-
Chapter 5: Normal Probability Distributions 109
Chapter 5. Normal Probability Distributions
5-2 The Standard Normal Distribution
Using a Continuous Uniform Distribution. In Exercises 1-4, refer
to the continuous uniform distribution depicted in figure 5-2,
assume that a class length between 50.0 min and 52.0 min is
randomly selected, and find the probability that the given time is
selected.
1. 0.15 3.00.5 )503.50( 0.5 minutes) 50.3 than less (class
===P
2. 0.5 10.5 51)-(520.5 minutes) 51.0an greater th (class
===P
3. 15.00.30.5 50.5)-(50.8 0.5 minutes) 50.8 and minutes
50.5between (class ===P
4. 65.01.3)0.5 (50.5)-(51.80.5 min) 51.8 andmin 50.5between
(class ===P
Using the Standard Normal Distribution. In Exercises 5-8, assume
that voltages in a circuit vary between 6 volts and 12 volts, and
voltages are spread evenly over the range of possibilities, so that
there is a uniform distribution. Find the probability of the given
range of voltage levels.
5. For a discrete probability distribution, P(x) =1. Since the
values on the x axis range from 6 to 12, this is a range of 6.0. To
get the closed area within the rectangle to be equal to 1, the
height of the rectangle has to be 1/6 = 0.167 and these are placed
adjacent to each other to cover all values in the full range of 6
to 12
P(voltage greater than 10 volts) = 333.03/16/2261)1012(
61
====
6. P (voltage less than 11 volts) = 833.06/5561)611(
61
===
7. P (voltage between 7 and 10 volts) = 500.02/16/3361)710(
61
====
8. P(voltage between 6.5 and 8.0 volts) =
250.04/16/5.15.161)5.68(
61
====
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110 Chapter 5: Normal Probability Distributions
Using the Standard Normal Distribution. In Exercises 9-28,
assume that the readings on scientific thermometers are normally
distributed with a mean of 0C and a standard deviation of 1.00C. A
thermometer is randomly selected and tested. In each case, draw a
sketch, and find the probability of each reading in degrees
Celsius.
9. Less than 0.25. The probability distribution of readings is a
standard normal distribution because the readings are normally
distributed with a mean of 0 and standard deviation of 1. We need
to find the area below z= 0.25. From Table A-2, this is 0.4013.
So, P(x < 0.25) = 0.4013.
10. Probability of a thermometer reading less than 2.75C, z=
2.75 Area below z of 2.75= 0.0030, P(x < 2.75) = 0.0030
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4
z=-0.25
Area found in Table A-2= 0.4013
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4
Area found in Table A-2= 0.0030
z=-2.75
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Chapter 5: Normal Probability Distributions 111
11. Probability of a thermometer reading less than 0.25C, z=
+0.25 Area below z of +0.25= 0.5987, P(x < +0.25) = 0.5987
12. Probability of a thermometer reading less than 2.75C, z=
+2.75 Area below z of +2.75= 0.9970, P(x < +2.75) = 0.9970
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4
Area found in Table A-2= 0.5987
z=0.25
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4
Area found in Table A-2= 0.9970
z=2.75
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112 Chapter 5: Normal Probability Distributions
13. Probability of a thermometer reading greater than 2.33C, z=
+2.33 Area below z of +2.33= 0.9901, P(x > +2.33) = 1 0.9901 =
0.0099
14. Probability of a thermometer reading greater than 1.96C, z=
+1.96 Area below z of +1.96= 0.9750, P(x > +1.96) = 1 0.9750 =
0.0250
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4
Area found in Table A-2= 0.9901
z=2.33
Area= 1- 0.9901= 0.0099
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4
Area found in Table A-2= 0.9750
z=1.96
Area = 1- 0.9750= 0.0250
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Chapter 5: Normal Probability Distributions 113
15. Probability of a thermometer reading greater than 2.33C, z=
2.33 Area below z of 2.33= 0.0099, P(x > 2.33) = 1 0.0099=
0.9901
16. Probability of a thermometer reading greater than 1.96C, z=
1.96 Area below z of 1.96= 0.0250, P(x > 1.96) = 1 0.0250=
0.9750
-4 -3.5
-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5
2 2.5 3 3.5 4
Area found in Table A-2= 0.0099
z=-2.33
Area= 1- 0.0099= 0.9901
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4
Area found in Table A-2= 0.0250
z= -1.96
Area= 1- 0.0250= 0.9750
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114 Chapter 5: Normal Probability Distributions
17. Probability of a thermometer reading between 0.5C and 1.5C,
between z= +0.50 and z= +1.50, Area below z of +1.50= 0.9332 and
area below z of +0.50= 0.6915 P(+0.50 < x< +1.50) = 0.9332
0.6915 = 0.2417
18. Probability of a thermometer reading between 1.5C and 2.5C,
between z= +1.50 and z= +2.50, Area below z of +2.50= 0.9938 and
area below z of +1.50= 0.9332 P(+1.50 < x < +2.50) = 0.9938
0.9332 = 0.0606
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4
z= 2.50
Total area up to z=2.50= 0.9938
z =1.50
Area= 0.9938-0.9332= 0.0606
Area found in Table A-2=0.9332
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4
Area found in Table A-2= 0.6915
z= 0.50
Total area up to z=1.50= 0.9332
z =1.50
Area= 0.9332-0.6915= 0.2417
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Chapter 5: Normal Probability Distributions 115
19. Probability of a thermometer reading between 2.00C and 1.0C,
z= 2.00 and z= 1.00 Area below z of 1.00 is 0.1587 and area below z
of 2.00 is 0.0228 P(2.00 < x < 1.00)= 0.1587 0.0228=
0.1359
20. Probability of a thermometer reading between 2.00C and
2.34C, z= +2.00 and z= +2.34 Area below z of +2.34 is 0.9904 and
area below z of +2.00 is 0.9772 P(+2.00 < x < +2.34)= 0.9904
0.9772= 0.0132
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4
Area found in Table A-2= 0.0228
z= -1.0
Total area up to z=-1.0 = 0.1587
z =-2.0
Area= 0.1587- 0.0228= 0.1359
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4
Area= 0.9904-0.9772= 0.0132
z= 2.34
Total area up to z=2.34=.9904
z =2.0
Area from table A-2= 0.9772
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116 Chapter 5: Normal Probability Distributions
21. Probability of a thermometer reading between 2.67C and
1.28C, z= 2.67 and z= +1.28 Area below z of +1.28 is 0.8997 and
area below z of 2.67 is 0.0038 P(2.67 < x < +1.28)= 0.8997
0.0038= 0.8959
22. Probability of a thermometer reading between 1.18C and
2.15C, z= 1.18 and z= +2.15 Area below z of +2.15 is 0.9842 and
area below z of 1.18 is 0.1190 P(1.18 < x < +2.15)= 0.9842
0.1190 = 0.8652
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4
Area= 0.8997-0.0038= 0.8959
z= 1.28
Total area up to z=2.34=0.8997
z=-2.67
Area from table A-2= 0.0038
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4
Area= 0.9842-0.1190= 0.8652
z= 2.15
Total area up to z=2.15= 0.9842
z=-1.18
Area from table A-2= 0.1190
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Chapter 5: Normal Probability Distributions 117
23. Probability of a thermometer reading between 0.52C and
3.75C, z= 0.52 and z= +3.75 Area below z of +3.75 is 0.9999 and
area below z of 0.52 is 0.3015 P(0.52 < x < +3.75)= 0.9999
0.3015 = 0.6984
24. Probability of a thermometer reading between 3.88C and
1.07C, z= 3.88 and z= +1.07 Area below z of +1.07 is 0.8577 and
area below z of 3.88 is 0.0001 P(3.88 < x < +1.07)= 0.8577
0.0001 = 0.8576
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4
Area= 0.9999-0.3015= 0.6984
z= 3.75
Total area up to z=3.75= 0.9999
z=-0.52
Area from table A-2= 0.3015
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4
Area= 0.8577-0.0001=0.8576
z= 1.07
Total area up to z=1.07= 0.8577
z=--3.88
Area from table A-2= 0.0001
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118 Chapter 5: Normal Probability Distributions
25. Probability of a thermometer reading greater than 3.57C, z=
+3.57 Area below z of +3.57=0.9999, P(x > +3.57) = 1 0.9999 =
0.0001
26. Probability of a thermometer reading less than -3.61C, z=
3.61 Area below z of 3.61= 0.0002, P(x < 3.61) = 0.0001
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4
z= 3.57
Area from table A-2= 0.9999
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4
Area= 1-0.0001= 0.9999
z= -3.61
Area= 1-0.9999= 0.0001
Area from Table A-2= 0.0001
-
Chapter 5: Normal Probability Distributions 119
27. Probability of a thermometer reading greater than 0C, z=
0.00 Area below z of 0.00= 0.5000, P(x > 0.00) =1 0.5000=
0.5000
28. Probability of a thermometer reading less than 0C, z= 0.00
Area below z of 0.00= 0.5000, P(x < 0.00) = 0.5000
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4
Area= 1-0.5000=0.5000
z= 0
Area from Table A-2=0.5000
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4
z= 0
Area from Table A-2= 0.5000
-
120 Chapter 5: Normal Probability Distributions
Basis for Empirical Rule. In Exercises 29-32, find the indicated
area under the curve of the standard normal distribution, then
convert it to a percentage and fill in the blank. The results form
the basis for the empirical rule introduced in Section 2-5.
29. About 68.26% of the area is between z = 1 and z = +1 (or
within one standard deviation of the mean). Since the area below z=
1.00 is 0.1587 the area between the mean and z= 1.00 is 0.5000
0.1587 = 0.3413, then the total area between z= 1.00 and z= +1.00
is
2 0.3413= 0.6826, converted to a percentage is 0.6826 100%
68.26%
30. About 95.44% of the area is between z= 2 and z= +2 (or
within two standard deviation of the mean). Since the area below z=
2.00 is 0.0228 the area between the mean and z= 2.00 is 0.5000
0.0228 = 0.4772, then the total area between z= 2.00 and z= +2.00
is
2 0.4772= 0.9544, converted to a percentage is 0.9544 100%=
95.44%
31. About 99.74% of the area is between z= 3 and z = +3 (or
within three standard deviation of the mean). Since the area below
z= -3.00 is 0.0013 the area between the mean and z= 3.00 is 0.5000
0.0013 = 0.4987, then the total area between z= 3.00 and z= +3.00
is
2 0.4987= 0.9974, converted to a percentage is 0.9974 100%=
99.74%
32. About 99.98%of the area is between z= 3.5 and z = +3.5 (or
within 3.5 standard deviation of the mean). Since the area below z=
3.50 is 0.0001 the area between the mean and
z= -3.50 is 0.5000 0.0001 = 0.4999, then the total area between
z= 3.50 and z= +3.50 is 2 0.4999= 0.9998, converted to a percentage
is 0.9998 100%= 99.98%
Finding Probability. In Exercises 33-36, assume that the
readings on the thermometers are normally distributed with a mean
of 0C and a standard deviation of 1.00C. Find the indicated
probability, where z is the reading in degrees.
33. P (1.96 < z 2.575) = 1 (Area below z= 2.575) = 1 0.0050 =
0.9950
36. P (1.96< z < 2.33) = (Area below z= +2.33) (Area below
z= +1.96) = 0.9901 0.9750= 0.0151
Finding Temperature Values. In Exercises 37-40, assume that the
readings on the thermometers are normally distributed with a mean
of 0C and a standard deviation of 1.00C. A thermometer is randomly
selected and tested. In each case, draw a sketch, and find the
temperature reading corresponding to the given information.
37. 0.90 in the body of the table corresponds to a z score of
+1.28. So, the 90th percentile is the temperature reading of +
(1.28 ) = 0 + (1.28 1.00) = 1.28C.
-
Chapter 5: Normal Probability Distributions 121
38. 0.20 in the body of the table corresponds to a z score of
0.84. So, the 20th percentile is the temperature reading of + (0.84
) = 0 + (0.84 1.00) = 0.84C.
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 z=
-0.84
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 z=
1.28
Area from Table A-2=0.9000
Area from Table A-2= 0.2000
-
122 Chapter 5: Normal Probability Distributions
39. 0.05 in the body of the table corresponds to a z score of
1.645. So, the 5th percentile is the temperature reading of +
(1.645 ) = 0 + (1.645 1.00) = 1.645C.
40. 0.03 in the body of the table corresponds to a z score of
1.88. This is the lower cutoff point. 1 0.03= 0.97 in the body of
the table corresponds to a z score of +1.88. This is the higher
cutoff point. Thus, thermometers with reading lower than 1.88 C or
higher than +1.88 C would be rejected and thermometers between 1.88
would not be rejected. In practice, values of 1.88 or +1.88 would
probably be rejected in this case since it indicates the lowest and
highest 3% would be rejected.
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 z=
-1.88
Area from Table A-2= 0.0300
z= -1.88
Area between 1.88= 0.9700
Total area up to z=1.88= 0.9700
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 z=
-1.645
Area from Table A-2= 0.0500
-
Chapter 5: Normal Probability Distributions 123
41. a. The percentage of data that are between one standard
deviation from the mean corresponds to the area between 1.00z and
+1.00z scores. This area is 68.26%.
b. The percentage of data that are between 1.96 standard
deviations from the mean corresponds to the area between 1.96z and
+1.96z scores. This area is 95.00%.
c. The percentage of data that are between three standard
deviations from the mean corresponds to the area between 3.00z and
+3.00z scores. This area is 99.74%.
d. The percentage of data that are between one standard
deviation below the mean and two standard deviations above the mean
one corresponds to the area between 1.00z and +2.00z scores. This
is 0.9772 0.1587 = 0.8185. This area is 81.85%
e. The percentage of data that are more than two standard
deviations away from the mean corresponds to 1 Area between 2.00z
and +2.00z scores = 1 0.9544 = 0.0456 or 4.56%.
5-3 Applications of Normal Distributions
IQ Scores. In Exercises 1-8, assume that adults have IQ scores
that are normally distributed with a mean of 100 and a standard
deviation of 15 (as on the Wechsler test). (Hint: Draw a graph in
each case.)
1. The IQ of 115 is converted to a z score as follows:
00.11515
15100115
+==
=
=
xz
Referring to Table A-2, z = +1.00 corresponds to an area of
0.8413, so P(IQ < 115) = 0.8413
2. The IQ of 131.5 is converted to a z score as follows:
10.215
5.3115
1005.131+==
=
=
xz
Referring to Table A-2, z = +2.10 corresponds to an area of
0.9821, so P(IQ > 131.5) = 1 0.9821= 0.0179
100z
0
x(IQ) 115
1
Area = 0.8413
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124 Chapter 5: Normal Probability Distributions
3. The IQs of 90 and 110 are converted to a z scores as
follows:
67.01510
15100110
,67.01510
1510090
+==
=
==
=
=
=
xz
xz
Referring to Table A-2, z = 0.67 corresponds to an area of
0.2514 and z = +0.67 corresponds to an area of 0.7486, so P(90 <
IQ < 110) = 0.7486 0.2514 = 0.4972
100z
0
x(IQ) 131.5
2.1
Area below z= 2.10= 0.9821
Area= 1-0.9821= 0.0179
100z
0
x(IQ) 110
0.67
Area= 0.2514 Area= 0.7486-0.2514= 0.4972
90
-0.67
Total area up to z= 0.67= 0.7486
-
Chapter 5: Normal Probability Distributions 125
4. The IQs of 110 and 120 are converted to a z scores as
follows:
33.11520
15100120
,67.01510
15100110
+==
=
=+==
=
=
xz
xz
Referring to Table A-2, z = +0.67 corresponds to an area of
0.7486 and z = +1.33 corresponds to an area of 0.9082, so P(110
< IQ
-
126 Chapter 5: Normal Probability Distributions
7. The IQ score separating the top 15% from the others is the
same score that separates the bottom (100 15) % from the others100
15 = 85. We find 0.85 in the body of the table and find the
corresponding z score. The z score for a cumulative area of 0.85 =
1.04
( ) 6.1156.15100)1504.1(100 =+=+=+= zx The IQ score separating
the top 15% from the others = 115.6
8. The IQ score separating the top 55% from the others is the
same score that separates the bottom (100 55) % from the others.100
55 = 45. We find 0.45 in the body of the table and find the
corresponding z score. The z score for a cumulative area of 0.45 =
0.13
( ) ( ) 05.9895.1100)1513.0(100 =+=+=+= zx The IQ score
separating the top 55% from the others = 98.05
100z
0
x(IQ)
0.84
Area= 0.80
112.6
100z
0
x(IQ)
1.04
Area = 0.85
115.6
Area = 0.15
-
Chapter 5: Normal Probability Distributions 127
9. Body Temperature a. 6.100 0.62, ,20.98 === x
87.362.04.2
62.02.986.100
z +==
=
=
x
From Table A-2, P(Temperature < 100.6) = P (z< +3.87) =
0.9999. P(Temperature> +3.87) = P(z > +3.87) = 1 0.9999 =
0.0001 This corresponds to 0.01%. Yes, this percentage suggests
that the cutoff of 100.6C is appropriate. b. Since we want 5% of
the people to exceed the required temperature, we use (100 5)%to
find the area to
the left of the cutoff line first. This corresponds to an area
0.95. From Table A-2, this corresponds to a z score of +1.645. ( )
22.9902.12.98)62.0645.1(2.98 =+=+=+= zx Thus, 5% of the people will
exceed 99.2C
10. Lengths of Pregnancies, 15 ,268 == a. x = 308. We are to
find P(Pregnancy> 308days). We find P(Pregnancy < 308) and
subtract it from 1.
67.21540
15268308
+==
=
=
xz
From the Table, P(z < 2.67) = P(pregnancy < 308) = 0.9962
P(pregnancy > 308) =1 0.9962 = 0.0038 This result shows that is
highly unlikely for a pregnancy to last 308 days or more. Therefore
it is more likely
that her husband is not responsible for her pregnancy, but there
is no proof one way or the other. b. If premature babies are in the
lower 4%, we find the cutoff time for the area 0.04. ( )
75.24125.26268)1575.1(268 ==+=+= zx So, the length that separates
premature babies from normal ones is 242 days.
11. Designing Helmets, 1 ,6 == To find the cutoff points for the
smallest 2.5% and the largest 2.5%, we find the z scores for the
areas 0.025 and
(1 0.025) or 0.975. From the table, these are 1.96 and +1.96
respectively.
( )( ) 896.796.16)196.1(6
404.496.16)196.1(6=+=++=+=
==+=+=
zx
zx
The minimum and maximum head breadths are 4 inches and 8 inches
respectively.
100z
0
x(IQ)
-0.13
Area= 0.45
98.05
Area= 0.55
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128 Chapter 5: Normal Probability Distributions
12. CD Player Warranty, 4.1 ,1.7 ==
a. 64.04.19.0
4.11.78
,0.8 +=====
xzx .
The area for this z score is 0.7389. So the probability that a
CD player will have a replacement time less than 8 years is
0.7389
b. We need to find the cutoff point for the upper 2%. So, we
find the z score for an area of (10.02) or 0.98. This corresponds
to z= + 2.05. 97.987.21.7)4.105.2(1.7)( =+=+=+= zx
Therefore, the time length of the warranty should be 10
years.
Heights of Women. In Exercises 13-16, assume that heights of
women are normally distributed with a mean given by = 63.6 in. and
a standard deviation given by = 2.5 in. (based on data from the
National Health Survey). In each case, draw a graph.
13. Beanstalk Club Height Requirement
56.25.24.6
5.26.6370
,5.2 ,6.63 +======
xz
This corresponds to a probability of 0.9948. So, 99.48% of the
women have height < 70 in. Therefore (100 99.48) or 0.52% of the
women meet the requirement of being at least 70in. in height.
14. Height Requirement for Women Soldiers We need to find the z
scores and areas for 58 in. and 80 in.
56.65.24.16
5.26.6380
24.25.26.5
5.26.6358
+==
=
=
=
=
=
=
xz
xz
The areas for these z scores are 0.0125 and 0.9999 respectively.
The probability of being between these heights is 0.9999 0.0125 =
0.9874. So, 98.74% of women meet this requirement. Not many women
are being denied entry into the army due to height.
63.6z
2.56
70
Height(in) )
Area = 0.9948 Area = 0.0052
-
Chapter 5: Normal Probability Distributions 129
15. Height Requirement for Rockettes We need to find the z
scores and areas for 66.5 in. and 71.5 in.
16.35.29.7
5.26.635.7116.1
5.29.2
5.26.635.66
+==
=
=+==
=
=
xz
xz
The areas for these z scores are 0.8770 and 0.9992 respectively.
The probability of being between these heights is 0.9992 0.8770 =
0.1222. The probability of meeting this new height is 0.1222. Only
12.22% of women meet this requirement. Yes, it seems that the
height of the Rockettes is well above the mean.
63.6 z
Height(in) )
Area= 0.0125
Total Area up to z= 6.56= 0.9999
-2.24 6.56
80
0
58
Area=0.9874
63.6z
Height(in) )
Area = 0.8770
Total Area up to z= 3.16= 0.9992
3.16 1.16
66.5
0
71.5
Area = 0.1222
-
130 Chapter 5: Normal Probability Distributions
16. Height Requirement for Rockettes To find the cutoffs for the
shortest 20% and the tallest 20%, we need to find to find the z
scores corresponding
to the areas 0.20 and (1 0.20) or 0.80. From the Table, these z
values are 0.84 and +0.84. We then use the formula:
( )( ) 7.651.26.63)5.284.0(6.63
5.611.26.63)5.284.0(6.63=+=+=+=
==+=+=
zx
zx
So, the new minimum and maximum allowable heights are 61.5 in.
and 65.7 in. respectively.
17. Birth Weights, 495 ,3420 == To find the cutoff weights for
the lightest 2% we need to find to find the z score corresponding
to the area 0.02.
From the Table, the z score is -2.05. We then use the formula: (
) 25.240575.10143420)49505.2(3420 ==+=+= zx . Therefore, the weight
of 2405g separates the lightest 2% of American babies from the
others.
63.6z
Height(in) )
-0.84 0.84
61.5
0
65.7
3420 z
Weight(g)
-2.05
2405
0
Area = 0.02
-
Chapter 5: Normal Probability Distributions 131
18. Birth Weights, 500 ,3570 == To find the cutoff weights for
the lightest 2% we need to find to find the z scores corresponding
to the areas
0.02. From the Table, this z is 2.05. We then use the formula: (
) 254510253570)50005.2(3570 ==+=+= zx . Therefore, the weight of
2545g
separates the lightest 2% of Norwegian babies from the others.
This result is not very different from the result in Exercise 17.
Its a difference of 140g.
19. Units of Measurement, 29 ,143 == a. z scores are measured in
units of number of standard deviations from the mean, but they do
not possess the
units of the original variable b. The mean will be 0, the
standard deviation will be 1, and the distribution will be normal
since the original
distribution is normal. z scores have the same shape of
distribution as does the original variable distribution; converting
to z scores does not result in a normal distribution of z scores if
the original distribution was not normally distributed
c. After converting to kg., the distribution will be normal
since the original distribution is normal, 1 lb= 0.4536 kg
deviation standard kg15.31kg 29) (0.4536 lb29mean kg64.86kg 143)
(0.4536 lb143
===
===
20. Using Continuity Correction a. 105 ,15 100 === x
33.0155
15100105
+==
=
=
xz
So, P(IQ < 105) = 0.6293. Therefore, P(IQ >105) = 1 0.6293
= 0.3707 b. We will replace 105 with an interval of 104.5 and
105.5. Because we want the probability of a score greater
than 105, we want the area bounded by the interval including the
area to the right. We convert 104.5 to a z score
30.015
5.415
1005.104+==
=
=
xz
So P(IQ104.5) = 1 0.6179=0.3821. P(IQ >105, adjusted for
continuity) = 0.3821 c. The results from (a) and (b) are nearly the
same. There is very little difference.
3570z
Weight(g)
-2.05
2545
0
Area= 0.02
-
132 Chapter 5: Normal Probability Distributions
5-4 Sampling Distributions and Estimators
1. Survey of Voters No, we cannot assume that the survey was
done incorrectly because the value of a statistic varies from
sample
to sample due to sampling variability. In this example, the
values for the sample proportion are different because of sampling
variability. A variation of 49% and 51% would seem to happen by
chance relatively often.
2. Sampling Distribution of Cholesterol Levels The sampling
distribution is a distribution of all possible means of the
cholesterol levels of any 40 randomly
selected women.
3. Sampling Distribution of Body Temperatures No, the histogram
will not show the shape of a sampling distribution of sampling
means. It will show the
distribution of individual values within one sample. A sampling
distribution will show a distribution of all possible means of
similar samples with the same sample size.
4. Sampling Distribution of Survey Results a. The 52% is a
statistic because it gives the value for one sample. b. The
sampling distribution suggested by the data is the distribution of
the proportions of all possible samples
of 1038 randomly selected people. c. I would feel more confident
if the sample size were 2000 because larger sample sizes tend to
have greater
representation of the population and they tend to have lower
error.
5. Phone Center Selecting samples with replacement, there will
be 32= 9 equally likely samples.
Sample Number a. Sample Sample Mean, x b. Probability
1 10,10 10.0 1/9 2 10, 6 8.0 1/9 3 10, 5 7.5 1/9 4 6, 10 8.0 1/9
5 6, 6 6.0 1/9 6 6, 5 5.5 1/9 7 5, 10 7.5 1/9 8 5, 6 5.5 1/9 9 5, 5
5.0 1/9
Sum of Sample Means =x 63.0
Mean of statistic values ==
9
x 7.0
Population parameter ==
++=
321
35610 7.0
Sampling Distribution Sample
Mean, x Probability
10.0 1/9 8.0 2/9 7.5 2/9 6.0 1/9 5.5 2/9 5.0 1/9
-
Chapter 5: Normal Probability Distributions 133
b. The probability of each sample is 1/9. The distribution of
sample means is bi-modal and somewhat flat.
c. Mean of sample statistics= 0.7963
9===
x d. Yes, the mean of the sampling distribution is equal to the
mean of the population of the three values. Yes,
these means are always equal, but only if every possible sample
is included.
6. Telemarketing Selecting samples with replacement, there will
be 42= 16 equally likely samples.
Sampling Distribution
Sample Mean, x Probability
11 1/16 10 2/16 9 1/16 7 2/16 6 4/16 5 2/16 3 1/16 2 2/16 1
1/16
b. The sampling distribution is of the 16 sample means, each of
which has a probability of occurring. It has one mode and it is
symmetrical.
Sample Number a. Sample
Sample Mean, x
c. Probability
1 1, 1 1 1/16 2 1, 11 6 1/16 3 1, 9 5 1/16 4 1, 3 2 1/16 5 11,
11 11 1/16 6 11, 1 6 1/16 7 11, 9 10 1/16 8 11, 3 7 1/16 9 9, 9 9
1/16
10 9, 1 5 1/16 11 9, 11 10 1/16 12 9, 3 6 1/16 13 3, 3 3 1/16 14
3, 1 2 1/16 15 3, 11 7 1/16 16 3, 9 6 1/16
Sum of Sample Means
=x 96.0 Mean of statistic values
==16
x 6.0
Population parameter ==
+++=
424
439111
6.0
-
134 Chapter 5: Normal Probability Distributions
c. Mean of sample statistics= 0.61696
16===
x d. Yes, the mean of the sampling distribution is equal to the
mean of the population of the four values.
Yes, these means are always equal, but only if every possible
sample is included.
7. Heights of L.A. Lakers Selecting samples with replacement,
there will be 52= 25 equally likely samples.
Sample Number a. Sample Sample Mean, x Probability
1 85, 85 85.0 1/25 2 85, 79 82.0 1/25 3 85, 82 83.5 1/25 4 85,
73 79.0 1/25 5 85, 78 81.5 1/25 6 79, 79 79.0 1/25 7 79, 85 82.0
1/25 8 79, 82 80.5 1/25 9 79, 73 76.0 1/25
10 79, 78 78.5 1/25 11 82, 82 82.0 1/25 12 82, 85 83.5 1/25 13
82, 79 80.5 1/25 14 82, 73 77.5 1/25 15 82, 78 80.0 1/25 16 73, 73
73.0 1/25 17 73, 85 79.0 1/25 18 73, 79 76.0 1/25 19 73, 82 77.5
1/25 20 73, 78 75.5 1/25 21 78, 78 78.0 1/25 22 78, 85 81.5 1/25 23
78, 79 78.5 1/25 24 78, 82 80.0 1/25 25 78, 73 75.5 1/25
Sum of Sample Means =x 1985
Mean of statistic values ==
25
x 79.4
Population parameter ==
++++=
5397
57873827985 79.4
-
Chapter 5: Normal Probability Distributions 135
Sampling Distribution
Sample Mean, x Probability
85.0 1/25 83.5 2/25 82.0 3/25 81.5 2/25 80.5 2/25 80.0 2/25 79.0
3/25 78.5 2/25 78.0 1/25 77.5 2/25 76.0 2/25 75.5 2/25 73.0
1/25
b. The probability of each sample occurring is 1/25. The
sampling distribution of means consists of the 25 sample means with
their corresponding probabilities. It has more than one mode and it
is not symmetrical.
c. The means of the sampling distribution is 4.7925
1985==
=
n
x d. Yes, the mean of the sampling distribution is equal to the
mean of the population of the five heights listed
above. Yes, these means are always equal as long as every
possible sample is included.
-
136 Chapter 5: Normal Probability Distributions
8. Genetics, p(F)= 3/4= 0.75, q= 0.25 Selecting samples with
replacement, there will be 42= 16 equally likely samples.
M=Mike(male)=0, A=Anna(female)=1, B=Barbara(female)=1,
C=Chris(female)=1
Sample Number a. Sample
Proportion of Females
(Sample Mean) Probability
1 M,M= 0, 0 0.0 1/16 2 M,A= 0, 1 0.5 1/16 3 M,B= 0, 1 0.5 1/16 4
M,C= 0, 1 0.5 1/16 5 A,A= 1, 1 1.0 1/16 6 A,M= 1, 0 0.5 1/16 7 A,B=
1, 1 1.0 1/16 8 A,C= 1, 1 1.0 1/16 9 B,B= 1, 1 1.0 1/16
10 B,M= 1, 0 0.5 1/16 11 B,A= 1, 1 1.0 1/16 12 B,C= 1, 1 1.0
1/16 13 C,C= 1, 1 1.0 1/16 14 C,M= 1, 0 0.5 1/16 15 C,A= 1, 1 1.0
1/16 16 C,B= 1, 1 1.0 1/16
Sum of Sample Means =x 12.0
Mean of statistic values ===
1612
16x
0.75
Population parameter ==
+++=
43
41110 0.75
Sampling Distribution
Sample Mean, x
Probability
0.0 1/16 0.5 6/16 1.0 9/16
b. The probability of each proportion is 1/16. The sampling
distribution of proportions consists of the 16 sample proportions
with their corresponding probabilities of 1/16. The distribution
has one mode and is clearly not symmetrical.
c. The mean of the sampling distribution is 75.01612
==
=
n
x d. The mean of the sampling distribution is equal to the
population proportion of females. Yes, the mean of
the sampling distribution of proportions always equals the
population proportion as long as every possible sample is
included.
-
Chapter 5: Normal Probability Distributions 137
9. Quality Control Selecting samples with replacement, there
will be 52= 25 equally likely samples. D1= 1, D2= 1, A1= 0, A2=0,
A3=0
Sample Number a. Sample
Sample Mean x
Probability
1 D1, D1= 1, 1 1.0 1/25 2 D1, D2= 1, 1 1.0 1/25 3 D1, A1= 1, 0
0.5 1/25 4 D1, A2= 1, 0 0.5 1/25 5 D1, A3= 1, 0 0.5 1/25 6 D2, D2=
1, 1 1.0 1/25 7 D2, D1= 1, 1 1.0 1/25 8 D2, A1= 1, 0 0.5 1/25 9 D2,
A2= 1, 0 0.5 1/25
10 D2, A3= 1, 0 0.5 1/25 11 A1, A1= 0, 0 0.0 1/25 12 A1, A2= 0,
0 0.0 1/25 13 A1, A3= 0, 0 0.0 1/25 14 A1, D1= 0, 1 0.5 1/25 15 A1,
D2= 0, 1 0.5 1/25 16 A2, A2= 0, 0 0.0 1/25 17 A2, A3= 0, 0 0.0 1/25
18 A2, D1= 0, 1 0.5 1/25 19 A2, D2= 0, 1 0.5 1/25 20 A2, A1= 0, 0
0.0 1/25 21 A3, A3= 0, 0 0.0 1/25 22 A3, D1= 0, 1 0.5 1/25 23 A3,
D2= 0, 1 0.5 1/25 24 A3, A1= 0, 0 0.0 1/25 25 A3, A2= 0, 0 0.0
1/25
Sum of Sample Means =x 10.0
Mean of statistic values ===
2510
25x
0.40
Population parameter
==
++++=
52
500011
0.40
Sampling Distribution Sample
Mean, x Probability
0.0 9/25 0.5 12/25 1.0 4/25
b. The sampling distribution consists of the 25 proportions and
their corresponding probabilities of 1/25 each. The sampling
distribution has one mode, but it is not symmetrical.
-
138 Chapter 5: Normal Probability Distributions
c. The mean of the sampling distribution is 40.02510
==
=
n
x d. Yes, the mean of the sampling distribution is equal to the
population proportion of defects. Yes, the mean
of the sampling distribution of proportions always equals the
population proportion as long as every possible sample is
included.
10. Women Senators a. From a random sample, these results were
obtained: D, R, D, D, D. b. The proportion of democrats is 4/5=
0.80. c. The proportion from part b is a statistic because it is
the proportion in a particular sample. d. No, the sample proportion
(4/5 = 0.8) does not equal the population proportion (10/13 = 0.77)
No random
sample of size 5 can equal the population proportion because the
proportions in the samples must be multiples of 0.2. The
possibilities are: 0.0, 0.2, 0.4, 0.6, 0.8, 1.0. The population
proportion (0.77) is not equal to any of these.
e. If all possible samples of size 5 are listed, then the mean
of all the sample proportions will be equal to population
proportion.
11. Mean Absolute Deviation From Table 5-2, x= 1, 2, 5, = 2.67
Population Mean Absolute Deviation, see this formula in Section
2-5.
56.1367.4
333.267.067.1
3)67.25()67.22(67.21
==
++=
++=
n
xx
Sample Number a. Sample
Sample Mean x
Absolute Deviation
2)( 21 xxd =
1 1, 1 1.0 0.0 2 1, 2 1.5 0.5 3 1, 5 3.0 2.0 4 2, 1 1.5 0.5 5 2,
2 2.0 0.0 6 2, 5 3.5 1.5 7 5, 1 3.0 2.0 8 5, 2 3.5 1.5 9 5, 5 5.0
0.0
89.098
====
n
ddMAD
Since MAD = 0.89 1.56 (the population absolute mean deviation)
the mean absolute deviation is not a good estimate of the
population mean absolute deviation.
-
Chapter 5: Normal Probability Distributions 139
12. Median as an Estimator
Sample Number Sample Mean ( )x Median Probability
1 1,1,1 1.00 1 1/27 2 1,1,2 1.33 1 1/27 3 1,1,5 2.33 1 1/27 4
1,2,1 1.33 1 1/27 5 1,5,1 2.33 1 1/27 6 1,2,5 2.67 2 1/27 7 1,5,2
2.67 2 1/27 8 1,2,2 1.67 2 1/27 9 1,5,5 3.67 5 1/27
10 2,2,2 2.00 2 1/27 11 2,2,1 1.67 2 1/27 12 2,2,5 3.00 2 1/27
13 2,1,2 1.67 2 1/27 14 2,5,2 3.00 2 1/27 15 2,1,5 2.67 2 1/27 16
2,5,1 2.67 2 1/27 17 2,1,1 1.33 1 1/27 18 2,5,5 4.00 5 1/27 19
5,5,5 5.00 5 1/27 20 5,5,1 3.67 5 1/27 21 5,5,2 4.00 5 1/27 22
5,1,5 3.67 5 1/27 23 5,2,5 4.00 5 1/27 24 5,1,2 2.67 2 1/27 25
5,2,1 2.67 2 1/27 26 5,1,1 2.33 1 1/27 27 5,2,2 3.00 2 1/27
7.22772
===
n
xxx 5.227
68===
n
MdnxMdn
In this case, the mean of the sample means and the mean of the
sample medians both are not equal to the population mean. Only the
mean of the sample means is equal to the population mean. The mean
of the medians is negatively biased. We conclude that the mean of
the sample mean a better estimate of the population mean than the
mean of the medians.
-
140 Chapter 5: Normal Probability Distributions
5-5 The Central Limit Theorem
Using the Central Limit Theorem. In Exercises 1-6, assume that
mens weights are normally distributed with a mean given by . = 172
lb and a standard deviation given by = 29 lb (based on data from
the National Health Survey).
1. a. P(x < 167)
.17.029
529
172167
=
=
=
xz From Table A-2, P(z < 0.17) = 0.4325.
There is a 0.4325 probability that an individual man will weigh
less than 167 lb. b. P( x < 167)
03.1833.45
629
5
3629
172167=
=
=
=
=
n
xz
. From Table A-2, P(z < 1.03) = 0.1515.
There is a 0.1515 probability that a group of 36 men will have a
mean weight less than 167 lb.
2. a. P(x > 180)
.28.0298
29172180
+==
=
=
xz From Table A-2, P(z < +0.28) = 0.6103.
Therefore, P(z > +0.28) = 1 0.6103 = 0.3897. There is a
0.3897 probability that an individual man will weigh more than 180
lb.
b. P( x > 180)
76.29.2
8
1029
8
10029
172180+===
=
=
n
xz
. From Table A-2, P(z < +2.76) = 0.9971.
Therefore, P(z > 0.28) = 1 0.9971 = 0.0029. There is a 0.0029
probability that a group of 100 men will have a mean weight more
than 180 lb.
3. a. P(170 < x < 175) 10.0
293
29172175
z ,07.029
229
172170+==
=
==
=
=
=
xxz .
From Table A-2, P(z < 0.07) = 0.4721 and P (z < +0.10) =
0.5398. The difference is 0.5398 0.4721 = 0.0677. There is a 0.0677
probability that an individual man will weigh between 170 lb and
175 lb
b. P(170 < x < 175)
55.0625.3
2
829
2
6429
172170=
=
=
=
=
n
xz
83.0625.33
829
3
6429
172175+===
=
=
n
xz
From Table A-2, P(z < 0.55) = 0.2912, and P(z < +0.83) =
0.7967. The difference is 0.7967 0.2912 = 0.5055. There is a 0.5055
probability that a group of 64 men will have a mean weight between
170 lb and 175 lb
4. a. P(100 < x < 165) 24.0
297
29172165
z ,48.22972
29172100
=
=
=
==
=
=
=
xxz .
From Table A-2, P(z < 2.48) = 0.0066 and P(z < 0.24) =
0.4052.
-
Chapter 5: Normal Probability Distributions 141
The difference is 0.4052 0.0066 = 0.3986. There is a 0.3986
probability that an individual man will weigh between 100 lb and
165 lb
b. P(100 < x < 165)
34.22222.372
929
72
8129
172100=
=
=
=
=
n
xz
17.2222.3
7
929
7
8129
172165=
=
=
=
=
n
xz
From Table A-2, P(z < 22.34) ~ 0.0001, and, P (z < 2.17) =
0.0150. The difference is 0.0150 0.0001 = 0.0149. There is a 0.0149
probability that a group of 81 men will have a mean weight between
100 lb and 165 lb
5. a. P( x > 160)
07.280.512
529
12
2529
172160=
=
=
=
=
n
xz
From Table A-2, P(z < 2.07) = 0.0192 Therefore P(z > 2.07)
= 1 0.0192 = 0.9808. There is a 0.9808 probability that a group of
25 men will
weigh more than 160 lb. b. The central limit theorem can be used
in part (a) because the original distribution is a normal
distribution
and we assume the sampling distribution would be normal even
though the sample size is less than 30.
6. a. P(160 < x < 180)
55.05.14
8
229
8
429
172180 z
83.050.14
12
229
12
429
172160
+===
=
=
=
=
=
=
=
n
x
n
xz
.
From Table A-2, P (z < 0.83) = 0.2033 and P(z < +0.55) =
0.7088. The difference is 0.7088 0.2033 = 0.5055. There is a 0.5055
probability that a group of 4 men will have a
mean weight between 160 lb and 180 lb. b. The central limit
theorem can be used in part (a) because the original distribution
is a normal distribution
we assume the sampling distribution would be normal even though
the sample size is less than 30.
7. Redesign of Ejection Seats, = 143, = 29 a. P(140 < x <
211) 34.2
2968
29143211
z ,10.029
329
143140+==
=
==
=
=
=
xxz
From Table A-2, P(z < 0.10) = 0.4602 and P(z < +2.34) =
0.9904. The difference is 0.9904 0.4602 = 0.5302. There is a 0.5302
probability that an individual woman will
weigh between 140 lb and 211 lb. b. P(140 < x < 211)
-
142 Chapter 5: Normal Probability Distributions
07.14833.468
62968
3629
143211 z
62.0833.43
629
3
3629
143140
+===
=
=
=
=
=
=
=
n
x
n
xz
.
From Table A-2, P (z < 0.62) = 0.2676 and P(z < +14.07) ~
0.9999. The difference is 0.9999 0.2676= 0.7323. There is a 0.7323
probability that a group of 36 women will have a mean weight
between 140 lb and 211 lb
c. The results from part (a) are more important because the
seats will be occupied by individual women, and not by groups of
women.
8. Designing Motorcycle Helmets, = 6, = 1 a. P(x < 6.2)
.2.012.0
162.6
==
=
=
xz From Table A2, P(z < 0.2) = 0.5793.
There is a 0.5793 probability that an individual man will have a
head breadth less than 6.2 in.
b. 0.21.02.0
101
2.0
1001
62.6+===
=
=
n
xz
.
From Table A-2, P(z < +2.0) = 0.9772. There is a 0.9772
probability that a group of 100 men will have a mean head breadth
less than 6.2 in.
c. The results from (b) above are for a group of men. Since the
helmets are to be used by one man alone at a time, the results of
(a) are more appropriate for the production manager to use.
9. Designing a Roller Coaster, = 14.4, = 1 a. P(x > 16.0)
.26.2707.06.1
414.11
6.1
21
4.1416+===
=
=
n
xz
From Table A-2, P (z < +2.26) = 0.9881. Therefore, P (z >
2.26) = 1 0.9881 = 0.0119. The probability that the mean of the 2
men is greater than 16 in. is 0.0119.
b. No, most riders will be able to fit since the probability of
both riders having a mean hip breadth of greater than 16in. is very
low.(0.0119). Yes, this design appears to be acceptable.
10. Uniform Random-Number Generator, = 0.5, = 0.289 .42.2
0289.007.0
10289.0
07.0
100289.0
50.057.0+===
=
=
n
xz
From Table A-2, P(z < +2.42) = 0.9922. Therefore, P(z >
2.42) = 1 0.9922 = 0.0078. The probability of getting 100 numbers
with a mean greater than 0.57 is 0.0078. It would be unusual to
generate
100 such numbers and get a mean of greater than 0.57. This is
because the probability of this occurring is very low (0.0078).
11. Blood Pressure, = 114.8, = 13.1 a. .92.1
1.132.25
1.138.114140
+==
=
=
xz
From Table A-2, P(z < +1.92) = 0.9726.Therefore, P(z >
+1.92) = 1 0.9726 = 0.0274. There is a 0.0274 probability that an
individual woman will have a systolic blood pressure greater than
140.
-
Chapter 5: Normal Probability Distributions 143
b. .85.355.6
2.25
21.132.25
41.13
8.114140+===
=
=
n
xz
From Table A-2, P(z < +3.85)= 0.9999. Therefore, P(z >
3.85) = 1 0.9999= 0.0001. There is a 0.0001 probability that a
group of 4 women will have a mean systolic blood pressure greater
than 140.
c. The central limit theorem can be used in part (b) because the
original distribution is a normal distribution, even though the
sample size is less than 30.
d. No. Although the mean result for the 4 women is less than
140, the individual values could be above or below 140 due to
sampling variability.
12. Reduced Nicotine in Cigarettes, = 0.941, = 0.313 a.
.19.1
40313.0
941.0882.0=
=
=
n
xz
From Table A-2, P(z < 1.19) = 0.1170.
There is a 0.1170 probability of randomly selecting 40
cigarettes with a mean of 0.882 g or less. b. Based on the results,
the amount of nicotine seems to be lower. This is because it is
very unlikely to select a
group of 40 cigarettes with a mean nicotine level of less than
0.882 if the mean and standard deviation have not changed.
Therefore, it is likely that these values have changed as the
company claims.
13. Elevator Design, = 172, = 29, n= 16, P = 0.975 We first find
the z score for the area P= 0.975 from the body of table A-2.This
corresponds to z = +1.96. We then use the formula:
( )21.18621.14172
25.796.11720.4
2996.11721629
*96.1172
=+
=+=
+=
+=
+=
nzx
.
To get the total value for 16 men, 186.21 16 = 2979.4. This is
the maximum total allowable weight if we want a 0.975 probability
of this weight not being exceeded with 16 men.
14. Seating Design, = 14.4, = 1, n = 18, P= 0.975 a. We first
find the z score for the area P= 0.975 from the body of table A-2.
This corresponds to z = +1.96. We then use the formula:
86.1446.04.14
236.096.14.14243.4196.14.14
18196.14.14
=+
=+=
+=
+=
+=
nzx
.
To get the total value for 18men, = 14.86 18 = 267.48 in. This
is the minimum length of the bench if we want a 0.975 probability
that it will fit the combined hips of 18 men.
b. Using the result in (a) would be wrong because we actually
want to build a bench for 18 male college football player are most
probably bigger in size than normal men.
15. Correcting for a Finite Population, = 143, = 29, N=120, n =
8 a. If we do not want to exceed this limit, we need to find the
probability of the 8 of them having a total weight
less than 1300 lb. A total capacity of 1300 lb for the 8 women
means 1300/8 = 162.5 lb per woman on average.
96.194.95.19
970.025.105.19
941.025.105.19
119112
828.229
5.19
11208120
829
1435.162
1
==
=
=
=
=
=
NnN
n
xz
From Table A-2, P(z < +1.96) = 0.975.
-
144 Chapter 5: Normal Probability Distributions
The probability of their total weight not exceeding 1300lb =
0.9750. b. We first find the z score for the area P= 0.9900 from
the body of Table A-2. This corresponds to z = +2.33.
We then use the formula:
18.16618.23143970.0*25.1034.2143
941.0828.22934.2143
11208120
82933.2143
1=+=+
=+=
+=
+=N
nNn
zx
To get the total value for 8 women, = 166.18 8 = 1329 lb. This
is the maximum allowable weight of passengers in the elevator if we
want a 0.99 probability that the elevator will not be
overloaded.
16. Population Parameters, 2, 3, 6, 8, 11, 18
a. 0.8648
618118632
==
+++++=
=
Nx
x 2 3 6 8 11 18 x= 48 x -6 -5 -2 0 3 10 (x )= 0
(x )2 36 25 4 0 9 100 (x )2= 174
( ) 385.56
1742==
=
Nx
b.
Sample Number Samples (without
replacement) Sample
mean, x xx ( xx )2
1 2, 3 2.5 -5.5 30.25 2 2, 6 4.0 -4.0 16.00 3 2, 8 5.0 -3.0 9.00
4 2, 11 6.5 -1.5 2.25 5 2, 18 10.0 2.0 4.00 6 3, 6 4.5 -3.5 12.25 7
3, 8 5.5 -2.5 6.25 8 3, 11 7.0 -1.0 1.00 9 3, 18 10.5 2.5 6.25
10 6, 8 7.0 -1.0 1.00 11 6,11 8.5 0.5 0.25 12 6, 18 12.0 4.0
16.00 13 8, 11 9.5 1.5 2.25 14 8, 18 13.0 5.0 25.00 15 11, 18 14.5
6.5 42.25
= 120.00 0.0 174.00
c. Mean of sample means, 0.815
120===
x
xn
x
d. Mean and standard deviation, See part (c). for the mean of
8.0
Standard deviation of sample means, ( ) 406.360.11
1500.1742
===
=
x
xn
x
e. By comparing the result in part (a) with the result in part
(c), we see that 8== x
-
Chapter 5: Normal Probability Distributions 145
x
NnN
n
==
===
=
406.38944.0808.3
8.0808.354
414.1385.5
1626
2385.5
1
Value is the same as the result found in part (d).
5-6 Normal as Approximation to Binomial
Using Normal Approximation. In Exercises 1-8, the given values
are discrete. Use the continuity correction and describe the region
of the normal distribution that corresponds to the indicated
probability. For example, the probability of more than 20 girls
corresponds to the area of the normal curve described with this
answer: the area to the right of 20.5.
1. The probability of more than 15 males with blue eyes
corresponds to the area of the normal curve to the right of 15.5,
P(x > 15)= Pc(x > 15.5)
2. The probability of at least 24 students understanding
continuity correction corresponds to the area of the normal curve
to the right of 23.5, P(x > 23)= Pc(x > 23.5)
3. The probability of fewer than 100 bald eagles sighted in a
week corresponds to the area of the normal curve to the left of
99.5, P(x < 100)= Pc(x < 99.5)
4. The probability that the number of working vending machines
in the United States is exactly 27 corresponds to the area of the
normal curve between 26.5 and 27.5, P(x =27)=
Pc(26.5 < x < 27.5)
5. The probability of no more than 4 students absent in a
biostatistics class corresponds to the area of the normal curve to
the left of 4.5, P(x 4)= Pc(x < 4.5)
6. The probability that the number of Canada geese residing in
one pond is between 15 and 20 inclusive corresponds to the area of
the normal curve between 14.5 and 20.5, P(15 x 20)= Pc(14.5 < x
< 20.5)
7. The probability that the number of rabbit offspring is
between 8 and 10 inclusive corresponds to the area of the normal
curve between 7.5 and 10.5, P(8 x 10)= Pc(7.5 < x < 10.5)
8. The probability of exactly 3 American elm trees with Dutch
elm disease corresponds to the area of the normal curve between 2.5
and 3.5, P(x= 3)= Pc(2.5 < x < 3.5)
Using Normal Approximation. In Exercises 9-12, do the following.
(a) Find the indicated binomial probability by using Table A-1. (b)
If np 5 and nq 5, also estimate the indicated probability by using
the normal distribution as an approximation to the binomial
distribution; if np < 5 or nq < 5, then state that the normal
approximation is not suitable.
9. a. n = 14, p = 0.5, From Table A-1, P(9) = 0.122. b. Normal
approximation np= nq= 14 0.5= 7 (both 5, normal approximation is
justified)
34.1871.15.2
871.175.9
80.0871.15.1
871.175.8
8711535.05.01475014
==
=
=
==
=
=
====
===
xz
xz
..npq.np
-
146 Chapter 5: Normal Probability Distributions
z = 0.80 corresponds to a probability of 0.7881 z = 1.34
corresponds to a probability of 0.9099 P(9) from Normal
Approximation= 0.9099 0.7881 = 0.1218 (very good approximation to
0.122)
10. a. n = 12, p = 0.8, From Table A-1, P (7) = 0.053 b.
.., nq..np 422012698012 ==== nq = 2.4 which is 55.5) P (girls
> 55) = P (z > +1.1) = 1 0.8643 = 0.1357 No, since P(girls
> 55) is greater than 0.05, it is not unusual to get more than
55 girls out of 100 births.
14. Probability of at Least 65 Girls, np= 50, nq= 50 (both 5,
normal distribution justified)
925
5.145
505645255.05.0100
505010050100
.
.
x
z
npq.np
., pn
+==
=
=
====
===
==
P(x < 65), finding Pc(x < 64.5) P (girls 65) = P (z >
+2.9) = 1 0.9981 = 0.0019 Yes, since P(girls 65) is less than 0.05,
it is unusual that there would be 65 or more girls out of 100
births.
-
Chapter 5: Normal Probability Distributions 147
15. Probability of at Least Passing, np= 50, nq= 50 (both 5,
normal distribution justified)
9155.9
550559
52550501005050100
answer) falseor (true50100
.
.
x
z
..npq.np ., pn
+==
=
=
====
===
==
P(x 60), finding Pc(x > 59.5) P (score 60) = P (z > +1.9)
= 1 0.9713 = 0.0287 No, since P(score 60) is less than 0.05, it is
unusual to get a score of at least 60 by guessing
16. Multiple-Choice Test, np= 5, nq= 20 (both 5, normal
distribution justified)
.
.
x
z..
x
z
npq .np
., pn
75225.5
25510251
25.2
2552
248.02.02552025
correct) is options 5 ofout (one2025
+==
=
==
=
=
=
====
===
==
P(3 < x < 10), finding Pc(2.5 < x < 10.5) P (z<
1.25) = 0.1056, P (z < +2.75) = 0.9970 P (-1.25 < z <
+2.75) = 0.9970 0.1056 = 0.8914
17. Mendels Hybridization Experiment, np= 145, nq= 435 (both 5,
normal distribution justified)
.
.
.
x
z
.npq.np
., pn
62043.105.6
43101455151
431075.10875.025.0580145250580
250580
+==
=
=
====
===
==
P(x 152), finding Pc(x > 151.5) P (z at least 0.62) = P(z
> +0.62) = 1 0.7324 = 0.2676 No, there is no evidence that the
Mendelian rate of 25% is wrong because it is not unusual to get 152
yellow
pods out of 580 seedlings, p= 0.2676
18. Cholesterol-Reducing Drug, np= 16.4, nq= 846.6 (both 5,
normal distribution justified)
.
.
.
.
..
x-
z
....npq..np
., pn
520014
1032014
39716518014091698100190863
3971601908630190863
+==
==
====
===
==
P(x 19), finding Pc(x > 18.5) P(z at least 0.52) = P (z >
+0.52) = 1 0.6985 = 0.3015 It is not unusual to have 19 people with
flu symptoms (P= 0.3015). Therefore, the flu symptoms are
probably
not due to taking the drug.
-
148 Chapter 5: Normal Probability Distributions
19. Probability of at Least 50 Color-Blind Men, np= 54, nq= 546
(both 5, normal distribution justified)
64.001.7
5.401.7
545.49-z
01.714.4991.009.0600npq540.09600np
0.09,p ,600n
=
=
==
====
===
==
x
P(x 50), finding Pc(x > 49.5) P(z at least 0.64) = P(z >
0.64) = 1 0.2611 = 0.7389
It is quite likely to have 50 color blind men among this group
of 600 men (P= 0.7389). However, the researchers cannot be very
confident since there is still quite some chance of not getting up
to 50 men.
20. Cell Phones and Brain Cancer, np= 143, nq= 419,952 (both 5,
normal distribution justified)
.
.
..
x
z
.npq ..np
., pn
61095.1133.7
9511831425135
951178.142999660.0000340.0420095831420003400420095
0003400420095
=
=
=
=
====
===
==
P(x 135), finding Pc(x < 135.5) P (z < 0.61) = 0.2709
It is not unusual to have 135 or fewer cases of brain cancer in
the population (P= 0.2709). Therefore, the media reports that cell
phones cause brain cancer are not supported by the evidence.
21. Identifying Gender Discrimination, np= 31, nq= 31 (both 5,
normal distribution justified)
4129373
599373
315219373515505062
3150625062
.
.
.
.
.
x
z
....npq .np
., pn
=
=
=
=
====
===
==
P(x 21), finding Pc(x < 21.5) P (z < 2.41) = 0.0080
It is unusual to have 21 female employees out of 62 new
employees being hired assuming no gender discrimination. (P=
0.0080) These results support the charge of gender discrimination
taking place.
22. Blood Group, np= 180, nq= 220 (both 5, normal distribution
justified)
.
.
.
.
.
x
z
...npq.np
., pn
350959
53959
180517695999550450400
180450400450400
=
=
=
=
====
===
==
P(x 177), finding P(x > 176.5) P(z < 0.35)= 0.3632, P(z
> 0.35) = 1 0.3632= 0.6368
It is not unusual to have at least 177 Group O donors in this
group of 400 people. The pool may be sufficient, however this pool
may not be sufficient because the probability is not high (P =
0.6368).
-
Chapter 5: Normal Probability Distributions 149
23. Acceptance Sampling, np= 5, nq= 45 (both 5, normal
distribution justified)
.
.
.
.
.
x
z
....npq .np
., pn
651122
53122
55112254901050
510501050
=
=
=
=
====
===
==
P(x 2), finding P(x > 1.5) P(z < 1.65)= 0.0495, P(z >
1.65) = 1 0.0495= 0.9505 Yes, this plan would detect defects at the
10% level about 95% of the time.
24. Car Crashes, np= 170, nq= 330 (both 5, normal distribution
justified)
79.259.105.29
59.101705.199
z
59.102.11266.034.0500npq 17034.0500np
34.0p ,500n
+==
=
=
====
===
==
x
P(x 200), finding P(x > 199.5) P(z < +2.79)= 0.9974, P(z
> +2.79) = 1 0.9974 = 0.0026 The probability of having 40 %(
200) of 500 men having accidents is very low (p< 0.05) when the
true
probability is 0.34. Therefore, the claim that the accident rate
in New York City is higher than 34% is supported by the evidence in
this result.
25. Cloning Survey, np= 506, nq= 506 (both 5, normal
distribution justified)
.
..x
z
...npq.np
., pn
80249115
5.3949115
5065.900911525350501012
506501012501012
+==
=
=
====
===
==
P(x 900), finding P(x > 900.5) P (z < +24.80) 0.9999, P(z
> +24.80) = 1 0.9999 = 0.0001 The probability of having 89%
(900) of 1012 people in a sample assuming a general probability of
0.5 is very
low. Yes, this evidence supports the claim that a majority of
people are opposed to cloning
5-7 Assessing Normality (Note: In Section 5-7 all graphics were
generated using SPSS)
Interpreting Normal Quantile Plots. In Exercises 1-4, examine
the normal quantile plot and determine whether it depicts data that
have a normal distribution.
1. The data are not normally distributed since the data plot
dots depart from being a straight line that follows the normal
quantile plot that is expected if the data are normally
distributed.
2. The data are not normally distributed since the data plot
dots depart from being a straight line that follows the normal
quantile plot that is expected if the data are normally
distributed.
3. The data are normally distributed since the data plot dots
are very close to a straight line that follows the normal quantile
plot that is expected if the data are normally distributed.
-
150 Chapter 5: Normal Probability Distributions
4. The data are normally distributed since the data plot dots
are very close to a straight line that follows the normal quantile
plot that is expected if the data are normally distributed.
Determining Normality. In Exercises 5-8, refer to the indicated
data set and determine whether the requirement of a normal
distribution is satisfied. Assume that this requirement is loose in
the sense that the population distribution need not be exactly
normal, but it must be a distribution that is basically symmetric
with only one mode.
5. BMI, Data Set 1 in Appendix B
The histogram above shows a distribution with one mode,
relatively symmetrical, and bell-shaped. It can be said to
approximate a normal distribution.
35.0030.0025.0020.00 15.00
BMIMales
12
10
8
6
4
2
0
Fre
quen
cy
-
Chapter 5: Normal Probability Distributions 151
6. Head Circumferences, Data Set 4 in Appendix B
The histogram above shows a distribution with one mode,
relatively symmetrical, and bell-shaped, except for two values in
the lower part of the range. While this distribution is not
perfectly symmetrical it could be considered to be approximately
normal.
42.5040.0037.50 35.00
HeadCircMales
20
15
10
5
0
Freq
uenc
y
-
152 Chapter 5: Normal Probability Distributions
7. Water Conductivity
The histogram above shows a distribution with one mode. However,
the distribution is not symmetrical and bell-shaped so it would not
be considered to be approximately normal.
60.0040.0020.00
WaterConductivity
12.5
10.0
7.5
5.0
2.5
0.0
Freq
uenc
y
-
Chapter 5: Normal Probability Distributions 153
8. Heights of Poplar Trees
The histogram above shows a distribution with one mode. However,
the distribution is not symmetrical or bell-shaped so it would not
be considered to be approximately normal.
14.0012.0010.008.006.004.00 2.00 0.00
PoplarTreeHgt
12
10
8
6
4
2
0
Freq
uenc
y
-
154 Chapter 5: Normal Probability Distributions
Generating Normal Quantile Plots. In Exercises 9-12, use the
data from the indicated exercise in this section. Use a
TI-83/84Plus calculator or software (such as SPSS, SAS, STATDISK,
Minitab. or Excel) capable of generating normal quantile plots (or
normal probability plots). Generate the graph, then determine
whether the data appear to come from a normally distributed
population. NOTE: The following Normal Quantile Plots, except those
in Exercises 15 and 16 were generated by SPSS. When using the SPSS
option for standardized or z scores, both axes are put into z score
units, not just the Y-axis.
9. From Exercise 5
420-2-4
Standardized Observed Value
4
2
0
-2
-4
Expe
cted
No
rmal
Val
ueNormal Q-Q Plot of BMIMales
The BMI data from Exercise 5 seems to come from a normal
distribution. Most of the points are very close to the straight
line.
-
Chapter 5: Normal Probability Distributions 155
10. From Exercise 6
20-2-4
Standardized Observed Value
4
2
0
-2
-4
Expe
cted
Nor
mal
Va
lue
Normal Q-Q Plot of HeadCircMales
The head circumference data from Exercise 6 seems to come from a
normal distribution. Most of the points, except for two of them,
are very close to the line.
-
156 Chapter 5: Normal Probability Distributions
11. From Exercise 7
420-2-4
Standardized Observed Value
4
2
0
-2
-4
Expe
cted
N
orm
al V
alue
Normal Q-Q Plot of WaterConductivity
The data on the conductivity variable are not normally
distributed. The points depart quite a bit from the straight
line.
-
Chapter 5: Normal Probability Distributions 157
12. From Exercise 8
420-2-4
Standardized Observed Value
4
2
0
-2
-4
Expe
cte
d N
orm
al V
alu
e
Normal Q-Q Plot of PoplarTreeHgt
This tree height data distribution is not normal. The points are
not close to the line. Also, there are some obvious outliers seen
in the plot.
13. Comparing Data Sets
420-2-4
Standardized Observed Value
4
2
0
-2
-4
Expe
cted
No
rmal
Va
lue
Normal Q-Q Plot of HgtWomen
43210-1-2-3
Standardized Observed Value
4
2
0
-2
-4
Expe
cted
No
rmal
Va
lue
Normal Q-Q Plot of CholestWomen
The distribution for height appears to be normal, but the
distribution for cholesterol does not appear to be normal. This
could be because cholesterol levels depend on diet and many other
human behaviors in different ways that do not yield normally
distributed results while height is a more natural variable less
influenced by human behaviors.
-
158 Chapter 5: Normal Probability Distributions
14. Comparing Data Sets
543210-1-2
Standardized Observed Value
4
2
0
-2
-4
Expe
cte
d No
rmal
Val
ueNormal Q-Q Plot of SystBPWomen
420-2-4
Standardized Observed Value
4
2
0
-2
-4
Expe
cte
d No
rmal
Val
ue
Normal Q-Q Plot of ElbowBrdthWomen
Systolic blood pressure does not appear to have a distribution
that approximates a normal distribution, but the distribution of
elbow breadth could approximate a normal distribution. This could
be because systolic blood pressure levels depend on diet and other
human behaviors that do not yield normally distributed results
while elbow breadth is a more natural variable less influenced by
human behaviors.
Constructing Normal Quantile Plots. In Exercises 15 and 16, use
the given data values and identify the corresponding z scores that
are used for a normal quantile plot, then construct the normal
quantile plot and determine whether the data appear to be from a
population with a normal distribution.
15. Heights of L.A. Lakers Sorting the data by order gives us
73, 78, 79, 82, 85 n = 5, 1/2n, 3/2n, 5/2n, 7/2n, 9/2n = 0.1, 0.3,
0.5, 0.7, 0.9 Corresponding z scores, using Table A-2 for these
areas are: 1.28, 0.52, 0.00, +0.52, and +1.28 We now pair the
sorted heights with their corresponding z scores: (73, 1.28) (78,
0.52) (79, 0) (82, +0.52) (85, +1.28) We plot these (x,y)
coordinates to get the normal quantile plot.
-
Chapter 5: Normal Probability Distributions 159
NOrmal Q-Q Plot for Laker's Height
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Observed Score (Height)
Expe
cte
d N
orm
al Va
lue
70 74 76 78 80 82 84 86
This distribution looks like it approximates a normal
distribution.
16. Monitoring Lead in Air Sorting the data by order gives us
0.42, 0.48, 0.73, 1.10, 1.10, 5.40 n= 6, 1/2n, 3/2n, 5/2n, 7/2n,
9/2n, 11/2n = 0.083, 0.167, 0.417, 0.583, 0.750, 0.917
Corresponding z scores by using Table A-2 for these areas are:
1.38, 0.67, 0.21, +0.21, +0.67 and +1.39 We now pair the sorted
heights with their corresponding z scores: (0.42, -1.38)
(0.48,-0.67) (0.73, 0.21) (1.10, 0.21) (1.10, 0.67) (5.40, 1.39) We
plot these (x,y) coordinates to get the normal quantile plot.
-
160 Chapter 5: Normal Probability Distributions
Normal Q-Q PLot for Lead in Air
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Observed Value (Lead in Air)
Expe
cte
d St
anda
rdiz
ed
Valu
e
0 1 2 3 4 5 6
The distribution of the data clearly is not normal.
17. Using Standard Scores
No, the transformation to z scores involves subtracting a
constant and dividing by a constant, so the plot of the (x,z,)
points will always be a straight line, regardless of the nature of
the distribution.
-
Chapter 5: Normal Probability Distributions 161
18. Lognormal Distribution
3210-1-2
Standardized Observed Value
2
1
0
-1
-2
Expe
cted
Nor
mal V
alu
e
Normal Q-Q Plot of PhoneTime
210-1-2
Standardized Observed Value
2
1
0
-1
-2
Expe
cted
No
rmal
Va
lue
Normal Q-Q Plot of LogPhoneTime
The above distribution on the left is clearly not normal.
However, the distribution on the right, after the log (x + 1)
transformation is much closer to being a normal distribution. This
illustrates that at times a transformation can provide a
distribution much closer to a normal distribution than the original
distribution has.
Review Exercises
1. High Cholesterol Levels, = 178.1, = 40.7 a. P(x > 260)
01.27.409.81
7.401.178260
+==
=
=
xz
P(x > 260)= P(z > + 2.01), Using Table A-2, P(z <
+2.01)= 0.9778 P(z > +2.01)= 1 P(z +2.01)= 1 0.9778= 0.0222 b.
P(170 < x < 200)
54.07.409.21
7.401.17820020.0
7.401.8
7.401.178170
+==
=
==
=
=
=
xz
xz
P(z < +0.54)= 0.7054, P(z < -0.20)= 0.4207 P(170 < x
< 200)= P(-0.20 < z < +0.54)= 0.7054 0.4207= 0.2847 c.
P(170 < x < 200), with n= 9
61.157.139.21
37.409.21
97.40
1.178200 z
60.057.131.8
37.401.8
97.40
1.178170
+===
=
=
=
=
=
=
=
n
x
n
xz
.
From Table A-2, P (z < 0.60)= 0.2743 and P(z < +1.61)=
0.9463. The difference is 0.9463 0.2743= 0.6720. There is a 0.6720
probability that a group of 9 men will have a mean cholesterol
level between 170 mg/dL and 200 mg/dL
d. The top 3% is equivalent to bottom 97%. From Table A-2, the
area 0.97 corresponds to a z score of +1.88 254.640.7)1.88(178.1
)(z =++=+= x Therefore, the cutoff for men should be a cholesterol
level of 254.6
-
162 Chapter 5: Normal Probability Distributions
2. Babies at Risk, a. = 3420, = 495
46.24951220
49534202200
z =
=
=
=
x
P( z < 2.46) = 0.0069. Therefore, 0.69% of babies are in at
risk category. If the Chicago hospital has 900 births, we expect
0.69 % of the 900 to be at risk 6.21 babies would be at
risk. b. Lowest 2%. From Table A-2, the area 0.02 corresponds to
a z score of 2.05 4052495)(-2.053420 )(z =+=+= x The cutoff weight
for the lowest 2% is 2405 g. c. P( x > 3700)
26.275.123
280
4495280
16495
34203700-z +===
==
n
x
From Table A-2, P(z < 2.26) = 0.9881.Therefore, P(z >2.26)
= 1 0.9881 = 0.0119. The probability that 16 newborn babies will
have mean weight greater than 3700 is 0.0119.
d. P(3300 < x < 3700) with n= 49
96.371.70
280
7495280
49495
34203700-z
70.171.70
120
7495
120
49495
34203300-z
+===
==
=
=
=
==
n
x
n
x
From Table A-2, P(z < 1.70) = 0.0446, and, P(z < +3.96) =
0.9999. P(3300 < x < 3700)= P(z < +3.96) P(z < 1.70)=
0.9999 0.0446 = 0.9553. There is a 0.9553 probability that a group
of 49 babies will have a mean birth weight between 3300 g and
3700 g.
3. Blue Genes, since np= 25 and nq= 75, both > 5, use of
normal approximation to a binomial distribution, with continuity
correction, is justified
P(x 19), find Pc(x < 19.5)
27.133.4
5.54.33
255.1933.475.1875.025.0100
250.25100 0.25 p 100,
=
=
=
=
====
===
==
xz
npq
npn
From Table A-2, the area below a z score of 1.27 is 0.1020.
Since P= 0.1020 > 0.05, it would not be considered to be unusual
to have 19 or fewer offspring with blue eyes out of 100 births.
4. Marine Corps Height Requirements for Men, = 69, = 2.8 a. P(64
< x < 78)
21.38.2
98.26978
z 79.18.25
8.26964
z +==
=
==
=
=
=
xx
From Table A-2, the area below a z score of 1.79 is 0.0367 and
for a z score of +3.21 is 0.9993. P(64 < x < 78)= P(z <
+3.21) P(z < 1.79)= 0.9993 0.0367= 0.9626 Therefore 96% of men
meet this requirement so not many men (only about 3.7%) are denied
entry into the
Marines because of their height.
-
Chapter 5: Normal Probability Distributions 163
b. The shortest 2% corresponds to an area of 0.02 which
corresponds to a z score of 2.05. The tallest 2% corresponds to an
area of 0.98 which corresponds to a z score of +2.05
74.7474.569)8.205.2(69)(26.6374.569)8.205.2(69)(
=+=+=+=
==+=+=
zx
zx
The new minimum and maximum heights would be 63.3 in. and 74.7
in. c. P( x > 68) with n= 64
86.235.01
88.21
648.2
6968z =
=
=
=
=
n
x
The area below a z score of 2.86 is 0.0021. P(z > 2.86) = 1
0.0021= 0.9979 The probability of randomly drawing a sample of 64
with a mean height greater than 68 in. is 0.9979.
5. Sampling Distributions a. With a sample size of 100, which is
considered a large sample size, we would expect the distribution
of
sample means to be normally distributed regardless of the shape
of distribution from which the samples are drawn. The basis for
making this claim is the Central Limit Theorem.
b. The standard deviation of the sample means is referred to as
the standard error of the mean. If = 512 and samples are of size,
n= 100, it is found as:
2.5110512
100512
====
nx
c. With a sample size of 1200, which is considered a very large
sample size, we would expect the distribution of sample proportions
from x/n to be normally distributed even though the original
distribution is a binomial distribution. The basis for making this
claim is the Central Limit Theorem.
6. Gender Discrimination, n= 20, p= 0.30, q= 0.70 np= 6, nq= 14
(since both 5, a normal distribution approximation is
justified)
71.1049.2
5.3049.2
65.2z
)5.2(),2(049.22.470.030.020npq
630.020np
=
=
=
=
-
164 Chapter 5: Normal Probability Distributions
7. Testing for Normality, From Data Set 6 in Appendix B, Bear
Neck Size From the graphs below, the distribution is approximately
normal. The histogram, with a normal distribution superimposed on
it, has one mode and is roughly bell-shaped and the normal quantile
plot has most of the points on the straight line.
8. Testing for Normality, From Data Set 12 in Appendix B,
Pre-Exercise with No Stress From the graphs below, the distribution
is approximately normal. The histogram, with a normal distribution
superimposed on it, has one mode is roughly bell-shaped and the
normal quantile plot has most of the points on the straight
line.
Cumulative Review Exercises
1. Eye Measurement Statistics Ordered scores: 55 59 62 63 66 66
66 67 in mm a. Sample Mean
35.00 30.00 25.00 20.00 15.00 10.00
BearNeckSize
10
8
6
4
2
0
Freq
ue
ncy
4 20-2-4
Standardized Observed Value
4
2
0
-2
-4
Expe
cted
Nor
mal
Val
ue
Normal Q-Q Plot of BearNeckSize
2 1 0-1-2Standardized Observed Value
2
1
0
-1
-2
Expe
cted
N
orm
al V
alu
e
Normal Q-Q Plot of PreExrcsSystBP
140.00 130.00120.00 110.00 100.00 90.00
PreExrcsSystBP
5
4
3
2
1
0
Freq
ue
ncy
-
Chapter 5: Normal Probability Distributions 165
0.638
5048
5566666362596667==
+++++++==
n
xx
b. Since there are a even number of scores, the median is the
middle point between the two middle, Median, x~ = (63+66)/2 =
64.5
c. The mode is the number that occurs the most frequent = 66
(occurs 3 times) d. Standard deviation 31876504 2 == xx
( )
21.4714.17
714.1756992
78254016255008
)18(8)504()31876(8
)1(2
2222
===
==
=
=
=
ss
nn
xxns
e. 95.021.44
21.46359
=
=
=
=
s
xxz
f. 6 of the 8 numbers are greater than 59, 6/8= 0.75 or 75% g.
Assuming a normal distribution, the area below a z score of 0.95,
P(z < 0.95)= 0.1711 P (z > 0.95) = 1 0.1711= 0.8289. This
corresponds to 82.89% h. This data set is ratio level of
measurement since there are equal intervals of measurement and
there is a
natural staring point at zero. i. The exact un-rounded distances
are continuous data that can be any value on the continuum.
2. Left-Handedness, p= 0.10 a. This is a binomial distribution
with p= 0.1. Probability of 3 out of a sample of 3 being left
handed P(L1) = 0.1, P(L2)= 0.1, P(L3)= 0.1 P(all three are L)=
P(L1) P(L2) P(L3)= 0.13= 0.001 b. P(at least 1 person left-handed)=
1 P(no lefthanders)= 1 P( N1) P(N2) P(N3) = 1 (0.9 0.9 0.9)=
1 0.729= 0.271 c. The sample size of 3 is too small, np= 0.3
< 5, np 5 is not satisfied. d. In a group of 50 people, the mean
number of left handed people would be 0.51.050 === np e. Standard
deviation, 121.25.49.01.050 ==== npq f. P(x > 8)
41.1121.23
121.258
==
=
=
xz
Area below a z score of 1.42 is 0.9207.Therefore, P(x > 8) =
1 0.9207= 0.0793. Since P= 0.0793 > 0.05, it would not be
considered an unusual result to get 8 lefthanders out of 50
subjects.