132
The Practice of Statistics for AP*, 4/e
Chapter 6
Section 6.1
Check Your Understanding, page 344:
1.
We are looking for the probability that the student gets either an A or a B.
This probability is
0.42
0.26
0.68.
+
=
2.
We are looking for
(
)
2
0.02
0.10
0.12.
P X
<
=
+
=
3.
This histogram is left skewed.
This means that higher grades are more likely, but that there are a few
lower grades.
Check Your Understanding, page 349:
1.
(
)
(
)
(
)
(
)
0 0.3
1 0.4
2 0.2
3 0.1
1.1.
X
µ
=
+
+
+
=
The long-run average, over many Friday mornings, will
be about 1.1 cars sold.
2.
(
) (
)
(
) (
)
(
) (
)
(
) (
)
2
2
2
2
2
0
1.1
0.3
1
1.1
0.4
2
1.1
0.2
3
1.1
0.1
0.89.
X
σ
=
−
+
−
+
−
+
−
=
So
0.943.
X
σ
=
On
average, the number of cars sold on a randomly selected Friday will differ from the mean (1.1) by 0.943
cars sold.
Exercises, page 353:
6.1
(a)
If you toss a coin 4 times, the sample space is {HHHH, HHHT, HHTH, HTHH, THHH, HHTT,
HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT}.
All of these outcomes are
equally likely and so have probability
1
0.0625.
16
=
To find the probability of X taking on any specific
number, count the number of outcomes with exactly this number of heads and multiply by 0.0625.
For
example, there are 4 ways to get exactly 3 heads so
(
)
(
)
3
4 0.0625
0.25.
P X
=
=
=
This leads to the
following distribution:
Value
0
1
2
3
4
Probability
0.0625
0.25
0.375
0.25
0.0625

Chapter 6: Random Variables
133
(b)
The histogram shows that this distribution is symmetric with a center at 2.
(c)
(
)
(
)
3
1
4
1
0.0625
0.9375.
P X
P X
≤
= −
=
= −
=
There is a 93.75% chance that you will get three or
fewer heads on 4 tosses of a fair coin.
6.2
(a)
If we roll two 6-sided dice, the sample space is {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2),
(2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1),
(5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}.
All of these are equally likely, so the
probability of any one outcome is
1
.
36
To find the probability of X taking on any specific number, count
the number of outcomes where the sum of the dice is exactly this number and multiply by 0.0278.
For
example, there are 4 ways to get a sum of 5 (the outcomes (1,4), (2,3), (3,2), and (4,1))
so
(
)
1
4
3
4
.
36
36
P X
=
=
This leads to the following distribution:
Value
2
3
4
5
6
7
8
9
10
11
12
Prob
1
36
2
36
3
36
4
36
5
36
6
36
5
36
4
36
3
36
2
36
1
36
(b)
The histogram shows that the distribution is symmetric about a center of 7.
(c)
(
)
(
)
1
2
3
6
1
5
5
1
4
1
1
1
.
36
36
36
36
6
6
P T
P T
≥
= −
≤
= −
+
+
= −
= −
=
This means that about five-sixths
of the time, when you roll a pair of 6-sided dice, you will have a sum of 5 or more.

134
The Practice of Statistics for AP*
, 4/e
6.3
(a) “At least one nonword error” is the event
{
}
1
X
≥
or
{
}
0 .
X
>
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.
6.4
(a)
“Plays with at most two toys” is the event
{
}
2
X
≤
or
{
}
3 .
X
<
P(X
≤
2
) = (X = 0
) + P(X
= 1
) +
P(X
= 2
)
= 0
.03 + 0.16 + 0.30
= 0
.49.
(b)
The event {X > 3} is “the child plays with more than three
toys.”
(
)
(
)
(
)
3
4
5
0.17
0.11
0.28.
P X
P X
P X
>
=
=
+
=
=
+
=
(
)
(
)
3
1
2
1
0.49
0.51.
P X
P X
≥
= −
≤
= −
=
6.5
(a)
All of the probabilities are between 0 and 1 and they sum to 1 so this is a legitimate probability
distribution.
(b) This is a right skewed distribution with the largest amount of probability on the digit 1.

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