7.3

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Chapter
37
The Normal
Probability
Distribution
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Pearson Prentice Hall. All rights reserved
© 2010 Pearson Prentice Hall. All©rights
Section 7.3 Applications of the Normal Distribution
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7-2
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7-3
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7-4
EXAMPLE
Finding the Probability of a Normal Random
Variable
It is known that the length of a certain steel rod is normally
distributed with a mean of 100 cm and a standard deviation of
0.45 cm.* What is the probability that a randomly selected steel
rod has a length less than 99.2 cm?
99.2  100 

P( X  99.2)  P  Z 

0.45 

 P  Z  1.78
 0.0375
Interpretation: If we randomly selected 100 steel rods, we would expect about 4
of them to be less than 99.2 cm.
*Based
upon information obtained from Stefan Wilk.
© 2010 Pearson Prentice Hall. All rights reserved
7-5
EXAMPLE
Finding the Probability of a Normal Random
Variable
It is known that the length of a certain steel rod is normally
distributed with a mean of 100 cm and a standard deviation of
0.45 cm. What is the probability that a randomly selected steel
rod has a length between 99.8 and 100.3 cm?
100.3  100 
 99.8  100
P(99.8  X  100.3)  P 
Z 

0.45 
 0.45
 P  0.44  Z  0.67 
 0.4186
Interpretation: If we randomly selected 100 steel rods, we would expect about 42
of them to be between 99.8 cm and 100.3 cm.
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7-6
EXAMPLE
Finding the Percentile Rank of a Normal Random
Variable
The combined (verbal + quantitative reasoning) score on the GRE is normally
distributed with mean 1049 and standard deviation 189.
(Source: http://www.ets.org/Media/Tests/GRE/pdf/994994.pdf.)
The Department of Psychology at Columbia University in New York requires a
minimum combined score of 1200 for admission to their doctoral program.
(Source: www.columbia.edu/cu/gsas/departments/psychology/department.html.)
What is the percentile rank of a student who earns a combined GRE score of
1300?
The area under the normal curve is a probability, proportion, or percentile.
Here, the area under the normal curve to the left of 1300 represents the
percentile rank of the student.
Area left of 1300 = Area left of (z = 1.33)
= 0.91 (rounded to two decimal places)
Interpretation: The student scored at the 91st percentile. This means the student
scored better than 91% of the students who took the GRE.
7-7
© 2010 Pearson Prentice Hall. All rights reserved
EXAMPLE
Finding the Proportion Corresponding to a Normal
Random Variable
It is known that the length of a certain steel rod is normally
distributed with a mean of 100 cm and a standard deviation of
0.45 cm. Suppose the manufacturer must discard all rods less
than 99.1 cm or longer than 100.9 cm. What proportion of rods
must be discarded?
The proportion is the area under the normal curve to the left of 99.1 cm plus
the area under the normal curve to the right of 100.9 cm.
Area left of 99.1 + area right of 100.9 = (Area left of z = -2) + (Area right of z = 2)
= 0.0228 + 0.0228
= 0.0456
Interpretation: The proportion of rods that must be discarded is 0.0456. If the company
manufactured 1000 rods, they would expect to discard about 46 of them.
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7-8
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7-9
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7-10
EXAMPLE
Finding the Value of a Normal Random Variable
The combined (verbal + quantitative reasoning) score on the GRE is normally
distributed with mean 1049 and standard deviation 189.
(Source: http://www.ets.org/Media/Tests/GRE/pdf/994994.pdf.)
What is the score of a student whose percentile rank is at the 85th percentile?
The z-score that corresponds to the 85th percentile is the z-score such that the
area under the standard normal curve to the left is 0.85. This z-score is 1.04.
x = µ + zσ
= 1049 + 1.04(189)
= 1246
Interpretation: The proportion of rods that must be discarded is 0.0456. If the company
manufactured 1000 rods, they would expect to discard about 46 of them.
© 2010 Pearson Prentice Hall. All rights reserved
7-11
EXAMPLE
Finding the Value of a Normal Random Variable
It is known that the length of a certain steel rod is normally distributed with
a mean of 100 cm and a standard deviation of 0.45 cm. Suppose the
manufacturer wants to accept 90% of all rods manufactured. Determine
the length of rods that make up the middle 90% of all steel rods
manufactured.
z1 = -1.645 and z2 = 1.645
Area = 0.05
Area = 0.05
x1 = µ + z1σ
= 100 + (-1.645)(0.45)
= 99.26 cm
x2 = µ + z2σ
= 100 + (1.645)(0.45)
= 100.74 cm
Interpretation: The length of steel rods that make up the middle 90% of all steel rods
manufactured would have lengths between 99.26 cm and 100.74 cm.
© 2010 Pearson Prentice Hall. All rights reserved
7-12
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