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2-7
Relative Class Frequencies
Class frequencies can be converted to relative class
frequencies to show the fraction of the total number
of observations in each class.
A relative frequency captures the relationship between
a class total and the total number of observations.
2-8
EXAMPLE – Creating a Frequency
Distribution Table
Ms. Kathryn Ball of AutoUSA
wants to develop tables, charts,
and graphs to show the typical
selling price on various dealer
lots. The table on the right
reports only the price of the 80
vehicles sold last month at
Whitner Autoplex.
2-9
Constructing a Frequency Table -
Example
Step 1: Decide on the number of classes.
A useful recipe to determine the number of classes (k) is
the “2 to the k rule.” such that 2k > n.
There were 80 vehicles sold. So n = 80. If we try k = 6, which
means we would use 6 classes, then 26 = 64, somewhat less
than 80. Hence, 6 is not enough classes. If we let k = 7, then 27
128, which is greater than 80. So the recommended number of
classes is 7.
Step 2: Determine the class interval or width.
The formula is: i (H-L)/k where i is the class interval, H is
the highest observed value, L is the lowest observed value,
and k is the number of classes.
($35,925 - $15,546)/7 = $2,911
Round up to some convenient number, such as a multiple of 10
or 100. Use a class width of $3,000
2-10
Step 3: Set the individual class limits
Step 4: Tally the vehicle selling prices
into the classes.
Step 5: Count the number of items in
each class.
Constructing a Frequency Table - Example
2-11
Relative Frequency Distribution
To convert a frequency distribution to a relative frequency
distribution, each of the class frequencies is divided by the
total number of observations.
2-12
Graphic Presentation of a Frequency
Distribution
The three commonly used graphic forms are:
Histograms
Frequency polygons
Cumulative frequency distributions
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2-13
Histogram
HISTOGRAM A graph in which the classes are marked on the
horizontal axis and the class frequencies on the vertical axis. The
class frequencies are represented by the heights of the bars and
the bars are drawn adjacent to each other.
2-14
Frequency Polygon
A frequency polygon
also shows the shape
of a distribution and is
similar to a histogram.
It consists of line
segments connecting
the points formed by
the intersections of the
class midpoints and the
class frequencies.
2-15
Histogram Versus Frequency Polygon
Both provide a quick picture of the main characteristics of the
data (highs, lows, points of concentration, etc.)
The histogram has the advantage of depicting each class as a
rectangle, with the height of the rectangular bar representing
the number in each class.
The frequency polygon has an advantage over the histogram. It
allows us to compare directly two or more frequency
distributions.
2-16
Cumulative Frequency Distribution
Selling Prices
($ 000) Frequency
Less than
Class
Cumulative
Frequency
15 up to 18 8 Less than 15 0
Less than 18 8
18 up to 21 23 Less than 21 31
21 up to 24 17 Less than 24 48
24 up to 27 18 Less than 27 66
27 up to 30 8 Less than 30 74
30 up to 33 4 Less than 33 78
33 up to 36 2 Less than 36 80
Total 80
2-17
Cumulative Frequency Distribution
2-18
Cumulative Frequency Distribution
More than Class Cum. Freq.
More than 14 80 [80]
More than 17 72 [80-8]
More than 20 49 [72-23]
More than 23 32 [49-17]
More than 26 14 [32-18]
More than 29 06 [14-8]
More than 32 02 [6-4]
More than 35 00 [2-2]
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2-19
Cumulative Frequency Distribution
14 17 20 23 26 29 32 35
80
70
60
50
40
30
20
10
●
●
●
●
●
● ● ●
2-20
From the following data set prepare:
(a) a data array of ascending order,
(b) a frequency distribution with six classes
under exclusive and inclusive method, and
(c) a cumulative as well as a relative frequency
distributions
4.9, 3.4, 4.7,3.4, 5.5, 3.8, 4.0, 5.5, 4.1, 5.5,
4.1, 4.2, 4.3, 2.0, 4.7, 4.8, 4.9, 5.5, 3.8, 4.1
Example -1
2-21
Data array in ascending order:
2.0, 3.4, 3.4, 3.8, 3.8, 4.0, 4.1, 4.1, 4.1, 4.2,
4.3, 4.7, 4.7, 4.8, 4.9, 4.9, 5.5, 5.5, 5.5, 5.5
Solution: Example -1
2-22
Solution- Exclusive Method
Classes
Tally
Marks Frequency
Cumulative
Frequency
Relative
Frequency
2.0-2.6 | 1 1 0.05
2.6-3.2 - 0 1+0=1 0.00
3.2-3.8 || 2 1+2=3 0.10
3.8-4.4 ||||| ||| 8 3+8=11 0.40
4.4-5.0 ||||| 5 11+5=16 0.25
5.0-5.6 |||| 4 16+4=20 0.20
Total 20 20 1.0000
2k > n, 26 = 64 > 20
i= (H─L)/k= (5.5-2.0)/6=0.58. Round up the 10th decimal to 0.6
2-23
Solution- Inclusive Method
Classes
Tally
Marks Frequency
Cumulative
Frequency
Relative
Frequency
2.0-2.5 | 1 1 0.05
2.6-3.1 - 0 1+0=1 0.00
3.2-3.7 || 2 1+2=3 0.10
3.8-4.3 ||||| ||| 8 3+8=11 0.40
4.4-4.9 ||||| 5 11+5=16 0.25
5.0-5.5 |||| 4 16+4=20 0.20
Total 20 20 1.0000
2k > n, 26 = 64 > 20
i= (H─L)/k= (5.5-2.0)/6=0.58. Round up the 10th decimal to 0.6
2-24
The Orange County Transportation Commission is concerned
about the speed motorists are driving on a section of the main
highway. Here are the speeds of 45 motorists: (See the table in
the next page.)
Use these data to construct relative frequency distribution using
5 equal intervals and 11 equal intervals. The US Transport
Department (TD) reports that, nationally, no more than 10
percent motorists exceed 55 mph.
a) Do Orange County motorists follow the TD’s report about
national driving patterns?
b) Which distribution did you use to answer part (a)?
c) The US TD has determined that the safest speed for this
highway is more than 36 but less than 59 mph. What
distribution helped you answer this question?
Exercise Problem-2 (Levin and Rubin)
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2-25
Table:
Exercise Problem-2 (Levin and Rubin)
2-26
Solution: Inclusive Method
Classes Tally Marks Freq. Relative
Frequency
15-25 ||| 3 0.067
26-36 |||| 4 0.089
37-48 ||||| ||||| || 12 0.267
48-58 ||||| ||||| ||||| ||| 18 0.400
59-69 ||||| ||| 8 0.177
Total 45 1.0000
2k > n, 25 = 32 > 20 ???
i= (H─L)/k= (69-15)/5=10.8. Now round up to 11
2-27
Solution: Inclusive Method
Classes Tally Marks Frequency R. Frequency
15-19 ||| 3 .0667
20-24 - 0 .0000
25-29 || 2 .0444
30-34 || 2 .0444
35-39 ||||| 5 .1111
40-44 ||| 3 .0667
45-49 ||||| ||||| | 11 .2444
50-54 ||| 3 .0667
55-59 ||||| ||| 8 .1778
60-64 ||||| 5 .1111
65-69 ||| 3 .0667
Total 45 1.0000
2k > n, 211 = 2048 > 20; OK
i= (H─L)/k= (69-15)/11=4.91. Now round up to 5 2-28
a) No, we can see from either distribution that
more than 10% of the motorists drive at 55
mph or more. 5 intervals: over 17.77%; 11
intervals: 35.56%)
b) Either can be used; the 11-interval distribution
gives a more precise answer
c) The five-interval distribution shows that
66.67% of the motorists drive between 37 and
58 mph, inclusive.