Question: Complete the frequency distribution table and then find the mean, median, and mode of the grouped data.
Class limits
[Scores] f x̄m fx̄m lcb < cf
41 - 45 1
36 - 40 8
31 - 35 8
26 - 30 14
21 - 25 7
16 - 20 2
n = _____
Σfx̄m = _____
1. **State the problem:**
We are given grouped data with class limits and frequencies $f$. We need to complete the frequency distribution table by finding the class midpoints $\bar{x}_m$, the product $f \times \bar{x}_m$, the lower class boundaries (lcb), and the cumulative frequencies $< cf$. Then, find the mean, median, and mode of the grouped data.
2. **Calculate class midpoints $\bar{x}_m$:**
The midpoint of each class is the average of the lower and upper class limits.
$$\bar{x}_m = \frac{\text{Lower limit} + \text{Upper limit}}{2}$$
| Class limits | $f$ | $\bar{x}_m$ |
|--------------|-----|-------------|
| 41 - 45 | 1 | $\frac{41+45}{2} = 43$ |
| 36 - 40 | 8 | $\frac{36+40}{2} = 38$ |
| 31 - 35 | 8 | $\frac{31+35}{2} = 33$ |
| 26 - 30 | 14 | $\frac{26+30}{2} = 28$ |
| 21 - 25 | 7 | $\frac{21+25}{2} = 23$ |
| 16 - 20 | 2 | $\frac{16+20}{2} = 18$ |
3. **Calculate $f \times \bar{x}_m$:**
| Class limits | $f$ | $\bar{x}_m$ | $f \times \bar{x}_m$ |
|--------------|-----|-------------|-----------------------|
| 41 - 45 | 1 | 43 | $1 \times 43 = 43$ |
| 36 - 40 | 8 | 38 | $8 \times 38 = 304$ |
| 31 - 35 | 8 | 33 | $8 \times 33 = 264$ |
| 26 - 30 | 14 | 28 | $14 \times 28 = 392$ |
| 21 - 25 | 7 | 23 | $7 \times 23 = 161$ |
| 16 - 20 | 2 | 18 | $2 \times 18 = 36$ |
4. **Calculate total frequency $n$ and total $\Sigma f \bar{x}_m$:**
$$n = 1 + 8 + 8 + 14 + 7 + 2 = 40$$
$$\Sigma f \bar{x}_m = 43 + 304 + 264 + 392 + 161 + 36 = 1200$$
5. **Calculate lower class boundaries (lcb):**
Lower class boundary is the midpoint between the lower limit of a class and the upper limit of the previous class.
Since classes are continuous and adjacent, the gap between classes is 1, so:
$$\text{lcb of first class} = 15.5$$ (one half less than 16)
Then add 5 for each next class:
| Class limits | lcb |
|--------------|-------|
| 16 - 20 | 15.5 |
| 21 - 25 | 20.5 |
| 26 - 30 | 25.5 |
| 31 - 35 | 30.5 |
| 36 - 40 | 35.5 |
| 41 - 45 | 40.5 |
6. **Calculate cumulative frequency $< cf$:**
| Class limits | $f$ | $< cf$ |
|--------------|-----|--------|
| 16 - 20 | 2 | 2 |
| 21 - 25 | 7 | 2 + 7 = 9 |
| 26 - 30 | 14 | 9 + 14 = 23 |
| 31 - 35 | 8 | 23 + 8 = 31 |
| 36 - 40 | 8 | 31 + 8 = 39 |
| 41 - 45 | 1 | 39 + 1 = 40 |
7. **Calculate the mean:**
$$\text{Mean} = \frac{\Sigma f \bar{x}_m}{n} = \frac{1200}{40} = 30$$
8. **Calculate the median:**
Median class is the class where cumulative frequency $< cf$ just exceeds $\frac{n}{2} = 20$.
From the table, median class is 26 - 30 (since $< cf = 23$).
Median formula:
$$\text{Median} = l + \left(\frac{\frac{n}{2} - F}{f_m}\right) \times c$$
Where:
- $l = 25.5$ (lower class boundary of median class)
- $F = 9$ (cumulative frequency before median class)
- $f_m = 14$ (frequency of median class)
- $c = 5$ (class width)
Calculate:
$$\text{Median} = 25.5 + \left(\frac{20 - 9}{14}\right) \times 5 = 25.5 + \frac{11}{14} \times 5 = 25.5 + 3.93 = 29.43$$
9. **Calculate the mode:**
Mode class is the class with highest frequency, which is 26 - 30 with $f = 14$.
Mode formula:
$$\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times c$$
Where:
- $l = 25.5$ (lower class boundary of modal class)
- $f_1 = 14$ (frequency of modal class)
- $f_0 = 7$ (frequency of class before modal class)
- $f_2 = 8$ (frequency of class after modal class)
- $c = 5$ (class width)
Calculate:
$$\text{Mode} = 25.5 + \left(\frac{14 - 7}{2 \times 14 - 7 - 8}\right) \times 5 = 25.5 + \left(\frac{7}{28 - 15}\right) \times 5 = 25.5 + \frac{7}{13} \times 5 = 25.5 + 2.69 = 28.19$$
**Final answers:**
- $n = 40$
- $\Sigma f \bar{x}_m = 1200$
- Mean = $30$
- Median = $29.43$
- Mode = $28.19$