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Book Free DemoThe 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 \(145.85\).
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