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Unformatted text preview: 1.6 1.8 1.18 1.20 1.26 2.4 Quantitative data are measurements that are recorded on a meaningful numerical scale. Qualitative
data are measurements that are not numerical in nature; they can only be classiﬁed into one of a
group of categories. A population is a set of existing units such as people. objects, transactions, or events. A sample is a
subset of the units of a population. a. Since the members were asked to ﬁll out a questionnaire, the method of data collection is a
mail survey. The population surveyed is the 4000 members of the Institute of Management. The variable of interest is the expected change in the number of temporary employees by
2002. Since there were only 3 possible choices for answers to the question that are mnnumerical, the
variable is qualitative. Since fewer than half of the members in the sample indicated that they expected to increase the
number of temporary employees by 2002, we can infer that fewer than half of all members expect to increase the number of temporary employees by 2002.
The variable of interest to the researchers is the rating of highway bridges.
Since the rating of a bridge can be categorized as one of three possible valueI. it is qualitative. The data set analyzed is a population since all highway bridges in the U.S. were categorized. The data were collected observationally. Each bridge was observed In in mural setting. The pepulation of interest is the set of all adults in the Minhang District, a suburb of
Shanghai, China. The sample size was 3,423 + 3,593 = 7,016.
The study made inferences about all "people in China." Since pnly those in the Minhang District were sampled, the results may not be characteristic of
all Chinese people. The group of people surveyed was from a very small group of people in China.‘ The people in the Minhang District may be quite different from the population of
China in general. To construct a relative frequency table for the data, we must ﬁnd the relative frequency for
each Cruise line. To ﬁnd the relative frequency, divide the frequency by the total population
size, 1,591,560. The relative frequency for Canaveral is 152,240r’1,591,560 = .096. The rest
of the relative frequencies are found in a similar manner and are reported in the table. 2.8 2.12 ' (frutse Elite Number oi Passengers ﬁelanve Frequency
(fanaveral (Dolphmi 152.240 15224011591350 = .133 Carnival (Fantasy) 480,924 4809241 1591560 = .302
Disney (Magic) 73,504 7350411591560 =1 .046
Premier (Oceanic) 270,361 27036111591560 = .170
Royal Caribbean (Nordic Empress) 106.161 10616111591560 = .067
Sun Cruz Casinos 453,806 45380611591560 = .285
Sterling Cruises (New Yorker) 15.782 1578211591560 = .010
Topaz Int’l Shipping (Topaz) 28,820 2828011591560 = .018
Other 10.502 1050211591560 = .007 oral 1,591, 1. 1 b. The cruise ship with the highest relative frequency (.302) is Carnivai (Fantasy). This means  that Carnival handles approximately .302 of the total pepulation. The relative frequency bar chart is:
0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.00 ﬁfe fa dfqéeifczyd’” fag as Most of the books (63%) received a "favorableirecomrnended" review. About the same
percentage of books received the following reviews: "cautious or very little recommendation" (10%), "little or no preference" (9%), and "outstandingi’signiﬁcant contribution" (12%). Only
5% of the books received "would not recommend" reviews. If the top two categories are added together. the percent recommended is "15% (actually
slightly higher than "15%). This agrees with the study. __.. 1‘ fl Rainw firm
' is
. ..__l_ a .
..__1__ .5 as 4.5'55 as 113.5 12514.5 use
“mamas: L. 16 b. The highest proportion of test scores {.25} fell in the measurement class 15—95 The proportion of scores between 3.5 and 5.5 is .15 The proportion of test scores higher than 11.5 is .10 + .05 + .05 = .20 The proportion of the 100 Students who scored less than 5.5 is .05 + .15 = .20 Using MINKTAB, the stemandleaf display is: Stewardleaf of PAF N=17
Leaf Unit = 1.0 6 0 000009 8 ‘l 25 (2) 2 1+5 ? 3 13 5 4 0 4 5 4 6 2 3 F 05? The median is the middle number once the data are arranged in order. The data arranged in
order are: 0.0,0,0,0,9,12, l5. 24, 25, 31, 33. 40, 6'2, 70, 75. "IT. The middle number or the median is 24. 77+33+T5++31 4'53
17 l? = 27.82 The mean of the data is f 2 £3: =
H The number occurring most frequently is 0. The mode is 0. The mode corresponds to the smallest number. It does not seem to locate the center of the distribution. Borh the mean and the median are in the middle of the stem—andleaf display.
Thus, it appears that both of them locate the center of the data. “L: H
a ll H s = m = 1.1402 After adding 3 to each of the data points,
Range = 6 — 3 = 3
222 x2— (BX): 102—...
s2=_£—”—_5=1.3 = nwl 5—1 l m =1.1402 After subtracting 4 from each of the data points,
Range = —1—(—4)= 3 2 _ 2
2x2_ (Ex) 39 _(13)
52: '1 =____._L :13 3= 1.3 =1.1402
11—1 5—1 The range, variance. and standard deviation remain the same when any number is added to or
subtracted from each measurement in the data set. 2.60 From the printout, the mean is 40.0555556 and the standard deviation is 2.1770812. Both of
these measures are measured in the same units as the original data, which is miles per gallon. Since the sample mean is a good estimate of the population mean, the manufacturer should be
satisﬁed. The sample mean is 40.0555556 which is greater than 40. The range of the data set is 45 — 35 = 10. Using Chebyshev’s Rule, the range should cover
approximately 6 standard deviations. Thus, a good estimate of the standard deviation would
be 1016 = 1.67. Using the Empirical Rule, the range should cover approximately 4 standard
deviations. Thus, a good estimate of the standard deviation would be 1074 = 2.5 The given
standard deviation is 2.2 which is between these two estimates. Thus, it is a reasonable value. Using MINITAB, the frequency histogram is {the relative frequency histogram would have the
same shape): Histogram of [:1 It 36
Midpoint
35
36
37
38
39
40
41
42
(.3
u.
45 i I 'I' *i* tit ii ti lt Iﬁ'ﬁﬁ *‘kt
t**t*I* “111'
it i
I d—hNb‘ﬁ‘ObUU—‘J Yes, the data appear to be moundshaped. Because the data are moundshaped, we can use the Empirical Rule. We would expect
approximately 68% of the data within the interval Y i 3, approximately 95% of the data
within the interval 3 i 2.5, and approximately all of the data within the interval Y i 35. The interval E i s is 40.056 i 2.177 or (37.879, 42.233). Twentyseven of the observations
fall in this interval or 27136 = .75 or 75%. This number is a little larger than 68%. The interval E i 25 is 40.056 1 20.177) or (35.702, 44.410). Thirtyfour of the
observations fall in this interval or 34736 x .94 or 94%. This number is very close to 95%. The interval .Y i 35 is 40.056 i 3(2. 177) or (33.525, 46.587). Thirtysix of the observations
fail in this interval or 36736 = 1.00 Or 100%. This number is the same as all of the
observations. 2.72 Since the element 40 has a zscore of —2 and 90 has a zscore of 3, —2= 40—pand3=90_’u
U U
=20=40—p. =30=90—p
un—zamo =p+30=90 I» p, = 40 + 20
By substitution, 40 + 20 + 30 = 90
w 50 = 50
=9 o = 10 By substitution, p, = 40 + 2(10) = 60 Therefore, the population mean is 60 and the standard deviation is 10. 2.78 a. From the problem, .u. = 2.7 and a = .5 z= x ;“=zo=x—p=>x:p+zcr
Forz = 2.0, x = 2.? + 2.0(.5) = 3.7
For 2: = —1.0,x = 2.7  l.0(.5) = 2.2
For 2: = .5, x = 2.7 + .5(.5) : 2.95
Forz = —2.5,x = 2.7 — 2.5(.5) = 1.45
b. For 2 = —3.6,x = 2.7 —1.6(.5)= 1.9 e. If we assume the distribution of GPAs is approximately mound~shaped. we can use the
Empirical Rule. From the Empirical Rule, we know that z .025
or = 2.5 % of the students will have GPAs above
3.7 (with z = 2). Thus, the GPA corresponding
to summa cum laude (top 2.5%) will be greater
than 3.? (z > 2). We know that :16 or 16% of the students will have GPAs above 3.2 (z = 1). Thus, the
limit on GPAs for cum laude (top 16%) will be greater than 3.2 (2 > 1). We must assume the distribution is moundshaped. ...
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 Fall '07
 Safarzadeh

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