Lesson Video: Frequency Density and Histograms | Nagwa Lesson Video: Frequency Density and Histograms | Nagwa

# Lesson Video: Frequency Density and Histograms Mathematics

In this video, we will learn how to calculate frequency density and plot histograms.

17:58

### 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.

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