Lesson video

Hi everyone.

My name is Ms. Koo and I'm really excited to be learning with you today as we're looking at histograms. Histograms are fantastic graphical representations of data.

I hope you enjoy the lesson.

Let's make a start.

Hi everyone and welcome to this lesson on histograms with unequal bar width.

Under the unit, Graphical representations of data: cumulative frequency and histograms. And by the end of the lesson you'll be able to construct a histogram from a table.

Today's keywords consist of discreet data.

Discrete data is data that can only take distinct, specific values.

For example, shoe sizes or number of people.

We'll also be looking at the keyword continuous data and continuous data can take any value within a range.

For example, height, mass, and temperature.

We'll also be looking at a histogram.

And a histogram is a diagram consisting of rectangles, whose area is proportional to the frequency in each class and whose width is equal to the class interval.

We'll also be looking at the keyword frequency density.

Frequency density is proportional to the frequency per unit for the data in each class.

Often the multiplier is one, meaning that the frequency density is equal to the frequency divided by the class width.

Today's lesson will be broken into two parts.

We'll be looking at frequency density first and then moving on to constructing histograms with unequal bar width.

So let's look at frequency density.

Which city do you think has the densest population from the list, circa 2024? And explain why you think this.

Do you think it's Dhaka in Bangladesh, London in the United Kingdom, Vatican City in Italy or Valencia, Spain? Have a little think.

well done if you've got this.

Well, it's Dhaka in Bangladesh.

There are more people living per unit area in Dhaka than any other city listed.

Now, population density is the number of people living per unit area and Dhaka has 19,447 people per square kilometre.

London has 11,144 people per square kilometre.

The Vatican city only has 1,194 people per square kilometre and Valencia has 5,800 people per square kilometre.

But how do you think we calculate the population density? Have a little think.

Well, it's the total population divided by the land area.

So we can apply the same principle to frequency density.

Frequency density is proportional to the frequency per unit for data in each class.

So let's examine the frequency density for this data.

Here is a table.

What do you notice about the class widths for these masses? Have a little think.

Well hopefully you can spot they are unequal.

Now how would we calculate the frequency per unit? In other words, the frequency density.

Well, if we know the class width is five, because five subtract zero is five, we need to work out the frequency per unit.

Well, to do this, we know the frequency is 20.

So that means the frequency density is the frequency per unit of class width.

So it's simply 20 divided by five.

So our frequency density is four.

Now let's have a look at our next class.

The class width here is 10 because 15 subtract off five is 10.

Now we know the frequency is 20.

So the frequency density is the frequency per unit of class width.

So it's simply 20 divided by 10, which is two.

Lastly, we know the class width is 20 because 35 subtract off 15 is 20.

Now we also know the frequency is 20.

So what's our frequency density? Well the frequency density is the frequency per unit of class width.

So it's simply 20 divided by 20, which is one.

So now we've identified our frequency density.

Here's a table showing the time in minutes it takes pupils to complete a quiz.

I want you to work out the frequency density.

I'm going to give you a hint, add an extra column, class width and add another column, frequency density, as it centralises all that wonderful information in one place.

Well done, let's see how you got on.

Well, adding that extra column of class width, you can see we have a class width of 20, a class width of 10, and a class width of five.

Working out our frequency density, our frequency density must be two because 40 divided by 20 is two.

Next frequency density is 3.

5 because 35 divided by 10 is 3.

5.

The next frequency density is 4.

8 because 24 divided by five is 4.

8.

Really well done if you got this.

So in summary, we know frequency density is proportional to the frequency per unit for the data in each class.

Unless scaled, we say frequency density is equal to frequency divided by class width.

This is because we're finding the frequency per unit of class width.

I want us to have a look at another check.

Here's the same table showing the time it takes pupils to complete a quiz.

Which time interval shows that for every minute two pupils finish the quiz? Have a little think.

Well, it's our first time interval.

Zero is less than T, less than equal 20.

This is because the frequency density is two and the frequency density shows the frequency per unit of class width.

In other words, for every minute we expected two pupils to finish the quiz in that time interval.

Great work everybody.

So now it's time for your task.

Question one says, here's a table showing the time in minutes it takes for pupils to complete a puzzle cube.

I want you to work out the frequency density.

I'm going to give you a nice little hint and ask you to add on those extra columns.

Add a column of class width and add a column of frequency density.

It just centralises all that wonderful information.

Press pause as you'll need more time.

Well done, let's move on to question two.

Which time interval shows that for every two minutes nine pupils finish the puzzle? Have a little think, press pause as you need more time.

Well done, let's go through these answers.

Well, for question one, hopefully you've added that extra column class width and added these class widths 10, 20, 20, 10 and 60 and worked out our frequency densities to be 0.

5, 0.

95, 1.

7, 4.

5, and one third or 0.

3 recurring.

For the second part, which time interval shows that for every two minutes nine pupils finish the puzzle? Well the frequency density is 4.

5 when 50 is less than T, less than or equal to 60.

This means for every minute 4.

5 pupils finish the puzzle.

Therefore nine pupils finish every two minutes.

Really well done if you've got this.

Great work everybody.

Now it's time for the second part of our lesson constructing histograms with unequal bar widths.

Now frequency density is proportional to the frequency per unit for the data in each class.

This means the class zero, less than W, less than equal to five.

We can assume every unit of class width has a frequency of four parcels.

And you can see in our table here.

We have a frequency density of four for this particular class width.

So we can represent this on our histogram.

Here we have a unit of our class width.

This means four parcels are represented in this unit because our frequency density is four.

Now given that we have a class width of five, this means each unit of our class width represents four parcels.

So this unit of our class width represents another four parcels.

This unit from our class width represents another four parcels.

This unit of our class width represents another four parcels, and this unit of our class width represents another four parcels.

Therefore, if we assume each unit of our class width has four parcels and we have five units, how do you think we can find the frequency of the parcels using the histogram? This is a great question.

Have a little think.

Well it's simply five multiplied by our four, which is 20 and you can see that in our table.

In other words, we can work out the frequency using the class width and the frequency density.

The frequency is the area of the bar.

So this is a really nice little summary if you want to write it down.

The frequency is equal to the class width multiplied by the frequency density or the frequency is equal to the area of the bar.

So let's finish constructing the histogram.

Next we have frequency density of two for the class of five, less than W, less than or equal to 15.

So we know this unit from our class width represents two parcels.

This unit from our class width represents two parcels, so on and so forth.

So all of these units from our class width represent two parcels.

So let's check this frequency.

How do we find the frequency? Well, it's the area of the bar.

So we simply do 10 times two.

Yes, it confirms it's 20, but we already knew that from our table.

Now we have a frequency density of one for the class width 15 less than W, less than equal to 35.

And this means we have one parcel per unit class width.

Here we have one parcel, so on and so forth.

So you can see we have 20 units where each unit represents one parcel.

Let's work out that frequency using the area of the bar.

20 times one is 20 and we already knew that from our table.

So in summary, we know frequency density is proportional to the frequency per unit for the data in each class.

Unless scaled, we can say frequency density is equal to frequency divided by class width.

This is because we're finding the frequency per unit of class width.

From a histogram, we know frequency is the area of the bar and this is the same as frequency is equal to class width multiplied by our frequency density.

This is a great amount of information on the screen.

Please press pause if you want to copy it down.

So let's move on to a check.

Here we have a table and Sam has drawn this histogram showing the length of pencils.

The table shows we have a frequency of six pencils between the length of zero and 10 centimetres, not including zero, but including 10.

We have a frequency of nine pencils between the lengths of 10 to 15 centimetres, not including our 10 but including our 15.

And we have a frequency of six pencils with the length between 15 and 25 centimetres, not including our 15 but including our 25.

What I want you to do is explain where Sam has made the error.

Have a little look, press pause as you need more time.

Well done, let's see how you got on.

Well Sam has plotted the frequency, not the frequency density.

Remember we have unequal bar widths, so we must use our frequency density as our Y axis.

So now what I would like you to do is draw the correct histogram using the above table.

Use the squares in your book and the axis labelled as well.

A big hint would be to add those two extra rows, identify that class width and add that information there.

And then work out that frequency density.

See if you can give it a go.

Press pause if you need more time.

Well done, let's see how you got on.

Well adding that class width, we should have a class width of 10, five and 10.

Working out our frequency density, we should have 0.

6, 1.

8 and 0.

6.

Now we can draw our histogram.

So from our histogram you can see we have a bar width between zero and 10 centimetres with a height of 0.

6 because this is our frequency density.

The next bar is between 10 and 15 centimetres with a frequency density of 1.

8 and the final bar shows lengths between 15 and 25 centimetres with a frequency density of 0.

6.

Really well done if you got this.

Now Laura says, "What's the point of frequency density? Can't I just draw a histogram like I did before?" So here is a table and let's explain why she can't just draw a histogram plotted against frequency when we have unequal bar widths.

Here's our information and I've plotted it against the frequency.

Do you think this is a good representation of how the data is distributed? Have a little think.

Well it certainly does not clearly show the distribution of the masses of these parcels.

So when the class widths are unequal, we can see how the data is distributed more easily using frequency density.

And here we can see there are more parcels per kilogramme between 0 less than W, less than equal to five kilogrammes.

And we can also see there are fewer parcels between 15 and 30 kilogrammes per kilogramme.

The distribution of data is clearly seen using our frequency density when we have unequal class widths.

So why do you think we do not need to use the frequency density when the class intervals are the same? Have a little think.

Well this is because the frequency per unit class width is the same for each class.

Therefore we can see the distribution of data by simply looking at the frequency.

Great work, everybody.

So now it's time for your task.

I want you to match the data table with the histogram showing lengths of string.

Press pause as you'll need more time.

Great work, let's move on to question two.

I want you to draw the histogram showing the time it took some pupils to complete a quiz.

Big hint, add that extra column with class width and add that extra column with frequency density as it always helps.

And here's the graph and I've kindly scaled it for you too.

See if you can give it a go.

Press pause if you need more time.

Well done, let's move on to question three.

I want you to draw the histogram showing the length of cat tails.

Same again, add that extra column of class width and add that extra column of frequency density.

See if you can give it a go using the graph here.

Notice how I have not scaled it for you, so make sure you use the correct scale.

Well done, let's move on to question four.

Here's the histogram showing the times pupils walked or jogged 1.

5 kilometres.

A says which time interval shows more pupils completed the 1.

5 kilometre per minute? And I want you to explain.

Part B asks which time interval has the smallest frequency? And I want you to explain.

And part C says, which time interval shows that for every two minutes, three pupils finished the 1.

5 kilometre? And I'd like you to explain.

See if you can give it a go.

Press pause for more time.

Well done, let's see how you got on.

Well for question one, you should have made this pairing.

A should have been paired up with table two.

B should have been paired up with table one, and C should have been paired up with table three.

Well done, for question two, adding on that extra column of class width and frequency density, you should have got this.

Press pause if you need more time to mark.

And then converting it to our histogram, we should have got our histogram looking like this.

Well done.

For question three, same again, adding that extra column of class width and frequency density, you should have had these values.

Press pause if you need more time to mark.

And for our histogram, I have used this scale for my X and Y axis.

You should have the same distribution even if you chose a different scale.

Well done if you got this.

Lastly for question four, which time interval shows more pupils completed the 1.

5 kilometre per minute? And explain.

Well, it's 15 less than T, less than equal to 20 because the frequency density is the highest.

Next, B, which time interval has the smallest frequency? Well, it's between zero and five minutes.

Not including zero, but including five.

And this is because the frequency is the area of the bar and this bar has the smallest area.

And for C, which time interval shows that for every two minutes, three pupils finished the 1.

5 kilometre race? Well, it's between five less than T, less than or equal to 15 because the frequency density is 1.

5.

This means that for every minute in this class width 1.

5 pupils finished their 1.

5 kilometre race or walk.

Therefore, for every two minutes, three pupils finished their 1.

5 kilometre walk.

Great work if you've got this.

Fantastic work everybody.

So in summary, frequency density is the frequency divided by class width.

This is because we are finding the frequency per unit of class width.

And if the class widths are not equal, we should use the frequency density for the Y axis instead of the frequency.

Really well done everybody.

It was great learning with you.