CONSTRUCTION OF A FREQUENCY DISTRIBUTION
Statistics and ProbabilityCONSTRUCTION OF A FREQUENCY DISTRIBUTION
In THIS Lecture, we will discuss the frequency distribution of a continuous variable & the graphical ways of representing data pertaining to a continuous variable i.e. histogram, frequency polygon and frequency curve.
You will recall that in Lecture No. 1, it was mentioned that a continuous variable takes values over a continuous interval (e.g. a normal Pakistani adult maleâ€™s height may lie anywhere between 5.25 feet and 6.5 feet). Hence, in such a situation, the method of constructing a frequency distribution is somewhat different from the one that was discussed in the last lecture.
EXAMPLE:
Suppose that the Environmental Protection Agency of a developed country performs extensive tests on all new car models in order to determine their mileage rating. Suppose that the following 30 measurements are obtained by conducting such tests on a particular new car model.
EPA MILEAGE RATINGS ON 30 CARS (MILES PER GALLON) | ||
36.3 | 42.1 | 44.9 |
30.1 | 37.5 | 32.9 |
40.5 | 40.0 | 40.2 |
36.2 | 35.6 | 35.9 |
38.5 | 38.8 | 38.6 |
36.3 | 38.4 | 40.5 |
41.0 | 39.0 | 37.0 |
37.0 | 36.7 | 37.1 |
37.1 | 34.8 | 33.9 |
39.9 | 38.1 | 39.8 |
EPA: Environmental Protection Agency There are a few steps in the construction of a frequency distribution for this type of a variable.
CONSTRUCTION OF A FREQUENCY DISTRIBUTION Step1
Identify the smallest and the largest measurements in the data set.
In our example: | |
Smallest value (X0) | = 30.1, |
Largest Value (Xm) | = 44.9, |
Step2 |
Find the range which is defined as the difference between the largest value and the smallest value In our example:
Range =Xmâ€“X0 = 44.9 â€“ 30.1 = 14.8
Let us now look at the graphical picture of what we have just computed.
30.1 44.9
30 35 40 45
14.8
(Range)
Step3
Decide on the number of classes into which the data are to be grouped. (By classes, we mean small subintervals of the total interval which, in this example, is 14.8 units long.)There are no hard and fast rules for this purpose. The decision will depend on the size of the data. When the data are sufficiently large, the number of classes is usually taken between 10 and 20.In this example, suppose that we decide to form 5 classes (as there are only 30 observations).
Step4
Divide the range by the chosen number of classes in order to obtain the approximate value of the class interval i.e. the width of our classes. Class interval is usually denoted by h. Hence, in this example Class interval = h = 14.8 / 5 = 2.96 Rounding the number 2.96, we obtain 3, and hence we take h = 3. This means that our big interval will be divided into small subintervals, each of which will be 3 units long.
Step5
Decide the lower class limit of the lowest class. Where should we start from? The answer is that we should start constructing our classes from a number equal to or slightly less than the smallest value in the data. In this example, smallest value = 30.1 So we may choose the lower class limit of the lowest class to be 30.0.
Step6
Determine the lower class limits of the successive classes by adding h = 3 successively. Hence, we obtain the following table:
Class Number | Lower Class Limit | |
1 | 30.0 | |
2 | 30.0 + 3 | = 33.0 |
3 | 33.0 + 3 | = 36.0 |
4 | 36.0 + 3 | = 39.0 |
5 | 39.0 + 3 | = 42.0 |
Step7
Determine the upper class limit of every class. The upper class limit of the highest class should cover the largest value in the data. It should be noted that the upper class limits will also have a difference of h between them. Hence, we obtain the upper class limits that are visible in the third column of the following table.
Class Number | Lower Class Limit | Upper Class Limit |
1 | 30.0 | 32.9 |
2 | 30.0 + 3 = 33.0 | 32.9 + 3 = 35.9 |
3 | 33.0 + 3 = 36.0 | 35.9 + 3 = 38.9 |
4 | 36.0 + 3 = 39.0 | 38.9 + 3 = 41.9 |
5 | 39.0 + 3 = 42.0 | 41.9 + 3 = 44.9 |
Hence we obtain the following classes:
Classes |
30.0 â€“ 32.9 |
33.0 â€“ 35.9 |
36.0 â€“ 38.9 |
39.0 â€“ 41.9 |
42.0 â€“ 44.9 |
The question arises: why did we not write 33 instead of 32.9? Why did we not write 36 instead of 35.9? and so on.
The reason is that if we wrote 30 to 33 and then 33 to 36, we would have trouble when tallying our data into these classes. Where should I put the value 33? Should I put it in the first class, or should I put it in the second class? By writing 30.0 to 32.9 and 33.0 to 35.9, we avoid this problem. And the point to be noted is that the class interval is still 3, and not 2.9 as it appears to be. This point will be better understood when we discuss the concept of class boundaries â€¦ which will come a little later in todayâ€™s lecture.
Step8
After forming the classes, distribute the data into the appropriate classes and find the frequency of each class, in this example:
Class | Tally | Frequency |
30.0 â€“ 32.9 | || | 2 |
33.0 â€“ 35.9 | ||| | 4 |
36.0 â€“ 38.9 | |||| |||| |||| | 14 |
39.0 â€“ 41.9 | |||| ||| | 8 |
42.0 â€“ 44.9 | || | 2 |
Total | 30 |
This is a simple example of the frequency distribution of a continuous or, in other words, measurable variable.
CLASS BOUNDARIES:
The true class limits of a class are known as its class boundaries. In this example: It should be noted that the difference between the upper class boundary and the lower class boundary of any class is equal to the class interval h = 3.
Class Limit | Class Boundaries | Frequency |
30.0 â€“ 32.9 | 29.95 â€“ 32.95 | 2 |
33.0 â€“ 35.9 | 32.95 â€“ 35.95 | 4 |
36.0 â€“ 38.9 | 35.95 â€“ 38.95 | 14 |
39.0 â€“ 41.9 | 38.95 â€“ 41.95 | 8 |
42.0 â€“ 44.9 | 41.95 â€“ 44.95 | 2 |
Total | 30 |
32.95 minus 29.95 is equal to 3, 35.95 minus 32.95 is equal to 3, and so on. A key point in this entire discussion is that the class boundaries should be taken up to one decimal place more than the given data. In this way, the possibility of an observation falling exactly on the boundary is avoided. (The observed value will either be greater than or less than a particular boundary and hence will conveniently fall in its appropriate class).Next, we consider the concept of the relative frequency distribution and the percentage frequency distribution. Next, we consider the concept of the relative frequency distribution and the percentage frequency distribution. This concept has already been discussed when we considered the frequency distribution of a discrete variable. Dividing each frequency of a frequency distribution by the total number of observations, we obtain the relative frequency distribution. Multiplying each relative frequency by 100, we obtain the percentage of frequency distribution. In this way, we obtain the relative frequencies and the percentage frequencies shown below
Class Limit
30.0 â€“ 32.9
33.0 â€“ 35.9
36.0 â€“ 38.9
39.0 â€“ 41.9
42.0 â€“ 44.9
Frequency
2
4
14
8
2
30
Relative Frequency
2/30 = 0.067
4/30 = 0.133
14/30 = 0.467
8/30 = 0.267
2/30 = 0.067
%age Frequency
6.7
13.3
4.67
26.7
6.7
The term â€˜relative frequenciesâ€™ simply means that we are considering the frequencies of the various classes relative to the total number of observations. The advantage of constructing a relative frequency distribution is that comparison is possible between two sets of data having similar classes. For example, suppose that the Environment Protection Agency perform tests on two car models A and B, and obtains the frequency distributions shown below:
MILEAGE | FREQUENCY | |
Model A | Model B | |
30.0 â€“ 32.9 | 2 | 7 |
33.0 â€“ 35.9 | 4 | 10 |
36.0 â€“ 38.9 | 14 | 16 |
39.0 â€“ 41.9 | 8 | 9 |
42.0 â€“ 44.9 | 2 | 8 |
30 | 50 |
In order to be able to compare the performance of the two car models, we construct the relative frequency distributions in the percentage form:
MILEAGE | Model A | Model B |
30.032.9 | 2/30 x 100 = 6.7 | 7/50 x 100 = 14 |
33.035.9 | 4/30 x 100 = 13.3 | 10/50 x 100 = 20 |
36.038.9 | 14/30 x 100 = 46.7 | 16/50 x 100 = 32 |
39.041.9 | 8/30 x 100 = 26.7 | 9/50 x 100 = 18 |
42.044.9 | 2/30 x 100 = 6.7 | 8/50 x 100 = 16 |
From the table it is clear that whereas 6.7% of the cars of model A fall in the mileage group 42.0 to 44.9, as many as 16% of the cars of model B fall in this group. Other comparisons can similarly be made. Let us now turn to the visual representation of a continuous frequency distribution. In this context, we will discuss three different types of graphs i.e. the histogram, the frequency polygon, and the frequency curve.
HISTOGRAM:
A histogram consists of a set of adjacent rectangles whose bases are marked off by class boundaries along the Xaxis, and whose heights are proportional to the frequencies associated with the respective classes. It will be recalled that, in the last lecture, we were considering the mileage ratings of the cars that had been inspected by the Environment Protection Agency. Our frequency table came out as shown below:
Class Limit | Class Boundaries | Frequency |
30.0 â€“ 32.9 | 29.95 â€“ 32.95 | 2 |
33.0 â€“ 35.9 | 32.95 â€“ 35.95 | 4 |
36.0 â€“ 38.9 | 35.95 â€“ 38.95 | 14 |
39.0 â€“ 41.9 | 38.95 â€“ 41.95 | 8 |
42.0 â€“ 44.9 | 41.95 â€“ 44.95 | 2 |
Total | 30 |
In accordance with the procedure that I just mentioned, we need to take the class boundaries along the X axis We obtain
29.95 32.95 35.95 38.95 41.95 44.95 Miles pergallon
Now, as seen in the frequency table, the frequency of the first class is 2. As such, we will draw a rectangle of height equal to 2 units and obtain the following figure:
Y
14 12
10
Number of Cars
8 6 4 2 0 Miles per gallon
The frequency of the second class is 4. Hence we draw a rectangle of height equal to 4 units against the second class, and thus obtain the following situation:
Y 14
Number of Cars
12
10 8 6 4 2 0
Miles per gallon
The frequency of the third class is 14. Hence we draw a rectangle of height equal to 14 units against the third class, and thus obtain the following picture:
Y 14
Number of Cars
12
10 8 6 4 2 0
Miles per gallon
Number of Cars
8 6 4 2 0
Miles per gallon
This diagram is known as the histogram, and it gives an indication of the overall pattern of our frequency distribution.
FREQUENCY POLYGON:
A frequency polygon is obtained by plotting the class frequencies against the midpoints of the classes, and connecting the points so obtained by straight line segments. In our example of the EPA mileage ratings, the classes are
Class Boundaries |
29.95 â€“ 32.95 |
32.95 â€“ 35.95 |
35.95 â€“ 38.95 |
38.95 â€“ 41.95 |
41.95 â€“ 44.95 |
These midpoints are denoted by X. Now let us add two classes to my frequency table, one class in the very beginning, and one class at the very end.
Class Boundaries | Mid Point (X) | Frequency (f) |
26.95 â€“ 29.95 | 28.45 | |
29.95 â€“ 32.95 | 31.45 | 2 |
32.95 â€“ 35.95 | 34.45 | 4 |
35.95 â€“ 38.95 | 37.45 | 14 |
38.95 â€“ 41.95 | 40.45 | 8 |
41.95 â€“ 44.95 | 43.45 | 2 |
44.95 â€“ 47.95 | 46.45 |
The frequency of each of these two classes is 0, as in our data set, no value falls in these classes.
Class Boundaries | Mid Point (X) | Frequency (f) |
26.95 â€“ 29.95 | 28.45 | 0 |
29.95 â€“ 32.95 | 31.45 | 2 |
32.95 â€“ 35.95 | 34.45 | 4 |
35.95 â€“ 38.95 | 37.45 | 14 |
38.95 â€“ 41.95 | 40.45 | 8 |
41.95 â€“ 44.95 | 43.45 | 2 |
44.95 â€“ 47.95 | 46.45 | 0 |
Now, in order to construct the frequency polygon, the midpoints of the classes are taken along the Xaxis and the frequencies along the Yaxis, as shown below:
Y 14
Number of Cars
12
10 8 6 4 2 X0
Miles per gallon
It is wellknown that the term â€˜polygonâ€™ implies a manysided closed figure. As such, we want our frequency polygon to be a closed figure. This is exactly the reason why we added two classes to our table, each having zero frequency. Because of the frequency being zero, the line segment touches the Xaxis both at the beginning and at the end, and our figure becomes a closed figure.
16 14
2 0
Miles per gallon
And since this graph is touching the Xaxis, hence it cannot be called a frequency polygon (because it is not a closed figure)!
FREQUENCY CURVE: When the frequency polygon is smoothed, we obtain what may be called the frequency curve.
In our example:
Y 16
Number of Cars
14 12 10 8 6 4 2 0
Miles per gallon
March 28th, 2011 at 7:45 pm
very informative,, lectures are well detailed and easy to understand