**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