Question :
| Number of Letters | 1 – 4 | 4 – 7 | 7 – 10 | 10 – 13 | 13 – 16 | 16 – 19 |
| Number of Surnames | 6 | 30 | 40 | 16 | 4 | 4 |
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames ? Also, find the modal size of the surnames.
Solution :
Calculation of median :
First, we prepare the following table to compute the median :
| Number of Letter | Number of surnames (frequency) | cumulative frequency |
| 1 – 4 | 6 | 6 |
| 4 – 7 | 30 | 36 |
| 7 – 10 | 40 | 76 |
| 10 – 13 | 16 | 92 |
| 13 – 16 | 4 | 96 |
| 16 – 19 | 4 | 100 |
We have : n = 100, so \(n\over 2\) = 50
The cumulative frequency just greater than 50 is 76 and the corresponding class is (7 – 10). Thus, (7 – 10) is the median class such that
\(n\over 2\) = 50, l = 7, cf = 36, f = 40 and h = 3.
Substituting these values in the formula,
Median = l + (\({n\over 2} – cf\over f\))\(\times \) h
= 7 + (\(50 – 26\over 40\))(3) = 7 + 1.05 = 8.05
Calculation of mean :
Number of letters
Mid-value (\(x_i\))
Frequency (\(f_i\))
\(f_ix_i\)
1 – 4
2.5
6
15
4 – 7
5.5
30
165
7 – 10
8.5
40
340
10 – 13
11.5
16
184
13 – 16
14.5
4
58
16 – 19
17.5
4
70
Total
100
832
Mean = \(\sum f_ix_i\over \sum f_i\) = \(832\over 100\) = 8.32
Calculation of mode :
The class (7 – 10) has the maximum frequency. Therefore, this is the modal class.
Here, l = 7, h = 3, \(f_1\) = 40, \(f_0\) = 30 and \(f_2\) = 16
Now, let us substitute these values in the formula
Mode = l + (\(f_1 – f_0\over 2f_1 – f_0 – f_2\))(h) = 7 + \(10\over 34\) \(\times\) 3 = 7 + 0.88 = 7.88
