Test Bank For Probability And Statistics For Engineering
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And The Sciences 8th Ed by Jay L. Devore. vv vv vv vv vv vv vv vv vv
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
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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, Pulitzer
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c. John Anderson, Emily Black, Bill Carter, Kay Davis
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d. 2.58. 2.96, 3.51, 3.69
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PTS: v v v v 1
2. A Southern State University system consists of 23 campuses. An administrator wishes to make an
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inference about the average distance between the hometowns of students and their campuses. Describe and
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discuss several different sampling methods that might be employed. Would this be an enumerative or an
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analytic study? 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
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system and ask each student in the sample to report the distance from their hometown to campus.
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Alternatively, the sample could be generated by taking a stratified random sample by taking a simple
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random sample from each of the 23 campuses and again asking each student in the sample to report the
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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
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study is enumerative because there exists a finite, identifiable population of objects from which to sample.
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PTS: v v v v 1
3. A Michigan city divides naturally into ten district neighborhoods. How might a real estate appraiser select
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a sample of single-family homes that could be used as a basis for developing an equation to predict
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appraised value from characteristics such as age, size, number of bathrooms, and distance to the nearest
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school, and so 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
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sample by taking a simple random sample from each of the 10 district neighborhoods. From each of the
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homes in the sample the necessary variables would be collected. This would be an enumerative study
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because there exists a finite, identifiable population of objects from which to sample.
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, PTS: v v v v 1
4. An experiment was carried out to study how flow rate through a solenoid valve in an automobile‘s
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pollution-control system depended on three factors: armature lengths, spring load, and bobbin depth.
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Two different levels (low and high) of each factor were chosen, and a single observation on flow was
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made for each 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 vv vv vv vv v v vv v v
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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PTS: v v v v 1
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
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the second 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 vv
3H 56678 leaf: v v 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.
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In addition, the display is not symmetric and appears to be positively skewed. The spread of the data is
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.75 - .31 = .44, which is .44/.45 = .978 or about 98% of the typical value of .45. This constitutes a
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reasonably large amount of variation in the data. The data value .75 is a possible outlier.
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PTS: v v v v 1
6. Temperature transducers of a certain type are shipped in batches of 50.
vv A sample of 60 batches was
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selected, and the number of transducers in each batch not conforming to design specifications was
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determined, resulting in the following data:
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0 4 v v v v 2 v v 1 1 3 4 1 2 3 2 2 8 4 5 1 3 1
v v 1 v v v v 3
2 1 v v v v 2 v v 1 3 2 0 5 3 3 1 3 2 4 7 0 2 3
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5 0 v v v v 2 v v 1 0 6 4 2 1 6 0 3 3 3 6 1 2 3
v v 3 v v v v 2
, a. Determine frequencies and relative frequencies for the observed values of x = number of
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nonconforming transducers in a batch. vv vv vv vv
b. What proportion of batches in the sample has at most four nonconforming transducers? What
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proportion has fewer than four? What proportion has at least four nonconforming units?
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ANS:
a.
Number Nonconforming vv Relative Frequency Frequency vv
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 vv v v 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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PTS: v v v v 1
7. The number of contaminating particles on a silicon wafer prior to a certain rinsing process was determined
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for each wafer in a sample size 100, resulting in the following frequencies:
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Number of particles vv v v Frequency Number of particles vv v v 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
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four 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-
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2)/100 = vv
.97, or 97%. In a similar fashion, the proportion containing at least 6 particles is (100 – 1-2-3-12-11-
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15)/100 = 56/100 = .56, or 56%. vv vv vv vv vv vv
b. The proportion containing between 4 and 9 particles inclusive is (11+15+18+10+12+4)/100 = 70/100
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= .70, or 70%. The proportion that contain strictly between 4 and 9 (meaning strictly more than 4
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and strictly less than 9) is (15+ 18+10+12)/100= 55/100 = .55, or 55%.
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, PTS: v v v v 1
8. The cumulative frequency and cumulative relative frequency for a particular class interval are the sum
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of frequencies and relative frequencies, respectively, for that interval and all intervals lying below it.
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Compute 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
Frequency
vv Frequency
vv vv Relative
vv Frequency
60 vv – vv under vv65 3 .075 3 .075
65 vv – vv under vv70 6 .15 9 .225
70 vv – vv under vv75 7 .175 16 .40
75 vv – vv under vv80 1 .025 17 .425
80 vv – vv under vv85 12 .30 29 .725
85 vv – vv under vv90 7 .175 36 .90
90 vv – vv under vv95 2 .05 38 .95
95 vv – vv under vv100 2 .05 40 1.0
PTS: v v v v 1
9. Consider the following observations on shear strength of a joint bonded in a particular
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vv manner:
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.
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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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corresponding trimming percentage? How does the value of this
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vv and median?
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ANS:
a. The sum of the n = 11 data points is 514.90, so
vv vv vv = 514.90/11 = 46.81.
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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
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16.4 22.2
v v v v
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,
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value. The mean differs from the median because the largest sample observations are much further from
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the median than 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
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observations is 400.6. The trimmed mean
vv is 400.6/9 = 44.51. The trimming percentage is
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100(1/11) = 9.1%.
vv lies between the mean and median.
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PTS: v v v v 1
10. A sample of 26 offshore oil workers took part in a simulated escape exercise, resulting in the
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accompanying data 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
