Download 13. A psychologist determined that the number of sessions required

13. A psychologist determined that the number of sessions required to obtain the trust
of a new patient is either 1, 2, or 3. Let x be a random variable indicating the
number of sessions required to gain the patient’s trust. The following probability
function has been proposed.
f (x) =
x
6
for x = 1, 2,
or 3
a. Is this probability function valid? Explain.
b. What is the probability that it takes exactly 2 sessions to gain the patient’s
trust?
c. What is the probability that it takes at least 2 sessions to gain the patient’s
trust?
13. a. Yes, since f (x) 
b.
f (2) = 2/6
c.
f (2) + f (3)
0 for x = 1,2,3 and  f (x) = f (1) + f (2) + f (3)
= 1/6 + 2/6 + 3/6 = 1
= .333
= 2/6 + 3/6 = .833
23. According to a survey, 95% of subscribers to The Wall Street Journal Interactive
Edition have a computer at home. For those households, the probability
distributions for the number of laptop and desktop computers are given (The Wall
Street Journal Interactive Edition Subscriber Study, 1999).
Probability
Number of
Computers
Laptop
Desktop
0
.47
.06
1
.45
.56
2
.06
.28
3
.02
.10
a. What is the expected value of the number of computers per household for
each type of computer?
b. What is the variance of the number of computers per household for each
type of computer?
c. Make some comparisons between the number of laptops and the number of
desktops owned by the 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. A university found that 20% of its students withdraw without completing the
introductory 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. Airline passengers arrive randomly and independently at the passenger-screening
facility 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.
f (0) 
100 e10
 e10  .000045
0!
b. f (0) + f (1) + f (2) + f (3)
f (0) = .000045 (part a)
f (1) 
101 e10
 .00045
1!
Similarly, f (2) = .00225, f (3) = .0075
and f (0) + f (1) + f (2) + f (3) = .010245
c. 2.5 arrivals / 15 sec. period
f (0) 
Use  = 2.5
2.50 e2.5
 .0821
0!
d. 1 - f (0) = 1 - .0821
= .9179
49. Blackjack, or twenty-one as it is frequently called, is a popular gambling game
played in Las Vegas casinos. A player is dealt two cards. Face cards (jacks, queens,
and kings) and tens have a point value of 10. Aces have a point value of 1 or 11.
A 52-card deck contains 16 cards with a point value of 10 (jacks, queens, kings,
and tens) 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. A blackjack is a 10-point card and an ace for a value of 21. Use your answers to
parts (a), (b), and (c) to determine the probability that a player is dealt
blackjack. (Hint: Part (d) is not a hypergeometric problem. Develop your own
logical relationship as to how the hypergeometric probabilities from parts (a),
(b), and (c) can be combined to answer this question.)
49.
Parts a, b & c involve the hypergeometric distribution with N
and n = 2
a. r = 20, x = 2
= 52
 20   32 
  
2
0
(190)(1)
f (2)      
 .1433
1326
 52 
 
2
b. r
= 4, x = 2
 4  48 
  
2 0
(6)(1)
f (2)     
 .0045
1326
 52 
 
2
c. r
= 16, x = 2
16  36 
  
2
0
(120)(1)
f (2)     
 .0905
1326
 52 
 
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