HistogramsIB MYP Maths Extended: Revision notes

A histogram shows continuous data (such as time, mass or age) in classes. There are no gaps between the bars. Unlike a bar chart, a histogram can have classes of unequal width. When the widths are unequal, the height of the bar cannot be the frequency, or wide classes would look bigger than they are. Instead, the area of each bar represents the frequency, and the height is the frequency density. $\text{frequency}=\text{frequency density}\times\text{class width}$

$\text{frequency density}=\frac{\text{frequency}}{\text{class width}}$ Example: 21 customers waited between 2 and 5 minutes. The class width is $5-2=3$, so the frequency density is $\frac{21}{3}=7$. Add a column for the class width and another for the frequency density to your table. The class with the tallest bar is the one with the greatest frequency density, which is not always the class with the greatest frequency. For example, a class with frequency 30 and width 5 has a density of 6, which is lower than the density 7 of a class with frequency 21 and width 3.

1. Make a table with the class, frequency and class width.
2. Work out the frequency density for each class.
3. Draw a horizontal axis for the data (continuous, with a scale) and a vertical axis labelled frequency density.
4. Draw a bar for each class from its lower boundary to its upper boundary, with height equal to its frequency density. The bars touch.
5. Give the histogram a title and label the axes with units. Example: classes $0<t\le2$, $2<t\le5$, $5<t\le10$, $10<t\le20$ with frequencies $8,21,30,11$ have frequency densities $4,7,6,1.1$.

To find the frequency of a class, multiply its bar height by its width: $\text{frequency}=\text{frequency density}\times\text{class width}$. Example: classes with frequency densities $2.4$, $3.5$, $4.2$, $1.5$ and widths $10,10,20,30$ have frequencies $24,35,84,45$ and the total is $188$. The total of all the frequencies is the number of items in the data set, which you can use to check your work.

Sometimes a question asks about a range that covers only part of a class. Assume the data is evenly spread across the class, so the number you want is proportional to the part of the width. Example: the class $20\le a<40$ has frequency $84$. The number aged 20 to 30 is $\frac{10}{20}\times84=42$, or $4.2\times10=42$. Add the whole classes below or above the part. For under 30: $24+35+42=101$. Say that your answer is an estimate, because the real spread inside a class is not known.