Mode:
Let us recall what we learnt in class \(IX\) about the mode of ungrouped frequency distribution.
Mode is defined as the number which occurs most frequently in the given set of data. That is, the observation having the maximum number of frequencies is called mode.
Mode - Ungrouped frequency distribution:
The mode of an ungrouped frequency distribution can be determined if the value of an item having the maximum number of frequencies.
Example:
Find the mode of the following data.
| Height (in cm) | \(145\) | \(150\) | \(162\) | \(132\) | \(138\) |
| Number of students | \(5\) | \(6\) | \(12\) | \(25\) | \(3\) |
Solution:
Here, the number of students is the frequency. The maximum frequency is \(25\).
Therefore, the height having the maximum frequency is \(132\) \(cm\).
Hence, \(132\) is the mode.
Mode - Grouped frequency dsitribution:
The mode of the grouped frequency distribution can be determined using the formula:
Mode \(= l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h\)
The class interval with maximum frequency is called the modal class.
Where \(l\) is the lower limit of the modal class,
\(f_1\) is the frequency of the modal class,
\(f_0\) is the frequency of the class preceding the modal class,
\(f_2\) is the frequency of the class succeeding the modal class, and
\(h\) is the width of the class interval.
Example:
Find the mode of the following data:
| Class interval | \(130 - 140\) | \(140 - 150\) | \(150 - 160\) | \(160 - 170\) | \(170 - 180\) |
| Frequency | \(5\) | \(36\) | \(14\) | \(28\) | \(1\) |
Solution:
The maximum frequency is \(36\), and the modal class is \(140 - 150\).
The mode of the grouped frequency distribution can be determined using the formula:
Mode \(= l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h\)
Here, \(l = 140\), \(f_1 = 36\), \(f_0 = 5\), \(f_2 = 14\), \(h = 10\)
Substituting the known values in the above formula, we have;
Mode \(= 140 + \left(\frac{36 - 5}{2(36) - 5 - 14} \right) \times 10\)
\(= 140 + \left(\frac{36 - 5}{72 - 5 - 14} \right) \times 10\)
\(= 140 + (\frac{31}{53}) \times 10\)
\(= 140 + 0.585 \times 10\)
\(= 140 + 5.85\)
\(= 145.85\)
Therefore, the mode of the given data is \(145.85\).