Two missing frequencies
A table hides two frequencies. The total and the median are given.
- Classes 0-10 to 40-50. Frequencies 4, x, 12, 6 and y. The total n is 40. The median is 25.
- The total gives
4 + x + 12 + 6 + y = 40. Sox + y = 18. - The median 25 lies in 20-30. So
25 = 20 + ((20 − (4 + x)) ÷ 12) × 10. - That gives
6 = 16 − x, sox = 10. Theny = 18 − 10 = 8.
The rule
Use the total to link x and y. Use the median to find x. Then find y.
Here x and y are hidden frequencies, not class marks. x lies before the median class. y lies after that class.
The total n is 40, so n ÷ 2 = 20. The median 25 lies in 20-30. So 20-30 is the median class. That fixes three letters of the formula. l = 20 is its lower bound. f = 12 is its frequency. h = 10 is its width.
The class with frequency x comes before the median class. So x is part of cf. Here cf = 4 + x.
The class with frequency y comes after the median class. So y is not in cf.
| Part | Value here |
|---|---|
| n ÷ 2 | 20 |
| l, the class lower bound | 20 |
| cf, the total before | 4 + x |
| f, the frequency of the class | 12 |
| h, the class width | 10 |
Check with x = 10 and y = 8. The totals climb 4, 14, 26, 32, 40. The first total to reach 20 is 26. So 20-30 is the median class. The median is 20 + ((20 − 14) ÷ 12) × 10. That is 25.
Now try it. Pick the equation that finds x.
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