Cumulative frequency, quartiles and percentiles: Year 10 Extended statistics and the bell curve

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

From Lesson 1: a class has boundaries 20 and 40. What mid-interval value stands for it?
Answer
30. The midpoint is (20 + 40) ÷ 2 = 30.
From Year 8: find the median of 3, 8, 9, 12, 20, 21 and 25.
Answer
12. There are 7 values, and the 4th is 12.
From Year 9: what is 35% of 80?
Answer
28. 0.35 × 80 = 28.

Words used in this lesson

cumulative frequency: the running total of frequencies up to a class boundary. For example: a total of 7 up to the first boundary.
upper boundary: the boundary at the top end of a class, where its running total is plotted. For example: 40 for the class 30 ≤ x < 40.
percentile: the value below which a given percentage of the data lie. For example: 12 s, if 30% of the times are under 12 s (the 30th percentile).
interquartile range: the upper quartile minus the lower quartile. For example: IQR = Q₃ − Q₁.
outlier: a value beyond the fences, far from the rest of the data. For example: above Q₃ + 1.5 × IQR.
interpolate: estimate a value between two known points along a straight line. For example: halfway between two plotted points.

The rain gauge and the one wet day in ten

An observatory in Chennai records the rainfall on 80 days of a monsoon month and groups it in classes 10 mm wide. Plotting the running total at the top of each class gives a curve that starts at 0 and finishes at 80.

Predict first: where will the curve climb fastest? And how would you read the rainfall that was reached or exceeded on only one day in ten? Test your prediction with the slider.

Percentage of the 80 values 0

0%share of the values

0number of values

0rainfall value (mm)

0% of the 80 days is 0 days, reached at a rainfall of 0 mm.

The figure needs JavaScript. In words: 80 daily rainfall readings are grouped in classes of width 10 mm from 0 to 60 mm, and the cumulative frequencies at the upper boundaries 10, 20, 30, 40, 50 and 60 are 6, 20, 42, 60, 72 and 80. Reading 25% of 80, which is 20 days, gives 20 mm. Reading 75%, which is 60 days, gives 40 mm. Reading 50%, which is 40 days, gives about 29.1 mm by interpolating between (20, 20) and (30, 42).

Show me the steps

Slide to 25%. That is 25% of 80 = 20 days. The curve reaches a cumulative frequency of 20 at a rainfall of 20 mm, so the lower quartile is 20 mm.
Slide to 75%. That is 60 days, and the curve reaches 60 at 40 mm: the upper quartile. The interquartile range is 40 − 20 = 20 mm.
Slide to 50%. That is 40 days, which falls between the plotted points (20, 20) and (30, 42). Moving along that straight piece, the curve reaches 40 at about 29.1 mm: the median.
Slide to 90%. That is 72 days, and the curve reaches 72 at 50 mm. One day in ten had at least that much rain.

Cumulative frequency is a running total. Plot each total at the upper boundary of its class, because only then has every value in the class been counted, and begin at the lower boundary of the first class with a count of 0. Join the points.

To read a quartile or percentile: turn the share into a count first (the lower quartile is n ÷ 4, the median n ÷ 2, the upper quartile 3n ÷ 4, the P-th percentile P ÷ 100 of n), find that count on the vertical axis, and read across to the horizontal axis. Between two plotted points, move along the straight piece.

Outliers. A value is an outlier if it is below Q₁ − 1.5 × IQR or above Q₃ + 1.5 × IQR, where IQR = Q₃ − Q₁.

Plot at the upper boundary, not the midpoint. For the class 100 to under 120 the point goes at x = 120, with the running total up to and including that class. At the midpoint, 110, only part of the class has been counted.

Worked examples

1. Fish lengths (cm), 60 fish: 10 to under 20 has 5; 20 to under 30 has 17; 30 to under 40 has 24; 40 to under 50 has 11; 50 to under 60 has 3. Estimate the quartiles.

The running totals are 5, 22, 46, 57 and 60, plotted at 20, 30, 40, 50 and 60, with (10, 0) to start.
Q₁ is the 15th value. It lies between (20, 5) and (30, 22): 20 + (15 − 5) ÷ 17 × 10 ≈ 25.9.
The median is the 30th value, between (30, 22) and (40, 46): 30 + 8 ÷ 24 × 10 ≈ 33.3.
Q₃ is the 45th value, in the same piece: 30 + 23 ÷ 24 × 10 ≈ 39.6. The IQR is about 13.7.

2. Estimate the 90th percentile.

90% of 60 is 54. This is between (40, 46) and (50, 57).
40 + (54 − 46) ÷ 11 × 10 ≈ 47.3 cm.

3. Could any fish be an outlier?

1.5 × IQR ≈ 1.5 × 13.7 ≈ 20.6, so the fences are about 25.9 − 20.6 ≈ 5 and 39.6 + 20.6 ≈ 60.
Every fish is between 10 and 60 cm, so none is below 5 and none is above 60: no outliers.

Your go. The first step is given. For 40 values, the running totals at the boundaries 20, 30, 40 and 50 are 6, 18, 33 and 40, and the count is 0 at the boundary 10. Estimate the median.

n = 40, so the median is the 20th value, which lies between the points (30, 18) and (40, 33).
Now move along that piece from 18 to 20.

Check the answer

30 + (20 − 18) ÷ (33 − 18) × 10 = 30 + 1.33 … = 31.3. The 20th value is only a little way into the class, because the curve has to climb 2 of its 15 steps.

Check before you go on

1. The class 40 ≤ x < 50 has frequency 9, and 31 values are below 40. Which point is plotted on the cumulative frequency graph for this class?

(45, 40) (50, 40) (50, 9) (40, 40)

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. In a cumulative frequency table, 11 values are below 30 minutes and 27 values are below 40 minutes. How many values are in the class 30 to under 40?

3. A cumulative frequency curve is drawn for 60 values. At what cumulative frequency do you read across to find the lower quartile?

Level 2 · Building

4. A cumulative frequency graph of 40 values passes through (40, 18) and (50, 30), with a straight line between. Estimate the median, to one decimal place.

5. A cumulative frequency graph of 50 values passes through (30, 22) and (40, 34), with a straight line between. Estimate the percentage of the values that are greater than 38, to one decimal place.

Level 3 · Stretch

6. A set of values has Q₁ = 24 and Q₃ = 41. Any value above Q₃ + 1.5 × IQR is an outlier. What is that fence?

7. A cumulative frequency curve describes 120 values. Its median is 41 and its 90th percentile is 63. How many values lie between 41 and 63?

Something to think about

Two weather stations each have 80 days of rainfall and the same median, but station B's curve is much steeper in the middle than station A's. Without drawing anything, which station has the smaller interquartile range, and what does a steep part of a curve say about the data there?

One way to think about it

Station B has the smaller IQR. A steep part of the curve means that many values are packed into a narrow range of x: the running total climbs quickly. If the middle half of B's values are squeezed into a narrow stretch, Q1 and Q3 are close together. A shallow part means the values are spread thinly.

Exit ticket

8. A data set has Q₁ = 30 and Q₃ = 50. How many of the values 53, 72, 77 and 84 are outliers?

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