Histograms (Unequal Class Intervals)

Worked example

1. 1

   $5<t\le 15:\text{width}=15-5=10$ — Start with the bar you know. Find its width (top minus bottom), because its area, not its height, is the 30 people.

2. 2

   $\text{height}=30÷10=3$ — Frequency ÷ width gives this bar's frequency density. So the top of this bar stands for 3.

3. 3

   $30\text{ squares}=3,\text{so }1\text{ square}=0.1$ — This bar is 30 small squares tall. 3 ÷ 30 gives one square's value. Now the side has a scale.

4. 4

   $\text{heights: }2.4,3,2.4,1.6,0.3$ — Turn each bar's squares (24, 30, 24, 16, 3) into a height by multiplying by 0.1.

5. 5

   $0<t\le 5:2.4\times 5=12$ — Frequency = height × width, because the area of a bar is its frequency. Width is 5 − 0 = 5.

6. 6

   $15<t\le 30:2.4\times 15=36$ — Same height as the first bar, but 3 times as wide (30 − 15 = 15), so 3 times as many people.

7. 7

   $30<t\le 40:1.6\times 10=16$ — Height × width again. Width is 40 − 30 = 10.

8. 8

   $40<t\le 60:0.3\times 20=6$ — A low bar, but wide (60 − 40 = 20), so it still holds 6 people.

9. 9

   $12+30+36+16+6=100$ — Add every bar, including the 30 you were told, to get the total.

Worked example

1. 1

   $20<t\le 30:\text{width}=30-20=10$ — 20 cuts through the 15 to 30 bar. Only the part from 20 to 30 counts, so the width is 10, not 15.

2. 2

   $2.4\times 10=24$ — Height × width of just that part. We assume people are spread evenly across a bar, so this is an estimate.

3. 3

   $30<t\le 60:16+6=22$ — The bars after 30 lie wholly above 20 minutes, so take their full frequencies.

4. 4

   $24+22=46$ — Add the part bar and the whole bars: 46 people waited more than 20 minutes.

5. 5

   $P=\frac{46}{100}=\frac{23}{50}$ — Probability (or proportion) = how many ÷ total. As a percentage this is 46%.

Worked example

1. 1

   $0<h\le 10:8÷10=0.8$ — Width is 10 − 0 = 10. Frequency ÷ width gives the bar's height, the frequency density.

2. 2

   $10<h\le 15:12÷5=2.4$ — Width is only 5, so 12 plants make a tall bar: they are packed close together.

3. 3

   $15<h\le 25:30÷10=3$ — Width 10. This is the tallest bar, so pick a scale up the side that reaches at least 3.

4. 4

   $25<h\le 40:18÷15=1.2$ — Width 15. A wide class gets a shorter bar than its frequency suggests.

5. 5

   $\text{draw bars from }0\text{ to }40$ — Each bar runs exactly from the start to the end of its class, so the bars touch. Its height is the density you found.

Worked example

1. 1

   $15<t\le 30:\text{width}=30-15=15$ — The bar is drawn but the table is blank. Find the width first.

2. 2

   $2.4\times 15=36$ — Height × width gives the frequency, because a bar's area is its frequency. Write 36 in the table.

3. 3

   $40<t\le 60:\text{width}=60-40=20$ — This time the table has 6 but the bar is missing.

4. 4

   $6÷20=0.3$ — Frequency ÷ width gives the height. Draw the bar 0.3 high from 40 to 60.

5. 5

   $30<t\le 40:100-(12+30+36+6)=16$ — Neither is given, but the total is. Take every other class away from 100 to leave 16.

6. 6

   $16÷10=1.6$ — Now it is a normal “frequency, no bar” gap: divide by the width 10 and draw the bar.

Worked example

1. 1

   $0<t\le 5:20÷5=4$ — Frequency density of the bar you know: frequency ÷ width (5 − 0 = 5).

2. 2

   $4\to 6\text{ cm, so }1\to 1.5\text{ cm}$ — A density of 4 is drawn 6 cm tall. 6 ÷ 4 gives 1.5 cm for each unit of density.

3. 3

   $5<t\le 12:21÷7=3$ — Density of the other bar. Its width is 12 − 5 = 7.

4. 4

   $3\times 1.5=4.5\text{ cm}$ — Multiply its density by 1.5 cm to get its height.