The cumulative frequency curve shows the heights of 50 children who are members of a youth club. Using the data from the cumulative frequency curve, complete the following tables and draw a histogram to represent the data.
So the data has been presented on this cumulative frequency curve here. We can see that the height of the children measured in centimeters is on the horizontal axis. And then the cumulative frequency is given on the vertical axis.
The first table that we need to complete has ranges for the heights in the first column. And then the cumulative frequencies have been left blank. So we need to read these values from the cumulative frequency curve.
The first value in the table is for children whose heights are less than 120 centimeters. So to read this value from the curve, we find 120 on the horizontal axis, we go up to the curve itself, and then we go across to the vertical axis. We can see that the value here is one small square above five.
Now as five small squares on the vertical axis represents five, this means that one small square will represent one. So one small square above five is the value six. So the cumulative frequency of children whose heights are less than 120 is six.
The next cumulative frequency is for those children whose heights are less than 140 centimeters. So we find this value from the curve in the same way. We go up from 140 on the horizontal axis to the curve and then across to the vertical axis, where we see that the value is one small square above 15. So it is 16.
Remember that this is a cumulative frequency. So these 16 children whose heights are less than 140 centimeters include the six children whose heights are less than 120 centimeters. We can then read the remaining three cumulative frequencies from the graph in the same way, giving 26 for children whose heights are less than 150 centimeters, 42 for children whose heights are less than 175 centimeters, and 50 for children whose heights are less than 185 centimeters. Although we could’ve filled this value in without using the cumulative frequency curve, because we were told in the question that there are 50 children in total.
So we filled in the first of these two tables. And now let’s consider the second. In this table, the heights are given in intervals in the first column. For example, the first class is children whose heights are greater than or equal to 100 centimeters but less than 120 centimeters.
The first column we need to complete is the frequency column. So this will be just those children who belong in each class, not the cumulative frequency. For the first class, the cumulative frequency is actually the same as the frequency because we can see from our cumulative frequency diagram that the smallest height measured was 100. For the second class, well, remember we said that the cumulative frequency of 16 for children whose heights are less than 140 centimeters includes the six whose heights are also less than 120 centimeters.
So to work out the number of children, the frequency, whose heights are just between 120 and 140 centimeters, we subtract six from 16, which gives 10. In the same way, to work out the number of children whose heights are greater than equal to 140 but less than 150 centimeters, we subtract the previous cumulative frequency of 16 from the current cumulative frequency of 26. We can calculate the remaining two frequencies in the same way, 42 minus 26 and 50 minus 42, giving 16 and eight, respectively. It will be sensible to perform a quick check about frequencies at this point. If we add the five frequencies together, we do indeed get 50, the total number of children.
The next column we need to complete is the class width. This is the difference between the endpoints of each class, the upper endpoint minus the lower endpoint. For the first class then, that’s 120 minus 100, which is equal to 20. For the second class, it’s 140 minus 120, which is also equal to 20.
Don’t make the assumption though that all of the classes will be of the same width. For the next class, the width is 150 minus 140, which is equal to 10. It’s really important that you do calculate all of the class widths individually. And don’t make the incorrect assumption that they’ll all be the same. For the final two classes, the widths are 175 minus 150, which is 25, and 185 minus 175, which is 10.
The final column that we need to complete is the frequency density, which is calculated using the frequency and the class width. This is essentially a way of standardizing the frequencies to a common unit. So we calculate the frequency density by dividing the frequency of each class by that class width.
For the first class then, it’s six divided by 20, which is equal to 0.3. For the second class, it’s 10 divided by 20, which is equal to 0.5. We calculate the remaining three frequency densities in the same way. 10 divided by 10 is one, 16 divided by 25 is 0.64, and eight divided by 10 is 0.8.
So we’ve now answered the first part of the question asking us to complete both of the tables. And we now need to consider the second part, where we’re asked to draw a histogram to represent the data. We can see that the horizontal axis has been completed for us. We have the Heights measured in centimeters. But what about the vertical axis?
Well, the vertical axis on a histogram represents the frequency density. So we now need to decide the scale that we’re going to use. Our scale needs to go from zero, and it needs to include the largest value in our frequency density column, which is one. Looking carefully at the vertical axis, we can see that there are 12 large squares. So if we choose each large square to represent 0.1, this will be a sensible scale for our vertical axis. We can fill in these values in multiples of 0.1 all the way up to 1.1.
We can also see that, within each large square, there are five small squares. So if five small squares represent 0.1, then this means that one small square will represent 0.02 on our vertical axis.
Now we can begin plotting our histogram. We draw bars for each class. The first class includes heights from 100 to 120 centimeters. So these will be the endpoints for the first bar. The frequency density for this class is 0.3. So we can draw in the top of our first bar. We can then connect it to the horizontal axis.
Our second class has a lower endpoint of 120. So the second bar starts immediately after the end of the first bar. There must be no gaps in our histogram. The upper endpoint for this bar is 140. The frequency density for this bar is 0.5. So we’ve drawn our second bar.
Our third class covers heights from 140 to 150 centimeters and has a height of one or 1.0. For the next part, we just need to remind ourselves of the vertical scale. The frequency density is 0.64. And as one small square on our vertical axis represents 0.02, this will be two small squares above 0.6. The upper endpoint for this class is 175. So we draw in our fourth bar.
Our final class has an upper endpoint of 185. And the height of this bar is 0.8. So we have our completed histogram, where we’ve plotted the heights in each class against the frequency density of that class.
By using frequency density rather than frequency on the vertical axis of a histogram, this ensures that the frequency of each class is actually given by the area of that bar. This means that we can see visually by looking at a histogram which classes have the highest and lowest frequency.
So we’ve completed the problem. We used the information in the cumulative frequency curve to complete the two tables. And then we used the information in the table to draw a histogram of the heights of these 50 children.