TestBank ForProbabilityAnd Statistics For Engineering And
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The Sciences 8th Ed by Jay L. Devore. ll ll ll ll ll ll ll ll
Chapter 1 – Overview and Descriptive Statistics
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SHORT ANSWER
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1. Give one possible sample of size 4 from each of the following populations:
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a. All daily newspapers published in the United States
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b. All companies listed on the New York Stock Exchange
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c. All students at your college or university ll ll ll ll ll ll
d. All grade point averages of students at your college or university
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ANS:
a. Houston Chronicle, Des Moines Register, Chicago Tribune, Washington Post
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b. Capital One, Campbell Soup, Merrill Lynch, Pulitzerll ll ll ll ll ll
c. John Anderson, Emily Black, Bill Carter, Kay Davis
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d. 2.58. 2.96, 3.51, 3.69 ll ll ll
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2. A Southern State University system consists of 23 campuses. An administrator wishes to make an inference
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about the average distance between the hometowns of students and their campuses. Describe and discuss several
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different sampling methods that might be employed. Would this be an enumerative or an analytic study?
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Explain your reasoning.
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ANS:
One could take a simple random sample of students from all students in the California State University system and
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ask each student in the sample to report the distance from their hometown to campus. Alternatively, the sample
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could be generated by taking a stratified random sample by taking a simple random sample from each of the 23
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campuses and again asking each student in the sample to report the distance from their hometown to campus.
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Certain problems might arise with self reporting of distances, such as recording error or poor recall. This study is
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enumerative because there exists a finite, identifiable population of objects from which to sample.
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3. A Michigan city divides naturally into ten district neighborhoods. How might a real estate appraiser select a
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sample of single-family homes that could be used as a basis for developing an equation to predict appraised
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value from characteristics such as age, size, number of bathrooms, and distance to the nearest school, and so
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on? Is the study enumerative or analytic?
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ANS:
One could generate a simple random sample of all single family homes in the city or a stratified random sample by
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taking a simple random sample from each of the 10 district neighborhoods. From each of the homes in the
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sample the necessary variables would be collected. This would be an enumerative study because there exists a
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finite, identifiable population of objects from which to sample.
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4. An experiment was carried out to study how flow rate through a solenoid valve in an automobile‘s pollution-
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control system depended on three factors: armature lengths, spring load, and bobbin depth. Two different
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levels (low and high) of each factor were chosen, and a single observation on flow was made for each
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combination of levels.
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a. The resulting data set consisted of how many observations?
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b. Is this an enumerative or analytic study? Explain your reasoning.
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ANS:
a. Number observations equal 2 2 2=8 ll ll ll ll l l ll l l
b. This could be called an analytic study because the data would be collected on an existing
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process. There is no sampling frame.
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5. The accompanying data specific gravity values for various wood types used in construction .
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.41 .41 .42 .42. .42 .42 .42 .43 .44
.54 .55 .58 .62 .66 .66 .67 .68 .75
.31 .35 .36 .36 .37 .38 .40 .40 .40
.45 .46 .46 .47 .48 .48 .48 .51 .54
Construct a stem-and-leaf display using repeated stems and comment on any interesting features of the display.
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ANS:
One method of denoting the pairs of stems having equal values is to denote the stem by L, for ‗low‘ and the second
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stem by H, for ‗high‘. Using this notation, the stem-and-leaf display would appear as follows:
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3L 1 stem: tenths ll
3H 56678 leaf: l l hundredths
4L 000112222234
5L 144
5H 58
6L 2
6H 6678
7L
7H 5
The stem-and-leaf display on the previous page shows that .45 is a good representative value for the data. In
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addition, the display is not symmetric and appears to be positively skewed. The spread of the data is .75 - .31 =
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.44, which is .44/.45 = .978 or about 98% of the typical value of .45. This constitutes a reasonably large
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amount of variation in the data. The data value .75 is a possible outlier.
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6. Temperature transducers of a certain type are shipped in batches of 50. A sample of 60 batches was selected,
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and the number of transducers in each batch not conforming to design specifications was determined,
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resulting in the following data:
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0 4 l l l l 2 l l 1 1 3 4 1 2 3 2 2 8 4 5 1 3 1
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2 1 l l l l 2 l l 1 3 2 0 5 3 3 1 3 2 4 7 0 2 3
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5 0 l l l l 2 l l 1 0 6 4 2 1 6 0 3 3 3 6 1 2 3
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, a. Determine frequencies and relative frequencies for the observed values of x = number of
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transducers in a batch.
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b. What proportion of batches in the sample has at most four nonconforming transducers? What proportion
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has fewer than four? What proportion has at least four nonconforming units?
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ANS:
a.
Number Nonconforming Relative Frequency ll Frequency ll
0 0.117 7
1 0.200 12
2 0.217 13
3 0.233 14
4 0.100 6
5 0.050 3
6 0.050 3
7 0.017 1
8 0.017 1
1.001
The relative frequencies don’t add up exactly to 1because they have been rounded
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b. The number of batches with at most 4 nonconforming items is 7+12+13+14+6=52, which is a proportion of
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52/60=.867. The proportion of batches with (strictly) fewer than 4 nonconforming items is 46/60=.767.
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7. The number of contaminating particles on a silicon wafer prior to a certain rinsing process was determined for
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each wafer in a sample size 100, resulting in the following frequencies:
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Number of particles ll l l Frequency Number of particles ll l l Frequency
0 1 8 12
1 2 9 4
2 3 10 5
3 12 11 3
4 11 12 1
5 15 13 2
6 18 14 1
7 10
a. What proportion of the sampled wafers had at least two particles? At least six particles?
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b. What proportion of the sampled wafers had between four and nine particles, inclusive? Strictly between four
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and nine particles?
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ANS:
a. From this frequency distribution, the proportion of wafers that contained at least two particles is (100-1-2)/100 =
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.97, or 97%. In a similar fashion, the proportion containing at least 6 particles is (100 – 1-2-3-12-11-15)/100
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= 56/100 = .56, or 56%.
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b. The proportion containing between 4 and 9 particles inclusive is (11+15+18+10+12+4)/100 = 70/100 = .70,
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or 70%. The proportion that contain strictly between 4 and 9 (meaning strictly more than 4 and strictly less
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than 9) is (15+ 18+10+12)/100= 55/100 = .55, or 55%.
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8. The cumulative frequency and cumulative relative frequency for a particular class interval are the sum of
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frequencies and relative frequencies, respectively, for that interval and all intervals lying below it. Compute
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the cumulative frequencies and cumulative relative frequencies for the following data:
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75 89 80 93 64 67 72 70 66 85
89 81 81 71 74 82 85 63 72 81
81 95 84 81 80 70 69 66 60 83
85 98 84 68 90 82 69 72 87 88
ANS:
Class Frequency Relative Cumulative Cumulative
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ll Relative
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Frequency
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60 – under 65 ll ll ll 3 .075 3 .075
65 – under 70 ll ll ll 6 .15 9 .225
70 – under 75 ll ll ll 7 .175 16 .40
75 – under 80 ll ll ll 1 .025 17 .425
80 – under 85 ll ll ll 12 .30 29 .725
85 – under 90 ll ll ll 7 .175 36 .90
90 – under 95 ll ll ll 2 .05 38 .95
95 – under 100 ll ll ll 2 .05 40 1.0
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9. Consider the following observations on shear strength of a joint bonded in a particular manner:
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30.0 4.4 33.1 66.7 81.5 22.2 40.4 16.4 73.7 36.6 109.9
a. Determine the value of the sample mean. ll ll ll ll ll ll
b. Determine the value of the sample median. Why is it so different from the mean?
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c. Calculate a trimmed mean by deleting the smallest and largest observations. What is the
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ll corresponding trimming percentage? How does the value of this ll ll ll ll ll ll ll ll l l ll compare to the mean and ll ll ll ll
ll median?
ANS:
a. The sum of the n = 11 data points is 514.90, so
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b. The sample size (n = 11) is odd, so there will be a middle value. Sorting from smallest to largest: 4.4 16.4
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22.2
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30.0 33.1 36.6 40.4 66.7 73.7 81.5 109.9. The sixth value, 36.6 is the middle, or median, value. The
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mean differs from the median because the largest sample observations are much further from the median than
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are the smallest values.
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c. Deleting the smallest (x = 4.4) and largest (x = 109.9) values, the sum of the remaining 9 observations is
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400.6. The trimmed mean
ll is 400.6/9 = 44.51. The trimming percentage is 100(1/11) = 9.1%.
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lies between the mean and median.
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10. A sample of 26 offshore oil workers took part in a simulated escape exercise, resulting in the accompanying data
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on time (sec) to complete the escape:
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373 370 364 366 364 325 339 393
356 359 363 375 424 325 394 402
