Name: __________________________________________________________________________Date: _______________
Chapter 6 Test Review
Make sure to show all work when necessary for each problem.
1. (6.1) Consumer Banker Association released a report showing the lengths of automobile
leases for new automobiles. The results are as follows.
Lease Length in Months
Percent of Leases
Mid Point (x)
13-24
12.7%
18.5
25-36
37.1%
37-48
28.5%
30.5 42.5
49-60
21.5%
More than 60
0.2%
54.5 66.5
(a) Is this a valid probability distribution? Explain.
Yes it is. All the probabilities add to one.
(b) Use the midpoint of each class (call the midpoint of the last class 66.5 months) to
compute the expected lease term.
X
P(X)
xP(x)
x-u
(x-u)2
(x-u)2P(x)
18.5
.127
2.35
-19.3
372.49
47.31
30.5
.371
11.32
-7.13
50.84
18.86
42.5
.285
12.11
4.87
23.72
6.76
54.5
.215
11.72
16.87
284.6
61.19
66.5
.002
.13
28.87
833.48
1.67
=
11.65
135.79
u = 37.63
(c) What is the standard deviation?
(d) Sketch a graph of the probability distribution.
2. (6.2) For a binomial experiment, how many outcomes are possible for each trial? What
are the possible outcomes?
2 outcomes: success or failure
3. (6.2) In a binomial experiment, is it possible for the probability of success to change
from one trial to the next? Explain. It
is not possible because binomial
distributions are independent and repeated under
identical conditions.
4. (6.2) Jake has just been given a 10-question multiple-choice quiz in history class. Each
question has four answers, of which only one is correct. Since Jake has not attended class
recently, he doesn’t know any of the answers. Assuming Jake guesses on all 10 questions,
find the indicated probabilities.
n =10 p = .25 q = .75
Identify the following…
(a) What is the probability that he will answer all questions correctly? (use formula)
C10,10 .2510.750 = .00000095
(b) What is the probability that he will answer all questions incorrectly? (use formula)
C10,0 .250 .7510 = .056
(c) What is the probability that he will answer at least two of the questions correctly?
(Use the table)
.282 + .250 + .146 + .058 + .016 + .003 + 0 + 0 + 0 =
.755
(d) What is the probability that he will answer no more than half the questions correctly?
(Use the table)
.056 + .188 +.282 + .250 + .146 + .058 = .98
5. (6.3) According to Harper’s Index, 60% of all federal inmates are serving time for drug
dealing. A random sample of 9 federal inmates were selected. (Use the table)
(a) What is the probability that 6 or more are serving time for drug dealing?
n=9
p = .6
r = 6, 7, 8, 9
.251 + .161 + .060 + .010 = .482
(b) What is the probability that 8 or fewer are serving time for drug dealing?
1 – P(9) = 1 - .01 = .99
(c) What is the expected number of inmates serving time for drug dealing?
9 (.6) =
5.4
6. (6.3) USA Today reports that about 25% of all prison parolees become repeat offenders.
Alice is a social worker whose job is to counsel people on parole. Let us say success means
a person does become a repeat offender. Alice has been given a group of four parolees.
(a) Find the probability P(r) of r successes ranging from 0 to 4 and make a histogram for
the probability distribution.
r
0
1
2
3
4
P(
r)
.316
.422
.211
.047
.004
n=4
p = .25
(b) Describe the distribution. How is the distribution supported by the probability of
success?
Skewed right. Since the probability is small (p=.25), the mean is shifted to the right.
(c) What is the expected number of parolees in Alice’s groups who will not be repeat
offenders? What is the standard deviation?
u = .25 (4) = 1
σ = .25 (.75) (4) = . .87
7. The Orchard café has found that about 5% of the diners who make reservations don’t
show up. If 82 reservations have been made, how any diners can be expected to show up?
.95 (82) = 77.9
Find the standard deviation of this distribution.
.05 (.95) (82)
.= .
197