100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows: 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.
Step-by-Step Solution
Key Concept: For grouped data, median, mean and mode are obtained using the formulas:<br>- Median: \(L + \frac{\frac{N}{2} - C_f}{f}\,h\)<br>- Mean: \(\bar{x}=\frac{\sum f_i x_i}{N}\) where \(x_i\) are class mid‑points.<br>- Mode (grouped): \(L + \frac{f_1-f_0}{(2f_1-f_0-f_2)}\,h\) where \(f_1\) is the frequency of the modal class, \(f_0\) and \(f_2\) are the frequencies of the preceding and succeeding classes respectively.
1. Tabulate the data
| Class (letters) | Frequency (f) | Mid‑point (x) |
|-----------------|--------------|--------------|
| 1 – 4 | 6 | 2.5 |
| 4 – 7 | 30 | 5.5 |
| 7 – 10 | 40 | 8.5 |
| 10 – 13 | 16 | 11.5 |
| 13 – 16 | 4 | 14.5 |
| 16 – 19 | 4 | 17.5 |
Total \(N = 100\).
2. Median
- Cumulative frequencies: 6, 36, 76, 92, 96, 100.
- Position of median = \(\frac{N}{2}=50\).
- The 50th observation lies in the class 7 – 10 (cumulative just exceeds 50).
- \(L = 7\) (lower limit of median class), \(h = 3\) (class width), \(f = 40\) (frequency of median class), \(C_f = 36\) (cumulative frequency before median class).
- \[\text{Median}= L+\frac{\frac{N}{2}-C_f}{f}\,h = 7+\frac{50-36}{40}\times3 = 7+\frac{14}{40}\times3 = 7+1.05 = 8.05\]
Hence, the median number of letters \(\approx 8.05\).
3. Mean
- Compute \(\sum f_i x_i\):
\(6\times2.5 = 15\)
\(30\times5.5 = 165\)
\(40\times8.5 = 340\)
\(16\times11.5 = 184\)
\(4\times14.5 = 58\)
\(4\times17.5 = 70\)
\(\sum f_i x_i = 15+165+340+184+58+70 = 832\).
- Mean \(\bar{x}=\frac{\sum f_i x_i}{N}=\frac{832}{100}=8.32\).
Hence, the average (mean) number of letters is \(8.32\).
4. Mode (grouped data)
- The modal class is the one with highest frequency: 7 – 10 (frequency \(f_1=40\)).
- Frequencies of adjacent classes: \(f_0 = 30\) (previous class), \(f_2 = 16\) (next class).
- Using the modal formula:
\[\text{Mode}= L+\frac{f_1-f_0}{(2f_1-f_0-f_2)}\,h\]
\[= 7+\frac{40-30}{(2\times40-30-16)}\times3 = 7+\frac{10}{34}\times3\]
\[= 7+0.2941\times3 = 7+0.8823 \approx 7.88\]
Hence, the modal size of the surnames is about \(7.88\) letters.
Answers
- Median \(\approx 8.05\) letters
- Mean \(= 8.32\) letters
- Mode \(\approx 7.88\) letters
Correct Answer: Median ≈ 8.05 letters, Mean = 8.32 letters, Mode ≈ 7.88 letters