Chapter 5 Solutions
5.1. (a)
n
=
1500 (the sample size).
(b)
The “Yes” count seems like the most reasonable
choice, but either count is defensible.
(c)
X
=
525 (or
X
=
975).
(d)
ˆ
p
=
525
1500
=
0
.
35 (or
ˆ
p
=
975
1500
=
0
.
65).
5.2.
n
=
200 (the sample size),
ˆ
p
=
40%
=
0
.
40, and
X
=
n
ˆ
p
=
80.
5.3.
We have 20 independent trials, each with probability of success (heads) equal to 0.5, so
X
has a
B
(
20
,
0
.
5
)
distribution.
5.4.
Assuming no multiple births (twins, triplets, quadruplets), we have four independent trials,
each with probability of success (type O blood) equal to 0.25, so the number of children
with type O blood has a
B
(
4
,
0
.
25
)
distribution.
5.5. (a)
For a
B
(
4
,
0
.
3
)
distribution,
P
(
X
=
0
)
=
0
.
2401 and
P
(
X
≥
3
)
=
0
.
0837.
(b)
For a
B
(
4
,
0
.
7
)
distribution,
P
(
X
=
4
)
=
0
.
2401 and
P
(
X
≤
1
)
=
0
.
0837.
(c)
The number of
“failures” in a
B
(
4
,
0
.
3
)
distribution has a
B
(
4
,
0
.
7
)
distribution. With 4 trials, 0 successes is
equivalent to 4 failures, and 3 or more successes is equivalent to 1 or fewer failures.
5.6. (a)
For a
B
(
100
,
0
.
5
)
distribution,
µ
ˆ
p
=
p
=
0
.
5 and
σ
ˆ
p
=
p
(
1
−
p
)
n
=
1
20
=
0
.
05.
(b)
No;
the mean and standard deviation of the sample count are both 100 times bigger. (That is,
ˆ
p
=
X
/
100, so
µ
ˆ
p
=
µ
X
/
100 and
σ
ˆ
p
=
σ
X
/
100.)
5.7. (a)
ˆ
p
has approximately a Normal distribution with mean 0.5 and standard
deviation 0.025, so
P
(
0
.
4
<
ˆ
p
<
0
.
6
)
=
P
(
−
2
<
Z
<
2
)
.
=
0
.
9544.
(b)
P
(
0
.
45
<
ˆ
p
<
0
.
55
)
=
P
(
−
1
<
Z
<
1
)
.
=
0
.
6826.
Note:
The answers given are from Table A or software. The 68–95–99.7 rule would
give the answers 0.95 and 0.68—quite reasonable answers, especially since this is an
approximation anyway. For comparison, the exact answers are P
(
0
.
4
<
ˆp
<
0
.
6
)
.
=
0
.
9431
or P
(
0
.
4
≤
ˆp
≤
0
.
6
)
.
=
0
.
9648, and P
(
0
.
45
<
ˆp
<
0
.
55
)
.
=
0
.
6318 or
P
(
0
.
45
≤
ˆp
≤
0
.
55
)
.
=
0
.
7287. (Notice that the “correct” answer depends on our
understanding of “between.”)
5.8. (a)
P
(
X
≥
3
)
=
(
4
3
)
0
.
52
3
0
.
48
+
(
4
4
)
0
.
52
4
.
=
0
.
3431.
(b)
If the coin were fair,
P
(
X
≥
3
)
=
(
4
3
)
0
.
5
3
0
.
5
+
(
4
4
)
0
.
5
4
=
0
.
3125.
5.9. (a)
Separate ﬂips are independent (coins have no “memory,” so they do not try to
compensate for a lack of tails).
(b)
Separate ﬂips are independent (coins have no “memory,”
so they do not get on a “streak” of heads).
(c)
ˆ
p
can vary from one set of observed data to
another; it is not a parameter.
169
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170
Chapter 5
Sampling Distributions
5.10. (a)
X
is a
count;
ˆ
p
is a proportion.
(b)
The given formula is the
standard deviation
for a
binomial
proportion
. The variance for a binomial count is
np
(
1
−
p
)
.
(c)
The rule of thumb
in the text is that
np
and
n
(
1
−
p
)
should both be at least 10. If
p
is close to 0 (or close to
1),
n
=
1000 might not satisfy this rule of thumb. (See also the solution to Exercise 5.18.)
5.11. (a)
A
B
(
200
,
p
)
distribution seems reasonable for this setting (even though we do not
know what
p
is).
(b)
This setting is not binomial; there is no fixed value of
n
.
(c)
A
B
(
500
,
1
/
12
)
distribution seems appropriate for this setting.
5.12. (a)
This is not binomial;
X
is not a count of successes.
(b)
A
B
(
20
,
p
)
distribution
seems reasonable, where
p
(unknown) is the probability of a defective pair.
(c)
This should
be (at least approximately) a
B
(
n
,
p
)
distribution, where
n
is the number of students in our
sample, and
p
is the probability that a randomlychosen student eats at least five servings of
fruits and vegetables.
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 Spring '07
 Guggenberger
 Normal Distribution, Standard Deviation

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