In previous year, we learn the classification of data into grouped and ungrouped frequency distribution. And also learnt the representation of data in pictorial form.
In this chapter, we shall learn the procedure of solving the measures of central tendency of grouped data and the concept of cumulative frequency distribution.
The measures of central tendency is the value that tends to cluster around the middle value of the given data.
The commonly used measures of central tendency are:
- Mean(or average)
- Median
- Mode
We shall learn about the various methods of solving the given data using the measures of central tendency in the upcoming topics.
Mean - Ungrouped frequency distribution:
The mean of the ungrouped frequency distribution can be determined using the formula:
\(\overline X = \frac{f_1 x_1 + f_2 x_2 + ... + f_n x_n}{f_1 + f_2 + ... + f_n}\) \(= \frac{\sum_{i=1}^{n} f_i x_i}{\sum_{i=1}^{n} f_i}\)
Example:
The height(in \(cm\)) of \(20\) students in a classroom are:
|
Height
\(x_i\)
|
130
|
135
|
140
|
155
|
163
|
165
|
177
|
189
|
196
|
100
|
| Number of students\(f_i\) | 1 | 2 | 1 | 2 | 1 | 3 | 2 | 2 | 2 | 4 |
Find the mean height of the \(20\) students.
Solution:
To find the value of \(f_ix_i\), multiply the value of \(x\) and \(f\) of each entry.
Consider for the mark \(130\). That is, \(130 \times 1 = 130\)
Similarly, for the mark \(135\), we have \(135 \times 2 = 270\) and so on.
Tabulating these values, we get:
|
Marks
\(x_i\)
|
Frequency
\(f_i\)
|
\(f_ix_i\) |
| \(130\) | \(1\) | \(130\) |
| \(135\) | \(2\) | \(270\) |
| \(140\) | \(1\) | \(140\) |
| \(155\) | \(2\) | \(310\) |
| \(163\) | \(1\) | \(163\) |
| \(165\) | \(3\) | \(495\) |
| \(177\) | \(2\) | \(354\) |
| \(189\) | \(2\) | \(378\) |
| \(196\) | \(2\) | \(392\) |
| \(100\) | \(4\) | \(400\) |
| Total | \(\sum f_i = 20\) | \(\sum f_ix_i = 3032\) |
Substituting the known values in the above formula, we get:
Mean \(\overline X = \frac{3032}{20}\) \(= 151.6\)
Therefore, the mean of the given data is \(151.6\).
Mean - Grouped frequency distribution:
In most situations, we usually consider a very large amount of data(like population census) for a purposeful study. In such cases, we may find it difficult to write the data in ungrouped data.
Hence, to simplify our work, we need to convert the given ungrouped data into grouped data.
Let us convert the given height(in \(cm\)) of the students in the classroom into grouped frequency data.
\(100\), \(165\), \(189\), \(155\), \(140\), \(196\), \(165\), \(135\), \(135\), \(100\), \(100\), \(100\), \(155\), \(165\), \(196\), \(189\), \(177\), \(163\), \(177\), \(130\).
Consider the frequency distribution table.
| Height (in cm) | \(100 - 120\) | \(120 - 140\) | \(140 - 160\) | \(160 - 180\) | \(180 - 200\) |
| Students | \(4\) | \(3\) | \(3\) | \(6\) | \(4\) |
The above frequency table shows that the data are grouped in class intervals.
Consider the interval \(140 - 160\). There are \(3\) students in the heights between \(140 - 160\) metres. In grouped frequency, individual observations are not available. Thus, we need to determine the value that indicates the particular interval. This value is called a midpoint or class mark. The midpoint can be determined using the formula:
Midpoint \(= \frac{UCL + LCL}{2}\)
Where \(UCL\) is the upper class limit and \(LCL\) is the lower class limit.
Example:
Consider the interval \(140 - 160\). Let us find the midpoint of this interval.
Here, \(UCL = 160\) and \(LCL = 140\)
Midpoint of \(140 - 160\) is \(\frac{160 + 140}{2} =\) \(\frac{300}{2}\) \(= 150\)
Therefore, the midpoint of the interval \(140 - 160\) is \(150\).
The mean of a grouped frequency distribution can be determined using any one of the following methods.
- Direct method
- Assumed mean method
- Step deviation method