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# Alpha Group Ch 8

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by

## Andrew Yanes

on 4 April 2016

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#### Transcript of Alpha Group Ch 8

Chapter 8 #32
Chapter 8 #32
STANDARD ERROR OF THE MEAN
Chapter 8 #32
a) What is the standard error of the mean?
2
FINDING THE Z VALUE OF WHEN THE POPULATION STANDARD DEVIATION IS KNOWN
Chapter 8 #32
Alpha Group Ch 8
CRA CDs Inc. wants the mean lengths of the "cuts" on a CD to be 135 seconds (2 minutes ans 15 seconds). This will allowthe disk jockeys to have plenty of time for commercials within each 10-minute segment. Assume thedistributionof 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.
a) What is the standard error of the mean? 2
b) What percentage of the sample means will be greater than 140 seconds? 0.0062
c) What percentage of the sample means will be greater than 128 but less than 140 seconds?
0.9938

Leo Legorburo
is the standard deviation
N
is the sample size
Leo Legorburo
= 8
N
= 16
2 = 8/sqrt16
the sample mean
the population mean
Standard error of the mean
Leo Legorburo
= 140
=
135
=
8
=
16
Leo Legorburo
Chapter 8 #32
1. 140-135/2
2. Z=2.5=.4938
3. .5-.4938=0.0062

What percentage of the sample means will be greater than 140 seconds? 0.0062
Leo Legorburo
Chapter 8 #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 special deviation of 40,000
a) If we select a random sample of 50 households, what is the standard error of the mean?
5657
b) What is the likelihood of selecting a sample with a mean at least \$112,00?
0.3632
c) What is the likelihood of selecting a mean of more than \$100,000?
0.9616
d) Find the likelihood of selecting a sample owith a mean of more than \$100,000 but less than \$112,000?
0.5984
Chapter 8 #34
=
\$40,000
50
= \$40,000
7.07
5657
A.
Chapter 8 #34
B)
What is the likelihood of selecting a sample with a mean of at least \$112,000
Use finding the Z value of x when the population standard deviation is known
=
112,000 - \$110,000
\$40,000/
50
=
2000
5657
z=0.35
0.5-0.1368 =
0.3632
z = 0.35 = 0.1368
Chapter 8 #34
C) What is the likelihood of selecting a sample with a mean of more than \$100,000?
z=
100,000 - 110,000
40000/
50
=
\$-10,000
=
1.77
5657
p= .4616 + .5 =
0.9616
Andrew Yanes
Chapter 8 #34
Find the likelihood of selecting a sample with a mean if more than 100,000 but less than \$112,000
z for 100,000 = -1.77
z for 112,000= 0.35
0.4646 + 0.1368 =
0.5984
Andrew Yanes
z for 1.77 = .4646
Chapter 8 #38
The mean amount purchased on a typical customer at churchills 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
a) What is the likelihood the sample mean is at least \$25,000?
0.0171
b) What is the likelihood the sample mean is greater than \$22.50 but less than \$25.00?
0.9037
Chapter 8 #38
25-23.5
5/
50
1.5
.71
=
2.12
z(2.12)=0.4830
0.5-0.4830= 0.0171
Kevin Zamora
A)
Chapter 8 #38
What is the likelihood the sample mean is greater than \$22.50 but less than \$25.00
Z =
22.50-23.50
5/
50
= 1.41
z(1.41)=0.0793
P(-1.41>Z>2.12)= 1-0.0793-0.0170 =
0.9037
Kevin Zamora
Leo Legorburo
Rolando Rodriguez
Rolando Rodriguez
Rolando Rodriguez
Kevin Zamora
What percent of the sample means will be greater than 128 but less than 140 seconds?
--
x = 128
mu = 135
standard deviation = 2
Charles English
128 - 135
----------
2
= 3.5 (z value)
Charles English
z value 3.5 = 0.9998

0.9998 - 0.0062 =
0.9938
Charles English
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