13.Apsychologistdeterminedthatthenumberofsessionsrequiredtoobtainthetrustofanew patientiseither1,2,or3.Letxbearandom variableindicatingthenumberofsessionsrequired to gain the patient’s trust.The following probability function has been proposed.

f(x) = forx = 1,2,or3

a.Is this probability function valid? Explain.

b.Whatistheprobabilitythatittakesexactly2sessionstogainthepatient’strust?

c.What is the probability that it takes at least 2 sessions to gain the patient’s trust?

13.a.Yes, since f (x)  0 for x = 1,2,3 and f (x) = f (1) + f (2) + f (3) = 1/6 + 2/6 + 3/6 = 1

b.f (2) = 2/6 = .333

c.f (2) + f (3) = 2/6 + 3/6 = .833

23.Accordingtoasurvey,95%ofsubscriberstoTheWallStreetJournalInteractiveEdition haveacomputerathome. Forthosehouseholds,theprobabilitydistributionsforthenumberoflaptopanddesktopcomputersaregiven(TheWallStreetJournalInteractiveEdition Subscriber Study, 1999).

Probability
Number of Computers / Laptop / Desktop
0 / .47 / .06
1 / .45 / .56
2 / .06 / .28
3 / .02 / .10

a.Whatistheexpectedvalueofthenumberofcomputersperhouseholdforeachtype of computer?

b.Whatisthevarianceofthenumberofcomputersperhouseholdforeachtypeof computer?

c.Makesomecomparisonsbetweenthenumberoflaptopsandthenumberofdesktops ownedbythe Journal’s subscribers.

23.a.Laptop: E (x) = .47(0) + .45(1) + .06(2) + .02(3) = .63

Desktop: E (x) = .06(0) + .56(1) + .28(2) + .10(3) = 1.42

b.Laptop: Var(x) = .47(-.63)2 + .45(.37)2 + .06(1.37)2 + .02(2.37)2 = .4731

Desktop: Var(x) = .06(-1.42)2 + .56(-.42)2 + .28(.58)2 + .10(1.58)2 = .5636

c.From the expected values in part (a), it is clear that the typical subscriber has more desktop computers than laptops. There is not much difference in the variances for the two types of computers.

35.Auniversityfoundthat20%ofitsstudentswithdrawwithoutcompletingtheintroductory statistics course.Assume that 20 students registered for the course this quarter.

a.Compute the probability that two or fewer will withdraw.

b.Compute the probability that exactly four will withdraw.

c.Compute the probability that more than three will withdraw.

d.Compute the expected number of withdrawals.

35.a.f (0) + f (1) + f (2) = .0115 + .0576 + .1369 = .2060

b.f (4) = .2182

c.1 - [ f (0) + f (1) + f (2) + f (3) ]= 1 - .2060 - .2054 = .5886

d. = n p = 20 (.20) = 4

43.Airlinepassengersarriverandomlyandindependentlyatthepassenger-screeningfacility at a major international airport.The mean arrival rate is 10 passengers per minute.

a.Compute the probability of no arrivals in a 1-minute period.

b.Compute the probability that three or fewer passengers arrive in a 1-minute period.

c.Compute the probability of no arrivals in a 15-second period.

d.Compute the probability of at least one arrival in a 15-second period.

43.a.

b.f (0) + f (1) + f (2) + f (3)

f (0) = .000045 (part a)

Similarly, f (2) = .00225, f (3) = .0075

and f (0) + f (1) + f (2) + f (3) = .010245

c.2.5 arrivals / 15 sec. period Use  = 2.5

d.1 - f (0) = 1 - .0821 = .9179

49.Blackjack,ortwenty-oneasitisfrequentlycalled,isapopulargamblinggameplayed inLasVegascasinos.Aplayerisdealttwocards.Facecards(jacks,queens,andkings) andtenshaveapointvalueof10.Aceshaveapointvalueof1or11.A52-carddeck contains16cardswithapointvalueof10(jacks, queens, kings, andtens)and four aces.

a.What is the probability that both cards dealt are aces or 10-point cards?

b.What is the probability that both of the cards are aces?

c.What is the probability that both of the cards have a point value of 10?

d.Ablackjackisa10-pointcardandanaceforavalueof21.Useyouranswerstoparts(a),(b),and(c)todeterminetheprobabilitythataplayerisdealtblackjack.(Hint: Part(d)isnotahypergeometricproblem.Developyourownlogicalrelationshipasto howthehypergeometricprobabilitiesfromparts(a),(b),and(c)canbecombinedto answer this question.)

49.Parts a, b & c involve the hypergeometric distribution with N = 52 and n = 2

a.r = 20, x = 2

b.r = 4, x = 2

c.r = 16, x = 2

d.Part (a) provides the probability of blackjack plus the probability of 2 aces plus the probability of two 10s. To find the probability of blackjack we subtract the probabilities in (b) and (c) from the probability in (a).

P (blackjack) = .1433 - .0045 - .0905 = .048