P(441
446 |
= 448 and
= 21)
Steps 1 & 2
: We need to calculate two Z's here.
Z
1
=
=
=
=
-2.3333
[Z-Probability =
.0098
]
Z
2
=
=
=
=
-.6667
[Z-Probability =
.2425
]
Step 3:
Interpretation: We want to calculate the area under the curve between 441 and 446 (Z
1
and Z
2
). We subtract the
smallest Z-Probability from the largest
Z-Probability.
.2525 - .0098
=
.2427
→ P(441
446 |
= 448 and
= 21) =
.2427
•
Z-Score
-2.3333
Z-Probability
0.0098
=NORMSDIST(-2.3333)
Z-Score
-0.6667
Z-Probability
0.2525
=NORMSDIST(-0.6667)

Module 10: Introduction to Sampling
A.
The Sampling Distribution of
(Sample Mean)
1. Population size (N) unknown
Example 2
:
P(441
446 |
= 448 and
= 21) =
.2427
•

Module 10: Introduction to Sampling
A.
The Sampling Distribution of
(Sample Mean)
2. Population size (N) known
Occasionally we'll know the size of the population (N) from which the sample (n) was drawn.
When that is the case, we
have to apply the Finite Correction Factor. The FCC allows us to be a little bit more precise as it incorporates more
information into our analysis. The goal in statistics is to be as accurate as possible, and to that end a good statistician will
never leave relevant information out of a study.
Therefore, the rule in this class is as follows: If you know N, you must apply the FCC.
The formula for the FCC:
FCC =
The FCC fits into the denominator of our Z-Formula:
Z =
The effect of the FCC is to make the Z-Score larger
•

Module 10: Introduction to Sampling
A.
The Sampling Distribution of
(Sample Mean)
2. Population size (N) known
Example 3
: The average age of a production company's hourly employees is 37.6 years with a standard deviation of 8.3 years. If
a random sample of 45 employees is taken, what is the probability that the ample will have an average age of less than 40
years? Suppose the production company has 350 hourly employees.
P(
40 |
= 37.6,
= 8.3, and N = 350)
Note that in this case we're given an additional piece of information: The population size (350). There we must apply the FCC
Steps 1 & 2
: Calculate the Z-Score using the new formula and use Excel to determine the Z-Probability:
Z =
= Z =
=
=
= 2.0749
[Z-Probability =
.9810
]
Step 3:
Interpretation: .9810 is the area under the curve to the left of 40. We want to determine the area to the
left of 40 P(X < 40).
P(
40 |
= 37.6,
= 8.3, and N = 350) =
.9810
•
Z-Score
2.0749
Z-Probability
0.9810
=NORMSDIST(2.0749)

Module 10: Introduction to Sampling
A.
The Sampling Distribution of
(Sample Mean)
2. Population size (N) known
Example 3
: P(
40 |
= 37.6,
= 8.3, and N = 350) =
.9810
•

Module 10: Introduction to Sampling
A.
The Sampling Distribution of
(Sample Mean)
2. Population size (N) known
Example 4
: The average cost of a one-bedroom apartment in a particular town is $750 per month with a standard deviation of
$125. If a random sample of 50 one-bedroom apartments is obtained, what is the probability the sample mean will be greater
than $730. Assume there are a total of 485 one-bedroom apartments in this town.
P(
$730 |
= $750,
= $125, and N = 485)
Steps 1 & 2
: Calculate the Z-Score using the new formula and use Excel to determine the Z-Probability:
Z =
= Z =
=
=
= -1.1934
[
Z-Probability =
.1164
]
Step 3:
Interpretation: .1164 is the area under the curve to the left of 730. We want to determine the area to the
right of 730
P(X > 730).
We subtract .1164 from 1.
1 - .1164 =
.8836
P(
$730 |
= $750,
= $125, and N = 485) =
.8836
•
Z-Score
-1.1934
Z-Probability
0.1164
=NORMSDIST(-1.1934)

Module 10: Introduction to Sampling
A.
The Sampling Distribution of

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- Spring '14
- DebraACasto
- Standard Deviation