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Frequency density: 140-150: 5, 150-155: 8, 155-160: 11, 160-180: 6

Step by step, with the reason for every step · Histograms with unequal class widths · Statistics

Step by step

The question

Question: draw the histogram for 140-150: 5, 150-155: 8, 155-160: 11, 160-180: 6.

frequency density: 140-150: 5, 150-155: 8, 155-160: 11, 160-180: 6

The classes are not the same width.

So the bars cannot just be as high as the frequency.

1 Class widths

A table: each class and its frequency.

The width of a class is its end minus its start: 10, 5, 5, 20.

2 Frequency density

Frequency density is the frequency divided by the width: 5 ÷ 10 = 0.5, 8 ÷ 5 = 1.6, 11 ÷ 5 = 2.2, 6 ÷ 20 = 0.3.

The rule for the height of each bar.

frequency density = frequencyclass width

Careful!

With different widths, a wide bar as high as its frequency would look too big.

Its area must be the frequency.

Careful: bar height = frequency density, not frequency

3 Draw the bars

Across: the classes.

Up: the frequency density.

Each bar goes across its whole class and up to its density: 0.5 for the first.

Now the area of each bar shows its frequency.

4 Check

Let's check: width times height gives back each frequency: 5, 8, 11, 6.

So the bars are right!

10 × 0.5 = 5, 5 × 1.6 = 8, 5 × 2.2 = 11, 20 × 0.3 = 6

Let's look back.

First: class widths.

Then: frequency density.

Then: draw the bars.

Last: check.

Answer
frequency density = frequencyclass width

Your turn

Try this one yourself, then check:

frequency density: 140-150: 2, 150-155: 20, 155-160: 10, 160-180: 10
Show the answer
heights 0.2, 4, 2, 0.5

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