Download Chapter 6 Normal Probability Distributions

Chapter 6
Normal Probability Distributions
6-2 The Standard Normal Distribution
1. The word “normal” as used when referring to a normal distribution does carry with it some of
the meaning the word has in ordinary language. Normal distributions occur in nature and
describe the normal, or natural, state of many common phenomena. But in statistics the term
“normal” has a specific and well-defined meaning in addition to its generic connotations of
being “typical” – it refers to a specific bell-shaped distribution generated by a particular
mathematical formula.
2. Bell-shaped describes a smooth, symmetric distribution that has its highest point in the center
and tapers off to zero in either direction.
3. A normal distribution can be centered about any value and have any level of spread. A
standard normal distribution has a center (as measured by the mean) of 0 and has a spread (as
measured by the standard deviation) of 1.
4. The notation zα indicates the z score that has an area of α to its right. When α = 0.05, for
example, z0.05 indicates the z score with an area of 0.05 to its right.
0.5
0.4
f(x)
5. The height of the rectangle is 0.5.
Probability corresponds to area, and the
area of a rectangle is (width)·(height).
P(x>124.0) = (width)·(height)
= (125.0−124.0)(0.5)
= (1.0)(0.5)
= 0.50
0.3
0.2
0.1
0.0
123.0
123.5
124.0
x (volts)
124.5
125.0
123.0
123.5
124.0
x (volts)
124.5
125.0
6. The height of the rectangle is 0.5.
0.5
0.4
f(x)
Probability corresponds to area, and the
area of a rectangle is (width)·(height).
P(x<123.5) = (width)·(height)
= (123.5−123.0)(0.5)
= (0.5)(0.5)
= 0.25
0.3
0.2
0.1
0.0
The Standard Normal Distribution SECTION 6-2
0.5
0.4
f(x)
7. The height of the rectangle is 0.5.
Probability corresponds to area, and the
area of a rectangle is (width)·(height).
P(123.2<x<124.7) = (width)·(height)
= (124.7−123.2)(0.5)
= (1.7)(0.5)
= 0.75
137
0.3
0.2
0.1
0.0
123.5
124.0
x (volts)
124.5
125.0
123.0
123.5
124.0
x (volts)
124.5
125.0
0.5
0.4
f(x)
8. The height of the rectangle is 0.5.
Probability corresponds to area, and the
area of a rectangle is (width)·(height).
P(124.1<x<124.5) = (width)·(height)
= (124.5−124.1)(0.5)
= (0.4)(0.5)
= 0.20
123.0
0.3
0.2
0.1
0.0
NOTE: For problems 9-16, the answers are re-expressed (when necessary) in terms of items that
can be read directly from Table A-2. In general, this step is omitted in subsequent exercises and
the reader is referred to the accompanying sketches to se how the indicated probabilities and z
scores relate to Table A-2. “A” is used to denote the tabled value of the area to the left of the
given z score. As a crude check, always verify that
A>0.5000 corresponds to a positive z score and z>0 corresponds to an A >0.5000
A<0.5000 corresponds to a negative z score and z<0 corresponds to an A < 0.5000
9. P(z<0.75) = 0.7734
10. P(z >-0.75) = 1 – P(z<-0.75)
= 1 – 0.2266
= 0.7734
11. P(-0.60<z<1.20) = P(z<1.20) – P(z<-0.60)
= 0.8849 – 0.2743
= 0.6106
12. P(-0.90<z<1.60) = P(z<1.60) – P(x<-0.90)
= 0.9452 – 0.1841
= 0.7611
13. For A = 0.9798, z = 2.05.
14. For A = 0.2546, z = -0.66
15. If the area to the right of z is 0.1075, A = 1 – 0.1075 = 0.8925.
For A = 0.8925, z = 1.24.
16. If the area to the right of z is 0.9418, A = 1 – 0.9418 = 0.0582.
For A = 0.0582, z = -1.57.
138
CHAPTER 6 Normal Probability Distributions
NOTE: The sketch is the key to Exercises 17-36. It tells whether to subtract two Table A-2
probabilities, to subtract a Table A-2 probability from 1, etc. For the remainder of chapter 6,
THE ACCOMPANYING SKETCHES ARE NOT TO SCALE and are intended only as aids to help
the reader understand how to use the table values to answer the questions. In addition, the
probability of any single point in a continuous distribution is zero – i.e., P(x=a) = 0 for any
single point a. For normal distributions, therefore, this manual ignores P(x=a) and uses
P(x>a) = 1 – P(x<a).
17.P(z<-1.50) = 0.0668
18. P(z<-2.75) = 0.0030
0.0668
0.0030
<-----------1.50
<----------0
-2.75
Z
19. P(z<1.23) = 0.8907
0
Z
20. P(z<2.34) = 0.9904
0.8907
0.9904
<-------------------------------0
1.23
<-------------------------------0
Z
21. P(z>2.22)
= 1 – 0.9868
= 0.0132
2.34
Z
22. P(z>2.33)
= 1 – 0.9901
= 0.0099
0.9868
0.9901
<-------------------------------0
2.22
<-------------------------------Z
0
2.33
Z
The Standard Normal Distribution SECTION 6-2
23. P(z>-1.75)
= 1 – 0.0401
= 0.9599
24. P(z>-1.96)
= 1 – 0.0250
= 0.9750
0.0401
0.0250
<-----------1.75
<----------0
0
Z
26. P(1.00<z<3.00)
= 0.9987 – 0.8413
= 0.1574
<----------------------------------|
<----------------------------------|
0.8413
0.9987
0.6915
0.8413
<---------------------------0
-1.96
Z
25. P(0.50<z<1.00)
= 0.8413 – 0.6915
= 0.1498
0.50 1.00
<---------------------------Z
27. P(-3.00<z<-1.00)
= 0.1587 – 0.0013
= 0.1574
0
1.00 3.00
Z
28. P(-1.00<z<-0.50)
= 0.3085 – 0.1587
= 0.1498
<--------------|
<--------------|
0.1587
0.3085
0.0013
0.1587
<--------3.00 -1.00
139
<-------0
Z
-1.00 -0.50
0
Z
140
CHAPTER 6 Normal Probability Distributions
29. P(-1.20<z<1.95)
= 0.9744 – 0.1151
= 0.8593
30. P(-2.87<z<1.35)
= 0.9099 – 0.0021
= 0.9078
<--------------------------------|
0.9744
<--------------------------------|
0.9099
0.1151
0.0021
<-----------1.20
<----------0
1.95
Z
31. P(-2.50<z<5.00)
= 0.9999 – 0.0062
= 0.9937
0
1.35
Z
32. P(-4.50<z<1.00)
= 0.8413 – 0.0001
= 0.8412
<--------------------------------|
0.9999
<--------------------------------|
0.8413
0.0062
0.0001
<-----------2.50
-2.87
<----------0
5.00
Z
33. P(z<3.55) = 0.9999
-4.50
0
1.00
Z
34. P(z>3.68)
= 1 – 0.9999
= 0.0001
0.9999
0.9999
<-------------------------------0
3.55
<-------------------------------Z
0
3.68
Z
The Standard Normal Distribution SECTION 6-2
35. P(z>0.00)
= 1 – 0.5000
= 0.5000
36. P(z<0.00) = 0.5000
0.5000
0.5000
<----------------------
<----------------------
0
0
Z
37. P(-1<z<1) = 0.8413 – 0.1587 = 0.6826
about 68%
<--------------------------------|
0.9772
0.1587
0.0228
<-----------
<----------0
1.00
Z
39. P(-3<z<3) = 0.9987 – 0.0013 = 0.9974
About 99.7%
-2.00
0
2.00
Z
40. P(-3.5<z<3.5) = 0.9999 – 0.0001 = 0.9998
about 99.98%
<--------------------------------|
0.9987
<--------------------------------|
0.9999
0.0013
0.0001
<-----------3.00
Z
38. P(-2<z<2) = 0.9772 – 0.0228 = 0.9544
about 95%
<--------------------------------|
0.8413
-1.00
141
<----------0
3.00
Z
-3.50
0
3.50
Z
142
CHAPTER 6 Normal Probability Distributions
41. For z0.05, the cumulative area is 0.9500.
The closest entry is 0.9500,
for which z = 1.645
42. For z0.01, the cumulative area is 0.9900.
The closest entry is 0.9901.
for which z = 2.33.
0.9500
<-------------------------------0
0.9900
<--------------------------------
0.05
1.645
43. For z0.10, the cumulative area is 0.9000.
The closest entry is 0.8997,
for which z = 1.28.
0
1.28
2.33
Z
44. For z0.02, the cumulative area is 0.9800.
The closest entry is 0.9798,
for which z = 2.05
0.9000
<--------------------------------
0
Z
0.01
0.9800
<--------------------------------
0.10
0
Z
45. P(-1.96<z<1.96)
= 0.9750 – 0.0250
= 0.9500
2.05
0.02
Z
46. P(z<1.645) = 0.9500
<--------------------------------|
0.9750
0.9500
<--------------------------------
0.0250
<-----------1.96
0
1.96
Z
0
1.645
Z
The Standard Normal Distribution SECTION 6-2
47. P(z<-2.575 or z>2.575)
= P(z<-2.575) + P(z>2.575)
= 0.0050 + (1 – 0.9950)
= 0.0050 + 0.0050
= 0.0100
48. P(z<-1.96 or z>1.96)
= P(z<-1.96) + P(z>1.96)
= 0.0250 + (1 – 0.9750)
= 0.0250 + 0.0250
= 0.0500
<--------------------------------|
0.9950
<--------------------------------|
0.9750
0.0050
0.0250
<-----------2.575
143
<----------0
2.575
Z
49. For P95, the cumulative area is 0.9500.
The closest entry is 0.9500,
for which z = 1.645
-1.96
0
1.96
Z
50. For P1, the cumulative area is 0.0100.
The closest entry is 0.0099.
for which z = -2.33.
<--------------------------------|
0.9500
0.0100
<----------0
1.645
51. For the lowest 2.5%, the cumulative
area is 0.0250 (exact entry in the table),
indicated by z = -1.96.
By symmetry, the highest 2.5%
[= 1–.9750] are above z = 1.96.
0
Z
52. For the highest 0.5%, the cumulative
area is 0.9950 (exact special entry),
indicated by z = 2.575.
By symmetry, the lowest 0.5%
are below z = -2.575.
<--------------------------------|
0.9750
<--------------------------------|
0.9950
0.0250
0.0050
<-----------1.96
-2.33
Z
<----------0
1.96
Z
-2.575
0
2.575
Z
144
CHAPTER 6 Normal Probability Distributions
53. Rewrite each statement in terms of z, recalling that z is the number of standard deviations a
score is from the mean.
a. P(-2<z<2)
= 0.9772 – 0.0228
= 0.9544 or 95.44%
b. P(z<-1 or z >1)
= P(z<-1) + P(z>1)
= 0.1587 + (1 – 0.8413)
= 0.1597 + 0.1587 = 0.3174 or 31.74%
<--------------------------------|
0.9772
<--------------------------------|
0.8413
0.0228
0.1587
<-----------2.00
<----------0
2.00
Z
c. P(z<-1.96 or z > 1.96)
= P(z<-1.96) + P(z>1.96)
= 0.0250 + (1 – 0.9750)
= 0.250 + 0.250 = 0.0500 or 5.00%
1.00
Z
<--------------------------------|
0.9987
0.0250
0.0013
<-----------
<----------0
1.96
Z
e. P(z<-3 or z>3)
= P(z<-3) + P(z>3)
= 0.0013 + (1 – 0.9987)
= 0.0013 + 0.0013 = 0.0026 = 0.26%
<--------------------------------|
0.9987
0.0013
<-----------3.00
0
d. P(-3<z<3)
= 0.9987 – 0.0013
= 0.9974 or 99.74%
<--------------------------------|
0.9750
-1.96
-1.00
0
3.00
Z
-3.00
0
3.00
Z
The Standard Normal Distribution SECTION 6-2
145
54. Since area = 1 for the entire distribution,
(width) ⋅ (height) = 1
(2 3) ⋅ (height) = 1
height = 1/(2 3)
= 0.2887
a. P(-1<x<1)
= (width)·(height)
= (2)(0.2887)
= 0.5774
b. P(-1.00<z<1.00)
= P(z<1.00) – P(z<-1.00)
= 0.8413 – 0.1587
= 0.6826
<--------------------------------|
0.8413
0.30
0.25
f(x)
0.20
0.15
0.10
0.05
0.1587
0.00
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
<-----------1.00
x
0
1.00
Z
c. Yes. The error is 0.6826 – 0.5774 = 0.1052 = 10.52% as a straight probability error, and
(0.6826 – 0.5774)/0.5774 = 0.1052/0.5774 = 18.2% as a relative error.
55. The sketches are the key. They tell what
probability (i.e., cumulative area) to look
up when reading Table A-2 “backwards.”
They also provide a check against gross
errors by indicating whether a score is
above or below zero.
a. P(z<a) = 0.9599
a = 1.75
(see the sketch at the right)
b. P(z>b) = 0.9722
P(z<b) = 1- 0.9972
= 0.0028
b = -2.00
0.9599
0
1.75
Z
c. P(z>c) = 0.0668
P(z<c) = 1 – 0.0668
= 0.9332
c = 1.50
0.9332
0.0228
<--------------------------------
<-----------2.00
<--------------------------------|
0
Z
0
1.50
Z
146
CHAPTER 6 Normal Probability Distributions
d. Since P(z<-d) = P(z>d) by symmetry
and ΣP(z) = 1,
P(z<-d) + P(-d<z<d) + P(z>d) = 1
P(z<-d) + 0.5878 + P(z<-d) = 1
2·P(z<-d) + 0.5878 = 1
2 ·P(z<-d) = 0.4122
P(z<-d) = 0.2061
-d = -0.82
d = 0.82
<--------------------------------|
0.7939
<--------------------------------|
0.5478
0.2061
0.4522
<-----------0.82
e. Since P(z<-e) = P(z>e) by symmetry
and ΣP(z) = 1,
P(z<-e) + P(-e<z<e + P(z>e) = 1
P(z<-e) + 0.0956 + P(z<-e) = 1
2·P(z<-e) + 0.0956 = 1
2·P(z<-e) = 0.9044
P(z<-e) = 0.4522
-e = -0.12
e = 0.12
<----------0
0.82
Observe that 0.7939 – 0.2061 = 0.5878,
as given in the problem.
Z
-0.12
0
0.12
Z
Observe that 0.5478 – 0.4522 = 0.0956,
as given in the problem.
56. The minimum and maximum values are 123.0 and 125.0 respectively.
The range is 125.0 – 123.0 = 2.0 volts.
μ = (minimum + maximum)/2 = (123.0 + 125.0)/2 = 248.0/2 – 124.0 volts
σ = range/ 12 = 2/ 12 = 0.58 volts
6-3 Applications of Normal Distributions
1. A normal distribution can have any mean and any positive standard deviation. A standard
normal distribution has mean 0 and standard deviation 1 – and it follows a “nice” bell-shaped
curve. Non-standard normal distributions can follow bell-shaped curves that are tall and thin,
or short and fat.
2. All normal distributions are continuous, symmetric and unimodal.
a. The area under the curve for any continuous probability distributions is 1.
b. In any symmetric distribution, the median equals the mean. And so the median is 100.
c. In any symmetric unimodal distribution, the mean and median and mode are equal. And so
the mode is 100.
3. For any distribution, converting to z scores using the formula z = (x-μ)/σ produces a sameshaped distribution with mean 0 and standard deviation 1.
4. No. Since the distribution of random digits does not follow a normal distribution, the methods
of this section are not relevant. Because random digits follow a discrete uniform distribution,
in which each digit is equally likely, probability questions can be answered using the methods
of chapter 4. When selecting a random digit, there are 5 possibilities [0,1,2,3,4] less than 5 and
P(x<5) = 5/10 = 0.5.
Applications of Normal Distributions SECTION 6-3
147
5. P(x<120) = P(z<1.33)
= 0.9082
6. P(x>80) = P(z>-1.33)
= 1 – P(z<-1.33)
= 1 – 0.0918 = 0.9082
7. P(90<x<115) = P(-0.67<z<1.00)
= P(z<1.00) – P(z<-0.67)
= 0.8413 – 0.2514
= 0.5899
8. P(75<x<110) = P(-1.67<z<0.67)
= P(z<0.67) – P(z<-1.67)
= 0.7486 – 0.0475
= 0.7011
9. The z score with 0.6 below it is z = 0.25 [closest entry 0.5987].
x = μ + zσ
= 100 + (0.25)(15)
= 100 + 3.75 = 103.75, rounded to 103.8
10. The z score with 0.8 above is the z score with 0.2 below it; z = -0.84 [closest entry 0.2005].
x = μ + zσ
= 100 + (-0.84)(15)
= 100 – 12.6 = 87.4
11. The z score with 0.95 above is the z score with 0.05 below it; z = -1.645 [bottom of table].
x = μ + zσ
= 100 + (-1.645)(15)
= 100 – 24.675 = 75.325, rounded to 75.3
12. The z score with 0.99 below it is z = 2.33 [closest entry 0.9901].
x = μ + zσ
= 100 + (2.33)(15)
= 100 + 34.95 = 134.95, rounded to 135.0
13. normal distribution: μ = 100 and σ = 15
P(x<115)
= P(z<1.00)
= 0.8413
14. normal distribution: μ = 100 and σ = 15
P(x>131.5)
= P(z>2.10)
= 1 – 0.9821
= 0.0179
<--------------------------------|
0.8413
0.9821
<-------------------------------100
0
115
1.00
x
Z
100
0
131.5
2.10
x
Z
148
CHAPTER 6 Normal Probability Distributions
15. normal distribution: μ = 100 and σ = 15
P(90<x<110) = P(-0.67<z<0.67)
= 0.7486 – 0.2514
= 0.4972
16. normal distribution: μ = 100 and σ = 15
P(110<x<120) = P(0.67<z<1.33)
= 0.9082 – 0.7486
= 0.1596
<--------------------------------|
0.7486
<----------------------------------|
0.9082
0.7486
0.2514
<----------90
-0.67
<---------------------------100
0
110
0.67
17. normal distribution: μ = 100 and σ = 15
For P30, A = 0.3000 [0.3015] and z = -0.52.
and z = -0.52.
x = μ + zσ
= 100 + (-0.52)(15)
= 100 – 7.8 = 99.2
0.3000
x
Z
18. normal distribution: μ = 100 and σ = 15
For Q1, A = 0.2500 [0.2514]
and z = -0.67.
x = μ + zσ
= 100 + (-0.67)(15)
= 100 – 10.05 = 89.95, rounded to 90.0
0.2500
<----------?
-0.52
100 110 120
0 0.67 1.33
x
Z
<----------100
0
x
Z
19. normal distribution: μ = 100 and σ = 15
For Q3, A = 0.7500 [0.7486]
and z = 0.67.
x = μ + zσ
= 100 + (0.67)(15)
= 100 + 10.05 = 110.05, rounded to 110.1
?
-0.67
100
0
x
Z
20. normal distribution: μ = 100 and σ = 15
For the top 37%, A = 0.6300 [0.6293]
and z = 0.33.
x = μ + zσ
= 100 + (0.33)(15)
= 100 + 4.95 = 104.95, rounded to 105.0
<--------------------------------|
0.7500
0.6300
<-------------------------------100
0
?
0.67
x
Z
100
0
?
0.33
x
Z
Applications of Normal Distributions SECTION 6-3
21. a. normal distribution: μ = 69.0, σ = 2.8
P(x<72) = P(z<1.07)
= 0.8577 or 85.77%
b. normal distribution: μ = 63.6, σ = 2.5
P(x<72) = P(z<3.36)
= 0.9996 or 99.96%
<--------------------------------|
<--------------------------------|
0.8577
69.0
0
72
1.07
0.9996
x
Z
c. No. It is not adequate in that 14% of the
men need to bend to enter, and they may
be in danger of injuring themselves if they
fail to recognize the necessity to bend.
63.6
0
x
Z
0.9800
69.0
0
22. a. normal distribution: μ = 69.0, σ = 2.8
P(x<51.6) = P(z<-6.21)
= 0.0001 or 0.01%
72
3.36
<--------------------------------|
d. For A = 0.9800 [0.9798],
and z = 2.05.
x = μ + zσ
= 69.0 + (2.05)(2.8)
= 69.0 + 5.7
= 74.7 inches
(See the sketch at the right)
?
2.05
x
Z
b. normal distribution: μ = 63.6, σ = 2.5
P(x<51.6) = P(z<-4.80)
= 0.0001 or 0.01%
0.0001
0.0001
<----------51.6
-6.21
149
<----------69.0
0
x
Z
51.6
-4.80
63.6
0
x
Z
150
CHAPTER 6 Normal Probability Distributions
c. Maybe. While it may not be convenient,
it presents no danger of injury because of
the obvious need for everyone to bend.
Considering the small size of the plane,
The door is probably as large as possible.
<--------------------------------|
0.6000
d. For A = 0.6000 [0.5987],
and z = 0.25
x = μ + zσ
= 69.0 + (0.25)(2.8)
= 69.0 + 0.7
= 69.7 inches
(See the sketch at the right.)
69.0
0
23. a. normal distribution: μ = 69.0, σ = 2.8
P(x>74) = P(z>1.79)
= 1 – 0.9633
= 0.0367 or 3.67%
b. normal distribution: μ = 63.6, σ = 2.5
P(x>70) = P(z>2.56)
= 1 – 0.9948
= 0.0052 or 0.52%
0.9633
0.9948
<-------------------------------69.0
0
74.0
1.79
x
Z
?
0.25
<-------------------------------x
Z
63.6
0
70.0
2.56
x
Z
c. No. The requirements are not equally fair for men and women, since the percentage of men
that are eligible is so much larger than the percentage of women who are eligible.
24. a. normal distribution: μ = 69.0, σ = 2.8
For the tallest 4%, A = 0.9600 [0.9599]
and z = 1.75.
x = μ + zσ
= 69.0 + (1.75)(2.8)
= 69.0 + 4.9
= 73.9 inches
0.9600
0.9600
<-------------------------------69.0
0
?
1.75
b. normal distribution: μ = 63.6 σ = 2.5
For the tallest 4%, A = 0.9600 [0.9599]
and z = 1.75.
x = μ + zσ
= 63.6 + (1.75)(2.5)
= 63.6 + 4.4
= 68.0 inches
<-------------------------------x
Z
63.6
0
?
1.75
x
Z
Applications of Normal Distributions SECTION 6-3
25. normal distribution: μ = 63.6 σ = 2.5
a. P(58<x<80) = P(-2.24<z<6.56)
= 0.9999 – 0.0125
= 0.9874 or 98.74%
No. Only 1 – 0.9874 = 0.0126 = 1.26%
of the women are not eligible because
of the height requirements.
<--------------------------------|
0.9999
0.0125
<----------58
-2.24
b. For the shortest 1%, A = 0.0100 [0.0099]
and z = -2.33.
x = μ + zσ
= 63.6 + (-2.33)(2.5)
= 63.6 – 5.8
= 57.8 inches
0.0100
63.6
0
80
6.56
x
Z
For the tallest 2%, A = 0.9800 [0.9798]
and z = 2.05.
x = μ + zσ
= 63.6 + (2.05)(2.5)
= 63.6 + 5.1
= 68.7 inches
0.9800
<----------?
-2.33
151
<-------------------------------63.6
0
x
Z
26. normal distribution: μ = 69.0, σ = 2.8
a. P(64<x<80) = P(-1.79<z<3.92)
= 0.9999 – 0.0367
= 0.9632 or 96.32%
No. Only 1 – 0.9632 = 0.0368 = 3.68%
of the men are not eligible because of
the height requirements.
63.6
0
?
2.05
x
Z
<--------------------------------|
0.9999
0.0367
<----------64
-1.79
69.0
0
80
3.92
x
Z
152
CHAPTER 6 Normal Probability Distributions
b. For the shortest 3%, A = 0.0300 [0.0301]
and z = -1.88.
x = μ + zσ
= 69.0 + (-1.88)(2.8)
= 69.0 – 5.3
= 63.7 inches
0.0300
0.9600
<----------?
-1.88
<-------------------------------69.0
0
x
Z
27. normal distribution: μ = 3570 and σ = 500
a. P(x<2700)
= P(z<-1.74)
= 0.409 or 4.09%
0.0409
69.0
0
?
1.75
x
Z
b. For the lightest 3%, A = 0.0300 [0.0301]
and z = -1.88
x = μ + zσ
= 3570 + (-1.88)(500)
= 3570 – 940 = 2630 g
0.0300
<----------2700
-1.74
For the tallest 4%, A = 0.9600 [0.9599]
and z = 1.75.
x = μ + zσ
= 69.0 + (1.75)(2.8)
= 69.0 + 4.9
= 73.9 inches
<----------3570
0
?
-1.88
x
Z
3570
0
x
Z
c. Not all babies below a certain birth weight require special treatment. The need for special
treatment is determined at least as much by developmental considerations as by weight
alone. Also, the birth weight identifying the bottom 3% is not a static figure and would have
to be updated periodically – perhaps creating unnecessary uncertainty and inconsistency.
28. normal distribution: μ = 172 and σ = 29
a. P(x<174) = P(z<0.07)
= 0.5279
b. 3500/140 = 25 men
c. 3500/174 = 20 men
d. Historically, the mean weight of
American adults has been increasing
over time.
<--------------------------------|
0.5279
172
0
174
0.07
x
Z
Applications of Normal Distributions SECTION 6-3
29. normal distribution: μ = 98.29 and σ = 0.62
a. P(x>100.6)
= P(z>3.87)
= 1 – 0.9999
= 0.0001
Yes. The cut-off is appropriate in that
there is a small probability of saying
that a healthy person has a fever, but
many with low grade fevers may
erroneously be labeled healthy.
0.9999
100.6
3.87
b. For the highest 5%, A = 0.9500
and z = 1.645.
x = μ + zσ
= 98.20 + (1.645)(0.62)
= 98.20 + 1.02
= 99.22 °F
0.9500
<-------------------------------98.20
0
153
<-------------------------------x
Z
30. normal distribution: μ = 14.4 and σ = 1.0
For P99, A = 0.9900 [0.9901]
and z = 2.33.
x = μ + zσ
= 14.4 + (2.33)(1.0)
= 14.4 + 2.3
= 16.7 inches
98.20
0
x
Z
<--------------------------------|
0.9900
14.4
0
31. normal distribution: μ = 268 and σ = 15
a. P(x>308)
= P(z>2.67)
= 1 – 0.9962
= 0.0038
The result suggests that an unusual
event has occurred – but certainly not
an impossible one, as about 38 of every
10,000 pregnancies can be expected to
last as long.
?
1.645
?
2.33
x
Z
0.9962
<-------------------------------268
0
308
2.67
x
Z
154
CHAPTER 6 Normal Probability Distributions
b. For the lowest 4%, A = 0.0400 [0.0401]
and z = -1.75.
x = μ + zσ
= 268 + (-1.75)(15)
= 268 – 26
= 242 days
0.0400
<----------?
-1.75
32. normal distributions
women: μ = 22.7 and σ = 1.0
For P5, A = 0.0500 and z = -1.645.
x = μ + zσ
= 22.7 + (-1.645)(1.0)
= 22.7 – 1.64
= 21.06, rounded to 21.1 inches
268
0
x
Z
men: μ = 23.5 and σ = 1.1
For P95, A = 0.9500 and z = 1.645.
x = μ + zσ
= 23.5 + (1.645)(1.1)
= 23.5 + 1.81
= 25.31, rounded to 25.3 inches
<--------------------------------|
0.9500
0.0500
<----------?
-1.645
22.7
0
x
Z
23.5
0
?
1.645
x
Z
The minimum and maximum sitting distances are 21.1 inches and 25.3 inches respectively.
33. a. The distribution and summary statistics,
obtained using statistical software, are
given below. The data appear to have a
distribution that is approximately normal.
n = 40
pressure frequency
x = 118.9
90- 99
1
s = 10.46
100-109
110-119
120-129
130-139
140-149
150-159
4
17
12
5
0
1 .
40
NOTE: Using rounded σ=10.5 in part (b)
gives x5 = 101.6 and x95 = 136.2.
b. normal distribution: μ = 118.9, σ = 10.46
For P5, A = 0.0500 and z = -1.645.
For P95, A = 0.9500 and z = 1.645.
x5 = μ + zσ = 118.9 + (-1.645)(10.46)
= 118.9 – 17.2 = 101.7 mm
x95 = μ + zσ = 118.9 + (1.645)(10.46)
= 118.9 + 17.2 = 136.1 mm
<--------------------------------|
0.9500
0.0500
<----------?
-1.645
110.8
0
?
1.645
x
Z
Applications of Normal Distributions SECTION 6-3
34. a. The distribution and summary statistics,
obtained using statistical software, are
given below. The data appear to have a
distribution that is approximately normal.
n = 115
duration frequency
x = 220.1
0- 49
1
s = 86.0
50- 99
100-149
150-199
200-249
250-299
300-349
350-399
400-449
8
18
23
25
19
11
8
2 .
115
155
b. normal distribution: μ = 220.1, σ = 86.0
For Q1, A = 0.2500 and z = -0.67.
For Q2, A = 0.5000 and z = 0.00.
For Q3, A = 0.7500 and z = 0.67.
x1 = μ + zσ = 220.1 + (-0.67)(86.0)
= 220.1 – 57.6 = 162.5 days
x2 = μ + zσ = 220.1 + (0.00)(86.0)
= 220.1 + 0.0 = 220.1 days
x3 = μ + zσ = 220.1 + (0.67)(86.0)
= 220.1 + 57.6 = 277.7 days
<--------------------------------|
0.7500
0.5000
<--------------------0.2500
<----------?
-0.67
220.1
0
?
0.67
x
Z
35. a. The z scores are always unit free. Because the numerator and the denominator of the
fraction z = (x-μ)/σ have the same units, the units will divide out.
b. For a population of size N, μ = Σx/N and σ2 = Σ(x-μ)2/N.
As shown below, μ = 0 and σ = 1 will be true for any set of z scores – regardless of the
shape of the original distribution.
Σz = Σ(x-μ)/σ = (1/σ)[Σ(x-μ)]
= (1/σ)[Σx – Σμ]
= (1/σ)[Nμ – Nμ] = (1/σ)[0] = 0
2
Σz = Σ[(x-μ)/σ ]2= (1/σ)2[Σ(x-μ)2]
= (1/σ2)[Nσ2] = N
μZ = (Σz)/N = 0/N = 0
σ2Z = Σ(z-μZ)2/N = Σ(z-0)2/N = Σz2/N = N/N = 1 and σZ = 1
The re-scaling from x scores to z scores will not affect the basic shape of the distribution –
and so in this case the z scores will be normal, as was the original distribution.
36. normal distribution: μ = 100 and σ = 15
a. P(x>103) = P(z>0.20)
= 1 – 0.5793
= 0.4207
b. PC(x>103) = P(x>103.5)
= P(z>0.23)
= 1 – 0.5910
= 0.4090
c. There is a difference of 0.4207 – 0.4090 = 0.0117 when the correction is not used.
This absolute probability error of 0.0117 is a relative error of 0.0117/0.4090 = 2.9%.
The difference is small, but not negligible.
156
CHAPTER 6 Normal Probability Distributions
37. normal distribution: μ = 25 and σ = 5
a. For a population of size N, μ = Σx/N and σ2 = Σ(x-μ)2/N.
As shown below, adding a constant to each score increases the mean by that amount but
does not affect the standard deviation. In non-statistical terms, shifting everything by k units
does not affect the spread of the scores. This is true for any set of scores – regardless of the
shape of the original distribution. Let y = x + k.
μY = [Σ(x+k)]/N
= [Σx + Σk]/N
= [Σx + Nk]/N
= (Σx)/N + (Nk)/N
= μX + k
σ2Y = Σ[y – μY]2/N
= Σ[(x+k) – (μX + k)]2/N
= Σ[x – μX]2/N
= σ2X and so σY = σX
If the teacher adds 50 to each grade,
new mean = 25 + 50 = 75
new standard deviation = 5 (same as before).
b. No. Curving should consider the variation. Had the test been more appropriately
constructed, it is not likely that every student would score exactly 50 points higher. If the
typical student score increased by 50, we would expect the better students to increase by
more than 50 and the poorer students to increase by less than 50. This would make the
scores more spread out and would increase the standard deviation.
c. For the top 10%, A = 0.9000 [0.8997] and z = 1.28.
x = μ + zσ
= 25 + (1.28)(5) = 25 + 6.4 = 31.4
For the bottom 70%, A = 0.7000 [0.6985] and z = 0.52.
x = μ + zσ
= 25 + (0.52)(5) = 25 + 2.6 = 27.6
For the bottom 30%, A = 0.3000 [0.3015] and z = -0.52.
x = μ + zσ
= 25 + (-0.52)(5) = 25 – 2.6 = 22.4
For the bottom 10%, A = 0.1000 [0.1003] and z = -1.28.
x = μ + zσ
= 25 + (-1.28)(5) = 25 – 6.4 = 18.6
This produces the grading scheme given at the right.
A: higher than 31.4
B: 27.6 to 31.4
C: 22.4 4o 27.6
D: 18.6 to 22.4
E: less than 18.6
d. The curving scheme in part (c) is fairer because it takes into account the variation as
discussed in part (b). Assuming the usual 90-80-70-60 letter grade cut-offs, for example, the
percentage of A’s under the scheme in part (a) with μ = 75 and σ = 5 is
P(x>90) = 1 – P(x<90)
= 1 – P(z<3.00)
= 1 – 0.9987
= 0.0013 or 0.13%
This is considerably less than the 10% A’s under the scheme in part (c) and reflects the fact
that the variation in part (a) is unrealistically small.
Applications of Normal Distributions SECTION 6-3
38. normal distributions
a. For P67, A = 0.6700 and z = 0.44.
xSAT = μ + zσ
= 1518 + (0.44)(325)
= 1518 + 143
= 1661
xACT = μ + zσ
= 21.1 + (0.44)(4.8)
= 21.1 + 2.1
= 23.2
<--------------------------------|
0.6700
1518
21.1
0
?
?
0.44
SAT
ACT
1518
21.1
0
1900
?
1.175
SAT
ACT
Z
b. zSAT = (x-μ)/σ
= (1900 – 1518)/325
= 1.175
xACT = μ + zσ
= 21.1 + (1.175)(4.6)
= 21.1 + 5.6
= 26.7
Z
39. Work with z scores to answer the question using a standard normal distribution. This is a
four-step process as follows.
Step 1. Find Q1 and Q3.
For Q1, A = 0.2500 [0.2514] and z = -0.67.
For Q3, A = 0.7500 [0.7486] and z = 0.67.
Step 2. Find the IQR = Q3 – Q1, and then determine (1.5)(IQR)
IQR = Q3 – Q1 = 0.67 – (-0.67) = 1.34
(1.5)(IQR) = (1.5)(1.34) = 2.01
Step 3. Find the lower and upper limits L and U beyond which outliers occur.
L = Q1 – (1.5)(IQR) = -0.67 – 2.01 = -2.68
U = Q3 + (1.5)(IQR) = 0.67 + 2.01 = 2.68
Step 4. Find P(outlier) = P(z<L) + P(z>U).
P(outlier) = P(z<-2.68 or z>2.68)
= P(z<-2.68) + P(z>2.68)
= 0.0037 + (1 – 0.9963)
= 0.0037 + 0.0037
= 0.0074
0.9963
<-------------------------------0.0037
<-----------2.68
0
2.68
Z
157
158
CHAPTER 6 Normal Probability Distributions
6-4 Sampling Distributions and Estimators
1. Given a population distribution of scores, one can take a sample of size n and calculate any
of several statistics. A sampling distribution is the distribution of all possible values of
such a particular statistic.
2. No. When dealing with one sample of size n, a collection of individual scores, a person is
investigating the distribution of the individual scores. In order to investigate the distribution of
the means, a person needs a collections of means – which are typically obtained from multiple
samples of size n.
3. An unbiased estimator is one whose expected value is the true value of the parameter which it
estimates. The sample mean is an unbiased estimator of the population mean because the
expected value (or mean) of its sampling distribution is the population mean.
4. In most situations the population is large enough so that repeated sampling will not alter the
selection probabilities – and this is equivalent to sampling with replacement. Assuming such
is the case allows the use of the simpler mathematical formulas involving independent events
instead of the more complicated formulas involving dependent events.
5. No. The students at New York University are not necessarily representative (by race, major,
etc.) of the population of all U.S. college students.
6. a. Yes. The sample man is an unbiased estimator of the population mean.
b. No. The sample median is not an unbiased estimator of the population median.
c. Yes. The sample proportion is an unbiased estimator of the population proportion.
d. Yes. The sample variance is an unbiased estimator of the population variance.
e. No. The sample standard deviation is not an unbiased estimator of the population standard
deviation..
f. No. The sample range is not an unbiased estimator of the population range.
7. The sample means will have a distribution that is approximately normal. They will tend to
form a symmetric, unimodal and bell-shaped distribution around the value of the population
mean.
8. The sample proportions will have a distribution that is approximately normal. They will tend
to form a symmetric, unimodal and bell-shaped distribution around the value of the population
proportion.
9. a. The medians of the 9 samples are given in column 2
at the right. The sampling distribution of the
median is given in columns 3 and 4 at the right.
b. The population median is 3. The mean of the
sample medians is Σ x ·P( x ) = 45/9 = 5.
c. In general the sample medians do not target the
value of the population median. For this reason,
sample median is not a good estimator of the
population median.
sample
2,2
2,3
2,10
3,2
3,3
3,10
10,2
10,3
10,10
x
2
2.5
6
2.5
3
6.5
6
6.5
10
x
P( x ) x ·P( x )
2
1/9
2/9
2.5 2/9
5/9
3
1/9
3/9
6
2/9
12/9
6.5 2/9
13/9
10 1/9
10/9
9/9
45/9
Sampling Distributions and Estimators SECTION 6-4
159
NOTE: Section 5-2 defined the mean of a probability distribution of x’s as μx = Σ[x·P(x)]. If the
variable is designated by the symbol y, then the mean of a probability distribution of y’s is
μy = Σ[y·P(y)]. In this section, the variables are statistics – like x and p̂ . In such cases, the
formula for the mean may be adjusted – to μ x = Σ[ x ·P( x )] and μ p̂ = Σ[ p̂ ·P( p̂ )]. In a similar
manner, the formula for the variance of a probability distribution may also be adjusted to match
the variable being considered.
10. a. The standard deviations of the 9 samples
sample s2
s
s
P(s) s·P(s)
are given in column 3 at the right. The
2,2
0
0
0
3/9 0.000
sampling distribution of the standard
2,3
0.5
0.707
0.707 2/9 0.157
deviation is given in columns 4 and 5 at
2,10
32
5.657
4.950 2/9 1.100
3,2
0.5
0.707
5.657 2/9 1.257
the right.
3,3
0
0
9/9 2.514
b. Since the values 2,3,10 are considered a
3,10 24.5
4.950
population, the population variance is
2
2
2
2
2
10,2
32
5.657
σ = Σ(x-μ) /N = (3 + 2 +5 )/3 = 38/3.
10,3
24.5 4.950
The population standard deviation is
10,10
0
0
38/3 = 3.559. The mean of the sample
standard deviations is Σs·P(s) = 2.514.
c. In general the sample standard deviations do not target the value of the population standard
deviation. For this reason, the sample standard deviation is not a good estimator of the
population standard deviation.
11. a. The variances of the 9 samples are given in column
2 at the right. The sampling distribution of the
variance is given in columns 3 and 4 at the right.
b. Since the values 2,3,10 are considered a
population, the population variance is
σ2 = Σ(x-μ)2/N = (32 + 22 +52)/3 = 38/3. The mean
of the sample variances is Σs2·P(s2) = 114/9 = 38/3.
c. The sample variance always targets the value of the
population variance. For this reason, the sample
variance is a good estimator of the population
variance.
12. a. The means of the 9 samples are given in column 2
at the right. The sampling distribution of the
mean is given in columns 3 and 4 at the right.
b. The population mean is 5. The mean of the
sample means is Σ x ·P( x ) = 45/9 = 5.
c. The sample mean always targets the value of the
population mean. For this reason, the sample mean
is a good estimator of the population mean.
sample s2
2,2
0
2,3
0.5
2,10
32
3,2
0.5
3,3
0
3,10 24.5
10,2
32
10,3
24.5
10,10
0
sample
2,2
2,3
2,10
3,2
3,3
3,10
10,2
10,3
10,10
s2
P(s2) s2·P(s2)
0
3/9
0/9
0.5 2/9
1/9
24.5 2/9
49/9
32
2/9
64/9
9/9 114/9
x
x
2
2.5
6
2.5
3
6.5
6
6.5
10
2
2.5
3
6
6.5
10
P( x ) x ·P( x )
1/9
2/9
2/9
5/9
1/9
3/9
2/9
12/9
2/9
13/9
1/9
10/9
9/9
45/9
160
CHAPTER 6 Normal Probability Distributions
The information below and the box at the right apply to
Exercises 13-16.
original population of scores in order: 46 49 56 58
summary statistics: N= 4 Σx = 209 Σx2 = 11017
μ = Σx/N
= 209/4 = 52.25
population x = (x2+x3)/2
= (49+56)/2 = 52.5
population R = xn – x1
= 58 – 46 = 12
σ2 = [Σ(x-μ)2]/N
= [(-6.25)2 + (-3.25)2 + (3.75)2 + (5.75)2]/4
= [39.0625 + 10.5625 + 14.0625 + 33.0625]/4
= 96.75/4 = 24.1875
sample
46,46
46,49
46,56
46,58
49,46
49,49
49,56
49,58
56,46
56,49
56,56
56,58
58,46
58,49
58,56
58,58
x
x
46.0
47.5
51.0
52.0
47.5
49.0
52.5
53.5
51.0
52.5
56.0
57.0
52.0
53.5
57.0
58.0
46.0
47.5
51.0
52.0
47.5
49.0
52.5
53.5
51.0
52.5
56.0
57.0
52.0
53.5
57.0
58.0
x
P( x )
1/16
2/16
1/16
2/16
2/16
2/16
2/16
1/16
2/16
1/16
16/16
13. a. The sixteen possible samples are given in the “samples”
column of the box preceding this exercise
b. The sixteen possible means are given in column 2 of the
box preceding this exercise. The sampling distribution of
the mean is given in the first two columns at the right.
c. The population mean is 52.25. The mean of the
sample means is Σ x ·P( x ) = 836.0/16 = 52.25. They
are the same.
d. Yes. The sample mean always targets the value of the
population mean. For this reason, the sample mean
is a good estimator of the population mean.
46.0
47.5
49.0
51.0
52.0
52.5
53.5
56.0
57.0
58.0
14. a. The sixteen possible samples are given in the “samples”
column of the box preceding Exercise 13.
b. The sixteen possible medians are given in column 3 of the
box preceding Exercise 13. The sampling distribution of
the median is given in the first two columns at the right.
c. The population median is 52.5. The mean of the
sample medians is Σ x ·P( x ) = 836.0/16 = 52.25. They
are not the same.
d. No. The sample medians do not always target the value
of the population median. For this reason, the sample
median is not a good estimator of the population median.
46.0
47.5
49.0
51.0
52.0
52.5
53.5
56.0
57.0
58.0
x
P( x )
1/16
2/16
1/16
2/16
2/16
2/16
2/16
1/16
2/16
1/16
16/16
R
0
3
10
12
3
0
7
9
10
7
0
2
12
9
2
0
s2
0.0
4.5
50.0
72.0
4.5
0.0
24.5
40.5
50.0
24.5
0.0
2.0
72.0
40.5
2.0
0.0
x ·P( x )
46.0/16
95.0/16
49.0/16
102.0/16
104.0/16
105.0/16
107.0/16
56.0/16
114.0/16
58.0/16
836.0/16
x ·P( x )
46.0/16
95.0/16
49.0/16
102.0/16
104.0/16
105.0/16
107.0/16
56.0/16
114.0/16
58.0/16
836.0/16
15. a. The sixteen possible samples are given in the “samples” column of the box preceding
Exercise 13.
Sampling Distributions and Estimators SECTION 6-4
b. The sixteen possible ranges are given in column 4 of the
box preceding Exercise 13. The sampling distribution of
the range is given in the first two columns at the right.
c. The population range is 12. The mean of the
sample ranges is ΣR·P(R) = 5.375. They are not the same.
d. No. The sample ranges do not always target the value
of the population range. For this reason, the sample
range is not a good estimator of the population range.
16. a. The sixteen possible samples are given in the “samples”
column of the box preceding this exercise
b. The sixteen possible variances are given in column 5 of the
box preceding Exercise 13. The sampling distribution of
the variance is given in the first two columns at the right.
c. The population mean is 24.1875. The mean of the
sample variances is Σs2·P(s2) = 387.0/16 = 24.1875. They
are the same.
d. Yes. The sample variance always targets the value of the
population variance. For this reason, the sample variance
is a good estimator of the population variance.
17. Let p̂ be the symbol for the sample proportion. The 9
possible sample proportions are given in column 2
at the right. The sampling distribution of the
proportion is given in columns 3 and 4 at the right.
The population proportion of odd numbers is 1/3.
The mean of the sample proportions is Σ p̂ ·P( p̂ ) =
3.0/9 = 1/3. They are the same. The sample
proportion always targets the value of the population
proportion. For this reason, the sample proportion is
a good estimator of the population proportion.
sample
2,2
2,3
2,10
3,2
3,3
3,10
10,2
10,3
10,10
18. Let p̂ be the symbol for the sample proportion. The 8
possible sample proportions are given in column 2
at the right. The sampling distribution of the
proportion is given in columns 3 and 4 at the right.
The population proportion of girl births is 0.5
The mean of the sample proportions is Σ p̂ ·P( p̂ ) =
12/24 = 0.5. They are the same. The sample
proportion always targets the value of the population
proportion. For this reason, the sample proportion is
a good estimator of the population proportion.
sample
bbb
bbg
bgb
bgg
gbb
gbg
ggb
ggg
R
0
2
3
7
9
10
12
P(R)
4/16
2/16
2/16
2/16
2/16
2/16
2/16
16/16
R·P(R)
0/16
4/16
6/16
14/16
18/16
20/16
24/16
86/16
s2
0.0
2.0
4.5
24.5
40.5
50.0
72.0
P(s2)
4/16
2/16
2/16
2/16
2/16
2/16
2/16
16/16
p̂
p̂ P( p̂ ) p̂ ·P( p̂ )
0.0
0.5
0.0
0.5
1.0
0.5
0.0
0.5
0.0
p̂
0
1/3
1/3
2/3
1/3
2/3
2/3
1
0.0
0.5
1.0
4/9
4/9
1/9
9/9
s2·P(s2)
0.0/16
4.0/16
9.0/16
49.0/16
81.0/16
100.0/16
144.0/16
387.0/16
0.0/9
2.0/9
1.0/9
3.0/9
p̂ P( p̂ ) p̂ ·P( p̂ )
0 1/8
1/3 3/8
2/3 3/8
1 1/8
8/8
161
0/24
3/24
6/24
3/24
12/24
162
CHAPTER 6 Normal Probability Distributions
19. a. The sampling distribution of the proportion is
given in columns 5 and 6 of the table at the
right.
b. The mean of the sampling distribution is
Σ p̂ ·P( p̂ ) = 12.0/16 = 0.75.
c. The population proportion of females is
3/4 = 0.75. Yes, the two values are the same.
The sample proportion always targets the
value of the population proportion. For this
reason, the sample proportion is a good
estimator of the population proportion.
20. Identify the 2 defective chips as x and y, and
identify the 3 acceptable chips as a, b and c.
a. The sampling distribution of the proportion is
given in columns 5 and 6 of the table at the
right.
b. The mean of the sampling distribution is
Σ p̂ ·P( p̂ ) = 10.0/25 = 0.40.
c. The population proportion of defectives is
2/5 = 0.40. Yes, the two values are the same.
The sample proportion always targets the
value of the population proportion. For this
reason, the sample proportion is a good
estimator of the population proportion.
pair
mm
ma
mb
mc
am
aa
ab
ac
pair
xx
xy
xa
xb
xc
yx
yy
ya
yb
yc
p̂
0.0
0.5
0.5
0.5
0.5
1.0
1.0
1.0
p̂
1.0
1.0
0.5
0.5
0.5
1.0
1.0
0.5
0.5
0.5
pair
bm
ba
bb
bc
cm
ca
cb
cc
pair
ax
ay
aa
ab
ac
bx
by
ba
bb
bc
cx
cy
ca
cb
cc
p̂
p̂
0.5
1.0
1.0
1.0
0.5
1.0
1.0
1.0
0.0
0.5
1.0
p̂
p̂
0.5
0.5
0.0
0.0
0.0
0.5
0.5
0.0
0.0
0.0
0.5
0.5
0.0
0.0
0.0
P( p̂ ) p̂ ·P( p̂ )
1/16 0.0/16
6/16 3.0/16
9/16 9.0/16
16/16 12.0/16
P( p̂ ) p̂ ·P( p̂ )
0.0 9/25 0.0/25
0.5 12/25 6.0/25
1.0 4/25 4.0/16
25/25 10.0/25
21. Use of the formula for the given values is shown below. The resulting distribution agrees
with, and therefore describes, the sampling distribution for the proportion of girls in a family
of size 2.
P(x) = 1/[2(2-2x)!(2x)!]
P(x=0) = 1/[2(2)!(0)!] = 1/[2·2·1] = 1/4
P(x=0.5) = 1/[2(1)!(1)!] = 1/[2·1·1] = 1/2
P(x=1) = 1/[2(0)!(2)!] = 1/[2·1·2] = 1/4
22. The MAD of the population
is 3.33, as illustrated by the
table below.
x |x-μ|
2
3
3
2
10 5
15 10
μ = Σx/N = 15/3 = 5
MAD = (Σ|x-μ|)/N
= 10/3 = 3.33
sample x
2,2 2.0
2,3 2.5
2,10 6.0
3,2 2.5
3,3 3.0
3,10 6.5
10,2 6.0
10,3 6.5
10,10 10.0
|x- x |’s MAD
0,0
0
0.5,.5 0.5
4,4
4.0
0.5,.5 0.5
0,0
0
3.5,3.5 3.5
4,4
4.0
3.5,3.5 3.5
0,0
0
16.0
MAD P(MAD) MAD·P(MAD)
0
0.5
3.5
4.0
3/9
2/9
2/9
2/9
9/9
0/9
1/9
7/9
8/9
16/9
Sampling Distributions and Estimators SECTION 6-4
163
The 9 equally likely samples and their MAD’s are given in the above box on the left. Each
sample has a probability of 1/9. The sampling distribution of the MAD’s, a list of each sample
MAD and its total probability is given in the above box at the right.
Since μMAD = Σ[MAD·P(MAD)] = 16/9 = 1.78 ≠ 3.33, the sample MAD is not a good estimator
(i.e., is not an unbiased estimator) of the population MAD.
6-5 The Central Limit Theorem
1. The standard error of a statistic is the standard deviation of its sampling distribution. For any
given sample size n, the standard error of the mean is the standard deviation of the distribution
of all possible sample means of size n. Its symbol and numerical value are σ x = σ/ n .
2. To use the normal distribution for the distribution of the sample means for samples with n<30,
the original population must have a normal distribution.
3. The subscript x is used to distinguish the mean and standard deviation of the sample means
from the mean and standard deviation of the original population. This produces the notation
μx = μ
σ x = σ/ n
4. No. The distribution of the sample incomes will have the same shape as the original
distribution – which is positively skewed because of the lower bound of zero and the unlimited
upper bound. The distribution of the means of samples of size n>30 will have a normal
distribution, but the individual n>30 values will have the same shape as the original
population.
5. a. normal distribution
μ = 1518
σ = 325
P(x<1500)
= P(z<-0.06)
= 0.4761
0.4761
<----------1500
-0.06
1518
0
x
Z
1518
0
_
x
Z
b. normal distribution,
since the original distribution is so
μ x = μ = 1518
σ x = σ/ n = 325/ 100 = 32.5
P( x <1500)
= P(z<-0.55)
= 0.2912
0.2912
<----------1500
-0.55
164
CHAPTER 6 Normal Probability Distributions
6. a. normal distribution
μ = 1518
σ = 325
P(x>1600)
= P(z>0.25)
= 1 – 0.5987
= 0.4013
0.5987
<-------------------------------1518
0
1600
0.25
x
Z
b. normal distribution,
since the original distribution is so
μ x = μ = 1518
σ x = σ/ n = 325/ 64 = 40.625
P( x >1600)
= P(z>2.02)
= 1 – 0.9783
= 0.0217
7. a. normal distribution
μ = 1518
σ = 325
P(1550<x<1575)
= P(0.10<z<0.18)
= 0.5714 – 0.5398
= 0.0316
0.9783
<-------------------------------1518
0
1600
2.02
<----------------------------------|
0.5714
0.5398
<---------------------------1518 1550 1575
0 0.10 0.18
b. normal distribution,
since the original distribution is so
μ x = μ = 1518
σ x = σ/ n = 325/ 25 = 65
P(1550< x <1575)
= P(0.49<z<0.88)
= 0.8106 – 0.6879
= 0.1227
_
x
Z
x
Z
<----------------------------------|
0.8106
0.6879
<---------------------------1518 1550 1575
0 0.49 0.88
_
x
Z
c. Since the original distribution is normal, the Central Limit Theorem can be used in part (b)
even though the sample size does not exceed 30.
The Central Limit Theorem SECTION 6-5
8. a. normal distribution
μ = 1518
σ = 325
P(1440<x<1480)
= P(-0.24<z<-0.12)
= 0.4522 – 0.4052
= 0.0470
<--------------|
0.4522
0.4052
<-------1440 1480 1518
-0.24 -0.12 0
b. normal distribution,
since the original distribution is so
μ x = μ = 1518
σ x = σ/ n = 325/ 16 = 81.25
P(1440< x <1480)
= P(-0.96<z<-0.47)
= 0.3192 – 0.1685
= 0.1507
165
x
Z
<--------------|
0.3192
0.1685
<-------1440 1480 1518
-0.96 -0.47 0
_
x
Z
c. Since the original distribution is normal, the Central Limit Theorem can be used in part (b)
even though the sample size does not exceed 30.
9. a. normal distribution
μ = 172
σ = 29
P(x>180)
= P(z>0.28)
= 1 – 0.6103
= 0.3897
0.6103
<-------------------------------172
0
180
0.28
x
Z
b. normal distribution,
since the original distribution is so
μ x = μ = 172
σ x = σ/ n = 29/ 20 = 6.48
P( x >180)
= P(z>1.23)
= 1 – 0.8907
= 0.1093
0.8907
<-------------------------------172
0
180
1.23
_
x
Z
c. Yes. A capacity of 20 is not appropriate when the passengers are all adult men, since a
10.93% probability of overloading is too much of a risk.
166
CHAPTER 6 Normal Probability Distributions
10. a. normal distribution
μ = 100
σ = 15
P(x>133)
= P(z>2.20)
= 1 – 0.9861
= 0.0139
0.9861
<-------------------------------100
0
133
2.20
x
Z
b. normal distribution,
since the original distribution is so
μ x = μ = 100
σ x = σ/ n = 15/ 9 = 5
P( x >133)
= P(z>6.60)
= 1 – 0.9999
= 0.0001
0.9999
<-------------------------------100
0
133
6.60
_
x
Z
c. No. Even though the mean score is 133, some of the individual scores may be below 131.5.
11. a. normal distribution
μ = 172
σ = 29
P(x>167)
= P(z>-0.17)
= 1 – 0.4325
= 0.5675
0.4325
<----------167
-0.17
172
0
x
Z
b. normal distribution,
since the original distribution is so
μ x = μ = 172
σ x = σ/ n = 29/ 12 = 8.372
P( x >167)
= P(z>-0.60)
0.2743
= 1 – 0.2743
<----------= 0.7257
167
172
c. No. It appears that the 12 person capacity
-0.60
0
could easily exceed the 2004 lbs –
especially when the weight of clothes and
equipment is considered. On the other hand, skiers may be lighter than the general
population – as the skiing may not be an activity that attracts heavier persons.
_
x
Z
The Central Limit Theorem SECTION 6-5
167
12. a. normal distribution
μ = 268
σ = 15
P(x<260)
= P(z<-0.53)
= 0.2981
0.2981
<----------260
-0.53
268
0
x
Z
268
0
_
x
Z
b. normal distribution,
since the original distribution is so
μ x = μ = 268
σ x = σ/ n = 15/ 25 = 3
P( x <260)
= P(z<-2.67)
= 0.0038
c. Yes. It is very unlikely to experience a
mean that low by chance alone, and the
effects of the diet on pregnancy should be
a matter of concern.
13. a. normal distribution
μ = 114.8
σ = 13.1
P(x>140)
= P(z>1.92)
= 1 – 0.9726
= 0.0274
0.0038
<----------260
-2.67
0.9726
<-------------------------------114.8
0
140
1.92
x
Z
b. normal distribution,
since the original distribution is so
μ x = μ = 114.8
σ x = σ/ n = 13.1/ 4 = 6.55
P( x >140)
= P(z>3.85)
0.9999
= 1 – 0.9999
<-------------------------------_
= 0.0001
140
x
114.8
c. Since the original distribution is normal,
Z
0
3.85
the Central Limit Theorem can be used in
part (b) even though the sample size does
not exceed 30.
d. No. The mean can be less than 140 when one or more of the values is greater than 140.
168
CHAPTER 6 Normal Probability Distributions
14. a. normal distribution
μ = 6.0
σ = 1.0
P(x<6.2)
= P(z<0.20)
= 0.5793
<--------------------------------|
0.5793
6.0
0
b. normal distribution,
since the original distribution is so
μ x = μ = 6.0
6.2
0.20
x
Z
<--------------------------------|
0.9772
σ x = σ/ n = 1.0/ 100 = 0.10
P( x <6.2)
= P(z<2.00)
= 0.9772
6.0
0
6.2
2.00
_
x
Z
c. Probabilities concerning means do not apply to individuals. It is the information from
part (a) that is relevant, since the helmets will be worn by one man at a time – and that
indicates that the proportion of men with head breadth greater than 6.2 inches is
1 – 0.5793 = 0.4207 = 42.07%.
15. a. normal distribution
μ = 69.0
σ = 2.8
P(x<72)
= P(z<1.07)
= 0.8577
<--------------------------------|
0.8577
69.0
0
b. normal distribution,
since the original distribution is so
μ x = μ = 69.0
72
1.07
x
Z
<--------------------------------|
0.9999
σ x = σ/ n = 2.8/ 100 = 0.28
P( x <72)
= P(z<10.17)
= 0.9999
69.0
0
72
10.17
_
x
Z
The Central Limit Theorem SECTION 6-5
169
c. The probability in part (a) is more relevant. Part (a) deals with individual passengers, and
these are the persons whose safety and comfort need to be considered. Part (b) deals with
group means – and it is possible for statistics that apply “on the average” to actually describe
only a small portion of the population of interest.
d. Women are generally smaller than men. Any design considerations that accommodate
larger men will automatically accommodate larger women.
16. a. normal distribution
μ = 0.8565
σ = 0.0518
P(x>0.8535)
= P(z>-0.06)
= 1 – 0.4761
= 0.5239
0.4761
<----------0.8535
-0.06
0.8565
0
x
Z
0.8565
0
_
x
Z
b. normal distribution,
since the original distribution is so
μ x = μ = 8565
σ x = σ/ n = 0.0518/ 465 = 0.00240
P( x >0.8535)
= P(z>-1.25)
= 1 – 0.1056
= 0.8944
0.1056
<----------0.8535
-1.25
c. Whether the bags contain the labeled amount cannot be answered from the one bag
examined. If each bag is filled with exactly 465 candies, then the company needs to make
some adjustments – because then the probability a bag contains less than the claimed weight
is 1 – 0.8944 = 0.1056 [and since 0.1056 > 0.05, it would not be unusual to get a bag with
less than the claimed weight]. In practice, however, the bags are filled by volume or weight
– and not by candy count [and several bags would have to be weighed to get an estimate of
the bag-to-bag variability].
17. a. normal distribution
μ = 143
σ = 29
P(140<x<211)
= P(-0.10<z<2.34)
= 0.9904 – 0.4602
= 0.5302
<--------------------------------|
0.9904
0.4602
<----------140
-0.10
143
0
211
2.34
x
Z
170
CHAPTER 6 Normal Probability Distributions
b. normal distribution,
since the original distribution is so
μ x = μ = 143
σ x = σ/ n = 29/ 36 = 4.833
P(140< x <211)
= P(-0.62<z<14.07)
= 0.9999 – 0.2676
= 0.7323
<--------------------------------|
0.9999
0.2676
<----------140
-0.62
143
0
211
14.07
_
x
Z
c. The information from part (a) is more relevant, since the seats will be occupied by one
woman at a time.
18. a. normal distribution
μ = 5.670
σ = 0.062
P(5.550<x<5.790)
= P(-1.94<z<1.94)
= 0.9738 – 0.0262
= 0.9476
<--------------------------------|
0.9738
0.0262
<----------5.550
-1.94
5.670
0
5.790
1.94
x
Z
If 0.9476 of the quarters are accepted, then 1 – 0.9476 = 0.0524 of the quarters are rejected.
For 280 quarters, we expect (0.0524)(280) = 14.7 of them to be rejected.
b. normal distribution,
since the original distribution is so
μ x = μ = 5.670
σ x = σ/ n = 0.062/ 280 = 0.00371
P(5.550< x <5.790)
= P(-32.39<z<32.39)
= 0.9999 – 0.0001
= 0.9998
<--------------------------------|
0.9999
0.0001
<----------5.550
-32.39
5.670
0
5.790
32.39
c. Probabilities concerning means do not apply to individuals. It is the information from
part (a) that is relevant, since the vending machine deals with quarters one at a time.
_
x
Z
The Central Limit Theorem SECTION 6-5
171
19. normal distribution,
by the Central Limit Theorem
μ x = μ = 12.00
σ x = σ/ n = 0.09/ 36 = 0.015
P( x >12.29)
= P(z>19.33)
= 1 – 0.9999
= 0.0001
0.9999
<-------------------------------12.00
0
12.29
19.33
_
x
Z
Yes. The results suggest that the Pepsi cans
are being filled with more than 12.00 oz of
product. This is undoubtedly done on purpose, to minimize probability of producing cans with
less than the stated 12.00 oz of product.
20. normal distribution,
by the Central Limit Theorem
μ x = μ = 98.6
σ x = σ/ n = 0.62/ 107 = 0.0599
P( x <98.2)
= P(z<-6.67)
= 0.0001
0.0001
<-----------
Yes. Since the probability of obtaining the
given results is so small if the mean were
truly 98.6 °F, the data suggest that such a
value is not the true mean body temperature.
21. a. normal distribution
μ = 69.0
σ = 2.8
The z score with 0.9500 below it is 1.645.
x = μ + zσ
= 69.0 + (1.645)(2.8)
= 69.0 + 4.6
= 73.6 inches
b. normal distribution,
since the original distribution is so
μ x = μ = 69.0
σ x = σ/ n = 2.8/ 100 = 0.28
The z score with 0.9500 below it is 1.645
x = μ x + zσ x
= 69.0 + (1.645)(0.28)
= 69.0 + 0.5
= 69.5 inches
98.2
-6.67
_
x
Z
98.6
0
<--------------------------------|
0.9500
69.0
0
?
1.645
x
Z
<--------------------------------|
0.9500
69.0
0
?
1.645
_
x
Z
172
CHAPTER 6 Normal Probability Distributions
c. The probability in part (a) is more relevant. Part (a) deals with individual passengers, and
these are the persons whose safety and comfort need to be considered. Part (b) deals with
group means – and it is possible for statistics that apply “on the average” to actually describe
only a small portion of the population of interest.
22. a. Yes. Since n/N = 36/160 = 0.225 > 0.05, use the finite population correction factor.
b. normal distribution,
since the original distribution is normal
μ x = μ = 8.5
σx =
σ
n
N-n
N-1
3.96 160-36 3.96 124
=
= 0.583
36 160-1
36 159
P( x >10.0)
= P(z>2.57)
= 1 – 0.9949
= 0.0051
=
0.9949
<-------------------------------8.5
0
10.0
2.57
_
x
Z
23. a. Yes. Since n/N = 12/210 = 0.0571 > 0.05, use the finite population correction factor.
b. For 12 passengers, the 2100 lb limit implies a mean weight of 2100/12 = 175 lbs.
c. normal distribution,
since the original distribution is normal
μ x = μ = 163
σx =
σ
n
N-n
N-1
32 210-12 32 198
=
= 8.991
0.9082
12 210-1
12 209
<-------------------------------_
P( x >175)
x
163
175
= P(z>1.33)
Z
0
1.33
= 1 – 0.9082
= 0.0918
d. The best approach is systematic trial and error. The above calculations indicate that for
n = 12, P(Σx<2100) = P( x <2100/n) = P( x < 175) = 0.9082.
Repeat the above calculations for
n = 11, 10, etc. until P(Σx<2100) = P( x <2100/n) reaches 0.9990.
*For n = 11, x = 2100/11 = 190.91
normal distribution,
since the original distribution is normal
μ x = μ = 163
=
σx =
=
σ
n
N-n
N-1
32 210-11 32 199
=
= 9.415
11 210-1
11 209
The Central Limit Theorem SECTION 6-5
P( x <190.91)
= P(z<2.96)
= 0.9985
Since 0.9985 < 0.999, try n = 10.
173
<--------------------------------|
0.9985
163
0
190.91
2.96
_
x
Z
*For n = 10, x = 2100/10 = 210
normal distribution,
since the original distribution is normal
μ x = μ = 163
σx =
σ
n
N-n
N-1
<--------------------------------|
0.9999
32 210-10 32 200
=
= 9.899
10 210-1
10 209
P( x <210)
= P(z<4.75)
= 0.9999
Since 0.9999 > 0.999, set n = 10.
=
163
0
210
4.75
_
x
Z
NOTE: The z score with 0.999 below it is 3.10. To find n exactly, solve the following
equation to get n = 10.9206. Since n must be a whole number, round down to set n = 10.
x - μx
z=
σx
3.10 =
2100/n - 163
32 210-n
n 210-1
24. a. The box below gives the calculations
to determine μ=5.0 and σ=3.5590 for
the original population of N=3 values.
x
2
3
10
15
x-μ (x-μ)2
-3
9
-2
4
5
25
0
38
μ = Σx/N = 15/3 = 5.0
σ2 = Σ(x-μ)2/N = 38/3 = 12.6667
σ = 3.5590
b. The 6 equally like samples of size n=2 are
listed in the first column in the box below.
c. The values of x are listed in the second
column in the box below.
sample
2,3
2,10
3,2
3,10
10,2
10,3
x
2.5
6.0
2.5
6.5
6.0
6.5
30.0
x - μ x ( x - μ x )2
-2.5
1.0
-2.5
1.5
1.0
1.5
0.0
6.25
1.00
6.25
2.25
1.00
2.25
19.00
174
CHAPTER 6 Normal Probability Distributions
d. Since there are 6 equally likely samples, the desired values can be determined using the
box above on the right as follows.
μ x = Σ x /6 = 30/6 = 5.0
σ x 2 = Σ( x - μ x )2/6 = 19/6 = 3.1667
σ x = 3.1667 = 1.7795
e. μ = 5 = μ x
σ
n
N-n
3.5590 3-2
=
= 1.7795 = σ x
N-1
3-1
2
6-6 Normal as Approximation to Binomial
1. Assuming the true proportion of households watching 60 Minutes remains relative constantly
during the two years of sampling, the two years of sample proportions represent samples from
the same population. Since the sampling distribution of the proportion approximates a normal
distribution, the histogram depicting the two years of sample proportions should be
approximately normal – e.g., bell-shaped.
2. IQ’s are measured in whole numbers, but the normal distribution is continuous. In a
continuous representation, an IQ of x=107 corresponds to 106.5<x<107.5.
3. No. With n=4 and p=0.5, the requirements that np ≥ 5 and nq ≥ 5 are not met.
4. binomial problem: n = 100 and p = 0.2
μ = np = (100)(0.2) = 20
σ2 npq = (100)(0.2)(0.8) = 16; σ = 4
The μ=20 indicates that a person guessing would be expected to get about 20 correct answers.
The σ=4 is the standard deviation for the number of correct answers of a guessing person,
indicating that the typical number of correct answers would be about 4 away from 20.
5. The area to the right of 8.5.
In symbols, P(x>8) = PC(x>8.5).
6. The area to the right of 1.5.
In symbols, P(x ≥ 2) = PC(x>1.5).
7. The area to the left of 4.5.
In symbols, P(x<5) = PC(x<4.5).
8. The area between 3.5 and 4.5.
In symbols, P(x=4) = PC(3.5<x<4.5).
9. The area to the left of 15.5.
In symbols, P(x ≤ 15) = PC(x<15.5).
10. NOTE: This exercise is subject to interpretation. If “between” is taken literally, the (a) answer
is correct. If “between 12 and 16” is taken to mean “between 12 and 16 inclusive,” the (b)
answer is correct.
(a) The area between 12.5 and 15.5.
In symbols, P(12<x<16) = PC(12.5<x<15.5).
(b) The area between 11.5 and 16.5.
In symbols, P(12 ≤ x ≤ 16) = PC(11.5<x<16.5).
11. The area between 4.5 and 9.5.
In symbols, P(4 ≤ x ≤ 9) = PC(4.5<x<9.5).
12. The area between 23.5 and 24.5.
In symbols, P(x=24) = PC(23.5<x<24.5).
Normal as Approximation to Binomial SECTION 6-6
13. binomial: n=10 and p=0.5
a. from Table A-1, P(x=3) = 0.117
b. normal approximation appropriate since
np = 10(0.5) = 5.0 ≥ 5
nq = 10(0.5) = 5.0 ≥ 5
μ = np = 10(0.5) = 5.0
σ = npq = 10(0.5)(0.5) = 1.581
P(x=3)
= P(2.5<x<3.5)
= P(-1.58<z<-0.95)
= 0.1711 – 0.0571
= 0.1140
175
<--------------|
0.1711
0.0571
<-------2.5 3.5 5.0
-1.58 -0.95 0
x
Z
14. binomial: n=12 and p=0.8
a. from Table A-1, P(x=9) = 0.236
b. normal approximation not appropriate since
nq = 12(0.2) = 2.4 < 5
15. binomial: n=8 and p=0.9
a. from Table A-1, P(x ≥ 6) = 0.149 + 0.383 + 0.430 = 0.962
b. normal approximation not appropriate since
nq = 8(0.1) = 0.8 < 5
16. binomial: n=15 and p=0.4
a. from Table A-1,
P(x<3) = 0+ + 0.005 + 0.022 = 0.027
b. normal approximation appropriate since
np = 15(0.4) = 6.0 ≥ 5
nq = 15(0.6) = 9.0 ≥ 5
μ = np = 15(0.4) = 6.0
σ = npq = 15(0.4)(0.6) = 1.897
P(x<3)
= P(x<2.5)
= P(z<-1.84)
= 0.0329
0.0329
<----------2.5
-1.84
6.0
0
17. binomial: n=40,000 and p=0.03
normal approximation appropriate since
np = 40,000(0.03) = 1200 ≥ 5
nq = 40,000(0.97) = 3880 ≥ 5
μ = np = 40,000(0.03) = 1200
σ = npq = 40000(0.03)(0.97) = 34.117
0.9982
P(x ≥ 1300)
<-------------------------------= P(x>1299.5)
1200 1299.5
= P(z>2.92)
0
2.92
= 1 – 0.9982
= 0.0018
No. It is not likely that the goal of at least 1300 will be reached.
x
Z
x
Z
176
CHAPTER 6 Normal Probability Distributions
18. binomial: n=2822 and p=0.75
normal approximation appropriate since
np = 2822(0.75) = 2116.5 ≥ 5
nq = 2822(0.25) = 705.5 ≥ 5
μ = np = 2822(0.75) = 2116.5
σ = npq = 2822(0.75)(0.25) = 23.003
0.0075
P(x ≤ 2060)
<----------= P(x<2060.5)
2060.5 2116.5
= P(z<-2.43)
x
-2.43
= 0.0075
Z
0
Yes. Since 0.0075 ≤ 0.05, it would be
unusual to get 2060 Internet users or less if the true population proportion were really 0.75.
19. binomial: n=574 and p=0.50
normal approximation appropriate since
np = 574(0.50) = 287 ≥ 5
nq = 574(0.50) = 287 ≥ 5
μ = np = 574(0.50) = 287
σ = npq = 574(0.50)(0.50) = 11.979
P(x ≥ 525)
0.9999
<-------------------------------= P(x>524.5)
= P(z>19.83)
524.5
x
287
= 1 – 0.9999
Z
0
19.83
= 0.0001
Yes. Since the probability of getting at least 525 girls by chance alone is so small, it appears
that the method is effective and that the genders were not being determined by chance alone.
20. binomial: n=152 and p=0.50
normal approximation appropriate since
np = 152(0.50) = 76 ≥ 5
nq = 152(0.50) = 76 ≥ 5
μ = np = 152(0.50) = 76
σ = npq = 152(0.50)(0.50) = 6.164
P(x ≥ 127)
0.9999
<-------------------------------= P(x>126.5)
= P(z>8.19)
126.5
x
76
= 1 – 0.9999
Z
0
8.19
= 0.0001
Yes. Since the probability of getting at least 127 boys by chance alone is so small, it appears
that the method is effective and that the genders were not being determined by chance alone.
21. binomial: n=580 and p=0.25
normal approximation appropriate since
np = 580(0.25) = 145 ≥ 5
nq = 580(0.75) = 435 ≥ 5
μ = np = 580(0.25) = 145
σ = npq = 580(0.25)(0.75) = 10.428
Normal as Approximation to Binomial SECTION 6-6
177
b. P(x ≥ 152)
= P(x>151.5)
= P(z>0.62)
= 1 – 0.7324
= 0.2676
a. P(x=152)
= P(151.5<x<152.5)
= P(0.62<z<0.72)
= 0.7642 – 0.7324
= 0.0318
<----------------------------------|
0.7642
0.7324
0.7324
<--------------------------------
<---------------------------145 151.5 152.5
0
0.62 0.72
x
Z
145
0
151.5
0.62
x
Z
c. The part (b) answer is the useful probability. In situations involving multiple ordered
outcomes, the unusualness of a particular outcome is generally determined by the
probability of getting that outcome or a more extreme outcome.
d. No. Since 0,2676 > 0.05, Mendel’s result could easily occur by chance alone if the true rate
were really 0.25.
22. binomial: n=1002 and p=0.61
normal approximation appropriate since
np = 1002(0.61) = 611.22 ≥ 5
nq = 1002(0.39) = 390.78 ≥ 5
μ = np = 1002(0.61) = 611.22
σ = npq = 1002(0.61)(0.39) = 15.439
P(x ≥ 701)
0.9999
<-------------------------------= P(x>700.5)
x
= P(z>5.78)
611.22 700.5
Z
0
5.78
= 1 – 0.9999
= 0.0001
Since the probability of getting at least 701 actual voters by chance alone is so small, the result
suggests that the respondents were not truthful about their voting.
23. binomial: n=420,095 and p=0.000340
normal approximation appropriate since
np = 420,095(0.000340) = 142.83 ≥ 5
nq = 420,095(0.999660) = 419952.17 ≥ 5
μ = np = 420,095(0.000340) = 142.83
σ = npq = 420095(0.000340)(0.999660)
= 11.949
0.2709
<----------P(x ≤ 135)
= P(x<135.5)
142.83
135.5
x
= P(z<-0.61)
-0.61
Z
0
= 0.2709
To conclude that cell phones increase the likelihood of experiencing such cancers requires
x > 142.83. Since 135 < 142.83, these results definitely do not support the media reports.
178
CHAPTER 6 Normal Probability Distributions
24. binomial: n=100 and p=0.80
normal approximation appropriate since
np = 100(0.80) = 80 ≥ 5
nq = 100(0.20) = 20 ≥ 5
μ = np = 100(0.80) = 80
σ = npq = 100(0.80)(0.20) = 4
P(x=85)
= P(84.5<x<85.5)
= P(1.13<z<1.38)
= 0.9162 – 0.8707
= 0.0455
<----------------------------------|
0.9162
0.8707
<---------------------------80
0
84.5 85.5
1.13 1.38
x
Z
25. binomial: n=200 and p=0.06
normal approximation appropriate since
np = 200(0.06) = 12 ≥ 5
nq = 200(0.94) = 188 ≥ 5
μ = np = 200(0.06) = 12
σ = npq = 200(0.06)(0.94) = 3.359
0.2296
P(x ≥ 10)
<----------= P(x>9.5)
12
9.5
x
= P(z>-0.74)
-0.74
Z
0
= 1 – 0.2296
= 0.7704
Yes. Since there is a 77% chance of getting at least 10 universal donors, a pool of 200
volunteers appears to be sufficient. Considering the importance of the need, and the fact that
one can never have too much blood on hand, the hospital may want to use a larger pool to
further increase the 77% figure and/or determine how large a pool would be necessary to
increase the figure to, say, 95%.
26. binomial: n=80 and p=0.075
normal approximation appropriate since
np = 80(0.075) = 6 ≥ 5
nq = 80(0.925) = 74 ≥ 5
μ = np = 80(0.075) = 6
σ = npq = 80(0.075)(0.925) = 2.356
P(x ≤ 4)
0.2611
<----------= P(x<4.5)
= P(z<-0.64)
6
4.5
x
= 0.2611
-0.64
Z
0
P(batch accepted)
= 1 – P(batch rejected) = 1 – 0.2611 = 0.7389.
Yes. Not only does the 7.5% rate of defectives seem too high, but also the fact that only about
74% of the batches it ships will be accepted seems problematic.
Normal as Approximation to Binomial SECTION 6-6
27. binomial: n=100 and p=0.24
normal approximation appropriate since
np = 100(0.24) = 24 ≥ 5
nq = 100(0.76) = 76 ≥ 5
μ = np = 100(0.24) = 24
σ = npq = 100(0.24)(0.76) = 4.271
P(x ≥ 27)
0.7224
<-------------------------------= P(x>26.5)
= P(z>0.59)
26.5
24
= 1 – 0.7224
0
0.59
= 0.2776
No. Since 0.2776 > 0.05, 27 is not an unusually high number of blue M&M’s.
179
x
Z
28. binomial: n=784 and p=0.079
normal approximation appropriate since
np = 784(0.079) = 61.936 ≥ 5
nq = 784(0.921) = 722.064 ≥ 5
μ = np = 784(0.079) = 61.936
σ = npq = 784(0.079)(0.921) = 7.553
P(x ≥ 479)
0.9999
<-------------------------------= P(x>478.5)
= P(z>55.15)
x
61.936 478.5
= 1 – 0.9999
55.15
Z
0
= 0.0001
Yes. Since the probability of obtaining 479 or more such checks from normal honest
transactions by chance alone is so very strong, there is strong evidence to indicate that the
checks from the suspected companies do not follow the normal pattern and are likely
fraudulent.
29. binomial: n=863 and p=0.019
normal approximation appropriate since
np = 863(0.019) = 16.397 ≥ 5
nq = 863(0.981) = 846.603 ≥ 5
μ = np = 863(0.019) = 16.397
σ = npq = 863(0.019)(0.981) = 4.011
0.6985
<-------------------------------P(x ≥ 19)
= P(x>18.5)
x
18.5
16.397
= P(z>0.52)
0.52
Z
0
= 1 – 0.6985
= 0.3015
Since the 0.3015 > 0.05, 19 or more persons experiencing flu symptoms is not an unusual
occurrence for a normal population. There is no evidence to suggest that flu symptoms are an
adverse reaction to the drug.
180
CHAPTER 6 Normal Probability Distributions
30. binomial: n=57 and p=0.50
normal approximation appropriate since
np = 57(0.50) = 28.5 ≥ 5
nq = 57(0.50) = 28.5 ≥ 5
μ = np = 57(0.50) = 28.5
σ = npq = 57(0.50)(0.50) = 3.775
P(x ≤ 15)
= P(x<15.5)
= P(z<-3.44)
= 0.0003
0.0003
<----------15.5
-3.44
28.5
0
x
Z
No. Since 0.0003 ≤ 0.05, getting 15 or fewer false positives by chance alone would be an
unusual event. The polygraph does not appear to be making random guesses.
31. Let x = the number that show up.
binomial: n=236 and p=0.9005
normal approximation appropriate since
np = 236(0.9005) = 212.518 ≥ 5
nq = 236(0.0995) = 23.482 ≥ 5
μ = np = 236(0.9005) = 212.518
σ = npq = 236(0.9005)(0.0995) = 4.598
0.5832
<-------------------------------P(x>213)
= P(x>213.5)
x
212.518 213.5
0.21
= P(z>0.21)
Z
0
= 1 – 0.5832
= 0.4168
No. Since the 0.4168 > 0.05, overbooking is not an unusual event. It is a real concern for both
the airline and the passengers.
32. binomial: n=236 and p=0.50
normal approximation appropriate since
np = 213(0.50) = 106.5 ≥ 5
nq = 213(0.50) = 106.5 ≥ 5
μ = np = 213(0.50) = 106.5
σ = npq = 236(0.9005)(0.0995) = 7.297
P(x ≥ 122)
0.9803
<-------------------------------= P(x>121.5)
= P(z>2.06)
x
121.5
106.5
= 1 – 0.9803
0.21
Z
0
= 0.0197
No. Since the 0.0197 ≤ 0.05, having at last 122 men is an unusual event. The load does not
have to be adjusted very often.
Normal as Approximation to Binomial SECTION 6-6
181
33. Marc is placing 200(5) = $1000 in bets. To make a profit his return must be more than $1000.
In part (a) each win returns the original 5 plus 35(5), for a total of 5 + 175 = $180.
For a profit, he must win more than 1000/180 = 5.55 times – i.e., he needs at least 6 wins.
In part (b) each win returns the original 5 plus 1(5), for a total of 5 + 5 = $10.
For a profit, he must win more than 1000/10 = 100 times – i.e., he needs at least 101 wins.
a. binomial: n=200 and p=1/38
normal approximation appropriate since
np = 200(1/38) = 5.26 ≥ 5
nq = 200(37/38) = 194.74 ≥ 5
μ = np = 200(1/38) = 5.263
σ = npq = 200(1/38)(37/38) = 2.263
P(x ≥ 6)
0.5398
<-------------------------------= P(x>5.5)
= P(z>0.10)
x
5.263
5.5
= 1 – 0.5398
Z
0
0.10
= 0.4602
b. binomial: n=200 and p=244/495
normal approximation appropriate since
np = 200(244/495) = 98.59 ≥ 5
nq = 200(251/495) = 101.41 ≥ 5
μ = np = 200(244/495) = 98.586
σ = npq = 200(244/495)(251/495)
0.6064
= 7.070
<-------------------------------P(x ≥ 101)
x
98.586 100.5
= P(x>100.5)
Z
0
0.27
= P(z>0.27)
= 1 – 0.6064
= 0.3936
c. Since 0.4602 > 0.3936, the roulette game is the better “investment” – but since both
probabilities are less than 0.5000, he would do better not to play at all.
34. Let x = the number that show up.
binomial: n=unknown and p=0.9005
We need to find the n for which P(x ≤213) = PC(x<213.5) = 0.95.
From Table A-2, the z score below which 95% of the probability occurs is 1.645.
Solve to find the n for which 213.5 persons with reservations show up 95% of the time.
(x-μ)/σ = z
(213.5 - 0.9005n)/ n(0.9005)(0.0995) = 1.645
(213.5 - 0.9005n) = 1.645 n(0.9005)(0.0995)
45582.25 - 384.5135n + 0.8109n 2 = 0.2425n
0.8109n 2 - 384.7560n + 45582.25 = 0
n = [384.7560 ± (-384.7560) 2 - 4(.8109)(45582.25)]/2(0.8109)
= [384.7560 ± 13.6599]/1.6218
= 228.82 or 245.66
And so z = 1.645 when n=228.82. If n>228.02, then z<1.645 and the probability that
x ≤213 is less than 95% – and so we round down to n=228.
NOTE: Using the n=245.66 produces z = -1.645. This extraneous root was introduced then
182
CHAPTER 6 Normal Probability Distributions
both sides of the above equation were squared. The following less mathematical alternative
approach to the exercise is recommended: Follow the technique in Exercise 31, increasing n
and calculating P(x ≤213) until P(x ≤213) drops below 0.95.
n
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
μ = np
202.6
203.5
204.4
205.3
206.2
207.1
208.0
208.9
209.8
210.7
211.6
212.5
213.4
214.3
215.2
216.1
217.0
217.9
218.8
219.7
220.6
221.5
σ = npq
4.49
4.50
4.51
4.52
4.53
4.54
4.55
4.56
4.57
4.58
4.59
4.60
4.61
4.62
4.63
4.64
4.65
4.66
4.67
4.68
4.69
4.69
z = (213.5 - μ)/σ
2.42
2.22
2.01
1.81
1.61
1.41
1.21
1.01
0.81
0.61
0.41
0.21
0.02
-0.18
-0.37
-0.56
-0.76
-0.95
-1.14
-1.33
-1.52
-1.71
P(x<213.5)
0.9922
0.9868
0.9778
0.9649
0.9463
0.9207
0.8869
0.8438
0.7910
0.7291
0.6591
0.5832
0.5080
0.4286
0.3557
0.2877
0.2236
0.1711
0.1271
0.0918
0.0643
0.0436
And so for n=228 there is a 94.49% probability that x ≤213 and that everyone who shows up
with reservations will have a seat. For n>228 the probability falls below that acceptable
level – and continues decreasing as n (the number of accepted reservations) increases. The
rows in bold face type indicate the correct solution (n=228), the probability of everyone being
accommodated in Exercise 31 (0.5832), and the limits on the extraneous solution (z = -1.645
and n = 245.66).
35. a. binomial: n=4 and p=0.350
P(x≥1) = 1 – P(x=0)
= 1 – [4!/(4!0!)](0.350)0(0.650)4
= 1 – 0.1785
= 0.8215
b. binomial: n=56(4)=224 and p=0.350
normal approximation appropriate since
np = 224(0.350) = 78.4 ≥5
nq = 224(0.650) = 145.6 ≥5
μ = np = 224(0.350) = 78.4
σ = npq = 224(0.350)(0.650) = 7.139
P(x≥56)
= P(x>55.5)
= P(z>-3.20)
= 1 – 0.0007
= 0.9993
0.0007
<----------55.5
-3.20
78.4
0
x
Z
Normal as Approximation to Binomial SECTION 6-6
183
c. Let H = getting at least one hit in 4 at bats.
P(H) = 0.8215 [from part (a)]
For 56 consecutive games, [P(H)]56 = [0.8125]56 = 0.0000165
d. The solution below employs the techniques and notation of parts (a) and (c).
for [P(H)]56 > 0.10
P(H) > (0.10)1/56
P(H) > 0.9597
for P(H) = P(x ≥ 1) > 0.9597
1 – P(x=0) > 0.9597
0.0403 > P(x=0)
0.0403 > [4!/(4!0!)]p0(1-p)4
0.0403 > (1-p)4
(0.0403)1/4 > 1 – p
p > 1 – (0.0403)1/4
p > 1 – 0.448
p > 0.552
36. In Excel, BINOMDIST(825,2000,.5,false) = 0. But using the normal approximation,
P(824.5<x<825.5) = P(-7.849<z<-7.804) = NORMSDIST(-7.804) – NORMSDIST(-7.849) gives
an answer.
6-7 Assessing Normality
1. A normal quantile plot can be used to determine whether sample data come form a normal
distribution. In theory, it compares the z scores for the sample data with the z scores for
normally distributed data with the same cumulative relative frequencies as the sample data. In
practice, it uses the sample data directly – since converting to z scores is a linear
transformation that re-labels the scores but does not change their distribution.
2. If sample data come from a normal distribution, the normal quantile plot should approximate a
straight line. If the points do not approximate a straight line, or if the points show a systematic
pattern, the conclusion is that the sample data do not come from a normal distribution.
3. Because the weights are likely to follow a normal distribution, on expects the points in a
normal quantile plot to approximate a straight line.
4. Identify outliers. If there is more than one outlier present, reject the idea that the data come
from a normal distribution.
5. No, the data do not appear to come from a population with a normal distribution. The points
are not reasonably close to a straight line.
6. Yes, the data appear to come from a population with a normal distribution. The points are
reasonably close to a straight line.
7. Yes, the data appear to come from a population with a normal distribution. The points are
reasonably close to a straight line.
8. No, the data do not appear to come from a population with a normal distribution. The points
form a S-shaped pattern, and not a straight line.
184
CHAPTER 6 Normal Probability Distributions
9. Yes, the data appear to come from a population with a normal distribution. The frequency
distribution and histogram indicate that the data is approximately bell-shaped.
25
20
frequency
duration (hours) frequency
0 – 49
1
50 – 99
8
100 – 149
18
150 – 199
23
200 – 249
25
250 – 299
19
300 – 349
11
350 – 399
8
400 – 449
2
115
15
10
5
0
0
100
200
300
400
duration (hours)
10. No, the data do not appear to come from a population with a normal distribution. The
frequency distribution and the histogram indicate the data are positively skewed.
frequency
67
71
57
52
42
16
5
2
312
80
70
60
frequency
number of flights
0
1
2
3
4
5
6
7
50
40
30
20
10
0
0
1
2
3
4
5
6
7
number of flights
11. No, the data do not appear to come from a population with a normal distribution. The
frequency distribution and the histogram indicate there is a concentration of data at the lower
end.
frequency
20
5
8
9
5
1
48
20
frequency
degree days
0 – 499
500 – 999
1000 – 1499
1500 – 1999
2000 – 2499
2500 – 2599
15
10
5
0
___________________________________________________________________
0
500
1000
1500
2000
degree days
2500
3000
Assessing Normality SECTION 6-7
12. Yes, the data appear to come from a population with a normal distribution. The frequency
distribution and the histogram indicate the data is approximately bell-shaped.
frequency
2
0
1
2
4
4
5
6
7
3
2
2
2
40
7
6
frequency
volts
124.0
124.1
124.2
124.3
124.4
124.5
124.6
124.7
124.8
124.9
125.0
125.1
125.2
5
4
3
2
1
0
____________________________________________________________________
124.0
124.2
124.4
124.6
NOTE: The normal quantile plots for Exercises 13-16 may be obtained
directly from many sources or constructed using Minitab for n scores
in C1 using the commands at the right, where (2n-1) and (2n) are the
actual values, and then plotting C1 on the x-axis and C4 on the y-axis.
See Exercises 19 and 20 for more detail on the mechanics of this
process, which is the five-step “manual construction” process given in
the text.
125.0
125.2
MTB> Sort C1 C1
MTB> Set C2
DATA> 1:(2n-1)/2
DATA> end
MTB> Let C3 = C2/(2n)
MTB> INVCDF C3 C4
3
2
z score
13. Yes. Since the points approximate a
straight line, the data appear to come
from a population with a normal
distribution. The gaps/groupings in
the durations may reflect the fact that
the times have to reflect whole
numbers of orbits or other physical
constraints.
124.8
generator voltage levels
1
0
-1
-2
-3
0
100
200
300
400
duration (hours)
3
2
1
z score
14. No. Since the points do not
approximate a straight line, the data
do not appear to come from a
population with a normal
distribution.
0
-1
-2
-3
0
1
2
3
4
number of flights
5
6
7
185
CHAPTER 6 Normal Probability Distributions
15. No. Since the points do not
approximate a straight line, the data
do not appear to com form a
population with a normal
distribution.
3
2
1
z score
186
0
-1
-2
-3
0
500
1000
1500
2000
2500
degree days
2
1
z score
16. Yes. Since the points approximate a
straight line, the data appear to come
from a population with a normal
distribution. The gaps/groupings are
caused by the discrete nature of the
data.
0
-1
-2
124.0
124.2
124.4
124.6
124.8
125.0
125.2
generator voltage levels
17. The two histograms are shown below. The heights (on the left) appear to be approximately
normally distributed, while the cholesterol levels (on the right) appear to be positively skewed.
Many natural phenomena are normally distributed. Height is a natural phenomenon unaffected
by human input, but cholesterol levels are humanly influenced (by diet, exercise, medication,
etc.) in ways that might alter any naturally occurring distribution.
20
12
15
Frequency
Frequency
10
8
6
4
10
5
2
0
0
58
60
62
64
height
66
68
0
200
400
600
cholesterol level
800
1000
Assessing Normality SECTION 6-7
187
18. The two histograms are given below. The systolic blood pressure levels (on the left) appear to
be positively skewed, while the elbow breadths (on the right) appear to be more normally
distributed. Many natural phenomena are normally distributed. Elbow breadth is a natural
phenomenon unaffected by human input, but blood pressure is humanly influenced (by diet,
exercise, medication, etc.) in ways that might alter any naturally occurring distribution.
14
8
10
Frequency
Frequency
12
8
6
4
6
4
2
2
0
0
90 100 110 120 130 140 150 160 170 180
5.4 5.7
6.0 6.3 6.6
systolic blood pressure
6.9 7.2
7.5 7.8 8.1
elbow width
1
2
3
4
5
127
129
131
136
146
0.10
0.30
0.50
0.70
0.90
-1.28
-0.52
0
0.52
1.28
z score
19. The corresponding z scores in the table below were determined following the five-step
“manual construction” procedure of the text.
(1) Arrange the n scores in order and place them in the x column.
(2) For each xi, calculate the cumulative probability using cpi = (2i-1)/2n for i = 1,2,…,n.
(3) For each cpi, find the zi for which
1.5
P(z<zi) = cpi for i = 1,2,…,n.
The resulting normal quantile plot indicates
1.0
that the data appear to come from a
0.5
population with a normal distribution.
0.0
i x cp
z .
-0.5
-1.0
125
130
135
140
145
braking distance (feet)
20. The corresponding z scores in the table
1
2
3
4
5
6
3
5
15
17
18
158
0.083
0.250
0.417
0.583
0.750
0.917
-1.38
-0.67
-0.21
0.21
0.67
1.38
z score
below were determined following the five-step “manual construction” procedure of the text.
(1) Arrange the n scores in order and place them in the x column.
(2) For each xi, calculate the cumulative probability using cpi = (2i-1)/2n for i = 1,2,…,n.
(3) For each cpi, find the zi for which
P(z<zi) = cpi for i = 1,2,…,n.
1.5
The resulting normal quantile plot indicates
1.0
that the data do not appear to come from a
0.5
population with a normal distribution.
i x
cp
z .
0.0
-0.5
-1.0
-1.5
0
20
40
60
80
100
120
number of satellites
140
160
188
CHAPTER 6 Normal Probability Distributions
21. a. Yes. Adding two inches to each height shifts the entire distribution to the right, but it does
not affect the shape of the distribution.
b. Yes. Changing to a different unit measure re-labels the horizontal axis, but it does not
change relationships between the data points or affect the shape of the distribution.
c. No. Unlike a linear transformation of the form f(x) = ax + b, the log function f(x) = log(x)
does not have the same effect on all segments of the horizontal axis.
22. This exercise is worked using ln(x+1) = loge(x+1). The same conclusion is reached for
log10(x+1) or any other base. Using the five-step “manual construction of a normal quantile
plot” given in the text, the corresponding z scores in the table below were determined
as follows.
(1) Arrange the n=20 scores in order and place them in the x column.
(2) For each xi, calculate ln(xi+1).
(3) For each xi, calculate the cumulative probability using cpi = (2i-1)/2n for i = 1,2,…,n.
(4) For each cpi, find the zi for which P(z<zi) = cpi for i = 1,2,…,n.
Since the cp and z columns are based only on i, they apply to both the x and the ln(x+1)
values. The resulting normal quantile plots indicate the x data do not appear to come from
a population with a normal distribution, but the ln(x+1) data do.
102
103
106
267
307
329
339
362
510
547
547
662
725
1065
1091
1169
1396
1894
3822
4337
ln(x+1) cp
4.63
4.64
4.67
5.59
5.73
5.80
5.83
5.89
6.24
6.31
6.31
6.50
6.59
6.97
7.00
7.06
7.24
7.55
8.25
8.38
0.025
0.075
0.125
0.175
0.225
0.275
0.325
0.375
0.425
0.475
0.525
0.575
0.625
0.675
0.725
0.775
0.825
0.875
0.925
0.975
z .
-1.96
-1.44
-1.15
-0.93
-0.76
-0.60
-0.45
-0.32
-0.19
-0.06
0.06
0.19
0.32
0.45
0.60
0.76
0.93
1.15
1.44
1.96
2
1
z score
x
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
0
-1
-2
0
1000
2000
3000
4000
x = (days to failure)
2
1
z score
i
0
-1
-2
4.5
5.0
5.5
6.0
6.5
7.0
7.5
ln(x+1) = ln(days to failure + 1)
NOTE: There may be a pattern of gaps and clustering that should be investigated
8.0
8.5
Statistical Literacy and Critical Thinking
189
Statistical Literacy and Critical Thinking
1. A normal distribution is one that is symmetric and bell-shaped. More technically, it is one that
can be described by the following formula, where μ and σ are specified values,
f(x) =
e
-
1
2σ 2
( x − μ )2
σ 2π
.
A standard normal distribution is a normal distribution that has a mean of 0 and a standard
deviation of 1. More technically, it is the distribution that results when μ = 0 and σ = 1 in the
above formula.
2. In statistics, a normal distribution is one that is symmetric and bell-shaped. This statistical use
of the term “normal” is not to be confused with the common English word “normal” in the
sense of “typical.”
3. Assuming there are no trends over time in the lengths of movies, and that there is mean length
that remains fairly constant from year to year, the sample means will follow a normal
distribution centered around that enduring mean length.
4. Not necessarily. Depending on how the survey was conducted, the sample may be a
convenience sample (if AOL simply polled its own customers) or a voluntary response sample
(if the responders were permitted decide for themselves whether or not to participate).
Chapter Quick Quiz
1. The symbol z0.03 represents the z score with 0.0300 below it. According to Table A-1 [closest
value is 0.0301], this is -1.88.
2. For n=100, the sample means from any distribution with finite mean and standard deviation
follow a normal distribution.
3. In a standard normal distribution, μ = 0 and σ = 1.
4. P(z>1.00) =1 – 0.8413 = 0.1587
5. P(-1.50<z<2.50) = 0.9938 – 0.0668 = 0.9270
6. P(x<115) = P(<1.00) = 0.8413
7. P(x>118) = P(z>1.20) = 1 – 0.8849 = 0.1151
8. P(88<x<112) = P(-0.80<z<0.80) = 0.7881 – 0.2119 = 0.5762
9. For n=25, μ x = μ = 100 and σ x = σ/ n = 15/ 25 = 3.
P( x <103) = P(z<1.00) = 0.8413
10. For n=100, μ x = μ = 100 and σ x = σ/ n = 15/ 100 = 1.5.
P( x >103) = P(z>2.00) = 1 – 0.9772 = 0.0228
190
CHAPTER 6 Normal Probability Distributions
Review Exercises
1. a. normal distribution: μ = 69.0 and σ = 2.8
P(x>75)
= P(z>2.14)
= 1 – 0.9838
= 0.0162 or 1.62%
0.9838
0.9999
<-------------------------------69.0
0
75
2.14
b. normal distribution: μ = 63.6 and σ = 2.5
P(x>75)
= P(z>4.56)
= 1 – 0.9999
= 0.0001 or 0.01%
<-------------------------------x
Z
63.6
0
75
4.56
x
Z
c. The length of a day bed appears to be adequate to meet the needs of all but the very tallest
men and women.
2. For the tallest 5%, A = 0.9500 and z = 1.645
x = μ + zσ
= 69.0 + (1.645)(2.8)
= 69.0 + 4.6
= 73.6 inches
0.9500
<-------------------------------69.0
0
3. a. normal distribution: μ = 69.0 and σ = 2.8
P(x>78)
= P(z>3.21)
= 1 – 0.9993
= 0.0007 or 0.07% of the men
0.9993
78
3.21
x
Z
normal distribution: μ = 63.6 and σ = 2.5
P(x>78)
= P(z>5.76)
= 1 – 0.9999
= 0.0001 or 0.01% of the women
0.9999
<-------------------------------69.0
0
?
1.645
<-------------------------------x
Z
63.6
0
78
5.76
x
Z
Review Exercises
191
b. For the tallest 1%, A = 0.9900 [0.9901]
and z = 2.33
x = μ + zσ
= 69.0 + (2.33)(2.8)
= 69.0 + 6.5
= 75.5 inches
0.9900
<-------------------------------69.0
0
4. normal distribution: μ = 63.6 and σ = 2.5
P(66.5<x<71.5)
= P(1.16<z<3.16)
= 0.9992 – 0.8770
= 0.1222 or 12.22%
?
2.33
x
Z
<----------------------------------|
0.9992
0.8770
<---------------------------63.6 66.5 71.5
0 1.16 3.16
x
Z
Yes. The minimum height requirement for Rockettes is greater than th mean height of all
women.
5. binomial: n=1064 and p=0.75
normal approximation appropriate since
np = 1064(0.75) = 798 ≥ 5
nq = 1064(0.25) = 266 ≥ 5
μ = np = 1064(0.75) = 798
σ = npq = 1064(0.75)(0.25) = 14.124
0.2296
<----------P(x ≤ 787)
= P(x<787.5)
787.5
798
x
= P(z<-0.74)
-0.74
Z
0
= 0.2296
Since the 0.2296 > 0.05, obtaining 787 plants
with long stems is not an unusual occurrence for a population with p=0.75. The results are
consistent with Mendel’s claimed proportion.
6. a. True. The sample mean is an unbiased estimator of the population mean.
See Section 6-4, Exercises 12 and 13.
b. True. The sample proportion is an unbiased estimator of the population proportion.
See Section 6-4, Exercises 17 and 18 and 19.
c. True. The sample variance is an unbiased estimator of the population variance.
See Section 6-4, Exercises 11 and 16.
d. Not true. See Section 6-4, Exercises 9 and 14.
e. Not true. See Section 6-4, Exercise 15.
192
CHAPTER 6 Normal Probability Distributions
7. a. normal distribution
μ = 178.1
σ = 40.7
P(x>260)
= P(z>2.01)
= 1 – 0.9778
= 0.0222
0.9778
<-------------------------------178.1
0
b. normal distribution
μ = 178.1
σ = 40.7
P(170<x<200)
= P(-0.20<z<0.54)
= 0.7054 – 0.4207
= 0.2847
σ x = σ/ n = 40.7/ 9 = 13.567
P(170< x <200)
= P(-0.60<z<1.61)
= 0.9463 – 0.2743
= 0.6720
x
Z
<--------------------------------|
0.7054
0.4207
<----------170
-0.20
c. normal distribution,
since the original distribution is so
μ x = μ = 178.1
260
2.01
178.1
0
200
0.54
x
Z
<--------------------------------|
0.9463
0.2743
<----------170
-0.60
178.1
0
200
1.61
_
x
Z
d. For the top 3%,
A = 0.9700 [0.9699] and z = 1.88
x = μ + zσ
= 178.1 + (1.88)(40.7)
= 178.1 + 76.5
= 254.6
0.9700
<-------------------------------178.1
0
?
1.88
x
Z
Review Exercises
193
8. binomial: n=40 and p=0.50
normal approximation appropriate since
np = 40(0.50) = 20 ≥ 5
nq = 40(0.50) = 20 ≥ 5
μ = np = 40(0.50) = 20
σ = npq = 40(0.50)(0.50) = 3.162
0.0778
P(x ≤ 15)
<----------= P(x<15.5)
20
= P(z<-1.42)
15.5
x
-1.42
= 0.0778
Z
0
Since the 0.0778 > 0.05, obtaining 15 or
fewer women is not an unusual occurrence for a population with p=0.75. No. There is not
strong evidence to charge gender discrimination.
9. a. For 0.6700 to the left, A = 0.6700 and z = 0.44.
b. For 0.9960 to the right, A = 1 – 0.9960 = 0.0040 and z = -2.65.
c. For 0.025 to the right, A = 1 – 0.0250 = 0.9750 and z = 1.96.
10. a. The sampling distribution of the mean is a normal distribution.
b. μ x = μ = 3420 grams
c. σ x = σ/ n = 495/ 85 = 53.7 grams
11. Adding 20 lbs of carry-on baggage for every male will increase the mean weight by 20 but will
not affect the standard deviation of the
weights.
a. normal distribution
μ = 192
σ = 29
P(x>195)
= P(z>0.10)
= 1 – 0.5398
0.5398
<-------------------------------= 0.4602
192
0
195
0.10
x
Z
b. normal distribution,
since the original distribution is so
μ x = μ = 192
σ x = σ/ n = 29/ 213 = 1.987
P( x >195)
= P(z>1.51)
0.9345
= 1 – 0.9345
<-------------------------------= 0.0655
195
192
Yes. Since 0.0655 > 0.05, being
0
1.51
overloaded under those circumstances
(i.e., all passengers are male and carrying
20 lbs of carry-on luggage) would not be unusual.
_
x
Z
194
CHAPTER 6 Normal Probability Distributions
12. For review purposes, this exercise is worked in detail using a frequency distribution, a
frequency histogram, and a normal quantile plot.
a. Using a frequency distribution and a frequency histogram.
frequency
1
5
8
6
20
9
8
7
frequency
weight (grams)
7.9500 – 7.9999
8.0000 – 8.0499
8.0500 – 8.0999
8.1000 – 8.1444
6
5
4
3
2
1
0
7.95
8.00
8.05
8.10
8.15
weight (grams)
b. Using a normal quantile plot. Using the five-step “manual construction of a normal quantile
plot” given in the text, the corresponding z scores in the table below were determined as
follows.
(1) Arrange the n=20 scores in order and place them in the x column.
(2) For each xi, calculate the cumulative probability using cpi = (2i-1)/2n for i = 1,2,…,n.
(3) For each cpi, find the zi for which P(z<zi) = cpi for i = 1,2,…,n.
x
cp
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
7.9817
8.0241
8.0271
8.0307
8.0342
8.0345
8.0510
8.0538
8.0658
8.0719
8.0775
8.0813
8.0894
8.0954
8.1008
8.1041
8.1072
8.1238
8.1281
8.1384
0.025
0.075
0.125
0.175
0.225
0.275
0.325
0.375
0.425
0.475
0.525
0.575
0.625
0.675
0.725
0.775
0.825
0.875
0.925
0.975
z .
-1.96
-1.44
-1.15
-0.93
-0.76
-0.60
-0.45
-0.32
-0.19
-0.06
0.06
0.19
0.32
0.45
0.60
0.76
0.93
1.15
1.44
1.96
2
1
z score
i
0
-1
-2
8.00
8.04
8.08
8.12
8.16
weight (grams)
Yes. The weights appear to come from a population that has a normal distribution. The
frequency distribution and the histogram suggest a bell-shaped distribution, and the points on
the normal quantile plot are reasonably close to a straight line.
Cumulative Review Exercises
195
Cumulative Review Exercises
1. Arranged in order, the values are: 125 128 138 159 212 235 360 492 530 900
summary statistics: n = 10
Σx = 3279
Σx2 = 1639067
a. x = (Σx)/n = 3279/10 = 327.9 or $327,900
b. x = (x5 + x6)/2 = (212 + 235)/2 = 223.5 or $223,500
c. s = 250.307 or $250,307 [the square root of the answer given in part d]
d. s2 = [n(Σx2) – (Σx)2]/[n(n-1)]
= [10(1639067) – (3279)2]/[10(9)]
= 5638829/90 = 62653.655556 or 62,653,655,556 dollars2
e. z = (235,000 – 327,900)/250,307 = -0.37
f. Ratio, since differences are meaningful and there is a meaningful zero.
g. Discrete, since they must be paid in hundredths of dollars (i.e., in whole cents).
2. a. A simple random sample of size n is a sample selected in such a way that every sample of
size n has the same chance of being selected.
b. A voluntary response sample is one for which the respondents themselves made the decision
and effort to be included. Such samples are generally unsuited for statistical purposes
because they are typically composed of persons with strong feelings on the topic and are not
representative of the population.
3. a. P(V1 and V2) = P(V1)·P(V2|V1) = (14/2103)(13/2102) = 0.0000412
b. binomial: n=5000 and p=14/2103
normal approximation appropriate since
np = 5000(14/2103) = 33.29 ≥ 5
nq = 5000(2089/2103) = 4966.71 ≥ 5
μ = np = 5000(14/2103) = 33.286
σ = npq = 5000(14/2103)(2089/2103)
= 5.750
P(x ≥ 40)
= P(x>39.5)
= P(z>1.08)
= 1 – 0.8599
= 0.1401
0.8599
<-------------------------------33.286
0
39.5
1.08
x
Z
c. No. Since 0.1401 > 0.05, 40 is not an unusually high number of viral infections.
d. No. To determine whether viral infections are an adverse reaction to using Nasonex, we
would need to compare to 14/2103 = 0.00666 rate for Nasonex users to the rate in the
general population – and that information is not given.
4. No. By not beginning the vertical scale at zero, the graph exaggerates the differences.
196
CHAPTER 6 Normal Probability Distributions
5. a. Let L = a person is left-handed.
P(L) = 0.10, for each random selection
P(L1 and L2 and L3) = P(L1)·P(L2)·P(L3)
= (0.10)(0.10)(0.10)
= 0.001
b. Let N = a person is not left-handed.
P(N) = 0.90, for each random selection
P(at least one left-hander) = 1 – P(no left-handers)
= 1 – P(N1 and N2 and N3)
= 1 – P(N1)·P(N2)·P(N3)
= 1 – (0.90)(0.90)(0.90)
= 1 – 0.729
= 0.271
c. binomial: n=3 and p=0.10
normal approximation not appropriate since
np = 3(0.10) = 3 < 5
d. binomial: n=50 and p=0.10
μ = np = 50(0.10) = 5
e. binomial problem: n=50 and p=0.10
σ = npq = 50(0.10)(0.90) = 2.121
f. There are two previous approaches that may be used to answer this question.
(1) An unusual score is one that is more than two standard deviations from the mean. Use
the values for μ and σ from parts (d) and (e).
z = (x – μ)/σ
z8 = (8 – 5)/2.121
= 1.41
Since 8 is 1.41<2 standard deviations from the mean, it would not be an unusual result.
(2) A score is unusual if the probability of getting that result or a more extreme result is less
than or equal to 0.05.
binomial: n = 50 and p = 0.10
normal approximation appropriate since
np = 50(0.10) = 5 ≥ 5
nq = 50(0.90) = 45 ≥ 5
Use the values for μ and σ
from parts (d) and (e).
P(x ≥ 8)
= P(x>7.5)
= P(z>1.18)
= 1 – 0.8810
0.8810
<-------------------------------= 0.1190
5
0
7.5
1.18
x
Z
Since 0.1190 > 0.05, getting 8 left-handers in a group of 50 is not an unusual event.