Find the missing frequency
Show on a big screenA table has one frequency hidden. The mean is given. Find the hidden number.
- Frequencies 2, 5, f, 3 and 6. The classes are 0-10 to 40-50. The mean is 28.
- The class marks are 5, 15, 25, 35 and 45. So
Σfx = 460 + 25fandn = 16 + f. - Mean = Σfx ÷ n. So
(460 + 25f) ÷ (16 + f) = 28. 460 + 25f = 448 + 28f. So12 = 3f, andf = 4.
The rule
Write the mean with f in it. Set it equal to the given mean. Solve for f.
Call the hidden frequency f. It is a number we do not know yet. In the table, x is the class mark. The letter f is in two places. It is in the total of f × x. It is also in n, the total frequency.
| Class | f | x | f × x |
|---|---|---|---|
| 0-10 | 2 | 5 | 10 |
| 10-20 | 5 | 15 | 75 |
| 20-30 | f | 25 | 25f |
| 30-40 | 3 | 35 | 105 |
| 40-50 | 6 | 45 | 270 |
| Total | 16 + f | 460 + 25f |
Solve it. Multiply both sides by 16 + f. Then 460 + 25f = 28 × (16 + f). Now 28 × 16 = 448, so 460 + 25f = 448 + 28f. Take 25f and 448 from both sides. That gives 12 = 3f, so f = 4.
Check with f = 4. Add: 16 + 4 = 20, and 460 + 25 × 4 = 560. Then n = 20 and Σfx = 560. And 560 ÷ 20 = 28.
More frequency in a class moves the mean toward that class's mark. If the mean already equals that mark, it does not move.
Now try it. Slide f until the mean matches.
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Try the lab to finish this lesson.