143
Chapter 7: Sampling Distributions and the Central Limit Theorem
7.1
a.
–
c.
Answers vary.
d.
The histogram exhibits a mound shape.
The sample mean should be close to 3.5 =
μ
e.
The standard deviation should be close to
σ
/
3 = 1.708/
3 = .9860.
f.
Very similar pictures.
7.2
a.
P
(
Y
= 2) =
P
(
W
= 6) =
p
(4, 1, 1) +
p
(1, 4, 1) +
p
(1, 1, 4) +
p
(3, 2, 1) +
p
(3, 1, 2)
=
p
(2, 3, 1) +
p
(2, 1, 3) +
p
(1, 3, 2)+
p
(1, 2, 3) +
p
(2, 2, 2) =
216
10
.
b.
Answers vary, but the relative frequency should be fairly close.
c.
The relative frequency should be even closer than what was observed in part b.
7.3
a.
The histogram should be similar in shape, but this histogram has a smaller spread.
b.
Answers vary.
c.
The normal curve should approximate the histogram fairly well.
7.4
a.
The histogram has a right–skewed shape.
It appears to follow
p
(
y
) =
y
/21,
y
= 1, …, 6.
b.
From the Stat Report window,
μ
= 2.667,
σ
= 1.491.
c.
Answers vary.
d.
i. It has a right–skewed shape.
ii. The mean is larger, but the std. dev. is smaller.
e.
i. sample mean = 2.667, sample std. dev = 1.491/
12 = .4304.
ii. The histogram is closely mound shaped.
iii. Very close indeed.
7.5
a.
Answers vary.
b.
Answers vary, but the means are probably not equal.
c.
The sample mean values cluster around the population mean.
d.
The theoretical standard deviation for the sample mean is 6.03/
5 = 2.6967.
e.
The histogram has a mound shape.
f.
Yes.
7.6
The larger the sample size, the smaller the spread of the histogram.
The normal curves
approximate the histograms equally well.
7.7
a.
–
b.
Answers vary.
c.
The mean should be close to the population variance
d.
The sampling distribution is not mound–shaped for this case.
e.
The theoretical density should fit well.
f.
Yes, because the chi–square density is right–skewed.
7.8
a.
σ
2
= (6.03)
2
= 36.3609.
b.
The two histograms have similar shapes, but the histogram generated from the smaller
sample size exhibits a greater spread.
The means are similar (and close to the value
found in part a).
The theoretical density should fit well in both cases.
c.
The histogram generated with n = 50 exhibits a mound shape.
Here, the theoretical
density is chi–square with
ν
= 50 – 1 = 49 degrees of freedom (a large value).
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View Full Document144
Chapter 7: Sampling Distributions and the Central Limit Theorem
Instructor’s Solutions Manual
7.9
a.
P
(
Y
–
μ

≤
.3) =
P
(–1.2
≤
Z
≤
1.2) = .7698.
b.
P
(
Y
–
μ

≤
.3) =
P
(–.3
n
≤
Z
≤
.3
n
) = 1 – 2
P
(
Z
> .3
n
).
For
n
= 25, 36, 69, and
64, the probabilities are (respectively) .8664, .9284, .9642, and .9836.
c.
The probabilities increase with
n
, which is intuitive since the variance of
Y
decreases
with
n
.
d.
Yes, these results are consistent since the probability was less than .95 for values of
n
less than 43.
7.10
a.
P
(
Y
–
μ

≤
.3) =
P
(–.15
n
≤
Z
≤
.15
n
) = 1 – 2
P
(
Z
> .15
n
).
For
n
= 9, the
probability is .3472 (a smaller value).
b.
For
n
= 25:
P
(
Y
–
μ

≤
.3) = 1 – 2
P
(
Z
> .75) = .5468
For
n
= 36:
P
(
Y
–
μ

≤
.3) = 1 – 2
P
(
Z
> .9) = .6318
For
n
= 49:
P
(
Y
–
μ

≤
.3) = 1 – 2
P
(
Z
> 1.05) = .7062
For
n
= 64:
P
(
Y
–
μ

≤
.3) = 1 – 2
P
(
Z
> 1.2) = .7698
c.
The probabilities increase with
n
.
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