Objective: Use the Normal Probability Distribution to make decisions about a population.
I. The female division of the company
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Female Data in inches
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Using
the STAT functions to enter data and calculate the sample mean |
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A. The
sample mean is 65.5 inches and the sample standard deviation is 3 inches.
These values will be used as
estimates for the population mean and population
standard deviation when calculating the standard (z) scores.
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Using
the library function "normalcdf"
on your TI-83 calculator. |
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1. To access the
"Distributions" on the TI-83, press the "2nd" button
followed by the "VARS" button. Note: For an x
score of negative infinity (less than problems), use 10 standard
deviations to the left of the mean.
For an
x score of positive infinity (greater than problems), use 10 standard
deviations to the right of the mean. |
B1. What percent of female adults are taller than 6 feet?
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The percent of female adults taller than 6 feet is about 1.5%. This matches the answer using the table.
B2. What percent of female adults are taller than 5 feet?
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About 96.7% of females are taller than 5 feet. This answer differs from the answer using the table by 0.1%.
B3. What percent of female adult heights are between 60 inches and 72 inches?
The percent of female adult heights between 60 inches and
72 inches is about 95.1%.
This answer matches the
answer using the table.
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Using
the library function "invNorm"
on your TI-83 calculator. |
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An alternate solution to finding the solution to Example 3 is to use the inverse cumulative normal probability function on the TI-83. 1. To access the
"Distributions" on the TI-83, press the "2nd" button
followed by the "VARS" button. |
C1. The bottom 8%
The shortest female height for leather pants is
approximately 5 ft 1 in.
This
answer agrees with the answer using the table.
C2. The upper 6%.
Since
the "InvNorm" program works to the left, use 0.94 in the program.
The tallest female height for leather pants is
approximately 5 ft 10 in.
This
answer agrees with the answer using the table.
II. The male division of the company
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Male Data in inches
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Using
the STAT functions to enter data and calculate the sample mean |
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A. The
sample mean is 68.5 inches and the sample standard deviation is 3.25
inches. These values will be used as
estimates for the population mean and population
standard deviation when calculating the standard (z) scores.
|
|
Using
the library function "normalcdf"
on your TI-83 calculator. |
|
1. To access the
"Distributions" on the TI-83, press the "2nd" button
followed by the "VARS" button. Note: For an x
score of negative infinity (less than problems), use 10 standard
deviations to the left of the mean.
For an
x score of positive infinity (greater than problems), use 10 standard
deviations to the right of the mean. |
B1. What percent of male adults are shorter than 6 feet?
![]() |
The percent of male adults shorter than 6 feet is about
85.9%.
This
answer differs from the answer using the table by 0.1%.
B2. What percent of male adults are shorter than 5 feet?
The percent of male adults shorter than 5 feet is about
0.4%.
This answer
agrees with the answer using the table.
B3. What percent of male adult heights are between 60 inches and 72 inches?
The percent of male adult heights between 60 inches and 72
inches is about 85.5%.
This answer
differs from the answer using the table by 0.1%.
|
|
Using
the library function "invNorm"
on your TI-83 calculator. |
|
An alternate solution to finding the solution to Example 3 is to use the inverse cumulative normal probability function on the TI-83. 1. To access the
"Distributions" on the TI-83, press the "2nd" button
followed by the "VARS" button. |
C1. The bottom 9%.
![]() |
The shortest male height for leather pants is
approximately 5 feet 4 inches.
This answer agrees with the answer using the table.
C2. The upper 7%.
Since the "InvNorm" program works to the left, use 0.93 in the
program.
![]() |
![]() |
The tallest male height for leather pants is approximately
6 feet 1 inch.
This
answer agrees with the answer using the table.
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