Chapter 7
7.1
(a) P(less than 3) = P(1 or 2) = 2/6 = 1/3.
(b)–(c) Answers will vary.
7.2
(a) The possible arrangements are: BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG.
Each
arrangement has probability 1/8.
(b) Three of the eight arrangements have two (and only two)
girls, so P(X = 2) = 3/8 = 0.375. (c) The distribution of X is shown in the table below.
Value of X
0
1
2
3
Probability
1/8
3/8
3/8
1/8
7.3
(a) “At least one nonword error” is the event {X
≥
1} or {X>0}.
P(X
≥
1) = 1
−
P(X<1) = 1
−
P(X=0) = 1
−
0.1 = 0.9.
(b) The event {X
≤
2}
is “no more than two nonword errors,” or “fewer
than three nonword errors.”
P(X
≤
2) = (X = 0) + P(X = 1) + P(X = 2) = 0.1 + 0.2 + 0.3 = 0.6.
P(X < 2) = P(X = 0) + P(X = 1) = 0.1 + 0.2 = 0.3.
7.4
The probability histograms are shown below. The distribution of the number of rooms is
roughly symmetric for owners (graph on the left) and skewed to the right for renters (graph on
the right).
The center is slightly over 6 units for owners and slightly over 4 for renters.
Overall,
renteroccupied units tend to have fewer rooms than owneroccupied units.
Number of rooms in owneroccupied units
Probability
10
9
8
7
6
5
4
3
2
1
0.25
0.20
0.15
0.10
0.05
0.00
Number of rooms in renteroccupied units
10
9
8
7
6
5
4
3
2
1
0.4
0.3
0.2
0.1
0.0
7.5
(a) “The unit has five or more rooms” can be written as {X
≥
5}.
P(X
≥
5) = P(X = 5) + P(X
= 6) + P(X=7) + P(X=8) + P(X=9) + P(X = 10) = 0.868.
(b) The event {X > 5} is “the unit has
more than five rooms.”
P(X > 5) = P(X = 6) + P(X = 7) + P(X=8) + P(X=9) + P(X = 10) =
0.658.
(c) A discrete random variable has a countable number of values, each of which has a
distinct probability (P(X = x)).
P(X
≥
5) and P(X > 5) are different because the first event
contains the value X = 5 and the second does not.
7.6
(a) P(T=2) = 1
−
0.37 = 0.63 and P (T=3) = 0.37×0.63 = 0.2331.
(b) P(T
≤
4) is the
probability that no more than two people will pass on your message.
()
( )
2
4
2
3
4
0.63
0.37
0.63
0.37
0.63
0.9493
PT
≤=
=+
==
+
×
+
×
±
.
7.7
(a) P(X < 0.49) = 0.49.
(b) P(X
≤
0.49) = 0.49.
Note: (a) and (b) are the same because
there is no area under the curve at any one particular point.
(c) P(X
≥
0.27) = 0.73.
(d) P(0.27
< X < 1.27) = P(0.27 < X < 1) = 0.73.
(e) P(0.1
≤
X
≤
0.2 or 0.8
≤
X
≤
0.9) = 0.1 + 0.1 = 0.2.
(f)
P(not [0.3
≤
X
≤
0.8]) = 1
−
0.5 = 0.5. Or P(0
≤
X < 0.3 or 0.8 < X
≤
1) = 0.3 + 0.2 = 0.5 (g) P(X
170
Chapter 7
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View Full Document= 0.5) = 0.
7.8
(a) P(0
≤
X
≤
0.4) = 0.4.
(b) P(0.4
≤
X
≤
1) = 0.6.
(c) P(0.3
≤
X
≤
0.5) = 0.2.
(d) P(0.3 < X
< 0.5) = 0.2.
(e) P(0.226
≤
X
≤
0.713) = 0.713
−
0.226 = 0.487.
(f) A continuous distribution
assigns probability 0 to every possible outcome. In this case, the probabilities in (c) and (d) are
the same because the events differ by 2 possible values, 0.3 and 0.5, each of which has
probability 0.
7.9
(a)
=
ˆ
(0
.
4
5
Pp
≥
)
0.45
0.4
0.024
PZ
−
⎛
≥
⎜
⎝⎠
⎞
⎟
)
=
P
(Z
≥
2.08) = 0.0188.
(b)
=
ˆ
.
3
5
<
0.35
0.4
0.024
−
⎛
≥
⎜
⎞
⎟
= P(Z <
−
2.08) = 0.0188.
(c)
ˆ
(0.35
0.45)
≤
≤
= P(
−
2.08
≤
Z
≤
2.08) =
0.9812
−
0.0188 = 0.9624.
7.10
Answers will vary.
For a sample of 400 observations from the N(0.4, 0.024) distribution,
there were 9 values below 0.35.
Thus, the relative frequency is 9/400 = 0.0225, which is close to
but slightly higher than the value from Exercise 7.9 (b).
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 Fall '08
 O
 Probability, Standard Deviation, Probability theory, Blaylock

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