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Chapter 8 Sampling Methods and the Central Limit Theorem
1. SR 8-3 The lengths of service of all the executives employed by Standard Chemicals are:
題目提供原始資料
Name
A
B
C
D
E
自行整理出母體分配
Years
20
22
26
24
28
年資 X
次數
機率
20
22
24
26
28
1
1
1
1
1
0.2
0.2
0.2
0.2
0.2
a. Using the combination formula, how many samples of size 2 are possible? 答 ： 有
5!
5 C 2 = 2! (5 - 2)! = 10 個可能樣本。
b. List all possible samples of 2 executives from the population and compute their means.
c. Organize the means into a sampling distribution.答：共 10 個樣本和 X 的機率分配見下表。
所有 n=2 樣本
X 的機率分配
sample#
x1
x2
X
X
次數
機率
1
20
22
21.0
21
1
0.1000
2
20
24
22.0
22
1
0.1000
3
20
26
23.0
23
2
0.2000
4
20
28
24.0
24
2
0.2000
5
22
24
23.0
25
2
0.2000
6
22
26
24.0
26
1
0.1000
7
22
28
25.0
27
1
0.1000
8
24
26
25.0
10
1.0000
9
24
28
26.0
10
26
28
27.0
d. Compare the population mean and the mean of the sample means.答：由母體分配計算出母體
平均數 μ =24，由 X 的機率分配算出 μ X =24，二者相等。
e. Compare the dispersion in the population with that in the distribution of the sample mean.答：母
體 全 距 =8 ， 樣 本 全 距 =6 。 母 體 標 準 差 σ ＝ 2.8284 ， 利 用 公 式 可 以 計 算 出
σX =
σ
n
N n 2.8284 5 - 2
=
= 2.0000 × 0.8660 = 1.7321。直接由 X 的機率分配也可計
N 1
5-1
2
算出同樣結果 σ X = 1.7321 。
1
f. Is the distribution of population values normally distributed (bell-shaped)?答：母體是不連續的
均等分配。
g. Is the distribution of the sample mean computed in part (c) starting to show some tendency
toward being bell-shaped?答：樣本平均數 X 的分配呈鐘形且對稱。
2. EX9 在 N=6 的母體中抽取 n=3 的樣本，共有 6 C 3 =
6!
3! (6 - 3)!
= 20 個可能樣本，作法和 SR3
完全相同。相較於樣本平均，母體較為分散，樣本平均變動在 1.33-4.0 之間，母體變動在 0-6
之間。)
3. EX12 Scrapper Elevator Company has 20 sales representatives who sell its product throughout
the United States and Canada. The number of units of units sold last month by each representative is
listed below. Assume these sales figures to be the population values.
2
3
2
3
3
4
2
4
3
2
2
7
3
4
5
3
3
3
3
5
a. Draw a graph showing the population distribution, and compute the mean of the population.
答：整理出母體機率分配如下表，計算可知母體平均數 3.3， σ =1.2288。
次數
機率
X
2
3
4
5
5
9
3
2
0.25
0.45
0.15
0.10
7
1
20
0.05
1
4. EX15 A normal population has a mean of 60 and a standard deviation of 12. You select a
random sample of 9. Compute the probability the sample mean is:
63  60
a. Greater than 63.答：z =
 0.75 故機率為 0.2266＝0.5000  0.2734。
12 / 9
56  60
b. Less than 56. 答： z 
 1 ，故機率為 0.1587＝0.5000  0.3413。
12 / 9
c. Between 56 and 63. 答：0.6147＝0.3413 + 0.2734.
5. EX18 According to an IRS study, it takes a mean of 330 minutes for taxpayers to prepare, copy,
and electronically file a 1040 tax form. This distribution of times follows the normal distribution
and the standard deviation is 80 minutes. A consumer watchdog agency selects a random sample of
40 taxpayers.
80
a. What is the standard error of the mean in this example?答： sx 
 12.649
40
2
b. What is the likelihood the sample mean is greater than 320 minutes? 答 ： z =
320  330
 0.79 ，故機率為 So probability is 0.7852, found by 0.2852 + 0.5000.
80 / 40
350  330
c. What is the likelihood the sample mean is between 320 and 350 minutes? z =
 1.58
80 / 40
So probability is 0.7281, found by 0.2852 + 0.4429.
d. What is the likelihood the sample mean is greater than 350 minutes? 0.0571, found by 0.5000 
0.4429.
6. EX22 In the Department of Education at UR University, student records suggest that the population of
students spends an average of 5.5 hours per week playing organized sports. The population’s standard
deviation is 2.2 hours per week. Based on a sample of 121 students, Healthy Lifestyles Incorporated (HLI)
would like to apply the central limit theorem to make various estimates.
a.
b.
Compute the standard error of the sample mean. 答：
=
2.2
=0.2
11
121
What is he chance that HLI will find a sample mean between 5 and 6 hours? 答：由 5 < X < 6
以及 z =
c.
2.2
X-μ
σ
n
可計算出 - 2.5 < z < 2.5 ，故 f (- 2.5 < z < 2.5) =2×(0.4983)=0.9876。
How strange would it be to obtain a sample mean greater than 6.5 hours? 答：由 z =
X-μ
σ
n
計算出 z
值＝5.0, 機率為 f (5 < z ) =0.00000029 。
8. EX31 Mattel Corporation Produces a remote-controlled car that requires three AA batteries.
The mean life of these batteries in this product is 35.0 hours. The distribution of the battery lives
closely follows the normal probability distribution with a standard deviation of 5.5 hours.
a. As a part of its testing program Sony tests samples of 25 batteries 答：The distribution will be
b.
normal.
What can you say about the shape of the distribution of the sample mean? 答：
c.
What is the standard error of the distribution of the sample mean? 答：  x 
d.
What proportion of the samples will have a mean useful life of more than 36 hours? 答：
36  35
z
 0.91 So probability is 0.1814, found by 0.5000 - 0.3186
5.5 25
e.
What proportion of the sample will have a mean useful life greater than 34.5 hours? 答：
34.5  35
z
 0.45 So probability is 0.6736, found by 0.5000 + 0.1736.
5.5 25
5.5
 1.1
25
f. What proportion of the sample will have a mean useful life between 34.5 and 36.0 hours? 答：
0.4922, found by 0.3186 + 0.1736
3
9. EX34 Information from the American Institute of Insurance indicate the mean amount of life
insurance per household in the United States is $110,000. This distribution follows the normal
distribution with a standard deviation of $40,000.
a. If we select a random sample of 50 households, what is the standard error of the mean?
sx 
40,000
 5657
50
b. What is the expected shape of the distribution of the sample mean? 答：The distribution will be
normal
c. What is the likelihood of selecting a sample with a mean of at least $112,000? 答：
112, 000 110, 000
z
 0.35 So probability is 0.3632, found by 0.5000 - 0.1368.
40, 000 50
d. What is the likelihood of selecting a sample with a mean of more than $100,00? 答：
100, 000 110, 000
z
 1.77 So probability is 0.9616, found by 0.5000 + 0.4616.
40, 000 50
e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than
$112,000. 答：0.5984, found by 0.4616 + 0.1368.
10. EX38 The mean amount purchased by a typical customer at Churchill’s Grocery Store is
$23.50.With a standard deviation of $5.00.Assume the distribution of amounts purchased follows
the normal distribution. For a sample of 50 customers, answer the following questions.
25  23.5
a. What is the likelihood the sample mean is at least $25.00? 答： z 
 2.12 So
5 50
probability is 0.0170, found by 0.5000  0.4830.
b. What is the likelihood the sample mean is greater than $22.50 but less than $25.00? 答：
22.5  23.5
z
 1.41 So probability is 0.9037, found by 0.4207 + 0.4830.
5 50
c. Within what limits will 90 percent of the sample means occur? 答：Between 22.33 and 24.67,
 5 

 50 
found by 23.50  1.65 
11. EX42 Human Resource Consulting (HRC) is surveying a sample of 60 firms in order to study
health care costs for a client. One of the items being tracked is the annual deductible that employees
must pay. The state Bureau of Labor reports the mean of this distribution is $502 with a standard
deviation of $100.
100
a. Compute the standard error of the sample mean for HRC. 答：  x 
 12.91
60
b. What is the chance HRC finds a sample mean between $477 and $527? 答 ：
477 502
527 502
z1 =
= 1.94 and z 2 =
= 1.94 So probability is 0.9476, found by 0.4738
100 60
100 60
+ 0.4738.
c. Calculate the likelihood that the sample mean is between $492 and $512? 答 ：
4
z1 =
492 502
100
60
= 0.77 and z 2 =
512 502
100
60
= 0.77 So probability is 0.5588, found by 0.2794
+ 0.2794
d. What is the probability the sample mean is greater than $550? 答： z =
550 502
100
60
= 3.72 So
probability is virtually 0.
12 EX45 Nike’s annual report says that the average American buys 6.5 pairs of sports shoes per
year. Suppose the population standard deviation is 2.1 and that a sample of 81 customers will be
examined next year.
2.1
a. What is the standard error of the mean in this experiment?答：  x 
 0.2333
81
b. What is the probability that the sample mean is between 6 and 7 pairs of sports shoes?答：
6 - 6.5
7 - 6.5
z1 =
= 2.14 and z 2 =
= 2.14 Probability is 0.9676, found by 0.4838 +
2.1 81
2.1 81
0.4838.
c. What is the probability that the difference between the sample mean and the population mean is
6.25 - 6.5
6.75 - 6.5
less than 0.25 pairs? 答： z1 =
= 1.07 and z 2 =
= 1.07 Probability is
2.1 81
2.1 81
0.7154, found by 0.3577 + 0.3577
d. What is the likelihood the sample mean is greater than 7 pairs? 答：Probability is 0.0162, found
by 0.5000 - 0.4838.
其他題目：
1. 請簡單說明三種抽樣方法，台灣的失業率統計是採用那一種抽樣方法？
2. （97 關務三等,7-19）有一母體為{1,2,3,4,5,6}。若由該母體抽出放回，連續抽出兩數。
(一).
試列出可能的樣本，並計算每一樣本之平均數。
(二).
樣本數為 2，樣本平均數之抽樣分配為何？答：
x
f x

1
1 36
1.5
2 36
2
3 36
2.5
4 36
3
5 36
3.5
6 36
4
5 36
4.5
4 36
5
3 36
5.5
2 36
6
1 36
請說明樣本平均數之抽樣分配的變異數與母體變異數之關係。答：
Var X  Var X   Var X 
n
2
(三).
 
3. (97 高考, 7-23) 若 XYZ 公司之每日股價變動 X 為一隨機變數其隨機分配如下
1
2
3
X =x
P( X = x)
0.4
0.4
0.2
(一).
試求隨機取二個樣本之平均數的抽樣分配。
(二).
根據(一)式，計算樣本平均數之期望值和標準差。答：E X ＝1.8， X ＝0.5291。
 
 
利用中央極限定理計算出樣本平均數之期望值和變異數。 E X   ＝1.8，
Var X  Var X   0.56  0.28 。
n
2
(三).
 
5