PROJECT:  NORMAL PROBABILITY DISTRIBUTION

Objective:  Use the Normal Probability Distribution to make decisions about a population.

I.  The female division

Female Data in inches

66.4

68.1

66.7

67.9

63.1

67.8

66.1

68.9

66.1

69.2

64.9

67.6

57.6

65.1

66.7

67.8

66.8

63.6

67.5

60.2

69.4

68.4

62.2

67.2

64.7

66.3

64.2

62.2

64.3

67.2

63.2

58.1

 

A.  Using the calculator to find the sample mean and standard deviation (rounded to tenths) :
       1.  Mean height of an adult female is 65.5 inches.
      2.  Standard deviation of an adult female is 3 inches.

B1.  What percent of female adults are taller than 6 feet?

Using a single page Standard Normal Probability Distribution Table starting at the mean and extending to z.

z = 2.17 has a probability of 0.4850
0.5000 – 0.4850 = 0.0150 = 1.5%

Using a two page Standard Normal Probability Distribution Table starting at z and extending to the left.
z = 2.17 has a probability of 0.9850
1.0000 – 0.9850 = 0.0150 = 1.5%

           About 1.5% of females are taller than 6 feet.

B2.  What percent of female adults are taller than 5 feet?

Using a single page Standard Normal Probability Distribution Table starting at the mean and extending to z.

z = – 1.83 has a probability of 0.4664
0.5000 + 0.4664 = 0.9664 ≈ 96.6%

Using a two page Standard Normal Probability Distribution Table starting at z and extending to the left.
z = – 1.83 has a probability of 0.0336
1.0000 – 0.0336 = 0.9664 ≈ 96.6%

           About 96.6% of females are taller than 5 feet.

B3.  What percent of female adult heights are between 60 inches and 72 inches?        

Using a single page Standard Normal Probability Distribution Table starting at the mean and extending to z.
            z = – 1.83 has a probability of 0.4664.

              z = 2.17 has a probability of 0.4850.

0.4850 + 0.4664 = 0.9514 ≈ 95.1%

Using a two page Standard Normal Probability Distribution Table starting at z and extending to the left.
 z = 2.17 has a probability of 0.9850
z = – 1.83 has a probability of 0.0336
0.9850 – 0.0336 = 0.9514 ≈ 95.1%

           The percent of female adult heights between 60 inches and 72 inches is  about 95.1%

C1.  The bottom 8%

Using a single page Standard Normal Probability Distribution Table starting at the mean and extending to z.
Look up 0.4200 in the table.


 x = 65.5 + z(3)
x = 65.5 + (–1.41)(3)
x = 61.27
Using a two page Standard Normal Probability Distribution Table starting at z and extending to the left.
Look up 0.0800 in the table.

  
x = 65.5 + z(3)
x = 65.5 + (–1.41)(3)
x = 61.27

             The shortest female height for leather pants is approximately 5 ft 1 in.

C2.  The upper 6%

Using a single page Standard Normal Probability Distribution Table starting at the mean and extending to z.
Look up 0.4400 in the table.


 x = 65.5 + z(3)
x = 65.5 + (1.555)(3)
x = 70.165
Using a two page Standard Normal Probability Distribution Table starting at z and extending to the left.
Look up 0.9400 in the table.


x = 65.5 + z(3)
x = 65.5 + (1.555)(3)
x = 70.165

             The tallest female height for leather pants is approximately 5 ft 10 in.

 

II.  The male division

Male Data in inches

68

65.5

68.1

72.5

65.4

71.2

67.7

73.5

67.7

65.1

65.3

65.5

72

73.2

62.5

77.2

70.5

66.7

67.5

70.2

67.4

71.8

65.1

67.2

66.3

69.3

67.7

67

73.8

66.5

66.1

68.6

A.  Using the calculator to find the sample mean and standard deviation (rounded to hundredths) :
      1.  Mean height of an adult male is 68.5 inches.
      2.  Standard deviation of an adult male is 3.25 inches.

B1.  What percent of male adults are shorter than 6 feet?

Using a single page Standard Normal Probability Distribution Table starting at the mean and extending to z. 

z = 1.08 has a probability of 0.3599.

0.5000 + 0.3599 = 0.8599 ≈ 86.0%

 

Using a two page Standard Normal Probability Distribution Table starting at z and extending to the left.
z = 1.08 has a probability of 0.8599 ≈ 86.0%

             The percent of male adults shorter than 6 feet is about 86%.            

B2.  What percent of male adults are shorter than 5 feet?

Using a single page Standard Normal Probability Distribution Table starting at the mean and extending to z. 

 

 

z = – 2.62 has a probability of 0.4956.

0.5000 – 0.4956 = 0.0044 ≈ 0.4%

Using a two page Standard Normal Probability Distribution Table starting at z and extending to the left.
z = – 2.62 has a probability of 0.0044  ≈ 0.4%

            The percent of male adults shorter than 5 feet is about 0.4%.

B3.  What percent of male adult heights are between 60 inches and 72 inches?

Using a single page Standard Normal Probability Distribution Table starting at the mean and extending to z.

        z = – 2.62 has a probability of 0.4956.

           z = 1.08 has a probability of 0.3599.

0.4956 + 0.3599 = 0.8555 ≈ 85.6%

Using a two page Standard Normal Probability Distribution Table starting at z and extending to the left.
 z = 1.08 has a probability of 0.8599.
z = – 2.62 has a probability of 0.0044
 0.8599 – 0.0044 = 0.8555 ≈ 85.6%

            The percent of male adult heights between 60 inches and 72 inches is about 85.6%.

C1.  The bottom 9%

Using a single page Standard Normal Probability Distribution Table starting at the mean and extending to z.
Look up 0.4100 in the table.


x = 68.5 + z(3.25)
x = 68.5 + (–1.34)(3.25)
x = 64.145
Using a two page Standard Normal Probability Distribution Table starting at z and extending to the left.
Look up 0.0900 in the table.


x = 68.5 + z(3.25)
x = 68.5 + (–1.34)(3.25)
x = 64.145

              The shortest male height for leather pants is approximately 5 feet 4 inches.

C2.  The upper 7%

Using a single page Standard Normal Probability Distribution Table starting at the mean and extending to z.
Look up 0.4300 in the table.


x = 68.5 + z(3.25)
x = 68.5 + (1.48)(3.25)
x = 73.31
Using a two page Standard Normal Probability Distribution Table starting at z and extending to the left.
Look up 0.9300 in the table.


x = 68.5 + z(3.25)
x = 68.5 + (1.48)(3.25)
x = 73.31

                The tallest male height for leather pants is approximately 6 feet 1 inch.

 

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