X 8.5) = 25 2 *V(X) = 25 2 *3.9936 = 2496 d. E(h(X)) = E(X 0.01X 2 ) = E(X) 0.01*E(X 2 ) = 16.38 0.01*272.298 = 13.65702

6. A company that produces fine crystal knows from experience that 10% of its goblets have cosmetic flaws and must be classified as seconds. * a. Among six randomly selected goblets, how likely is it that only on is a second? b. Among six randomly selected goblets, what is the probability that at least two are seconds? c. How many out six randomly selected goblets would we expect to be seconds? d. What is the standard deviation of the number of second goblets? This is binomial with n = 6 and p = 0.1 a. P(X = 1) = dbinom(1,6,0.1) = 0.3543 b. P(X 2) = 1 pbinom(1,6,0.1) = 0.1143 c. E(X) = 6*0.1 = 0.6 d. SD(X) = 6.1.9 = 0.7348

7. A geologist has collected 10 specimens of basaltic rock and 10 specimens of granite.* The geologist instructs a laboratory assistant to randomly select 15 of the specimens for analysis. a. What is the probability function of the number of granite specimens selected for analysis? b. What is the probability that all specimens of one of the two types of rock are selected for analysis? c. What is the probability that the number of granite specimens selected for analysis is within 1 standard deviation of its mean value? This is hypergeometric with X = number of granite specimens, m = 10, n = 10, k = 15 a. P(X = x) = mcx*nc(k-x)/(m+n)ck = 10Cx*10C(15-x)/20C15 b. P(X = 10 or Y = 10) = 10C10*10C5/20C15 + 10C5*10C10/20C15 = 0.03215 c. E(X) = 15(0.5) = 7.5 V(X) = 15(0.5)(0.5)(1-14/19) = 0.9868 Sd(X) = 0.9934 1 P(6.5 < X < 8.5) = P(7 X 8) = dhyper(7,10,10,15) + dyper(8,10,10,15) = 0.6966

8. The number of requests for assistance received by a towing service is a Poisson process with average of 4 requests per hour.* a. Compute the probability that exactly ten requests are received during a particular 2-hour period. b. If the operators of the towing service take a 30-min break for lunch, what is the probability that they do not miss any calls for assistance? c. How many calls would you expect during their break? a. X = number of requests in 2 hours; E(x) = 4*2 = 8; P(X = 10) = dpois(10,8) = 0.0993 b. X = number of calls in 30 min; E(X) = 4/2 = 2; P(X = 0) = dpois(0,2) = 0.1353 c. E(X) = 4/2 = 2

9. Section 4.7, problem 4 Huck and Jim are waiting for a raft. The number of rafts floating by over intervals of time is a Poisson process with a rate of λ = 0.4 rafts per day. They agree in advance to let the first raft go and take the second one that comes along. What is the probability that they will have to wait more than a week? Hint: If they have to wait more than a week, what does that say about the number of rafts in a period of 7 days? X = number of rafts that go by in one week; E(X) = 0.4*7 = 2.8 If they wait for the second one, we want to know what is the probability that none or one raft went by? P(X 1) = ppois(1,2.8) = 0.2311

10. Suppose that X and Y are independent random variable having the joint probability distribution Find: a) E(2X 3Y) b) E(XY) c) Cov(X, Y) y x f(x, y) 2 4 1 0.1 0.15 3 0.2 0.3 5 0.1 0.15 a) We need to first find E(X) and E(Y) X 2 4 f(x) 0.4 0.6 E(X) = 2*.4 + 4*.6 = 3.2 Y 1 3 5 f(y) 0.25 0.5 0.25 E(Y) = 1*0.25 + 3*0.5 +5*0.25 = 3 E(2X 3Y) = 2*3.2 + 3*3 = 15.4 b) E(XY) = 1*2*.1 + 1*4*.15 + 3*2*.2 + 3*4*.3 + 5*2*.1 + 5*4*.15 = 9.6 c) Cov(X,Y) = E(XY) E(X)*E(Y) = 9.6 3.2*3 = 0