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Chapter 2 Exercise Solutions
Several exercises in this chapter differ from those in the 4th edition. An “*” following the exercise number indicates that the description has changed (e.g., new values). A second exercise number in parentheses indicates that the exercise number has changed. For example, “2-16* (2-9)” means that exercise 2-16 was 2-9 in the 4th edition, and that the description alsodiffers from the 4th edition (in this case, asking for a time series plot instead of a digidot plot). New exercises are denoted with an “☺”. 2-1*. (a) x = ∑ xi n = (16.05 + 16.03 +
n i =1

+ 16.07 ) 12 = 16.029 oz

(b)
n

s=

i =1

∑ x − ∑ xi
2 i i =1

( )
n

2

n

n −1

=

(16.052 +

+ 16.07 2 ) − (16.05 + 12 − 1

+ 16.07) 2 12

= 0.0202 oz

MTB > Stat > BasicStatistics > Display Descriptive Statistics Descriptive Statistics: Ex2-1
Variable Ex2-1 Variable Ex2-1 N N* Mean 12 0 16.029 Maximum 16.070 SE Mean 0.00583 StDev 0.0202 Minimum 16.000 Q1 16.013 Median 16.025 Q3 16.048

2-2. (a) x = ∑ xi n = ( 50.001 + 49.998 +
n i =1

+ 50.004 ) 8 = 50.002 mm

(b)
n

s=

i =1

2 ∑ xi − ∑ xi i =1

( )
n

2

n

n −1

=

(50.0012 +

+50.0042 ) − (50.001 + 8 −1

+ 50.004) 2 8

= 0.003 mm

MTB > Stat > Basic Statistics > Display Descriptive Statistics Descriptive Statistics: Ex2-2
Variable Ex2-2 Variable Ex2-2 N N* Mean 8 0 50.002 Maximum 50.006 SE Mean 0.00122 StDev 0.00344 Minimum 49.996 Q1 49.999 Median 50.003 Q3 50.005

2-1

Chapter 2 Exercise Solutions 2-3. (a) x = ∑ xi n = ( 953 + 955 +
n i =1

+ 959 ) 9 =952.9 °F

(b)
n

s=

i =1

∑ x − ∑ xi
2 i i =1

( )
n

2

n

n −1

=

(9532 +

+ 9592 ) − (953 + 9 −1

+ 959) 2 9

= 3.7 °F

MTB > Stat > Basic Statistics > Display Descriptive Statistics Descriptive Statistics: Ex2-3
Variable Ex2-3 Variable Ex2-3 N N* Mean 9 0 952.89 Maximum 959.00 SE Mean 1.24 StDev 3.72 Minimum 948.00 Q1 949.50 Median 953.00 Q3 956.00

2-4. (a) Inranked order, the data are {948, 949, 950, 951, 953, 954, 955, 957, 959}. The sample median is the middle value. (b) Since the median is the value dividing the ranked sample observations in half, it remains the same regardless of the size of the largest measurement.

2-5.
MTB > Stat > Basic Statistics > Display Descriptive Statistics Descriptive Statistics: Ex2-5
Variable Ex2-5 Variable Ex2-5N N* Mean 8 0 121.25 Maximum 156.00 SE Mean 8.00 StDev 22.63 Minimum 96.00 Q1 102.50 Median 117.00 Q3 144.50

2-2

Chapter 2 Exercise Solutions 2-6. (a), (d)
MTB > Stat > Basic Statistics > Display Descriptive Statistics Descriptive Statistics: Ex2-6
Variable Ex2-6 Variable Ex2-6 N N* Mean 40 0 129.98 Maximum 160.00 SE Mean 1.41 StDev 8.91 Minimum 118.00 Q1 124.00 Median 128.00 Q3 135.25(b) Use √n = √40 ≅ 7 bins
MTB > Graph > Histogram > Simple
Histogram of Time to Failure (Ex2-6)
20

15

Frequency

10

5

0

112

120

128

136 Hours

144

152

160

(c)
MTB > Graph > Stem-and-Leaf Stem-and-Leaf Display: Ex2-6
Stem-and-leaf of Ex2-6 Leaf Unit = 1.0 2 11 89 5 12 011 8 12 233 17 12 444455555 19 12 67 (5) 12 88999 16 13 0111 12 13 33 10 13 10 13 677 713 7 14 001 4 14 22 HI 151, 160 N = 40

2-3

Chapter 2 Exercise Solutions 2-7. Use

n = 90 ≅ 9 bins

MTB > Graph > Histogram > Simple

Histogram of Process Yield (Ex2-7)
18 16 14 12

Frequency

10 8 6 4 2 0

84

88

92

96

Yield

2-4

Chapter 2 Exercise Solutions 2-8. (a)
Stem-and-Leaf Plot 2 12o|68 6 13*|3134 12 13o|776978 28 14*|3133101332423404 (15)14o|585669589889695 37 15*|3324223422112232 21 15o|568987666 12 16*|144011 6 16o|85996 1 17*|0 Stem Freq|Leaf

(b) Use

n = 80 ≅ 9 bins

MTB > Graph > Histogram > Simple
Histogram of Viscosity Data (Ex 2-8)
20

15

Frequency

10

5

0

13

14

15 Viscosity

16

17

Note that the histogram has 10 bins. The number of bins can be changed by editing the X scale. However, if 9 bins...