Cumulative Frequency Questions | Worksheets and Revision | MME

# Cumulative Frequency Curves Questions, Worksheets and Revision

Level 6-7

## Cumulative Frequency

Cumulative frequency is the number of times that anything up to and including that value (or group of values) appeared. You will need to be able to work out the cumulative frequency as well as use this to plot a cumulative frequency graph.

Make sure you are happy with the following topics before continuing.

## Plotting a Cumulative Frequency graph

Below is a frequency table of data compiled on a group of college students’ heights.

Step 1: Construct a cumulative frequency table for this data.

Calculating the cumulative frequency is just adding up the frequencies as you go along.

The first value is the first frequency value, we then add this to the second value to get the second cumulative frequency value

$13 + 33 = 46$

Continuing this, we get that

$46 + 35 = 81$

people were $180$ cm of shorter, is

$81 + 11 = 92$

Step 2: Using the cumulative frequency, plot a cumulative frequency graph.

The points plotted on your graph should be plotted at the end of each class,

i.e. the point which has cumulative frequency of $13$ should be plotted at $160$ on the height axis, and so on.

You should join up the plotted points with a smooth curve. It should end up looking like an elongated ‘S’ shape.

Level 6-7

## Example: Finding the Median and IQR

Using the cumulative frequency graph below, calculate the median and interquartile range.

[3 marks]

There are $92$ people in total, so the lower quartile, median, and upper quartile will be the $23$rd person, $46$th person, and $69$th person respectively. So, we find these points on the $y$-axis, and then draw a line across to the graph to find the corresponding heights on the $x$-axis. This is shown on the cumulative frequency graph below.

Here, we get

$Q_1 = 163, \,\, \text{ median } = 170, \,\,Q_3 = 176$

The interquartile range is therefore

$176-163=13$

Level 6-7

## GCSE Maths Revision Cards

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### Example Questions

a)  For the cumulative frequency column, we simply need to add up the frequencies as we move downwards in the table:

The first cumulative frequency box is simply the number $16$.

For the second cumulative frequency box, we need to add $24$ to the  $16$ from the previous cumulative frequency box.

$24+16=40$

For the third cumulative frequency box, we need to add $19$ to the  $40$ from the previous cumulative frequency box.

$19+40=59$

This process of adding the frequency total to the cumulative frequency total repeats until the table is complete.

The final table should look like this:

b)  For the cumulative frequency diagram, we need to plot the time in minutes along the $x$-axis and your cumulative frequency totals on the $y$-axis.

Then we need to plot each of the cumulative frequency figures with the corresponding class interval maximums.  In other words, for the time interval of $0 - 20$ minutes, you would go along the $x$-axis to $20$ minutes (the maximum in this time range) and go upwards to the cumulative frequency value of $16$.

Once you have plotted all points, including the origin $(0,0)$, join up all points with a smooth curve.  (A cumulative frequency graph is always a smooth curve which goes up.)

Your graph should look like this:

From the graph, we can see that the minimum number of hours was $0$ and the maximum was $12$.

We can also see from the graph that is represents exercising data from $72$ people since this is where the graph ends.

In order to draw a box plot, we need to know the following values:

a)  the minimum value

b) the maximum value

c) the median value

d)  the lower quartile

e)  the upper quartile

We have already established that the minimum value is $0$ and the maximum value is $72$.

The median value is the middle value.  Since there are $72$ values in total, then the median value is the 36th value (since $36$ is half of $72$).  If we go up the $y$-axis and locate the 36th value, go across to the line and then down, we can see that is corresponds to a value of $3.2$ hours.

The lower quartile is half-way between the first value and the median.  To work out which value is the lower quartile, find $\frac{1}{4}$ of the total number of values:

$72 \div4 = 18$

The lower quartile is therefore the value of the 18th term.  If we go up the $y$-axis and locate the 18th value, go across to the line and then down, we can see that is corresponds to a value of $1.6$ hours.

The upper quartile is half-way between the final value and the median values.  To work out which value is the upper quartile, simply find $\frac{3}{4}$ of the total number of values:

$\dfrac{3}{4}\times72 = 54$

The upper quartile is therefore the value of the 54th term.  If we go up the $y$-axis and locate the 54th value, go across to the line and then down, we can see that is corresponds to a value of $5$ hours.

The graph below illustrates the above:

Using this information, the resulting box plot will look like this:

a)  We know that there are $8$ cars with a value between $£0$ and $£10,000$[/latex], so we can insert $8$ in the $£0 - £10,000$ cumulative frequency box.

The next box is the $£0 - £15,000$ box.  We know that there are $8$ cars with a value of less than $£10,000$, and a further $12$ cars which have a value of more than $£10,000$ but less than $£15,000$, therefore $20$ cars have a value of between $£0$ and $£15,000$, so this is the next value we can insert.

Continue this process until all values are calculated and your cumulative frequency table should look as follows:

To draw the graph, we need to plot the cumulative frequency totals on the vertical axis (the cumulative frequency is always on the $y$-axis and the price in pounds on the $x$-axis.  However, since we are dealing with grouped data (the prices are price bands), we need to plot the cumulative frequency total against the highest value in the band.  So, the first point we plot (after plotting $(0,0)$, the origin) would be the cumulative frequency total of $8$ against $£10,000$ (the top value in the $£0 - £10,000$ band).  The next point we would plot would be the cumulative frequency value of $20$ against $£15,000$ (the highest value in the $£10,000 - £15,000$ band).

Your completed cumulative frequency graph should look as follows:

b)  We know that there are $48$ cars in the showroom in total (since the maximum value on the cumulative frequency table is $48$).  To find the median, we need to read the value of the car that corresponds to a cumulative frequency total of $24$ (half of $48$).  By locating the value of $24$ on the cumulative frequency axis, we can see that this corresponds to a value of approximately $£16,000$.

c)  For this question, we need to locate $£17,500$ on the $x$-axis and see what cumulative frequency total this corresponds to.  By drawing a line up from $£17,500$ until it touches the line and then drawing a horizontal line to the $y$-axis, we should hit a cumulative frequency total of approximately $26$

This means that $26$ cars have a value that is up to $£17,500$.  Therefore the remaining cars must have a value which is greater than $£17,500$.  Since there are $48$ cars in total then the number of cars which have a value of more than $£17,500$ is simply $48-26=22$ cars.

a)  In order to draw our cumulative frequency graph, we need to work out the cumulative frequency totals:

To represent this information on a graph, we need to plot the cumulative frequency totals on the vertical axis against snake length on the horizontal axis.  After plotting our first point of $(0,0)$, the next point is $(1,23)$; the key thing to remember is that since the snake length data is grouped, we need to plot the highest value in each length band (so for the $0$ metres – $1$ metre band, we would plot the corresponding cumulative frequency total against $1$ metre).  Once we have plotted all the points, we need to join them together with a smooth line, with an end result similar to the below:

b)  The interquartile range is calculated by subtracting the lower quartile from the upper quartile.

In this data set, the lower quartile is the length of the 40th snake ($40$ because $40$ is $\frac{1}{4}$ of $160$.  To find the length of the 40th snake, find $40$ on the vertical cumulative frequency axis and find the corresponding length on the horizonal axis.  The length of the 40th snake is approximately $1.5$ metres.

In this data set, the upper quartile is the length of the 120th snake ($120$ because $120$ is $\frac{3}{4}$ of $160$.  To find the length of the 120th snake, find $120$ on the vertical cumulative frequency axis and find the corresponding length on the horizonal axis.  The length of the 120th snake is approximately $3.5$ metres.

Therefore the interquartile ranges is $2$ metres.

c)  For this question, we need to work out the median snake length at Bob Exotic’s Snake Sanctuary.  Since there are $160$ snakes in total, the median snake length is the length of the 80th snake ($80$ because $80$ is $\frac{1}{2}$ of $160$.  The 80th snake has a length of approximately $2.6$m.

If at the other snake sanctuary, the median snake length is $1.78$ metres, then to calculate how much smaller as a percentage, we need to find out how much smaller the median snake is:

$2.6\text{ m } – 1.78\text{ m} = 0.82\text{ m}$

To work out a percentage increase or decrease, you need to remember the simply formula:

$\dfrac{ \text{ difference}}{\text{ original value}}\times 100$

$\dfrac{0.82\text{ m}}{2.6\text{ m}}\times 100=31.5\%$

(In this question, however, it may not be obvious what the original value is.  The question asks us to work out by what percentage these snakes are smaller than the snakes in Bob Exotic’s Snake Sanctuary.  Because of the word ‘than’, it is the length of the snakes at Bob Exotic’s Snake Sanctuary that we should consider as the ‘original’ value.  Words / phrases like ‘compared to’ or ‘than’ always indicate what we are working out a percentage of.)

### Worksheets and Exam Questions

#### (NEW) Cumulative Frequency Exam Style Questions - MME

Level 6-7 New Official MME

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