Review 1
2001.11.8
Chapter 1:
1. Elements, Variable, and Observations:
Example:
Table 1.1 (p. 5) in the textbook!!
25 elements (25 companies): Advanced Comm. Systems, Ag-Chem
Equipment Co.,…,Webco Industries
Inc..
5 variables : Exchange, Ticker Symbol, Annual Sales, Share Price,
Price/Earnings Ratio.
25 observations: (OTC, ACSC, 75.10, 0.32, 39.10), (OTC, AGCH,
321.10, 0.48, 23.40),…, (AMEX, WEB, 153.50, 0.88,
7.50).
2. Type of Data: Qualitative Data and Quantitative Data
(a) Qualitative data may be nonnumeric or numeric.
(b) Quantitative data are always numeric.
(c) Arithmetic operations are only meaningful with quantitative data.
Example (continue):
Qualitative variables: Exchange, Ticker Symbol.
Quantitative variables: Annual Sales, Share Price, Price/Earnings Ratio.
Chapter 2: Figure 2.22, p. 66.
1. Summarizing qualitative data:
 Frequency distribution, relative frequency distribution, and
percent frequency distribution.
 Bar plot and Pie plot.
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Example:
Below you are given the grades of 20 students.
D
C
E
B
B
B
A
D
B
C
B
E
C
B
C
B
B
D
B
C
Then,
Grades
Frequency
Relative
Frequency
Percent
Frequency
E
2
2/20=0.1
10
D
3
3/20=0.15
15
C
5
5/20=0.25
25
B
9
9/20=0.45
45
A
1
1/20=0.05
5
Total
20
1
100
2. Summarizing qualitative data:
 Frequency distribution, relative frequency distribution, percent
frequency distribution, cumulative frequency distribution,
cumulative relative frequency distribution, cumulative percent
frequency distribution
 Histogram, Ogive, and stem-and leaf display.
Example:
Suppose we have the following data:
30
79
59
65
40
64
52
53
57
39
61
47
50
60
48
50
58
67
Suppose the number of nonoverlapping classes is determined to be 5.
Please construct the frequency distribution table (including frequency,
percent frequency, cumulative frequency, and cumulative percent
frequency) for the data.
[solution:]
2
Approximat e class width 
79  30
 9.8
5

The class width is 10.
Thus,
Class
Frequency
2
3
7
5
1
30-39
40-49
50-59
60-69
70-79
Percent
Frequency
Cumulative
Frequency
(2/18)100=11
(3/18)100=17
(7/18)100=39
(5/18)100=28
(1/18)100=5
2
5
12
17
18
Cumulative
Percent
Frequency
11
28
67
95
100
Chapter 3: Key Formulas, pp. 128-129.
Example:
Suppose we have the following data:
Rent
420-439
440-459
460-479
480-499
500-519
Frequency
8
17
12
8
7
Rent
520-539
540-559
560-579
580-599
600-619
Frequency
4
2
4
2
6
What are the mean rent and the sample variance for the rent?
[solution:]
10
xg 
fM
i 1
i
70
i
, where f i is the frequency of class i M i is the midpoint of
class i and n is the sample size. Then,
Rent
420-439
440-459
fi
8
17
Mi
429.5
449.5
Rent
520-539
540-559
fi
4
2
Mi
529.5
549.5
3
460-479
12
469.5
560-579
4
569.5
480-499
8
489.5
580-599
2
589.5
500-519
7
509.5
600-619
6
609.5
Thus,
10
fM
i
i 1
i
 34525 and x g  34525  493.21 .
70
For the sample variance,
 f M
10
s g2 
i 1
i
 xg 
2
i
70  1

208234.287
 3017.89
69
Chapter 4:
 Tabular and Graphical Methods: Crosstabulation (qualitative
and quantitative data) and Scatter Diagram (only quantitative
data).
 Numerical Method: Covariance and Correlation Coefficient.
Chapter 5:
1. Multiple Step Experiments, Permutations, and Combinations:
Example:
How many committees consisting of 3 female and 5 male students can be
selected from a group of 5 female and 8 male students?
[solution:]
 5  8
5!
8!
     

 560
3
5
    3!2! 5!3!
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2. Event, Addition Law, Mutually Exclusive Events and Independent
Event:
Example:
Assume you are taking two courses this semester (S and C). The
probability that you will pass course S is 0.835, the probability that you
will pass both courses is 0.276. The probability that you will pass at least
one of the courses is 0.981.
(a) What is the probability that you will pass course C?
(b) Is the passing of the two courses independent event?
(c) Are the events of passing the courses mutually exclusive? Explain.
[solution:]
(a)
Let A be the event of passing course S and B be the event of passing
course C. Thus,
P( A)  0.835, P( A  B)  0.276, P( A  B)  0.981 .
 P( Ac  B)  P( A  B)  P( A)  0.981  0.835  0.146
 P( B)  P( A  B)  P( Ac  B)  0.276  0.146  0.422
.
(b)
P( A | B) 
P( A  B) 0.276

 0.654  P( A)  0.835
P( B)
0.422
Thus, events A and B are not independent. That is, passing of two courses
are not independent events.
(c)
Since P( A  B)  0.276  0 , events A and B are not mutually exclusive.
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