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Study Guide Chapter 8
42
Chapter 8: Sampling Methods and
the Central Limit Theorem
Question 32
32. CRA CDs Inc. wants the mean lengths of the “cuts” on a CD to be 135
seconds (2 minutes and 15 seconds). This will allow the disk jockeys to have
plenty of time for commercials within each 10-minute segment. Assume the
distribution of the length of the cuts follows the normal distribution with a
population standard deviation of 8 seconds. Suppose we select a sample of 16
cuts from various CDs sold by CRA CDs Inc.
Question 32
Question 32 Part A
32. CRA CDs Inc. wants the mean lengths of the “cuts” on a CD to be 135
seconds (2 minutes and 15 seconds). This will allow the disk jockeys to have
plenty of time for commercials within each 10-minute segment. Assume the
distribution of the length of the cuts follows the normal distribution with a
population standard deviation of 8 seconds. Suppose we select a sample of 16
cuts from various CDs sold by CRA CDs Inc.
Question 32 Part A
a) What is the standard error of the mean?
Chapter 8
Part A
Step 1: Population Standard Deviation= 8 seconds
Sample number of observations= 16
Step 1: Population Standard Deviation= 8 secon...
Cerrina Ellis
Chapter 8
Part A
Step 2: Formula=
Cerrina Ellis
Chapter 8
Part A
Step 3: Solve
Plug in : 8/sqr16= 2
Cerrina Ellis
Question 32 Part B
32.CRA CDs Inc. wants the mean lengths of the “cuts” on a CD to be 135
seconds (2 minutes and 15 seconds). This will allow the disk jockeys to have
plenty of time for commercials within each 10-minute segment. Assume the
distribution of the length of the cuts follows the normal distribution with a
population standard deviation of 8 seconds. Suppose we select a sample of 16
cuts from various CDs sold by CRA CDs Inc.
Question 32 Part B
b) What percent of the sample means will be greater than 140
seconds?
Step 1 - Formula Normal Distribution
Sample Mean
Population Mean
Sigma
Sample Size
Melanie Guzman
Step 2: Plug in
Sample Mean = 140
Population Mean = 135
sigma= 8
Sample size = 16
Melanie Guzman
Step 3 - Calculate
Answer: 0.0062
z≥2.5
z<-.25
Melanie Guzman
Question 32 Part C
32.CRA CDs Inc. wants the mean lengths of the “cuts” on a CD to be 135
seconds (2 minutes and 15 seconds). This will allow the disk jockeys to have
plenty of time for commercials within each 10-minute segment. Assume the
distribution of the length of the cuts follows the normal distribution with a
population standard deviation of 8 seconds. Suppose we select a sample of 16
cuts from various CDs sold by CRA CDs Inc.
Question 32 Part C
c)What percent of the sample means will be greater than 128
but less than 140 seconds?
Answer: 99.38%
Dimitrios Karnegis
Steps for solving the probability of occurence between two variables
-First we must remember the Z formula:
Z= (Xbar - mu)/ standard deviation
Steps for solving the probability of occurence between t...
-Then we solve for both variables: In this case 128, and 140.
-Finally we subtract the two variables for our final answer
Dimitrios Karnegis
Solving for the probability that the mean is lesser than 140 seconds
Find P( Xbar: Xbar = 140 )
P( ( Xbar - μ ) / σ
( 140 - 135 ) / 2 )
= P( Z > 2.5 )
= P( Z < -2.5 )
When Z=-2.5 its Probability is .4938
.5000-.4938
= 0.9938
Solving for the probability that the mean is lesser than 140 seconds
Dimitrios Karnegis
Solving for the probability that the mean is higher than 128 seconds
Find P( Xbar: Xbar = 128 )
P( ( Xbar - μ ) / σ
P( 128 - 135 ) / 2 )
= P( Z > -3.5 )
= P( Z < 3.5 )
When Z=3.5 its Probability is .4998
.5000- .4998
= 0.0002
Solving for the probability that the mean is higher than 1...
Dimitrios Karnegis
Finding the difference of probability between 128 seconds and 140 seconds
Find P( 128 < Xbar < 140 )
= P( ( 128 - 135 ) / 2 < Z < ( 140 - 135 ) / 2 )
= P( -3.5 < Z < 2.5 )
= P( Z < 2.5 ) - P( Z < -3.5 )
= 0.9938 - 0.0002
= 0.9936
Finding the difference of probability between 1...
Dimitrios Karnegis
Question 34
34. Information from the American Institute of Insurance indicates 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
Question 34
Question 34 Part A
34. Information from the American Institute of Insurance indicates 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
Question 34 Part A
a) If we select a random sample of 50 households, what is
the standard error of the mean?
Standard error for mean formula
Sample of 50 households, we round to nearest whole number.
40,000/ sqrt(50)
=
5657
40,000/ sqrt(50)
=
5657
Question 34 Part B
34. Information from the American Institute of Insurance indicates 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.
Question 34 Part B
b) What is the likelihood of selecting a sample with a mean
of at least $112,000?
Formula for distribution of sample mean
Plug in!
Z = (112,000-110,000)/
40000/sqrt(50)
=
0.3536
Check z-chart then answer
z-chart: 0.3536
=
0.1368
(look for 0.35)
0.50-1368 = [0.3632]
Check z-chart then answer
Question 34 Part C
34. Information from the American Institute of Insurance indicates 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.
Question 34 Part C
c) What is the likelihood of selecting a sample with a mean
of more than $100,000?
Same formula as last time
Plug in
z = (100,000-110,000)
/5656.85
= -1.77
Plug in
Z-value and answer.
z-value of 1.77 = 0.4616
.50 + .4616 = .9616
Z-value and answer.
Question 34 Part D
34. Information from the American Institute of Insurance indicates 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.
Question 34 Part D
d) Find the likelihood of selecting a sample with a mean of
more than $100,000 but less than $112,000.
Solution - very difficult
Add the results of the previous two questions.
0.1368 + 0.4616 = .5984
Solution - very difficult
Question 38
38. 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.
Question 38 Part A
38. 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.
Question 38 Part A
a) What is the likelihood the sample mean is at least $25.00?
Catherine Abascal
Chapter 8: #38 A
Solve for:
That the likilhood that the sample mean is $25.00
Chapter 8: #38 A
Catherine Abascal
Chapter 8: #38 A
Select Formula:
Chapter 8: #38 A
How to know it is a Z table: Cause it says NORMAL Distribution
Catherine Abascal
Input Numbers into Formula
(25-23.50)/
(5÷√50)
x=25
μ=23.50
σ=5
√n=50
(25-23.50)/
(5÷√50)
= 2.12
Catherine Abascal
Chapter 8: #38 A
Keep Solving:
0.5000 - 0.4830
=0.017
Chapter 8: #38 A
The probability of the sample mean of at least $25.00 is 0.017.
Catherine Abascal
Question 38 Part B
38. 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.
Question 38 Part B
b) What is the likelihood the sample mean is greater than
$22.50 but less than $25.00?
Chapter 8: Exercise 38 B
- What we are solving for...?
Sample mean is greater than 22.50 but less than $25.00
Chapter 8: Exercise 38 B
Elinor Faedo
Chapter 8: Exercise 38 B
Step 2: Select Fromula
x= 22.50
m= 23.50
o = 5
n=50
Chapter 8: Exercise 38 B
(22.50-23.50)/ 5/ l⁄ 50 = 1.41
Elinor Faedo
Chapter 8: Exercise 38 B
Sep 3: Solving Cont.
P (-1.41> Z > 2.12) = 0.0793-0.0170 = 0.9037
Elinor Faedo