22.12 Cumulative frequency tables and graphs (Extended)

Cumulative frequency scores

Some frequency tables have an extra column for cumulative frequency. It is a running total of the frequencies. In other words, cumulative frequency is the total of that row's frequency and all the other frequencies from the previous scores in the data set.

Worked example

Example 1

Sam recorded the number of pets owned by $10$10 people in his class. Here is the regular frequency table.

Number of Pets	Frequency
$0$0	$2$2
$1$1	$5$5
$2$2	$2$2
$3$3	$1$1

Now let's look at how we would calculate and add in a cumulative frequency column.  Remember, we add each frequency to the previous frequency total. The first value in the cumulative frequency table will be the same as the value in the frequency column (since there's no previous value to add it to).

Number of Pets	Frequency	Cumulative Frequency
$0$0	$2$2	$2$2
$1$1	$5$5	$2+5=7$2+5=7
$2$2	$2$2	$7+2=9$7+2=9
$3$3	$1$1	$9+1=10$9+1=10

Notice that the final value in the cumulative frequency column is the same as the total number of people that were surveyed? That's how we know we've got our frequency scores all right.

Practice question

Question 1

Cumulative frequency histograms

We've looked at frequency histograms as graphical representations of the distribution of data by plotting the frequencies of each individual score. In cumulative frequency histograms, we plot the cumulative frequency scores. As such, the columns in a cumulative frequency histogram continue to increase, with the last scores having the tallest column.

This is a graph of a cumulative frequency histogram.

Cumulative frequency ogive

This is a line that:

• starts at the bottom-left of the first column
• increases by the frequency of each class
• ends at the total sum of all scores
• is joined together by line segments joining the top right corner of the histogram columns
• can be superimposed on the cumulative frequency histogram, or can stand alone.

So adding the ogive to the cumulative frequency histogram above we would get the following graph:

The ogive by itself is below:

Finding the median from the ogive

We can find the median of the data set using the ogive by:

1. Finding the middle point on the cumulative frequency axis (half the total number of scores)
2. Drawing a horizontal line to the polygon and then a vertical line down to the horizontal axis
3. Observing which histogram column the line is in which gives the median score.

Finding quartiles and percentiles from the ogive

We can find the percentiles and quartiles of the data set using the ogive by:

1. Find the corresponding percentage of the total number of scores and find that number on the cumulative frequency axis. e.g. For the $20$20th percentile, find $20%$20% of the total number of scores. 
2. Drawing a horizontal line to the polygon and then a vertical line down to the horizontal axis
3. Observing which histogram column the line is in which gives the required score.

Worked example

Example 2

Consider the cumulative frequency histogram and ogive given below:

Use the graph to estimate: 
a) The median.

b) The $90$90th percentile.

c) The lower quartile.

d) The upper quartile.

e) The interquartile range.

 

 

a) 

Think: The median represents the $50%$50% mark of the data, because $50%$50% of the data lies above and below it. 

Do:  We need to find $50%$50% of the total number of scores, which according to the graph is $50$50. 

$50%$50% of $50$50	$=$=	$50%\times50$50%×50
	$=$=	$25$25

 

Now to find the median we draw a horizontal line from $25$25 on the vertical axis until it hits the ogive, then draw a line vertically down: 

We can see that the dashed line hits the horizontal axis in the column bounded by $20$20 and $25$25. So we take the average of these numbers to find the median.  Therefore the median is $22.5.$22.5. 

Note: If the column was just labelled with one number, rather than two numbers at the end points, then that one number would be the median.

b) 

Think: The $90$90th percentile is the $90%$90% mark of the data, because $90%$90% of the data lies below it. 

Do:  We need to find $90%$90% of the total number of scores, which according to the graph is $50$50. 

$90%$90% of $50$50	$=$=	$90%\times50$90%×50
	$=$=	$45$45

 

Now to find the percentile we draw a horizontal line from $45$45 on the vertical axis until it hits the ogive, then draw a line vertically down: 

We can see that the dashed line hits the horizontal axis in the column bounded by $30$30 and $35$35. So we take the average of these numbers to find the percentile.  Therefore the $90$90th percentile is $32.5.$32.5. 

c) 

Think: The lower quartile is the $25%$25% mark of the data, because $25%$25% of the data lies below it. 

Do:  We need to find $25%$25% of the total number of scores, which according to the graph is $50$50. 

$25%$25% of $50$50	$=$=	$25%\times50$25%×50
	$=$=	$12.5$12.5

 

Now to find the percentile we draw a horizontal line from $12.5$12.5 on the vertical axis until it hits the ogive, then draw a line vertically down: 

We can see that the dashed line hits the horizontal axis in the column bounded by $15$15 and $20$20. So we take the average of these numbers to find the lower quartile. Therefore the lower quartile is $17.5.$17.5. 

d) 

Think: The upper quartile is the $75%$75% mark of the data, because $75%$75% of the data lies below it. 

Do:  We need to find $75%$75% of the total number of scores, which according to the graph is $50$50. 

$75%$75% of $50$50	$=$=	$75%\times50$75%×50
	$=$=	$37.5$37.5

 

Now to find the percentile we draw a horizontal line from $37.5$37.5 on the vertical axis until it hits the ogive, then draw a line vertically down: 

We can see that the dashed line hits the horizontal axis in the column bounded by $25$25 and $30$30. So we take the average of these numbers to find the upper quartile.  Therefore the upper quartile is $27.5.$27.5. 

 

e) 

Think: The interquartile range is the difference between the upper quartile and lower quartile. 

Do:  

Interquartile range	$=$=	$27.5-17.5$27.5−17.5
	$=$=	$10$10

 

Practice questions

Question 2

Question 3

Cumulative frequency and grouped data

For grouped data cumulative frequency scores are calculated the same way by adding the cumulative frequency column in the frequency distribution table. The difference with grouped data is that when finding the median we can only estimate the value.

Worked example

Example 3

The frequency distribution table below shows the heights, in centimetres, of a group of children aged $5$5 to $11$11.

Child's height in cm	class centre	frequency	cumulative frequency
$91$91-$100$100	$95$95	$5$5	$5$5
$101$101-$110$110	$105$105	$22$22	$27$27
$111$111-$120$120	$115$115	$30$30	$57$57
$121$121-$130$130	$125$125	$31$31	$88$88
$131$131-$140$140	$135$135	$18$18	$106$106
$141$141-$150$150	$145$145	$6$6	$112$112

Use the table to answer the following questions:

1. How many children were in the group?
2. How many children had heights greater than $130$130 cm but less than or equal to $140$140 cm?
3. Which class interval contained the most children?
4. How many children had a height less than or equal to $120$120 cm?
5. How many children had a height greater than $130$130 cm?

Do:

1. The final cumulative frequency value tells us there were $112$112 children in the group. This is equal to the sum of the values in the frequencies column. 
2. The frequency column indicates there are $18$18 children with heights in the range $131-140$131−140.
3. The class interval with the highest frequency is $120-129$120−129.
4. The cumulative frequency of the $111-120$111−120 class interval, tells us that $57$57 children had a height less than or equal to $120$120 cm.
5. Here we can add the final two frequencies: $18+6=24$18+6=24. Alternatively, we could subtract the cumulative frequency of $88$88 (corresponding to the class interval containing the height $130$130 cm), from the total number of children in the group: $112-88=24$112−88=24.

Practice question

Question 4

Cumulative frequency graphs for grouped data

We can construct a cumulative frequency histogram and polygon for grouped data using the class interval on the horizontal axis. 

Worked example 

Example 4

a) The global life expectancy data from 2016 is shown in the frequency distribution table below. Construct a cumulative frequency histogram and ogive for the data set.

class interval	frequency	cumulative frequency
$51-54$51−54	$5$5	$5$5
$55-60$55−60	$10$10	$15$15
$61-64$61−64	$25$25	$40$40
$65-70$65−70	$26$26	$66$66
$71-74$71−74	$40$40	$106$106
$75-80$75−80	$49$49	$155$155
$81-84$81−84	$28$28	$183$183
Total	$183$183

Do: Plot the cumulative frequency for each class interval to get the height of each column. The columns should be ascending each time.

The cumulative frequency histogram and ogive are displayed together below:

b) Estimate the median life expectancy age using the graph.

Think: The median age is in the middle of the data set when the data is in ascending order. The number of scores altogether is 183. The median from the graph is the middle of 183 scores which is 92.5. 

Do: Starting at 92.5 along the vertical axis, draw a line from the vertical axis to the ogive and then a perpendicular line down to the horizontal axis. Estimate the value of the median by its position on the horizontal axis.

The median is approximately 73 years.

Practice question

Question 5