One table, many questions
A class register lists the learners absent on each of 20 days. The numbers are invented. Each day's number is one reading. What is the usual number absent?
| Days | Absent each day |
|---|---|
| 1 to 5 | 2, 0, 1, 4, 0 |
| 6 to 10 | 1, 3, 0, 2, 1 |
| 11 to 15 | 0, 1, 0, 2, 1 |
| 16 to 20 | 0, 1, 3, 0, 2 |
Loading lab…
Step 1 of 4
- Five columns, one for each value that appears.
- Column "0" has 7 marks. Its bar height is 7.
- Columns "1" and "2": heights 6 and 4.
- Columns "3" and "4": heights 2 and 1. All twenty counted.
The rule
List each value once. Write beside it how many times it appears.
That count is its frequency. The frequencies add up to 20: 7 + 6 + 4 + 2 + 1 = 20.
| Learners absent | Days |
|---|---|
| 0 | 7 |
| 1 | 6 |
| 2 | 4 |
| 3 | 2 |
| 4 | 1 |
Find every average from this table.
The mean shares the total equally. The median is the middle value in order. The mode is the value that appears most. The 20 readings add to 0 + 6 + 8 + 6 + 4 = 24.
Put the 20 readings in order. The middle two are the 10th and 11th. Seven 0s fill places 1 to 7. Six 1s fill places 8 to 13.
| Average | Working | Answer |
|---|---|---|
| Mean | (0×7 + 1×6 + 2×4 + 3×2 + 4×1) ÷ 20 = 24 ÷ 20 | 1.2 |
| Median | The 10th and 11th of the 20 readings, in order, are both 1 | 1 |
| Mode | The value with the most days | 0 |
Watch out!
Do not average the five values 0, 1, 2, 3, 4. That gives 2. Dividing by 5 ignores the frequencies.