1. Cumulative frequency tables
Running totals
- Cumulative frequency is a running total of the frequencies, so each entry counts everything up to and including that class
- It is used with grouped continuous data, where the individual values are not known
- Build the column by adding each class frequency to the total so far
- The final cumulative frequency must equal the total frequency, which is a built-in check on the arithmetic
- Each entry answers the question "how many are less than or equal to the end of this class"
| Time (t minutes) | Frequency | Cumulative frequency |
|---|---|---|
| 0 < t ⩽ 10 | 4 | 4 |
| 10 < t ⩽ 20 | 11 | 15 |
| 20 < t ⩽ 30 | 21 | 36 |
| 30 < t ⩽ 40 | 16 | 52 |
| 40 < t ⩽ 50 | 8 | 60 |
Worked example
Building a cumulative frequency table
The table shows the lengths of 40 fish. Work out the cumulative frequency for each class.
| Length (x cm) | Frequency |
|---|---|
| 20 < x ⩽ 30 | 4 |
| 30 < x ⩽ 40 | 9 |
| 40 < x ⩽ 50 | 14 |
| 50 < x ⩽ 60 | 8 |
| 60 < x ⩽ 70 | 5 |
Solution:
- Cumulative frequency is a running total, so each entry includes everything before it
- First class: 4
- Second: 4 + 9 = 13
- Third: 13 + 14 = 27
- Fourth: 27 + 8 = 35
- Fifth: 35 + 5 = 40
- The cumulative frequencies are 4, 13, 27, 35, 40
- The final entry must equal the total of all the frequencies, and 4 + 9 + 14 + 8 + 5 = 40 confirms it
- The values can only increase or stay the same; a column that falls anywhere has an arithmetic slip in it
2. Drawing the diagram
Plotting at the upper bound
- The single most important rule: each running total goes above the upper bound of its class, never above the midpoint
- The reason is that only at the end of a class can every value in it be guaranteed to have been counted
- So the table above gives the points (10, 4), (20, 15), (30, 36), (40, 52) and (50, 60)
- Add a starting point at the lower bound of the first class with a cumulative frequency of zero, here (0, 0)
- The syllabus asks for points to be clearly marked, for example as small crosses, and joined with a smooth curve
- The curve usually has a stretched S-shape, rising steeply where the frequencies are largest, and it never turns back down
3. Reading estimates off the curve
Median, quartiles and percentiles
- Everything read off a cumulative frequency diagram is an estimate, because the original values are unknown
- Find the total, n, from the highest point the curve reaches, at the top right
- Then work from the cumulative frequency axis across to the curve and down to the horizontal axis
| Statistic | Position on the cumulative frequency axis |
|---|---|
| Lower quartile | n ÷ 4 |
| Median | n ÷ 2 |
| Upper quartile | 3n ÷ 4 |
| p-th percentile | p ÷ 100 × n |
- Note that the median position here is n ÷ 2, not (n + 1) ÷ 2 as it is for a list of values, because the curve treats the data as spread smoothly
- The interquartile range is the upper quartile minus the lower quartile, both read off in the same way
- Draw the horizontal and vertical lines on the diagram, since they show where the reading came from
Exam tip
Write down both quartiles before you subtract
An interquartile-range question gives a mark for one correct quartile on its own, whether you read the quartiles off a curve or find them from a list. So write each one down as you get it (lower quartile = …, upper quartile = …) and only then subtract. A misreading of the other quartile, or a slip in the subtraction, still leaves you that mark.
Worked example
Quartiles from a cumulative frequency diagram
The table in section 1 records the times of 60 people, and a smooth cumulative frequency curve is drawn through its points. Use the curve to estimate the median, the quartiles and the interquartile range.
Solution:
- The total is n = 60, which is where the curve finishes at the top right
- The lower quartile is at 60 ÷ 4 = 15 on the cumulative frequency axis
- The median is at 60 ÷ 2 = 30
- The upper quartile is at 3 × 60 ÷ 4 = 45
- For each one, draw a horizontal line from that value across to the curve, then a vertical line down to the time axis and read the value
- The cumulative frequency reaches 15 exactly at the end of the second class, so the lower quartile is 20 minutes
- The reading for 30 falls between 20 and 30 minutes, giving a median of roughly 27 minutes
- The reading for 45 falls between 30 and 40 minutes, giving an upper quartile of roughly 35 minutes
- Interquartile range = upper quartile − lower quartile, so about 35 − 20 = 15 minutes
How many are above or below a value
- To find how many are below a given value, go up from that value on the horizontal axis to the curve, then across to read the cumulative frequency
- To find how many are above it instead, subtract that reading from the total
- Read the question carefully: "more than", "at least" and "fewer than" all need the same reading, but only "fewer than" is answered by the reading itself; for "more than" or "at least", subtract the reading from the total
Worked example
How many took longer than a given time
Using the same curve for 60 people, estimate how many took longer than 35 minutes.
Solution:
- Start at 35 on the horizontal time axis and go straight up to the curve
- Now read horizontally to the vertical scale and note the running total there
- 35 minutes is halfway between the plotted points (30, 36) and (40, 52), but the curve is starting to level off there, so it reads a little above the halfway value of 44: about 45
- That means about 45 people took 35 minutes or less
- The question asks for longer than 35 minutes, so subtract from the total
- 60 − 45 = about 15 people
- Always check whether the question wants the reading itself or the total minus the reading