Unit 16 Section 3 : Cumulative Frequency

Cumulative frequencies are easy to calculate from a frequency table. Cumulative frequency graphs can then be used to estimate the median of a set of data. In this section we also look at the idea of quartiles, the interquartile range and the semiinterquartile range.

When you have a set of n values, in order,

Lower quartile   =   th value
Median   =   th value
Upper quartile   =   th value
Interquartile range   =   upper quartile – lower quartile
Semi-interquartile range   =  

If the data is arranged in an ordered list, and the number of data values, n, is odd then the th value will be a single item from the list, and this will be the median. For example, if n = 95 the median will be the = 48th value. However, if n is even then will determine the two central values that must be averaged to obtain the median. For example, if n = 156 then = 78.5, which tells us that we must average the 78th and 79th values to get the median.

For large sets of data, we estimate the lower quartile, median and upper quartile using the th, th and th values. For example, if n = 2000 , then we would estimate the lower quartile, median and upper quartile using the 500th, 1000th and 1500th values.

Example 1

For the following set of data,

471839510
(a)

determine the median,

First list the values in order:

3 4 5 7 9 10 18

As there are 7 values, the median will be the = 4th value.
Median = 7.

(b)

calculate the interquartile range,

The lower quartile will be the = 2nd value.
Lower quartile = 4.

The upper quartile will be the = 6th value.
Upper quartile = 10.

The interquartile range = upper quartile – lower quartile = 10 – 4 = 6

The semi-interquartile range = = = 3

Example 2

(a)

Draw a cumulative frequency graph for the following data:

Height (cm)150 ≤ h < 155155 ≤ h < 160160 ≤ h < 165165 ≤ h < 170170 ≤ h < 175
Frequency42256325

From the data table we can see that there are no heights under 150 cm.

Under 155 cm there are the first 4 heights.

Under 160 cm there are the first 4 heights plus a further 22 heights that are between 155 cm and 160 cm, giving 26 altogether.

Under 165 cm we have the 26 heights plus the 56 that are between 160 cm and 165 cm, giving 82 altogether.

Continuing this process until every height has been counted gives the following cumulative frequency table.

Height (cm)Under 150Under 155Under 160Under 165Under 170Under 175
Cumulative Frequency00 + 4
= 4
4 + 22
= 26
26 + 56
= 82
82 + 32
= 114
114 + 5
= 119

The cumulative frequency graph can now be plotted using the points in the table, (150, 0), (155, 4), (160, 26), (165, 82), (170, 114) and (175, 119).

To obtain the cumulative frequency polygon, we draw straight line sections to join these points in sequence.

(b)

Estimate the median from the graph.

There are 119 values, so the median will be the = 60th value.

This can be read from the graph as shown above.

Median ≈ 163 cm.

(c)

Estimate the interquartile range from the graph.

The lower quartile will be given by the th value.

Lower quartile ≈ 160.5 cm.

The upper quartile will be given by the th value.

Upper quartile ≈ 166.5 cm.

Using these values gives:

Interquartile range = 166.5 – 160.5 = 6 cm

Example 3

Estimate the semi-interquartile range of the data illustrated in the following cumulative frequency graph:

The sample contains 15 values, so the lower quartile will be the = 4th value.
Similarly, the upper quartile will be the 12th value.
These can be obtained from the graph, as follows:

Lower quartile = 1.4 kg

Upper quartile = 3 kg

Interquartile range = 3 – 1.4 = 1.6 kg

Semi-interquartile range = 0.8 kg

Exercises

Question 1

Determine the median and interquartile range of the following set of data:

11859734814162
Median =
Interquartile range =
Question 2

Calculate the semi-interquartile range of this sample:

422632415233887138525327463259
Semi-interquartile range =
Question 3

In a sample, the semi-interquartile range is 14. The lower quartile is 5 less than the median. Determine the median if the upper quartile is 91.

Median =
Question 4

Below are the times, in minutes, spent on homework one evening by a group of students.

Time Spent (min)0 ≤ t < 1010 ≤ t < 2020 ≤ t < 3030 ≤ t < 4040 ≤ t < 50
Frequency3710154
(a)
Time (min)Under 0Under 10Under 20Under 30Under 40Under 50
Cumulative
Frequency
(b)

Draw a cumulative frequency polygon for this data.

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(c)

Use the polygon to estimate the median.

minutes
(d)

Use the polygon to estimate the semi-interquartile range.

minutes
Question 5

Estimate the median and interquartile range of the data illustrated in the following cumulative frequency graph:

Median =
Interquartile range =
Question 6

Use a cumulative frequency graph to estimate the median and interquartile range of the following data:

Cost (£)10 ≤ c < 1111 ≤ c < 1212 ≤ c < 1313 ≤ c < 1414 ≤ c < 15
Frequency8124021
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Median = £
Interquartile range = £
Question 7

A factory collected data on the time for which a particular type of candle would burn. The data is summarised in the following table:

Time (mins)0 ≤ t < 1010 ≤ t < 2020 ≤ t < 3030 ≤ t < 4040 ≤ t < 50
Frequency1212155
(a)

How do the mean and median compare?

Mean = minutes

Median = minutes

So

(b)

Determine the semi-interquartile range for the data.

minutes
Question 8

The number of passengers on a bus route was recorded over a period of time, to give the following data:

Number of PassengersFrequency
0 ≤ n < 103
10 ≤ n < 207
20 ≤ n < 3012
30 ≤ n < 4013
40 ≤ n < 5029
50 ≤ n < 6027

Determine the median and semi-interquartile range of the data.

Median = passengers (to the nearest integer)
Semi-interquartile range = passengers (to the nearest integer)
Question 9

The cumulative frequency graph shows the height of 150 Norway fir trees.

(a)

Use the graph to estimate the median height and the interquartile range of the Norway firs.

Median = m
Interquartile range = m
(b)

Which one of the following sketches of frequency diagrams shows the distribution of heights of the Norway firs?

Diagram

Question 10

40 students worked on a farm one weekend. The cumulative frequency graph shows the distribution of the amount of money earned. No one earned less than £15.

(a)

Read the graph to estimate the median amount of money earned.

Median = £
(b)

Estimate the percentage of students who earned less than £40.

(c)

Write down the value of the interquartile range.

£
Lower quartile = £26.40
Upper quartile = £37.10
Interquartile range = £10.70
(d)

30 of the students work on the farm another weekend later in the year. The tables which follow show the distribution of the amount of money earned by the students.

Money Earned (£)No. of Students
≥ 25 and < 301
≥ 30 and < 352
≥ 35 and < 403
≥ 40 and < 454
≥ 45 and < 5010
≥ 50 and < 557
≥ 55 and < 603
 
Money Earned (£)No. of Students
< 250
< 301
< 353
< 406
< 4510
< 5020
< 5527
< 6030

Draw a cumulative frequency graph using a copy of the axes below.

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(e)

State whether each of the following statements is true or false.

A. Three of the students earned less than £35 each.

B. The median amount earned is between £40 and £45.

The median corresponds to cumulative frequency 15, so lies in the interval £45 ≤ amount earned < £50. The median = £47.50.

C. Most of the 30 students earned more than £50 each.

Only 10 out of the 30 students earned £50 or more.