Histograms and Cumulative Frequency for the ESAT

Discrete and continuous data

Data is classified into two main types: discrete and continuous. Understanding the difference is vital for selecting the correct representation. Discrete data can only take certain fixed values. For example, the number of pupils in a class must be an integer (you cannot have $23.3$ pupils). Other examples include the number of cars in a car park and UK shoe sizes, such as $3$, $3.5$, $4$, and $4.5$.

Continuous data can take on any value within a range. For instance, the heights of pupils in a class are continuous because a height could be $165.75$ cm. Even if measurements are rounded for recording purposes, the underlying data remains continuous.

Class intervals

When data is grouped, we use inequalities to express class intervals. These intervals must be continuous so that the upper boundary of one class is exactly the lower boundary of the next class. It is essential that every data point belongs to exactly one interval. Class intervals do not have to be of equal width.

There is a distinction between class limits and class boundaries. The class limits are the values written in a frequency table. The class boundaries are the 'true' values. For example, if heights are measured to the nearest cm and grouped as $150$ cm to $154$ cm, the class boundaries are $149.5 \leq h < 154.5$. The lower boundary is the smallest number that rounds up to the lower limit, and the upper boundary is the smallest number that would round up to the next limit.

In a group of $150$ to $154$ cm, the limits are $150$ and $154$, but the boundaries are $149.5$ and $154.5$. If the next group is $170$ to $179$ cm, the class width is $10$ cm, showing that intervals can vary in width.

Histograms

Histograms are used to visualize the underlying shape of a data distribution, such as whether it is symmetrical, skewed, increasing, or decreasing. They follow specific rules:

1. The area of the bar, not the height, is proportional to the frequency.
2. Bar widths correspond to the class intervals (class width) and are bounded by the class boundaries.
3. Bars are drawn on a continuous, linear, horizontal scale with no spaces between them unless a class has zero frequency.
4. The vertical axis represents frequency density.

Frequency density is defined by the formula:

$Frequency\ Density = \frac{frequency}{class\ width}$

If the data is spread evenly on both sides of the centre, the distribution is described as symmetrical.

Cumulative frequency

Cumulative frequency curves plot running totals of data. They are used to make estimates of statistics such as the median, quartiles, and range. These curves can be used for both discrete and continuous data. The running total is always plotted against the upper class limit (or upper boundary). The resulting curve usually forms an elongated S shape. Points may be joined with a smooth curve or straight lines.

On the vertical axis, if the total frequency is $n$, the median is the horizontal value corresponding to $n/2$ on the vertical axis. The lower quartile corresponds to $n/4$ and the upper quartile to $3n/4$. The interquartile range is the difference between these two horizontal values. On a graph of marks, a point $(x, y)$ shows that $y$ individuals achieved $x$ marks or fewer.

Worked Example: Discrete vs Continuous Data

Consider the following data sets:

1. Length of a leaf: This is continuous because a leaf can take any value within a range.
2. Number of insects caught in a trap: This is discrete because the count must be an integer.
3. Mass of kittens: This is continuous because mass can take any value within a range.

Worked Example: Class Intervals for Continuous Data

Heights of tomato plants are measured to the nearest cm. We must determine the class boundaries.

Class limits (cm)	Class interval (cm)
50 to 59	$49.5 \leq h < 59.5$
60 to 69	$59.5 \leq h < 69.5$
70 to 79	$69.5 \leq h < 79.5$
80 to 89	$79.5 \leq h < 89.5$
90 to 109	$89.5 \leq h < 109.5$

Note that the upper boundary of the first class ($59.5$) becomes the lower boundary of the second class.

Worked Example: Drawing a Histogram

Using the plant data above, we calculate frequency density.

Class interval	Frequency	Class width	Frequency density
$49.5 \leq h < 59.5$	15	10	$15 / 10 = 1.5$
$59.5 \leq h < 69.5$	20	10	$20 / 10 = 2.0$
$69.5 \leq h < 79.5$	25	10	$25 / 10 = 2.5$
$79.5 \leq h < 89.5$	53	10	$53 / 10 = 5.3$
$89.5 \leq h < 109.5$	30	20	$30 / 20 = 1.5$

The histogram is drawn with frequency density on the vertical axis. If there is a large gap between $0$ and the first boundary, the axis can be compressed or start from the first boundary.

Alternatively, a zigzag line can show the scale compression:

Or the bars can simply start at the relevant boundary on a linear scale:

Worked Example: Cumulative Frequency

We catch insects in 65 traps. We need to calculate cumulative frequency and estimate the median.

Insects ($n$)	Class Interval	Freq	Cumulative Freq	Interval
1 to 5	$0 < n \leq 5$	5	5	$0 < n \leq 5$
6 to 10	$5 < n \leq 10$	8	13	$0 < n \leq 10$
11 to 15	$10 < n \leq 15$	15	28	$0 < n \leq 15$
16 to 20	$15 < n \leq 20$	20	48	$0 < n \leq 20$
21 to 25	$20 < n \leq 25$	14	62	$0 < n \leq 25$
26 to 30	$25 < n \leq 30$	3	65	$0 < n \leq 30$

We plot cumulative frequency against upper limits ($5$, $10$, $15$, etc.).

Join the points, starting from $(0, 0)$.

To find the median, divide the total ($65$) by $2$ to get $32.5$. Read across from $32.5$ on the vertical axis to the curve, then down to the horizontal axis. The median is approximately $16$.

To find traps with more than $17$ insects: read up from $17$ on the horizontal axis to find $36$ on the vertical axis. This means $36$ traps have $17$ or fewer insects. Thus, $65 - 36 = 29$ traps have more than $17$ insects.