9th Class Mathematics Statistics Classification of Data

         Classification of Data

In order to tabulate a large number of data, we use the frequency distribution table. Frequency distribution is of two types:

1.       Discrete Frequency Distribution

2.       Continuous Frequency Distribution

In discrete frequency distribution method the frequency distribution is carried out with the help of raw data using the tally marks. But in continuous frequency distribution the data is divided into small groups of class interval and corresponding frequency is found. The frequency of a data is defined as the number of times a data is repeated in the given collection of data. The small groups into which the given data is divided is known as its class interval. The difference between the upper limit and lower limit of a class interval is known as its class size.

The class mark is defined as the average of the upper limit and lower limit of a class interval. The cumulative frequency is defined as the sum of all previous frequencies of the class interval.  

Here, 0 - 5, 5 - 10, ---are class intervals.

The difference 5 - 0 = 5 is the class size.

The average of the class interval is called class mark i.e. \[\frac{0+5}{2}=\frac{5}{2}=2.5\]

The cumulative frequency of class 5 - 10 is 4 + 10 = 14

There are various methods of representation of data. It can be represented in the form of bar graph. Histogram, ogive curve, frequency polygon curve or pie chart.

            Frequency Distribution Table

The frequency distribution table for a certain given data can be as shown below: It consists of class interval, tally marks and frequency.

            Frequency Distribution Table for an Ungrouped Data    

  Construct a frequency distribution table for the following data: \[\text{5},\text{1},\text{ 3},\text{ 4},\text{ 2},\text{1},\text{ 3},\text{ 5},\text{ 4},\text{ 2},\text{1},\text{ 5},\text{1},\text{ 3},\text{ 2},\text{1},\text{ 5},\text{ 3},\text{ 3},\text{ 2}\]

Solution:

  The following are the marks obtained by 50 students in mathematics in their previous examination held in their school. Prepare a frequency distribution table for the data.

45, 68, 41, 87, 61, 44, 67, 30, 54, 8, 39, 60, 37, 50, 19, 6, 42, 29, 32, 61, 25, 77, 62, 98, 47, 36,15, 40, 9, 25, 34, 50, 61, 75, 51, 96, 20, 13, 18, 35, 43, 88, 25, 95, 68, 81, 29, 41, 45, 87

Solution:

To decide the length of the class interval and to take all the scores given in the problem. We find the largest value and the smallest value from the given marks scored by the various students. We can do this by merely going through all the scores. In the data given above the largest value is 98 and the smallest value is 8.  

Difference = Largest value - Smallest value                

                       \[=\text{98}-\text{8 }=\text{ 9}0\]

Since the difference between the largest and the smallest value is 90, so if we take class intervals of size 5 then it will be as shown : 0 - 5, 5 - 10, 10 - 15, ..., 95 100. On the other hand if we take the class interval of size 10 then it will be as shown: 0 - 10, 10 - 20, 20 - 30,...... 90 - 100. Or if we take the class interval of class size 15 then it becomes: 0 - 15, 15 - 30, ....,90 - 105. It is advisable not to over reduce the number of class intervals. If the class intervals are mentioned in the problem then we proceed as given in the problem.

For the above proceed we prepare the frequency distribution table by taking class intervals as: 0 - 10, 10 - 20, 20 - 30, 30 - 40, 40 - 50, 50 - 60, 60 - 70, 70 - 80, 80-90 and 90-100.  

            Frequency Distribution Table of the Marks taken by 50 Students in a Mathematics Test