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Chapter 4
4.1
Sample: A small portion of a population that, when analyzed, may provide information
concerning the entire population.
Population: The total set of observations being considered in a statistical procedure.
4.2
Continuous statistic:
-weight of 5 pound bag of flour
-millimeters separating a closed car door from its frame
-amount of cutting fluid used during a particular
machining operation
Discrete statistic:
-number of 5 pound bags of flour produced
-number of cars passing through a toll booth
-number of telephone calls received in one hour
4.5
Skewed to the left.
4.16
It is important to use both statistical measures and descriptive concepts when describing a
histogram. The statistical measures describe the distribution mathematically while the
descriptive concepts provide a visual picture of the distribution. Combining both creates a
greater understanding of the process.
4.19
The central tendency of the data refers to describing where the data is grouped in relation
to the desired target value. The central tendency is described by the mean, the mode and
the median.
4.20
Measures of dispersion refer to the spread of the data in the distribution or around the
desired target. Measures of dispersion are the range and the standard deviation.
4.21
The salary data for the engineer includes key information concerning the standard
deviation and the range (min, max). The size of the sample (n) is also significantly larger,
providing more information. This additional information gives validity to the average of
$48,298 and median of $45,750.
4.22
Mean = 8 Mode = 8, Median = 8
4.28
Mean = 119.8
Mode = 119.8
Median = 119.8
Standard Deviation = 0.2
Range = 0.8
1
These values confirm that the histogram has a narrow spread. The mean, mode
and median are equal. The distribution is symmetrical.
4.29
 = 5.32
R=9
4.34
The mean, mode, median, standard deviation, and the range work together to describe a
histogram by providing analytical information concerning where the data is located, i.e.,
the central tendency of the data and the dispersion of the data. This information allows
those investigating a process to understand the process’s accuracy and precision.
4.35
The distribution is slightly negatively skewed. The mean is less than the median which is
less than the mode. Without mathematical data, the range and standard deviation can not
be easily judged, though the distribution does lack a long tail indicative of a large
standard deviation or range. The kurtosis can not be judged because there are no other
curves to compare it with.
4.36
The middle curve has a significantly larger standard deviation than the acceptable curve
for the process. It is platykurtic. The curve on the right has a significantly smaller
standard deviation than the acceptable curve for the process. It is leptokurtic.
4.37
15 – 8
Z = ------------- = 2.33
3
Area = 0.9893 Percentage of calls lasting over 15 minutes => 1 – 0.9893 = 0.0107
1.1% of calls last over 15 minutes
15  12
=1
3
Area = 0.8413
84.13% wait less than 15 minutes
4.38
Z=
4.39
Z=
4.40
Z=
4.42
X-bar = 20.6
(0.25  0.917)
= 1.60
0.005
Area = 0.9452
1  0.9452 = 0.0548
5.48% above 90 mm
65  70 5
= 1.67

3
3
Area = 0.0475
Area = 1  0.0475 = 0.9525
95.25% acceptable
std dev = 1.3
Z=
19.7  20.6
= 0.69
1.3
2
Xi = 19.7
The amount of product produced that falls below the lower specification limit of 19.7
pounds is 24.51%.
4.43
X-bar = 44.795
std dev = 0.402
Z=
44.000  44.795
= 1.98
0.402
Xi = 44.000
The percentage of coils produced that falls below the minimum requirement of 44.000 is
2.4%.
4.44
X-bar = 50
std dev = 18
Z=
0  50
= 2.78
18
Xi = 0
Area = 0.0027
X-bar = 50
std dev = 18
Z=
90  50
= 2.22
18
Xi = 90
Area = 0.9868
The percentage of temperatures that falls between 0 and 90 degrees is equal to
0.9868  0.0027 = 0.9841 or 98.41%.
4.45
X-bar = 119.8
std dev = 0.2
Z=
119.7  119.8
= 0.5
0.2
Z=
120.3  119.8
= 2.5
0.2
Xi = 119.7
X-bar = 119.8
std dev = 0.2
Xi = 120.3
The amount of product produced that falls below the lower specification limit of 119.7
mm is 30.85%
The amount of product produced that falls above the upper specification limit of 120.3
pounds is 1  0.9938 = 0.0062 or 0.62%.
These results verify that the process is not centered correctly in order to meet the
specifications. The standard deviation is also large when compared to the narrow spread
of the process.
3
65  45
= 2.0
Area1 = 0.9772
10
35  45
Z=
= 1.0
Area2 = 0.1587
10
Area3 = 0.9773  0.1587 = 0.8186
81.86% of clean-ups take between 35 and 65 minutes.
4.46
Z=
4.47
Z=
4.48
Z=
4.50
Z=
4.51
25 – 23
Z = -------------- = 0.5
4
Xi  3000
= 2.575
50
X  3000
2.575 = i
50
Xi = 2871.25 foot-candles
1000  900
=2
50
Area1 = 0.9773
Area2 = 1  0.9773 = 0.0227
2.27% of batteries survive beyond 1000 days
11.41  11.85
= 0.33
1.33
Area = 0.3707
37.07% of parts will be below the lower specification limit of 11.41 inches.
Area = 0.6915
1 – 0.6915 = 0.3085 30.85% have hysteresis greater than 25.
4.52
Since  = 0.10, then Z(/2) = 1.645
1.645(0.0008)
0.0015 +/
= (0.0017, 0.0013)
32
The engineers know with 90% confidence that the interval values for the population
mean are (0.0017, 0.0013).
4.53
Since  = 0.05, then Z(/2) = 1.96
1.96(100)
1200 +/
= (1231, 1169)
40
The manufacturer knows with 95% confidence that the interval values for the population
mean are (1231, 1169).
4.54
Since  = 0.05, then t(/2) = 2.365
4
44.795 +/
2.365(0.402)
= (45.131, 44.458)
8
At a 95% confidence level, the interval for the population mean is (45.131, 44.458).
4.55
Since  = 0.01, then Z(/2) = 2.575
2.575(2)
351 +/- -------------- = (350, 352)
√50
The chef knows with 99% confidence that the interval values for the population mean are
(350, 352).
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