Download Solution for odd problems.

Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
CHAPTER 6
Section 6-1
6-1.
6-3.
6-5.
6-7.
No, usually not. For example, if the sample is {2, 3} the mean is 2.5 which is not an observation in the
sample.
No, usually not. For example, the mean of {1, 4, 4} is 3 which is not even an observation in the sample.
Yes. For example, {5, 5, 5, 5, 5, 5, 5}, the sample mean = 5, sample standard deviation = 0
Sample average:
n
x
x
i 1
n
i

592.035
 74.0044 mm
8
Sample variance:
8
x
i 1
 592.035
i
8
x
i 1
2
i
 43813.18031
2
 n 
  xi 
n
592.0352
2
xi   i 1 
43813.18031 

n
8
s 2  i 1

n 1
8 1
0.0001569

 0.000022414 (mm ) 2
7
Sample standard deviation:
s  0.000022414  0.00473 mm
The sample standard deviation could also be found using
n
  xi  x
s
i 1
n 1
2
8
where
  xi  x 
2
 0.0001569
i 1
Dot Diagram:
.. ...:
.
-------+---------+---------+---------+---------+---------diameter
73.9920
74.0000
74.0080
74.0160
74.0240
74.0320
There appears to be a possible outlier in the data set.
6-1
Applied Statistics and Probability for Engineers, 6th edition
6-9.
Sample average:
x
84817
 7068.1 yards
12
Sample variance:
12
x
i 1
 84817
i
19
x
i 1
600057949
2
i
2
 n 
  xi 
n
84817 2
2
xi   i 1 
600057949 

n
12
s 2  i 1

n 1
12  1
564324.92
2

 51302.265  yards 
11
Sample standard deviation:
s  51302.265  226.5 yards
The sample standard deviation could also be found using
n
s
 x
i 1
i
 x
2
n 1
where
12
 x
i 1
 x   564324.92
2
i
Dot Diagram: (rounding was used to create the dot diagram)
.
. : .. ..
:
:
-+---------+---------+---------+---------+---------+-----C1
6750
6900
7050
7200
7350
7500
6-2
December 31, 2013
Applied Statistics and Probability for Engineers, 6th edition
6-11.
December 31, 2013
Sample average:
x
351.8
 43.975
8
Sample variance:
8
x
i 1
 351.8
i
19
x
i 1
16528.403
2
i
2
 n 
  xi 
n
2
 i 1 

351.8
2
x

16528.043 
 i
n
8
s 2  i 1

n 1
8 1
1057.998

 151.143
7
Sample standard deviation:
s  151.143  12.294
The sample standard deviation could also be found using
n
s
 x
i 1
i
 x
2
n 1
where
8
 x
i 1
 x   1057.998
2
i
Dot Diagram:
.
. ..
.
..
.
+---------+---------+---------+---------+---------+------24.0
32.0
40.0
48.0
56.0
64.0
6-13.

6905
 5.44
1270
The value 5.44 is the population mean because the actual physical population of all flight times during the
operation is available.
6-15.
Sample average of exercise group:
6-3
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
n
x
x
i 1
i
n

3454.68
 287.89
12
Sample variance of exercise group:
n
x
i 1
i
 3454.68
n
x
i 1
 1118521.54
2
i
2
 n 
  xi 
n
 i 1 
2
(3454.68) 2
x

1118521
.
54

 i
n
12
s 2  i 1

n 1
12  1
123953.71

 11268.52
11
Sample standard deviation of exercise group:
s  11268.52  106.15
Dot Diagram of exercise group:
Dotplot of 6 hours of Exercise
150
200
250
300
6 hours of Exercise
Sample average of no exercise group:
n
x
x
i 1
n
i

2600.08
 325.010
8
Sample variance of no exercise group:
n
x
i 1
i
n
x
i 1
2
i
 2600.08
 947873.4
2
 n 
  xi 
n
(2600.08) 2
2
xi   i 1 
947873.4 

n
8
s 2  i 1

n 1
8 1
6-4
350
400
450
Applied Statistics and Probability for Engineers, 6th edition

December 31, 2013
102821.4
 14688.77
7
Sample standard deviation of no exercise group:
s  14688.77  121.20
Dot Diagram of no exercise group:
Dotplot of No Exercise
150
200
250
300
350
No Exercise
400
450
500
n
6-17.
x
x
i 1
n
i

57.47
 7.184
8
2
 n

  xi 
n
2
 i 1 

57.47 
2
x

412.853 
 i
0.00299
n
8
s 2  i 1


 0.000427
n 1
8 1
7
s  0.000427  0.02066
Examples: repeatability of the test equipment, time lag between samples, during which the pH of the
solution could change, and operator skill in drawing the sample or using the instrument.
6-19.
a)
x  65.86 F
s  12.16 F
Dot Diagram
: :
.
.
.
. ..
.: .: . .:..: .. :: .... ..
-+---------+---------+---------+---------+---------+-----temp
30
40
50
60
70
80
b) Removing the smallest observation (31), the sample mean and standard deviation become
x  66.86 F
s = 10.74 F
6-21.
Sample mean:
n
x
x
i 1
n
i

188.82
 4.7205
40
Sample variance:
6-5
Applied Statistics and Probability for Engineers, 6th edition
40
x
i 1
i
40
x
 188.82
i 1
2
i
December 31, 2013
907.157
2
 n 
  xi 
n
188.822
2
xi   i 1 
907.157 

15.8322
n
40
s 2  i 1


 0.405954
n 1
40  1
39
Sample standard deviation:
s  0.405954  0.637145
Dot Diagram:
6-23.
Dot Diagram for Unseeded clouds:
Dot Diagram for Seeded clouds:
The sample mean for unseeded clouds is 164.588 and the data is not centered about the mean. Two large
observations increase the mean. The sample mean for seeded clouds is 441.985 and the data is not centered
about the mean. The average rainfall when clouds are seeded is higher when compared to the rainfall when
clouds are not seeded. The amount of rainfall of seeded clouds varies widely when compared to amount of
rainfall for unseeded clouds.
6-24.
Sample mean:
6-6
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
n
x
x
i 1
n
i

3737.98
 52.6476
71
Sample variance:
71
71
 xi  3737.98
x
i 1
i 1
2
i
200899
2
 n 
  xi 
n
2
 i 1 

3737.98
2
x

200899 
 i
4102.98
n
71
s 2  i 1


 58.614
n 1
71  1
70
Sample standard deviation:
s  58.614  7.65598
Dot Diagram:
There appears to be an outlier in the data.
Section 6-2
6-25.
a) N = 30
Leaf Unit = 0.10
1
2
6
12
(7)
11
8
5
2
3
3
4
4
5
5
6
6
7
Variable
CondNum
Maximum
7.000
4
9
0134
556899
0012444
567
234
678
00
N
30
N*
Mean
0 5.328
SE Mean StDev
0.178 0.975
6-7
Minimum
3.420
Q1
4.538
Median
5.220
Q3
6.277
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
b) One particular bridge has poor rating of 3.4
c) Calculating the mean after removing the bridge mentioned above:
Variable
N
Mean SE Mean StDev Minimum
Q1
CondNum
29
5.394
0.171 0.922
3.970 4.600
Maximum
7.000
There is a little difference in between the two means.
Median
5.260
Q3
6.295
6-27.
A back-to-back steam-and-leaf display is useful when two data sets are to be compared. Therefore, for this
problem the comparison of the seeded versus unseeded clouds can be performed more easily with the backto-back stem and leaf diagram than a dot diagram.
6-29.
The median will equal the mode when the sample is symmetric with a single mode. The symmetry implies
the mode is at the median of the sample.
6-31.
Stem-and-leaf display for cycles to failure: unit = 100
1
1
5
10
22
33
(15)
22
11
5
2
1|2
represents 1200
0T|3
0F|
0S|7777
0o|88899
1*|000000011111
1T|22222223333
1F|444445555555555
1S|66667777777
1o|888899
2*|011
2T|22
Median = 1436.5, Q1 = 1097.8, and Q3 = 1735.0
No, only 5 out of 70 coupons survived beyond 2000 cycles.
6-33.
Stem-and-leaf display for yield: unit = 1
1
1
7
21
38
(11)
41
27
19
7
1
1|2
represents 12
7o|8
8*|
8T|223333
8F|44444444555555
8S|66666666667777777
8o|88888999999
9*|00000000001111
9T|22233333
9F|444444445555
9S|666677
9o|8
Median = 89.250, Q1 = 86.100, and Q3 = 93.125
6-35.
Sample median is at
70  1 = 35.5th observation, the median is 1436.5.
2
Modes are 1102, 1315, and 1750 which are the most frequent data.
6-8
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
n
Sample mean:
6-37.
x
x
i 1
i
n

98259
 1403.7
70
Do not use the total as an observation. There are 23 observations.
Stem-and-leaf of Billion of kilowatt hours
Leaf Unit = 100
(18)
5
3
2
2
1
1
1
1
0
0
0
0
0
1
1
1
1
N
= 23
000000000000000111
23
5
9
6
Sample median is at 12th = 38.43.
n
Sample mean:
x
x
i 1
n
i

4398.8
 191.0
23
Sample variance: s2 = 150673.8
Sample standard deviation: s = 388.2
6-39.
Stem-and-leaf display. Strength: unit = 1.0 1|2 represents 12
1
1
2
4
5
9
20
26
37
46
(13)
41
33
22
13
8
5
532|9
533|
534|2
535|47
536|6
537|5678
538|12345778888
539|016999
540|11166677889
541|123666688
542|0011222357899
543|011112556
544|00012455678
545|2334457899
546|23569
547|357
548|11257
i  0.5
 100  95  i  95.5  95th percentile is 5479
100
6-9
Applied Statistics and Probability for Engineers, 6th edition
6-41.
December 31, 2013
Stem-and-leaf display. Yard: unit = 1.0
Note: Minitab has dropped the value to the right of the decimal to make this display.
1
5
8
16
20
33
46
(15)
39
31
12
4
1
22 | 6
23 | 2334
23 | 677
24 | 00112444
24 | 5578
25 | 0111122334444
25 | 5555556677899
26 | 000011123334444
26 | 56677888
27 | 0000112222233333444
27 | 66788999
28 | 003
28 | 5
n
Sample Mean x 
x
i 1
100
i

n
x
i 1
i

100
26030.2
 260.3 yards
100
Sample Standard Deviation
100
x
i 1
i
100
 26030.2
x
and
i 1
2
i
6793512
2
 n

  xi 
n
2
 i 1 
(26030.2
2
x

6793512

 i
17798.42
n
100
s 2  i 1


n 1
100  1
99
2
 179.782 yards
and
s  179.782  13.41 yards
Sample Median
Variable
yards
N
100
Median
260.85
i  0.5
 100  90  i  90.5  90 th percentile is 277.2
100
6-10
Applied Statistics and Probability for Engineers, 6th edition
6-43.
Stem-and-leaf display. Rating: unit = 0.10
1
2
5
7
9
12
18
(7)
15
8
4
3
2
December 31, 2013
1|2
represents 1.2
83|0
84|0
85|000
86|00
87|00
88|000
89|000000
90|0000000
91|0000000
92|0000
93|0
94|0
95|00
Sample Mean
n
x
x
i 1
n
40
i

x
i 1
40
i

3578
 89.45
40
Sample Standard Deviation
40
40
 xi  3578
x
and
i 1
i 1
2
i
320366
2
 n 
  xi 
n
35782
2
xi   i 1 
320366 

n
40  313.9
s 2  i 1

n 1
40  1
39
 8.05
and
s  8.05  2.8
Sample Median
Variable
rating
N
40
Median
90.000
22/40 = 55% of the taste testers considered this particular Pinot Noir truly exceptional.
6-45.
Stem-and-leaf display. Height: unit = 0.10
Female Students| Male Students
0|61
1
00|62
3
00|63
5
0000|64
9
6-11
1|2 represents 1.2
Applied Statistics and Probability for Engineers, 6th edition
00000000|65
0000|66
00000000|67
00000|68
00|69
0|70
17
(4)
16
8
3
1
December 31, 2013
2
65|00
3
66|0
7
67|0000
17
68|0000000000
(15) 69|000000000000000
18
70|0000000
11
71|00000
6
72|00
4
73|00
2 74|0
1
75|0
The male engineering students are taller than the female engineering students. Also there is a slightly
wider range in the heights of the male students.
Section 6-3
6-12
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
6-47.
Frequency Tabulation for Exercise 6-23.Cycles
-------------------------------------------------------------------------------Lower
Upper
Relative
Cumulative Cum. Rel.
Class
Limit
Limit
Midpoint
Frequency Frequency Frequency
Frequency
-------------------------------------------------------------------------------at or below
.000
0
.0000
0
.0000
1
.000
266.667
133.333
0
.0000
0
.0000
2
266.667
533.333
400.000
1
.0143
1
.0143
3
533.333
800.000
666.667
4
.0571
5
.0714
4
800.000 1066.667
933.333
11
.1571
16
.2286
5 1066.667 1333.333 1200.000
17
.2429
33
.4714
6 1333.333 1600.000 1466.667
15
.2143
48
.6857
7 1600.000 1866.667 1733.333
12
.1714
60
.8571
8 1866.667 2133.333 2000.000
8
.1143
68
.9714
9 2133.333 2400.000 2266.667
2
.0286
70
1.0000
above 2400.000
0
.0000
70
1.0000
-------------------------------------------------------------------------------Mean = 1403.66
Standard Deviation = 402.385
Median = 1436.5
Frequency
15
10
5
0
500
750
1000
1250
1500
1750
number of cycles to failure
6-13
2000
2250
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
6-49.
Frequency Tabulation for Exercise 6-25.Yield
-------------------------------------------------------------------------------Lower
Upper
Relative
Cumulative Cum. Rel.
Class
Limit
Limit
Midpoint
Frequency Frequency Frequency
Frequency
-------------------------------------------------------------------------------at or below
77.000
0
.0000
0
.0000
1
77.000
79.400
78.200
1
.0111
1
.0111
2
79.400
81.800
80.600
0
.0000
1
.0111
3
81.800
84.200
83.000
11
.1222
12
.1333
4
84.200
86.600
85.400
18
.2000
30
.3333
5
86.600
89.000
87.800
13
.1444
43
.4778
6
89.000
91.400
90.200
19
.2111
62
.6889
7
91.400
93.800
92.600
9
.1000
71
.7889
8
93.800
96.200
95.000
13
.1444
84
.9333
9
96.200
98.600
97.400
6
.0667
90
1.0000
10
98.600
101.000
99.800
0
.0000
90
1.0000
above 101.000
0
.0000
90
1.0000
-------------------------------------------------------------------------------Mean = 89.3756
Standard Deviation = 4.31591
Median = 89.25
6-51.
Histogram 8 bins:
Histogram of Cycles to failure (8 bins)
18
16
14
Frequency
12
10
8
6
4
2
0
500
750
1000
1250
1500
1750
2000
Cycles to failure of aluminum test coupons
Histogram 16 bins:
6-14
2250
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
Histogram of Cycles to failure (16 bins)
14
12
Frequency
10
8
6
4
2
0
200
500
800
1100
1400
1700
2000
Cycles to failure of aluminum test coupons
2300
Yes, both of them give the same similar information
6-53.
Histogram
Histogram of Energy
20
Frequency
15
10
5
0
0
1000
2000
3000
Energy consumption
4000
The data are skewed.
6-55.
The histogram for the spot weld shear strength data shows that the data appear to be normally distributed
(the same shape that appears in the stem-leaf-diagram).
6-15
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
Frequency
20
10
0
5320 5340 5360 5380 5400 5420 5440 5460 5480 5500
shear strength
6-57.
Yes, the histogram of the distance data shows the same shape as the stem-and-leaf display in exercise 6-33.
Frequency
20
10
0
220 228 236 244 252 260 268 276 284 292
distance
Yes, the histogram of the wine rating data shows the same shape as the stem-and-leaf display.
7
6
5
Frequency
6-59.
4
3
2
1
0
85
90
95
rating
6-16
Applied Statistics and Probability for Engineers, 6th edition
6-61.
Frequency Tabulation for Bridge Condition
Bin
Frequency
<=3
0
(3,4]
3
(4,5]
9
(4,6]
10
(6,7]
8
(7,8]
0
>8
0
Mean
5.328333333
Median
5.22
Standard Deviation
6-63.
0.975337548
Frequency Tabulation for Cloud Seeding
Bin
Frequency
<200
32
(200, 400]
11
(400, 600]
2
(600, 800]
1
(800, 1000]
2
(1000, 1200]
0
>1200
4
6-17
December 31, 2013
Applied Statistics and Probability for Engineers, 6th edition
Mean
Median
Standard Deviation
December 31, 2013
303.2865
116.8
514.9998
Section 6-4
6-65.
a) Descriptive Statistics: Bridge Condition
Variable
N N*
Mean SE Mean
BrgCnd
30
0 5.328
0.178
Maximum
7.000
StDev
0.975
Median = 5.220
Lower Quartile = Q1= 4.538
Upper Quartile = Q3= 6.277
b)
c) No obvious outliers.
6-18
Minimum
3.420
Q1
4.538
Median
5.220
Q3
6.277
Applied Statistics and Probability for Engineers, 6th edition
6-67.
December 31, 2013
a) Descriptive Statistics for unseeded cloud data:
Variable
Mean SE Mean StDev Minimum
Unseeded 164.6
54.6 278.4
1.0
Q1
23.7
Median
44.2
Q3
183.3
Maximum
1202.6
Median = 44.2
Lower Quartile = Q1 = 23.7
Upper Quartile = Q3 = 1202.6
b) Descriptive Statistics for seeded data:
Variable Mean SE Mean StDev
Seeded
442
128
651
Minimum
4
Q1
79
Median
222
Q3
445
Maximum
2746
Median = 222
Lower Quartile = Q1 = 79
Upper Quartile = Q3 = 445
c)
d) The two data sets are plotted here. A greater mean and greater dispersion in the seeded data are seen.
Both plots show outliers.
6-69.
Descriptive Statistics
Variable
time
Variable
time
N
8
Mean
2.415
Minimum
1.750
Median
2.440
Maximum
3.150
TrMean
2.415
Q1
1.912
a)Sample Mean: 2.415, Sample Standard Deviation: 0.543
b) Box Plot: There are no outliers in the data.
6-19
StDev
0.534
Q3
2.973
SE Mean
0.189
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
Boxplot of Time
3.2
3.0
2.8
Time
2.6
2.4
2.2
2.0
1.8
1.6
6-71.
Descriptive Statistics
Variable
N
Temperat
9
Variable
Min
Temperat
948.00
Mean
952.44
Max
957.00
Median
953.00
Q1
949.50
Tr Mean
952.44
Q3
955.00
StDev
3.09
SE Mean
1.03
a) Sample Mean = 952.44, Sample Variance = 9.53, Sample Standard Deviation = 3.09
b) Median = 953; An increase in the largest temperature measurement does not affect the median.
c)
Boxplot of Temperature
957
956
Temperature
955
954
953
952
951
950
949
948
6-73.
Descriptive Statistics of O-ring joint temperature data
Variable
Temp
N
36
Variable
Temp
Mean
65.86
Minimum
31.00
Median
67.50
Maximum
84.00
TrMean
66.66
Q1
58.50
StDev
12.16
SE Mean
2.03
Q3
75.00
a) Median = 67.50, Lower Quartile: Q1 = 58.50, Upper Quartile: Q3 = 75.00
b) Data with lowest point removed
Variable
Temp
N
35
Mean
66.86
Median
68.00
6-20
TrMean
67.35
StDev
10.74
SE Mean
1.82
Applied Statistics and Probability for Engineers, 6th edition
Variable
Temp
Minimum
40.00
Maximum
84.00
December 31, 2013
Q1
60.00
Q3
75.00
The mean and median have increased and the standard deviation and difference between the upper and
lower quartile have decreased.
c)Box Plot: The box plot indicates that there is an outlier in the data.
Boxplot of Joint Temp
90
80
Joint Temp
70
60
50
40
30
6-75.
The box plot shows the same basic information as the stem and leaf plot but in a different format. The outliers that
were separated from the main portion of the stem and leaf plot are shown here separated from the whiskers.
Boxplot of Billion of kilowatt hours
1800
1600
Billion of kilowatt hours
1400
1200
1000
800
600
400
200
0
6-77.
This plot, as the stem and leaf one, indicates that the data fall mostly in one region and that the measurements
toward the ends of the range are more rare.
6-21
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
Boxplot of Weld Strength
5500
Weld Strength
5450
5400
5350
6-79.
We can see that the two distributions seem to be centered at different values.
Boxplot of Female, Male
76
74
72
Data
70
68
66
64
62
60
Female
6-81.
Male
All distributions are centered at about the same value, but have different variances.
6-22
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
Boxplot of High Dose, Control, Control_1, Control_2
3000
2500
Data
2000
1500
1000
500
0
High Dose
Control
Control_1
6-23
Control_2
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
Section 6-5
6-83.
Stem-and-leaf display for Force: unit = 1
3
6
14
18
(5)
17
14
10
3
1|2
represents 12
17|558
18|357
19|00445589
20|1399
21|00238
22|005
23|5678
24|1555899
25|158
Time Series Plot of Pull-off Force
260
250
Pull-off Force
240
230
220
210
200
190
180
170
4
8
12
16
20
Index
24
28
32
36
40
In the time series plot there appears to be a downward trend beginning after time 30. The stem-and-leaf plot does not
reveal this.
6-85.
Stem-and-leaf of Wolfer sunspot
Leaf Unit = 1.0
17
29
39
50
50
38
33
23
20
16
10
8
7
4
1
1
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
N
= 100
01224445677777888
001234456667
0113344488
00145567789
011234567788
04579
0223466778
147
2356
024668
13
8
245
128
4
The data appears to decrease between 1790 and 1835, the stem and leaf plot indicates skewed data.
6-24
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
Time Series Plot of Wolfer sunspot
160
140
Wolfer sunspot
120
100
80
60
40
20
0
1
20
30
40
50
Index
60
70
Stem-and-leaf of Number of Earthquakes
Leaf Unit = 1.0
2
6
13
22
38
49
(13)
48
37
25
20
12
10
8
6
4
2
1
0
0
1
1
1
1
1
2
2
2
2
2
3
3
3
3
3
4
80
N
90
= 110
67
8888
0001111
222333333
4444455555555555
66666666777
8888888889999
00001111111
222222223333
44455
66667777
89
01
22
45
66
9
1
Time Series Plot
Time Series Plot of Number of Earthquakes
45
40
Number of Earthquakes
6-87.
10
35
30
25
20
15
10
5
1900
1910
1921
1932
1943
1954
Year
1965
1976
1987
6-25
1998
2009
100
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
6-89.
a)
Time Series Plot of Anomaly
0.75
0.50
Anomaly
0.25
0.00
-0.25
-0.50
1880 1891
1903
1915
1927
1939 1951
Year
1963
1975
1987
1999
There is an increasing trend in the most recent data.
b)
Time Series Plot of CO2
380
CO2
360
340
320
300
1880 1891
1903
1915
1927
1939 1951
Year
1963
1975
1987
1999
c) The plots increase approximately together. However, this relationship alone does not prove a cause and
effect.
6-26
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
Section 6-6
6-91.
a)
Scatterplot of Price vs Taxes
45
Price
40
35
30
25
4
5
6
7
8
9
Taxes
As the taxes increase, the price tends to increase.
b) From computer software the correlation coefficient is 0.876
b) Values for y tend to increase as values for x1 increase. However, values for y tend to decrease as values
for x2 or x3 decrease.
6-93.
The pattern of the data indicates that the sample may not come from a normally distributed population or that the
largest observation is an outlier. Note the slight bending downward of the sample data at both ends of the graph.
Normal Probability Plot for 6-1
Piston Ring Diameter
ML Estimates - 95% CI
99
ML Estimates
95
90
Percent
80
70
60
50
40
30
20
10
5
1
73.99
74.00
74.01
Data
6-27
74.02
Mean
74.0044
StDev
0.0043570
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
6-95.
A normal distribution is reasonable for these data.
Normal Probability Plot for 6-5
Visual Accomodation Data
ML Estimates - 95% CI
99
ML Estimates
95
Mean
43.975
StDev
11.5000
90
Percent
80
70
60
50
40
30
20
10
5
1
0
10
20
30
40
50
60
70
80
Data
6-97.
The data appear to be approximately normally distributed. However, there are some departures from the
line at the ends of the distribution.
Normal Probability Plot for temperature
Data from exercise 6-13
99
ML Estimates
95
90
Percent
80
70
60
50
40
30
20
10
5
1
30
40
50
60
70
80
90
Data
6-99.
6-28
100
Mean
65.8611
StDev
11.9888
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
Normal Probability Plot for cycles to failure
Data from exercise 6-15
ML Estimates
99
Mean
1282.78
StDev
539.634
Percent
95
90
80
70
60
50
40
30
20
10
5
1
0
1000
2000
3000
Data
The data appear to be approximately normally distributed. However, there are some departures from the
line at the ends of the distribution.
Normal Probability Plot for concentration
Data from exercise 6-24
99
ML Estimates
95
90
Percent
80
70
60
50
40
30
20
10
5
1
25
35
45
55
65
75
Data
6-101.
6-29
85
95
Mean
59.8667
StDev
12.3932
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
Normal Probability Plot for 6-22...6-29
ML Estimates - 95% CI
99
Female
Students
95
Male
Students
90
Percent
80
70
60
50
40
30
20
10
5
1
60
65
70
75
Data
Both populations seem to be normally distributed. Moreover, the lines seem to be roughly parallel indicating that the
populations may have the same variance and differ only in the value of their mean.
Supplemental Exercises
6-103.
Based on the digidot plot and time series plots of these data, in each year the temperature has a similar
distribution. In each year, the temperature increases until the mid year and then it starts to decrease.
Dotplot of 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009
2000
2001
2002
2003
2004
2005
2006
2007
2008
2009
12.6
13.2
13.8
14.4
15.0
Global Temperature
6-30
15.6
16.2
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
Time Series Plot of 2000, 2001, 2002, 2003, 2004, 2005, 2006, ...
17
Variable
2000
2001
2002
2003
2004
2005
2006
2007
2008
2009
Global Temperature
16
15
14
13
12
1
2
3
4
5
6
7
Month
8
9
Stem-and-leaf of Global Temperature
Leaf Unit = 0.10
5
29
42
48
(11)
58
50
39
29
12
12
13
13
14
14
15
15
16
10
N
11
12
= 117
23444
555566666666677777888999
1222233344444
555566
12222333344
55566667
22333334444
5555566679
00000011111112222222333334444
Time-series plot by month over 10 years
Time Series Plot of Global Temperature
17
Global Temperature
16
15
14
13
12
1
12
24
36
48
60
Month
6-31
72
84
96
108
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
6-105.
The unemployment rate is steady from 200-2002, then it increases until 2004, decreases steadily from 2004
to 2008, and then increases again dramatically in 2009, where it peaks.
Time Series Plot of Unemployment
10
9
Unemployment
8
7
6
5
4
3
1
6-107.
13
26
39
52
65
78
Month in 1999 - 2009
91
104
117
a) Sample 1 Range = 4, Sample 2 Range = 4
Yes, the two appear to exhibit the same variability
b) Sample 1 s = 1.604, Sample 2 s = 1.852
No, sample 2 has a larger standard deviation.
c) The sample range is a relatively crude measure of the sample variability as compared to the sample
standard deviation because the standard deviation uses the information from every data point in the
sample whereas the range uses the information contained in only two data points - the minimum and
maximum.
6-109.
From the stem-and-leaf diagram, the distribution looks like the uniform distribution. From the time series
plot, there is an increasing trend in energy consumption.
Stem-and-leaf of Energy
Leaf Unit = 100
4
7
9
10
13
(3)
2
2
2
2
2
3
N
= 28
0011
233
45
7
888
001
6-32
Applied Statistics and Probability for Engineers, 6th edition
12
10
6
3
3
3
3
3
December 31, 2013
23
4455
667
888
Time Series Plot of Engergy
4000
Engergy
3500
3000
2500
2000
1982
1985
1988
1991
1994
Year
1997
2000
2003
2006
6-111.
15
sales
10
5
0
Index
10
20
30
40
50
60
70
80
90
There appears to be a cyclic variation in the data with the high value of the cycle generally increasing. The
high values are during the winter holiday months.
b) We might draw another cycle, with the peak similar to the last year’s data (1969) at about 12.7 thousand
bottles.
6-113.
a) Stem-and-leaf display for Problem 2-35: unit = 1
1
8
18
(7)
15
12
7
5
0T|3
0F|4444555
0S|6666777777
0o|8888999
1*|111
1T|22233
1F|45
1S|77
6-33
1|2
represents 12
Applied Statistics and Probability for Engineers, 6th edition
3
December 31, 2013
1o|899
b) Sample Average = 9.325, Sample Standard Deviation = 4.4858
c)
20
springs
15
10
5
Index
10
20
30
40
The time series plot indicates there was an increase in the average number of nonconforming springs during
the 40 days. In particular, the increase occurred during the last 10 days.
The golf course yardage data appear to be skewed. Also, there is an outlying data point above 7500 yards.
7500
7400
7300
yardage
6-115.
7200
7100
7000
6900
6800
6-34
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
6-117.
Normal Probability Plot for Temperature
99
ML Estimates
95
Mean
48.125
StDev
2.63490
90
Percent
80
70
60
50
40
30
20
10
5
1
40
45
50
55
Data
A normal distribution is reasonable for these data. There are some repeated values in the data that cause
some points to fall off the line.
Boxplot for problem 6-53
6-119.
285
Distance in yards
275
265
255
245
235
225
The plot indicates that most balls will fall somewhere in the 250-275 range. This same type of information could
have been obtained from the stem and leaf graph.
6-121.
6-35
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
Boxplot for Exercise 6-85
1100
1000
900
800
700
600
Trial 1
-
-
Trial 2
Trial 3
Trial 4
There is a difference in the variability of the measurements in the trials. Trial 1 has the most
variability in the measurements. Trial 3 has a small amount of variability in the main group of
measurements, but there are four outliers. Trial 5 appears to have the least variability without any
outliers.
All of the trials except Trial 1 appear to be centered around 850. Trial 1 has a higher mean value
All five trials appear to have measurements that are greater than the “true” value of 734.5.
The difference in the measurements in Trial 1 may indicate a “start-up” effect in the data.
There could be some bias in the measurements that is centering the data above the “true” value.
6-123.
a) Stem-and-leaf of Drowning Rate
N
= 35
Leaf Unit = 0.10
4
5
8
13
17
(1)
17
15
14
13
12
8
6
6
5
1
1
Trial 5
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
2788
2
129
04557
1279
5
59
9
1
0
0356
14
2
5589
3
Time Series Plots
6-36
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
Time Series Plot of Drowning Rate
22.5
20.0
Drowning Rate
17.5
15.0
12.5
10.0
7.5
5.0
1972 1975 1978 1981 1984 1987 1990 1993 1996 1999 2002
Year
b)
Descriptive Statistics: Drowning Rate
Variable
Drowning Rate
N
35
N*
0
Variable
Drowning Rate
Maximum
21.300
Mean
11.911
SE Mean
0.820
StDev
4.853
Minimum
5.200
Q1
8.000
Median
10.500
Q3
15.600
c) Greater awareness of the dangers and drowning prevention programs might have been effective.
d) The summary statistics assume a stable distribution and may not adequately summarize the data because
of the trend present.
6-125.
a) From computer software the sample mean = 34,232.05 and the sample standard deviation = 20,414.52
b)
6-37
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
There is substantial variability in mileage. There are number of vehicles with mileage near the mean, but
another group with mileage near or even greater than 70,000.
c)
Stem-and-leaf of Mileage
Leaf Unit = 1000
7
13
16
31
35
45
(13)
42
36
29
23
19
16
8
4
3
2
1
0
0
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
N
= 100
1223444
567889
013
555555677889999
1122
5667888999
0001112223334
667788
0012334
567899
1114
688
00022334
5677
2
7
3
5
d) For the given mileage of 45324, there are 29 observations greater and 71 observations less than this
value. Therefore, the percentile is approximately 71%.
6-127.
a)
Stem-and-leaf of Force
Leaf Unit = 1.0
1
1
13
27
33
(9)
26
23
16
12
11
7
5
2
1
1
1
1
1
2
2
2
2
2
3
3
3
3
N
= 68
0
444455555555
66666677777777
889999
000011111
222
4444445
6777
8
0011
22
444
67
b) The sample mean is 21.265 and the sample standard deviation is 6.422.
c)
6-38
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
Probability Plot of Force
Normal
99.9
Mean
StDev
N
AD
P-Value
99
Percent
95
90
21.26
6.422
68
2.035
<0.005
80
70
60
50
40
30
20
10
5
1
0.1
0
10
20
Force
30
40
There a number of repeated values for force that are seen as points stacked vertically on the plot. This is
probably due to round off of the force to two digits. There are fewer lower force values than are expected
from a normal distribution.
d) From the stem-and-leaf display, 9 caps exceed the force limit of 30. This is 9/68 = 0.132.
e) The mean plus two standard deviations equals 21.265 + 2(6.422) = 34.109. From the stem-and-leaf
display, 2 caps exceed this force limit of 30. This is 2/68 = 0.029.
f) In the following box plots Force1 denotes the subset of the first 35 observations and Force2 denotes the
remaining observation. The mean and variability of force is greater for the second set of data in Force2.
Boxplot of Force1, Force2
40
35
Data
30
25
20
15
10
Force1
Force2
g) A separate normal distribution for each group of caps fits the data better. This is to be expected when the
mean and standard deviations of the groups differ as they do here.
6-39
Applied Statistics and Probability for Engineers, 6th edition
December 31, 2013
Probability Plot of Force1, Force2
Normal
99
Variable
Force1
Force2
95
90
Mean StDev N
AD
P
18.67 4.395 36 1.050 0.008
24.19 7.119 32 0.639 0.087
Percent
80
70
60
50
40
30
20
10
5
1
10
15
20
25
Data
30
35
40
6-40
45