 Histograms Questions | Worksheets and Revision | MME

## Histograms

When displaying grouped data, especially continuous data, a histogram is often the best way to do it – specifically in cases where not all the groups/classes are the same width. Histograms are like bar charts with $2$ key differences:

• There are no gaps between the bars
• It’s the area (as opposed to the height) of each bar that tells you the frequency of that class.

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

## Frequency Density

In order to make this work, when drawing a histogram, we plot frequency density on the $y$-axis rather than frequency. The frequency density for each group is found using the formula:

$\text{frequency density} = \dfrac{\text{frequency}}{\text{class width}}$

Level 6-7

## Example 1: Drawing a Histogram

Below is a grouped frequency table of the lengths of $71$ pieces of string.

Construct a histogram of the data.

[4 marks] To construct a histogram, we will need the frequency density for each class. Dividing the frequency of the first class by its width, we get

$\text{frequency density } =\dfrac{8}{20-0} = 0.4$ Once we have calculated the frequency density with the remaining groups, then it is good to add a third column to the table containing the frequency density values, see the completed table.

Once this new column is completed, all that remains is to plot the histogram.

With lengths on the $x$-axis and frequency density on the $y$-axis, each bar that we draw will have width equal to its class width, and height equal to the relevant frequency density.

The resulting histogram is shown.

Level 6-7

## Example 2: Interpreting Histograms

Below is a histogram showing the times taken to complete a quiz.

$44$ people took between $0$ and $1.5$ minutes.

Work out how many people took between $3$ and $4$ minutes.

[4 marks] To answer this question, we’re going to use the information to work out how much $1$ small square of area is worth.

Between $0$ and $1.5$ minutes includes all of the first bar and some of the second. From $0$ to $1$ minutes there are $10\times 12 =120$ small squares, and from $1$ to $1.5$ there are $5\times 20=100$ small squares (marked on the graph below for clarity). So, in total there are $100+120=220$ small squares between $0$ and $1.5$ minutes, and the question tells us that this accounts for $44$ people. Therefore, $1$ person is equal to

$220 \div 44=5\text{ small squares}$.

Now, reading from the graph we get that there are $11 \times 10 = 110$ small squares between $3$ and $4$ minutes, so given that $5$ small squares is one person, there must be

$110 \div 5 = 22\text{ people}$

who took between $3$ and $4$ minutes to do the quiz.

Level 8-9

### Example Questions

In order to draw a histogram, we need to know the frequency density for each row of data. The frequency density is calculated by dividing each frequency by its associated class width.

This means that we need to create a new column on the data table for the frequency densities.

The first row of the table has a plant height from $0 - 10$cm and a frequency of $6$. As mentioned above, the frequency density is the frequency divided by the band width, so the frequency density for the first row can be calculated as follows:

$\text{Frequency density} = 6 \div10 = 0.6$

By repeating this process for the remaining four rows, our completed frequency density column will look like the one below: Now we are in a position to draw the histogram. The height will be on the the $x$-axis and the frequency density on the $y$-axis. Since the band widths are not consistent (the band width of the $20 - 24$ cm category is only $4$ cm whereas the band width for the $30 - 50$ cm category is $20$ cm), this means that the widths of the bars you draw will not be the same.

Your completed histogram should look like the one below: The number of values in each class is represented by the area of each bar (and not the height). We have been told that $54$ people can hold their breath for at least a minute, so this means that the area of the bars from $60$ seconds upwards represents $54$ people.

People who can hold their breath for $1$ minute or more is represented by the whole of the last bar ($70 - 100$ seconds) and the right-hand part of the second-to-last bar ($60 - 70$ seconds). To work out the area in these two bars, we simply need to count the small squares:

$(5 \times 15) + (15 \times 4) = 75 + 60 = 135$

This is illustrated in red on the histogram below.

If $135$ small squares represents $54$ people, we can work out how many people one small square represents:

If

$54\text{ people} = 135\text{ small squares}$

then:

$\text{1 person } = \dfrac{135}{54} = 2.5\text{ small squares}$

Now that we know that $1$ person is represented by $2.5$ small squares, we need to work out how many small squares there are between $20$ and $40$ seconds.

The number of small squares between $20$ and $40$ is:

$(5 \times 32) + (5 \times 20) = 160 + 100 = 260$

This is illustrated in green on the graph below.

Therefore, the number of people who can hold their breath for between $20$ and $40$ seconds is:

$\dfrac{260}{2.5} = 104\text{ people}$ a) Since we are taking data from the histogram, we can see the frequency density and the band width, but we need to work out how many riders (the frequency) rode for $30$ kilometres or less.

The key formula when we are dealing with histograms is:

$\text{Frequency density} = \dfrac{\text{frequency}}{\text{bandwidth}}$

If we need to work out the frequency, then we simply need to rearrange this formula:

If

$\text{Frequency density} = \dfrac{\text{frequency}}{\text{bandwidth}}$

then

$\text{Frequency} = \text{ frequency density}\times\text{ bandwidth}$

The number of riders (the frequency) who rode between $0$ and $20$ kilometres can be calculated as follows:

$4\times20 = 80\text{ riders}$

The number of riders (the frequency) who rode between $20$ and $30$ kilometres can be calculated as follows:

$10\times10 = 100\text{ riders}$

Therefore the number of riders who rode between $0$ and $30$ kilometres is:

$80+100=180\text{ riders}$

b) In order to work out the mean journey length, we need to work out how many riders there are in total. In order to do this, we need to work out how many riders rode between $0 – 20$ kilometres, $20 – 30$ kilometres, $30 – 54$ kilometres etc.

We already know from the previous question that $80$ riders rode between $0$ and $20$ kilometres and that a further $100$ riders rode between $20$ and $30$ kilometres.

In the $30 – 57$ kilometres category, we have a band width of $27$ kilometres and a frequency density of $2$, so the number of riders can be calculated as follows:

$27\times2 = 54\text{ riders}$

In the $57 – 70$ kilometres category, we have a band width of $13$ kilometres and a frequency density of $9$, so the number of riders can be calculated as follows:

$13\times9 = 117\text{ riders}$

In the $70 – 90$ kilometres category, we have a band width of $20$ kilometres and a frequency density of $6$, so the number of riders can be calculated as follows:

$20\times6 = 120\text{ riders}$

Although we now exactly how many riders rode in each distance category, we cannot know exactly how far each rider rode since we are dealing with grouped data. In the $0 – 20$ kilometres category, the $80$ riders could have cycled $1$ kilometre or $19$ kilometres. What we have to do is assume that the distance that each cyclist rode is the midpoint of each distance category (this is why this is an estimated mean and not an accurate mean).

The easiest thing for us to do is to tabulate our data, with one column for the midpoint of each distance category, another column for the frequency (number of riders) and another column for the midpoint multiplied by the frequency (this last column is to work out the total distance travelled by all the riders in that category combined because to work out the mean, we will need to divide the total distance travelled by all riders by the number of riders).

The tabulated data should look like the below: The total of the frequency column is the total number of riders. The total of the ‘midpoint multiplied by frequency column’ is the total distance travelled by all of the riders. Therefore the estimated mean can be calculated as follows:

$\text{Estimated mean} = 22678.5\text{ kilometres} \div \text471\text{ riders} \approx 48\text{ kilometres}$

a) In order to complete the rest of the histogram, we need to work out the frequency densities for the length categories which have not already been drawn on the histogram.

The frequency density for the $0 – 4$ cm length category can be calculated as follows:

$\text{Frequency density} = 32 \div 4 = 8$

The frequency density for the $10– 20$ cm length category can be calculated as follows:

$\text{Frequency density} = 22 \div 10 = 2.2$

The frequency density for the $20 – 40$ cm length category can be calculated as follows:

$\text{Frequency density} = 42 \div 20 = 2.1$

The frequency density for the $40 – 45$ cm length category can be calculated as follows:

$\text{Frequency density} = 30 \div 5 = 6$

The frequency density for the $55 – 70$ cm length category can be calculated as follows:

$\text{Frequency density} = 9 \div 15 = 0.6$

Now that we have worked out the frequency density for each length category, we can now plot them on the histogram, with a result similar to the below: b) For this part of the question, we need to fill in the gaps in the frequency column of the table. In order to do this, we will need to take a frequency density reading from the histogram for the $2$ length categories in question.

Reading from the histogram, we see that the frequency density for the $4 – 10$ cm category is $3.5$, and the frequency density for the $45 - 55$ cm category is $4.6$. All we need to do is rearrange the frequency density formula so that we can work out the frequency.

Since

$\text{Frequency density} = \dfrac{\text{frequency}}{\text{bandwidth}}$

then

$\text{Frequency} =\text{frequency density}\times\text{bandwidth}$

Therefore, the frequency for the $4 – 10$ cm length category can be calculated as follows:

$3.5\times6=21$

The frequency for the $45 – 55$ cm length category can be calculated as follows:

$4.6\times10=46$ a) The key piece of information in this question is that $15$ bags of flour weigh between $35$ and $40$ pounds. What we need to do is look and see what area of the histogram this represents. The area of the $35 – 40$ pounds bar (do not accidentally work out the area of the entire $30 – 40$ pounds bar!) can be calculated as follows:

$2.5\times30\text{ small squares} = 75\text{ small squares}$

We can therefore conclude that $15$ bags of flour is represented by $75$ small squares.

If

$15\text{ bags} = 75\text { small squares}$

then

$1\text{ bag} = 5\text { small squares}$

All we need to do now is work out how many small squares there are from $80$ pounds upwards.

Between $80$ and $95$ pounds there are $75$ small squares, and between $95$ and $100$ pounds, there are a further $125$ small squares, giving us a total of $200$ small squares.

Since $5$ small squares represents a single bag of flour, then $200$ squares represents $40$ bags of flour.

b) The answer to part a) can only be an estimate because we are dealing with grouped data. We have made the assumption that the number of bags that weigh between $80$ and $95$ pounds is $\frac{3}{5}$ of the number of bags of flour that weigh between $70$ and $95$ pounds. At one extreme, it is possible that all of these bags of flour are less than $80$ pounds and, at the other extreme, it is possible that they might all weigh more than $80$ pounds.

c) We know from the question that there are $185$ bags of flour in total. Therefore the median weight of a bag of flour is the weight of the $93^{\text{rd}}$ bag (since $93$ is the ‘mid-point’ of $185$). We will therefore need to work out which weight band the $93^{\text{rd}}$ bag of flour falls into. This is going to be difficult (impossible) at this stage since we do not know how many bags of flour are in the $30 – 40$ pound category, the $40 – 55$ pound category etc.

We know from the first question, that $15$ bags of flour weigh between $35$ and $40$ pounds. Since this is half of the total of the $30 – 40$ pound category, the number of bags between $30$ and $40$ pounds is:

$15\times2 = 30\text{ bags}$

In the $40 – 55$ pound category, the area is $1.5$ times the $30 – 40$ pound strip, so this represents:

$30\times1.5 = 45\text{ bags}$

So far we have accounted for the first $75$ bags of flour $(50+75=125)$ so haven’t reached the $93^{\text{rd}}$ bag of flour yet.

The $55 – 65$ pound category has the same width as the $30 – 40$ pound category. If we compare the area to the $30 – 40$ pound category, its area is $25$ small squares larger than the $30 – 40$ pound category. Therefore, once we know what an area of $25$ small squares represents, we can add this to $30$ (the number of bags represented by the $30 – 40$ pound category).

We know from the first question that $5$ small squares corresponds to $1$ bag, so $25$ small squares will correspond to $5$ bags.

Therefore the $55 – 65$ pound category corresponds to $35$ bags.

Since there are $30$ bags in the $30 – 40$ pound category and a further $45$ bags in the $40 – 55$ pound category, there are $75$ bags that have a weight between $30$ and $55$ pounds. Therefore the $55 – 65$ pound category accounts for the $76^{\text{th}}$ bag to the $110^{\text{th}}$ bag ($110$ since there are $75$ bags between $30$ and $55$ pounds and $35$ bags between $55$ and $65$ pounds). We are trying to locate the weight of the $93^{\text{rd}}$ bag, so we know it must be in the $55$ to $65$ pound weight category.

We are now in a position to calculate the estimated weight of the $93^{\text{rd}}$ bag (this is the hard bit!).

By subtracting the $75$ bags that weigh less than $55$ pounds from $93$, we can work out that the $93^{\text{rd}}$ bag will be the $18^{\text{th}}$ of the $35$ bags between $55$ and $65$ pounds. We can write this as $\frac{18}{35}$. So where in the weight category does this fall? It will fall $\frac{18}{35}$ of the way between $55 – 65$ pounds. Since this is a weight category of $10$ pounds, we will need to perform the following calculation:

$\dfrac{18}{35}\times10=5.14\text{ pounds}$

Since the category starts at $55$ pounds, then the weight of the median bag (the $93^{\text{rd}}$) bag is $55+5.14=60.14 \text{ pounds}$

(This last part seems complicated, but only because the fraction is not that easy. If there were $20$ bags in the $55 – 65$ pound category, and it was the $10^{\text{th}}$ bag in this category that represented the median, since the $10^{\text{th}}$ bag in the category is exactly half way through the $20$ bags in the category, then its estimated weight would simply be half way between $55$ and $65$ pounds, so would therefore have a weight of $60$ pounds.)

### Worksheets and Exam Questions

#### (NEW) Histograms Exam Style Questions - MME

Level 6-7 New Official MME

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