frequency distributions twenty five medtech students were given a blood test to determine their...
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FREQUENCY DISTRIBUTIONS
Twenty five medtech students were given a blood test to determine their blood type. The data set is:
A B B AB O
O O B AB B
B B O A O
A O O O AB
AB A O B A
As we can see, this is a little messy, so we decide to organize this into a frequency distribution:
Class Frequency
A FREQUENCY DISTRIBUTION is a table that shows the partition of data into classes or intervals and how many data values are in each class.
Essentially, a frequency distribution must have at least two columns: one for the classes or data groupings, and another for the frequency or the no. of data values belonging to the respective class.
Additional columns can be included when necessary or helpful.
Example: (Categorical frequency distribution)
Twenty five medtech students were given a blood test to determine their blood type. The data set is:
A B B AB O
O O B AB B
B B O A O
A O O O AB
AB A O B A
What is the variable in this study?
Blood type of a student.
What kind of variable is this?
Qualitative.
How do we construct a frequency dist. for this data set?
(1) Draw a frequency distribution table, including a column for tallies (which is helpful in counting the frequencies).
Class Tallies Frequency
(2) Identify the classes. Note that, for qualitative variables, a data value is a category by itself. Therefore, the classes are the data values: A, B, AB, O
A
B
AB
O
(3) Tally the data: one tick for a class if a particular data value belongs to it; and count the respective frequencies.
|||||
||||| ||||||| ||||
||||
5
79
4
This is called a categorical frequency distribution because the variable we are tabulating is categorical (qualitative).
(4) Another column for the relative frequencies of the classes is optional, but sometimes necessary.
Class Tallies Frequency
A ||||| 5
B ||||| || 7
AB ||||| |||| 9
O |||| 4
Rel. freq.
class f requency
relative f req.total no. of data
The relative frequency of a class is the percentage of the population which the class occupies.
0.20
0.28
0.36
0.16
Example: (Grouped frequency distribution)
A survey of the 50 wealthiest people by the Forbes Magazine yields the following data on the ages (in years, as of 2012) of these billionaires:
49 57 38 73 81 74 59 76 65 69
54 56 69 68 78 65 85 49 69 61
48 81 68 37 43 78 82 43 64 67
52 56 81 77 79 85 40 85 59 80
60 71 57 61 69 61 83 90 87 74
What is the variable in this study?
Age
What kind of variable is this?
Quantitative
How do we construct the frequency table for this data set?
(1) Draw a frequency distribution table, including a column for tallies (which is helpful in counting the frequencies).
Class Tallies Frequency
The data in this example is quantitative (numerical), so the classes must be intervals. (ex.: 34 – 40, 41 – 47, etc.)
For the class labelled n1 - n2,n1 is called the lower class limit, which is the lowest
data value that can be tallied in the class n1 - n2; andn2 is called the upper class limit, which is the highest
data value that can be tallied in the class n1 - n2
Also, there must always be 5 to 20 interval classes for a better display of the distribution of data.
As a rule, both class limits must have the same number of decimal places (no. of digits after the decimal point) as the data. For example,
if the data are: 6.2, 12.8. 10.5, … (one decimal place) the interval classes must be like: 5.3 – 8.3, 8.4 – 11.4 …
Always take note of the no. of decimal places in the data!
(2) Compute the range of the data.
highest lowest range data value data value
In the example, range = 90 – 37
= 53
(3) Choose the desired number of classes (5 – 20) and compute the class width
range
class widthno. of classes
The class width is the “gap” between any one lower class limit to the next lower class limit.
Let us choose to have 8 classes.
Then the class width is: 53 / 8
= 6.625The class width must also have the same number of
decimal places as the data. If the computed class width has a long decimal portion, keep the number of decimal places as that of the data (removing the other decimal places) and add 1 to the last digit.
Since our data have no (0) decimal places, the class width must also have no decimal places.
So, class width = 6.625 ® 6 (keeping 0 dec. places, removing
others)® 7 (adding 1 to the last digit ‘6’)
So, class width = 7
(4) For the first lower class limit, select any number close to the lowest data value, and add the class width the same number of times as the number of classes. These will be the lower limits of the classes.
In the example, the smallest data value is 38.
We choose the first lower class limit to be 35.
Then we add the class width (7) to the first lower class limit (35), until we get 8 classes.
Class Tallies Frequency
35 -
42 - 49 -
56 -
63 -
70 -
77 -
84 -
(5) For the corresponding upper class limit, we simply subtract 1 from last digit of the next lower class limit.
Class Tallies Frequency
35
42
49
56
63
70
77
84
- 41
- 48
- 55
- 62
- 69
- 76
- 83
The class limits must be mutually exclusive (i.e., they don’t overlap), so that each data value will belong to exactly one class only. Also, note that the classes are equal in width.
- 90
(6) Since the classes are intervals, we must insert a second column for the so-called class boundaries.
Class Boundaries
Tallies Frequency
35 - 41
42 – 48
49 – 55
56 – 62
63 – 69
70 – 76
77 - 83
84 - 90
As we can see, the classes have a gap of 1 unit (35-41, 42-48). The class boundaries simply connect these classes when we construct the graph of the frequency distribution.
The class boundaries are also intervals (n1 - n2), like the class limits, having one more decimal place than that of the data, connecting the classes halfway between adjacent class limits.
lower lower class boundary 0.0...5class limit
upper upper class boundary 0.0...5class limit
The ± 0.0..5 depends on the no. of decimal places in the data. Class boundaries must have one more decimal place, right?
If the data has no decimal places, use ± 0.5
If the data has 1 decimal place, use ± 0.05
If the data has 2 decimal places, use ± 0.005
For the class 35 – 41, (no decimal places)
lower class boundary = 35 – 0.5 = 34.5
upper class boundary = 41 + 0.5 = 41.5
So the corresponding class boundaries is 34.5 – 41.5
Class Boundaries
Tallies Frequency
35 - 41 34.5 – 41.5
42 – 48
49 – 55
56 – 62
63 – 69
70 – 76
77 - 83
84 - 90
Computing the class boundaries for the other classes
41.5 – 48.5 48.5 – 55.5 55.5 – 62.5 62.5 – 69.5 69.5 – 76.5 76.5 – 83.5 83.5 – 90.5
(7) Tally the data: one tick for a class if a particular data value belongs to it; and count the respective frequencies.
Class limits
ClassBoundari
es
Tally Frequency
35 – 41 34.5 – 41.5
||| 3
42 – 48 41.5 – 48.5
||| 3
49 – 55 48.5 – 55.5
|||| 4
56 – 62 55.5 – 62.5
||||| ||||| 10
63 – 69 62.5 – 69.5
||||| ||||| 10
70 – 76 69.5 – 76.5
||||| 5
77 – 83 76.5 – 83.5
||||| ||||| 10
84 – 90 83.5 – 90.5
||||| 5
This is called a grouped frequency distribution because the classes are not categories but groups or intervals.
Example: (Grouped frequency distribution)
In a study of one-way commuting distance of FEU students, a random sample of 60 students gives the ff. data (in kms):
13.2
47.8
10.5
3.7 16.4
20.1
17.9
40.3
4.5 2.8
7 25.3
8 21.4
19.6
15.1
3.2 17.8
14.2
6.3
12.2
45.8
1.4 8.2 4.1 16.7
11.2
18.5
23.2
12.4
6 2.5 15.2
13 7 15.6
46.2
12.5
9.3 18.7
34.2
13.5
41.6
28.1
36 17.2
24 27.6
29.5
9.2
14.6
26.1
10.6
24 37 31.2
8.2 16.8
12.2
16
Make a frequency distribution with 6 classes and initial lower class limit = 1.0
First, take note that the data has 1 decimal place!
(1) Compute the range.
range = 47.8 – 1.4 = 46.4
(2) Compute the class width.
class width = 46.4/6 = 7.73
® 7.7 (keeping 1 dec. place, removing others)® 7.8 (adding 1 to the last digit ‘7’.)
(3) Add the class width (7.8) to the initial lower class limit (1.0) until we get the no. of classes (6).
1.0
1.0 + 7.8 = 8.88.8 + 7.8 = 16.6
16.6 + 7.8 = 24.424.4 + 7.8 = 32.232.2 + 7.8 = 40.0
These numbers will be the lower class limits of the 6 classes.
Class Boundaries
Tallies Frequency
1.0 –
8.8 –
16.6 –
24.4 –
32.2 –
40.0 –
(4) For the corresponding upper class limit, we simply subtract 1 from last digit of the next lower class limit.
8.7
16.5 24.3 32.1 39.9 47.7
(5) Compute the class boundaries for each class.
For the class 1.0 – 8.7, (1 decimal place)
lower class boundary = 1 – 0.05 = 0.95
upper class boundary = 8.7 + 0.05 = 8.75
So the corresponding class boundaries is 0.95 – 8.75
0.95 – 8.75 8.75 – 16.55 16.55 – 24.35 24.35 – 32.15 32.15 – 39.95 39.95 – 47.85
(6) Tally the data: one tick for a class if a particular data value belongs to it; and count the respective frequencies.
Class Boundaries Tallies Frequency
1.0 – 8.7 0.95 – 8.75 14
8.8 – 16.5 8.75 – 16.55 19
16.6 – 24.3
16.55 – 24.35
13
24.4 – 32.1 24.35 – 32.15
6
32.2 – 39.9 32.15 – 39.95
3
40.0 – 47.8 39.95 – 47.85
5