Lesson video

In this lesson, we're going to learn how to find frequency from a histogram.

All the information we need to complete this table can be gathered from this histogram.

Because the frequency density is the height to each bar.

The class width is the width of each bar.

And the frequency is the area of each bar.

Let's complete the first column of the table.

Using the bars that we can see drawn against the horizontal axis.

So we have nought to five, five to 10, 10 to 20 and 20 to 30.

And this is where we find the class width.

So the difference between zero and five is five.

The difference between five and 10 is five.

10 and 20 is 10.

20 and 30 is ten.

Now the frequency density we can find by using the vertical axis.

So we have 1.

2 for the first bar.

The second bar is 2.

8.

The third bar is 1.

3 high, and the fourth bar is nought point three high.

And the frequencies can be found by multiplying the class width by the frequency density in each case.

So we have six, then 14, then 13, and then three.

And that's how we can find frequency from a histogram.

Here's a question for you to try.

Pause the video to complete the task and restart the video when you're finished.

Here are the answers.

In this example, we had to draw two of the bars and to do that we need the frequency density.

So in the table we have the frequencies of eight and 17.

We need to find the class width using the first column and then dividing frequency by class width gives us the frequency density.

This is the heights of the bars that we need to draw.

For those bars that are already pre-drawn, we can find the frequency density by looking at each height and then dividing by the class width, in order to find the frequency.

Here's another question for you to try.

Pause the video to complete the task and restart the video when you're finished.

Here are the answers.

Remember that the area of the bar is the frequency.

So you have the heights of the pre-drawn bars.

You just need to multiply by the class width, and this will give you the frequency to complete your table.

Sometimes we won't be given a scale on the vertical axis and we need to work that out so we can find the rest of the frequencies.

We have one frequency for the bar between 20 and 30.

We know that it's 22.

And we know the class with this ten because that's the difference between 20 and 30.

So the frequency density is 22 divided by 10, which is 2.

2.

That's the height of the bar between 20 and 30.

If we look at the scale, we can see that that equates to 11 increments.

So each small increment on the vertical axis represents nought point two.

And we have markings every five increments, so they must represent one.

So here we have a scale which is increasing by one each time.

Now we can go ahead and find the rest of the frequencies from a histogram.

Let's start by putting all the class widths in.

By finding the difference between the upper and lower bounds for each class.

Now the frequency densities are simply the heights of each bar, which we can read now because we have a vertical scale.

And finally, our frequency can be worked out by multiplying together our class width and our frequency density for each group.

So 10 multiplied by nought point six, five locks with 1.

4, five multiplied by 3.

4, 10 multiplied by 1.

5, and 20 multiplied by nought point seven.

Now we found all the information from this histogram.

Here's a question for you to try.

Pause the video to complete the task and restart the video when you're finished.

Here are the answers.

It doesn't matter which of the first three bars you use to work out the scale.

Probably the easiest one would be the third one, because it has a height of three.

Because we know the frequency density is three, from our table.

And then we can see that the scale on the vertical axis was increase by one each time, because that is three.

And therefore once we've got the scale, we can work out all the other frequency densities, and subsequently the frequencies and put those in the table.

Here's another question for you to try.

Pause the video to complete the task and restart the video when you're finished.

Here are the answers.

In this example you had to use the class interval of 60 to 70 to help here.

Because that's the only case where we have the frequency, as six in our table, and we have a bar drawn on as well.

So we can work out the frequency density by dividing six by 10.

That gives us nought point six.

So we know that fourth bar on the diagram represents not nought point six.

This is six small increments.

So each small increment must be nought point one.

Meaning each of the markings on the vertical scale must represent nought point five.

So that's increasing by nought point five each time, and we can fill in that scale.

And then that means that we can interpret the rest of the histogram.

That's all for this lesson.

Thank you for watching.