### Video Transcript

In this video, we’re going to talk
about frequency density and how we can calculate the frequency density and then use
it to draw a histogram. Let’s begin by thinking about what
a histogram looks like. You might think that it looks a
little bit like a bar chart, and you’d be right. But there are some important
differences. When we have a bar chart like the
one below, then we have frequency on the 𝑦-axis. In a histogram, on the 𝑦-axis, we
instead have frequency density. Before we talk a little bit more
about frequency density, there’s a few other points visually about a histogram.

The first thing is to notice that
often in a histogram, the bars are of unequal width. Secondly, we notice that there are
no gaps between the bars in a histogram. The bar chart, however, does have
gaps between the bars. In the bar chart, it’s very easy to
find the frequency by reading the value on the 𝑦-axis. So how can we find frequency from a
histogram? Well, in a histogram, we find the
frequency by finding the area of the bars. As each bar is a rectangle, then we
can find the area by multiplying the length by the width. But what are these lengths and
widths?

We can work out the base length
here by finding the class width. If, for example, we had the width
here going from two to four, then the class width would be two, since four minus two
is two. The other length of this bar, this
rectangle, is given by the frequency density. So therefore, if we’re given a
histogram and we want to calculate the frequency, we do this by calculating the
class width times the frequency density for each particular bar. So what happens if, instead, we’re
given a table and we need to actually draw the histogram? If we were drawing a bar chart,
we’d simply plot the frequency on the 𝑦-axis. But remember, we need to find the
frequency density.

The fact remains that the area of
the bars tells us the frequency. That means, in our first bar, the
area would be 12. We can work out one of the
dimensions of our rectangle by calculating the class width of this interval. In this example, three, four, five,
and six would fit in this class, so the class width would be four. We can see from the table that the
frequency, the area of this bar, must be 12. Therefore, the other length of this
bar, the frequency density, must be three, since four times three would give us
12.

We can put this into the form of a
formula. Frequency density is equal to
frequency divided by the class width. To draw a histogram, we need to
work out the frequency density for each data set in a given table. So whether we’re interpreting a
histogram that we’re given or drawing a histogram from a table, we’ll need to be
able to use and recall both of these two formulas. Let’s have a look in more detail at
some questions. And we’ll begin by calculating the
frequency density.

The frequency table provided shows
the ages of teenagers in a neighborhood. Find the missing density from the
table.

In this question, we’re looking at
frequency density rather than just frequency. We would find frequency density if
we wanted to go on and create a histogram from the data. Frequency density tells us the
ratio of a frequency of a class to its width. We can calculate frequency density
by using the formula that frequency density is equal to frequency over the class
width. It can often be helpful to add in a
row to our table to work out the class width for each interval.

Let’s take a look at our first
interval and work out what exactly we mean by this. 13 is less than 𝑎 is less than or
equal to 14. The age in years means that the age
must be greater than 13 but not equal to it. The second part of this inequality
𝑎 is less than or equal to 14 means that the age can be 14 or less than. So if 𝑎 must be bigger than 13 and
not equal to it and less than 14, it leaves us really only with one possible
value. And that’s the age of 14 years
old. 14 is larger than 13, and although
it isn’t less than 14, it can be equal to it. The class width here would simply
be one.

In the second inequality, we’re
told that the age must be greater than 14. So teenagers in this category could
not be 14. And the ages go up to and including
17. So the teenagers in this category
would either be 15, 16, or 17 years old. So the class width here would be
three. In our final column, we have ages
greater than 17 but not including it and less than or equal to 19 years old. The teenagers in this category
would either be 18 years old or 19 years old. So the class width here would be
two.

Let’s have a look at the frequency
densities that have already been calculated. In the first column, we can see
that the frequency density given is 21. The formula tells us that we take
the frequency of 21 and we divide by the class width, in this case, one. And 21 over one does indeed give us
21. In the second column, to find the
frequency density, we will take our frequency of 24 and divide by the class width of
three. And 24 divided by three does indeed
give us a frequency density of eight. To find the missing frequency
density in the final column, we take the frequency of 10 and divide by the class
width of two. 10 divided by two gives us five,
the answer for the missing frequency density.

In this question, we weren’t asked
to create a histogram from the data. But if we did, what would it look
like? When we create a histogram on our
𝑦-axis, we don’t have frequency like we would in a bar chart. But instead, we plot frequency
density. The age would be plotted on the
𝑥-axis. And we can see that each bar in
this histogram goes up to each individual frequency density. The bars are all next to each
other. And in a histogram, we would
calculate the frequency by finding the area of each bar. Histograms can be a very good way
for visualizing the distribution of values in data sets. In this question, we didn’t need to
draw the histogram just to give the missing frequency density of five.

Let’s have a look at another
example of how to find the frequency density.

The frequency table below
represents the times students take to walk to school. Find the frequency density for each
frequency class.

In this question, we’re asked for
the frequency density. The frequency density gives us the
ratio of the frequency of a class to its width. We might calculate frequency
density if we wish to create a histogram. We can use the formula that
frequency density equals the frequency divided by the class width. First, we’ll work out the class
width and then we’ll find the frequency density for each class. So in this first class interval of
students walking to school, we have the time is given as zero is less than 𝑡 is
less than or equal to four. This means that the time 𝑡 minutes
can’t be equal to zero. But it could be one, two minutes,
three minutes, or even four minutes.

This inequality here — 𝑡, time, is
less than or equal to four — means that the time could also be equal to four
minutes. We could say then that there are
four students that took either one, two, three, or four minutes to walk to
school. As there are one, two, three, four
options in our class width, the class width is four. In our second column, we have the
inequality four is less than 𝑡 is less than or equal to eight, which means that our
time in 𝑡 minutes does not include four. But it does include five, six,
seven, or eight, since the time is less than or equal to eight minutes. Therefore, the interval from four
to eight gives us a class width of four.

The final two columns also have a
class width of four. To find the frequency density for
each interval then, we can use the formula. In our first column, we have a
frequency of four divided by a class width of four. So our first frequency density will
be one. In the next column, we have a
frequency of 12 divided by a class width of four. And 12 divided by four is
three. In the next column, a frequency of
16 divided by the class width of four gives us a frequency density of four. In the final column, the frequency
of eight divided by the class width of four gives us the frequency density of
two. And so we have our answer for the
frequency density for each frequency class. That’s one, three, four, and
two.

If we wanted to create a histogram
from the given data, it would look like this. We plot the frequency density on
the 𝑦-axis rather than just the frequency. And along the 𝑥-axis, we have the
time in minutes. We remember that we can calculate
the frequency if we’re given a histogram by finding the area of each of the
bars. In this question, the class widths
were all the same at a width of four. But often we find in histograms the
bars will be of varying widths. In this question, however, we
didn’t need to draw the histogram, just to give the frequency densities, which we
found.

In the next question, we’ll compare
a number of histograms with a given frequency table.

The frequency table below gives the
heights of a group of students. Which of the following histograms
represents the given data?

In this question, we’re thinking
about histograms. When we have a histogram, on the
𝑦-axis, we have frequency density, rather than frequency, which we would have in a
bar chart. This is why if we have a quick look
at our options, on our first bar, the height doesn’t go up to 60 as this would be
simply representing the frequency. The frequency in a histogram is
found by calculating the areas of the bars. To create a histogram from the
given frequency table, the first thing we need to do is calculate the frequency
density for each class. We can find frequency density by
finding the frequency and dividing by the class width. And what exactly do we mean by
class width?

Well, if we take a look at our
first class, the height ℎ of this group of students, in the first interval, we have
the inequality that 150 is less than ℎ is less than or equal to 153 centimeters. In this interval, we would
therefore have students who are 151 centimeters, 152 centimeters, or 153
centimeters. We couldn’t have students here
exactly equal to 150 centimetres because of this inequality which doesn’t include
the equals to. The class width in this case would
be three. So to find the frequency density,
we would take our frequency of 60 and divide by the class width of three. And 60 divided by three is 20.

In the second class, we can find
our class width by having 156 subtract 153. In this case, the class width would
be three. The values in this interval would
be 154, 155, and 156. Easing the formula to find the
frequency density, we would take our frequency of 45, divide by the class width of
three. And we’d have a value of 15 for
this frequency density. In our next column, the class width
is a little bit more interesting. In this case, we can’t include the
value of 156 centimeters, since our inequality says that we can’t have an equals to
this value. This interval will go up to and
including the value of 157. So we just have one possible
value. Frequency density here is the
frequency of 30 divided by the class width of one, giving us a frequency density of
30.

The next class will have a class
width of five, since we have the values 158, 159, 160, 161, and 162. The frequency density is the
frequency of 100 divided by the class width of five, which is 20. In our final class interval, we’ll
have a frequency density of 15, as 45 divided by the class width of three gives us
15.

So let’s take a look at the
histogram given in option (A). Our first interval goes from 150 to
153. The frequency density is 20. The first histogram shows this bar
correctly. The interval of 153 to 156 has a
correct frequency density of 15. The third interval from 156 to 157,
however, in option (A) is not given correctly. Here, the frequency density would
be given as 15. We could assume that the person who
drew this histogram incorrectly worked out the class width as two. So we can rule out option (A) as
the histogram.

If we look at options (B) and (C),
both of these have the same correct two bars to begin with. If we take a close look at option
(B), the interval 156 to 157 we can see on the histogram is incorrectly given as the
interval from 156 to 159. So this bar is wrong. And we can rule out option (B) as
the histogram. We need to check that the other
bars in option (C) are completely correct. We can see that our third bar
correctly has a frequency density of 30. The fourth bar, which structures
correctly along from 157 to 162, does have a correct frequency density of 20. The final bar does correctly go
from 162 to 165 on our 𝑥-axis. And the frequency density is
correct at 15. We can therefore say that option
(C) is the histogram to represent the data in the frequency table.

We’ll now look at one final
question on histograms.

The following table represents the
speed of the members of a cycling club over a long-distance race. David drew a histogram to represent
this data. What is the mistake in this
histogram?

Here we have a table that shows us
the speed of members of a cycling club. We’re shown a histogram of this
data. And we can recall that in a
histogram, we don’t plot frequency on the 𝑦-axis. But instead we plot frequency
density, just like David did. In order to find the frequency
density, we calculate the frequency divided by the class width. In order to calculate the frequency
density of our first class interval, we could quickly subtract it from five to give
us three. We could also see that the values
six, seven, and eight miles per hour would be in this interval. We can’t include the value of five
miles per hour because of the inequality here that the speed has to be greater than
five.

As the class width is three here,
to find the frequency density, we take our frequency of 12 and divide by the class
width of three, which gives us a frequency density of four. If we take a quick look at the
histogram, we can see that David correctly drew the interval from five to eight. And the frequency density was
correct with a value of four. In the second column, we have the
inequality that eight is less than 𝑠 is less than or equal to 10, where 𝑠 is the
speed. In this case, we’d have the values
of nine and 10, giving us a class width of two. The frequency density would be the
frequency of 24 divided by the class width of two, which gives us a density of
12. This bar is correct on the
histogram.

We can find the next frequency
density by dividing our frequency by five, which was our class width, to give us
seven. This is correctly given on David’s
histogram. The final three frequency densities
can be calculated as four, one, and one. If we look at our histogram, we can
see that our fourth bar is correct and so is the final one. But this one is not. So what is the mistake? Well, he drew his interval going up
to a height of 10, which would be the frequency. We could give our answer that the
bar of the interval 20 is less than 𝑠 is less than or equal to 30 represents the
frequency instead of the frequency density.

We can now recap what we’ve learned
in this video. We saw that histograms are like bar
charts. Only instead histograms have
frequency density on the 𝑦-axis. We can find the frequency by
finding the area of the bars. To draw a histogram, we can use the
formula for frequency density. To calculate the frequency from a
histogram, we could use this formula for finding the frequency.