11.3 Diagrams for single-variable data and the histogram — Practise with help

Work each question closed book first. Check your arithmetic by adding the frequencies back to the total.

Warm-up: remember

Work out $8 + 14 + 11 + 7$ .
Find $25\%$ of $60$ .
Write $\dfrac{18}{24}$ in its simplest form.

Answers

$8 + 14 + 11 + 7 = 40$ .
$0.25 \times 60 = 15$ .
$\dfrac{18}{24} = \dfrac{3}{4}$ .

If question 3 was hard, revise equivalent fractions before you start.

Practise with help

1. A class of width 20 contains 15 values. Find the frequency density.

Hint

Frequency density is the frequency divided by the class width.

Next step

Work out $\dfrac{15}{20}$ and leave it as a decimal or a fraction.

Full working

\text{frequency density} = \frac{15}{20} = 0.75.

The bar is three-quarters of a unit high.

2. A histogram bar has frequency density 2.5 for the class $10$ – $30$ . Find the frequency in the class.

Hint

The class width is $30 - 10$ . Multiply the density by it.

Next step

The width is $20$ , so work out $2.5 \times 20$ .

Full working

\text{frequency} = 2.5 \times 20 = 50.

So 50 values lie in the class.

3. A histogram has frequency density 3 for the class $0$ – $8$ . Find the number of values between $2$ and $8$ .

Hint

The part from $2$ to $8$ has width $8 - 2 = 6$ .

Next step

Multiply the density 3 by the width 6.

Full working

\text{frequency} = 3 \times 6 = 18.

The area over the part of the bar from $2$ to $8$ gives 18 values.

4. Complete the frequency density column: class $0$ – $5$ with frequency 15, class $5$ – $15$ with frequency 25.

Hint

Find each class width first. They are not the same.

Next step

For the first class, $\dfrac{15}{5}$ ; for the second, $\dfrac{25}{10}$ .

Full working

The first width is 5, so the density is $\dfrac{15}{5} = 3$ . The second width is 10, so the density is $\dfrac{25}{10} = 2.5$ . The second class has a larger frequency but a shorter bar, because it is twice as wide.

5. A histogram of 80 values has frequency density 0.5 in the class $20$ – $40$ . Estimate the probability that a value lies in the class.

Hint

First turn the frequency density into a frequency.

Next step

The width is $20$ , so the frequency is $0.5 \times 20 = 10$ . Now divide by the total of 80.

Full working

\text{frequency} = 0.5 \times 20 = 10, \qquad P = \frac{10}{80} = \frac{1}{8} = 0.125.

The estimate is $\dfrac{1}{8}$ .

6. A student wants to show the number of hours spent watching television by 40 people, grouped into classes 0–1, 1–2, 2–3 and 3–5 hours. Explain why a histogram is the right choice, and why a bar chart with gaps would mislead.

Hint

Decide whether the variable is continuous or categorical.

Next step

Compare the class widths: the last class is twice as wide as the others.

Full working

Time is continuous, so the classes have no gaps and a histogram is the right diagram. The last class is twice as wide as the others, so its bar must be drawn with frequency density, not frequency. A bar chart with gaps would treat the classes as separate categories and would let the wide class look twice as common as it is.