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Histograms

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Summary

Histograms

In a nutshell

Histograms offer a graphical way of representing grouped continuous data. A histogram is a good visual representation of how the data is distributed in terms of location, shape and dispersion. A key feature of a histogram is that the yy-axis denotes the frequency density and thus the frequency is proportional to the area of a bar.



Frequency density

In a histogram, the frequency of each interval is proportional to the area of the corresponding bar.


frequency density

Maths; Representation of data; KS5 Year 12; Histograms


class width


As the area of a rectangle is given by:

Area=length×widthArea = length \times width​​


It follows that:

Frequencyfrequency density×class widthFrequency \propto frequency \ density \times class \ width​​


Therefore:

Frequency=k×frequency density×class width\boxed{Frequency = k\times frequency \ density \times class \ width}​​


Where kk​ is the constant of proportionality.


Note: In the case k=1k=1, the frequency is given by the area of the bar.



Histograms

In order to draw a histogram, the frequency density must be calculated using the above formula for each class interval, or for each row of the grouped frequency table. A histogram can then be plotted with the class interval on the xx-axis and frequency density on the yy-axis. The completed histogram shows the distribution of data and can be used for further analysis.


Example 1

The table shows the ages of students in an after-school Maths club. Given that the interval 6<x106 \lt x \le 10​ has frequency density 11​, find the other frequency densities and draw a histogram to show the distribution of students.


Age

Frequency

6<x106 \lt x \le 10​​
88​​
10<x1210 \lt x \le 12​​
1010​​
12<x1312 \lt x \le 13​​
44​​
13<x1813 \lt x \le 18​​
22​​


First, add columns to show the class width and frequency density. The class width is calculated by subtracting the lower bound from the upper bound for each class.


Age

class width

Frequency

Frequency density

6<x106 \lt x \le 10​​
44​​
88​​
11​​
10<x1210 \lt x \le 12​​
22​​
1010​​
​​​
12<x1312 \lt x \le 13​​
11​​
44​​
​​​
13<x1813 \lt x \le 18​​
55​​
22​​
​​​


Use the formula for frequency along with the information given in the first row to find the constant of proportionality.

Frequency=k×frequency density×class width8=k×1×4k=2\begin{aligned}Frequency &= k\times frequency \ density \times class \ width\\ 8&=k\times 1\times 4\\k&=2\end{aligned}​​


Rearrange the formula to make frequency density the subject.

Frequency=k×frequency density×class widthFrequency=2×frequency density×class widthFrequency density=frequency2×class width\begin{aligned}Frequency &= k\times frequency \ density \times class \ width \\Frequency &= 2\times frequency \ density \times class \ width\\ Frequency \ density&=\dfrac{frequency}{2\times class\ width}\end{aligned}​​


Use this formula to work out the remaining frequency densities.


Age

class width

Frequency

Frequency density

6<x106 \lt x \le 10​​
44​​
88​​
11​​
10<x1210 \lt x \le 12​​
22​​
1010​​
​​​102×2=2.5\dfrac{10}{2\times2}=2.5​​
12<x1312 \lt x \le 13​​
11​​
44​​
​​​42×1=2\dfrac{4}{2\times 1}=2​​
13<x1813 \lt x \le 18​​
55​​
22​​
22×5=0.2\dfrac{2}{2\times 5}=0.2​​​​


Plot the frequency density against age on a grid:


Maths; Representation of data; KS5 Year 12; Histograms



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