Drawing Histograms

Relate bar area, class width and frequency.

1. Frequency is represented by the area of a histogram bar.
2. Frequency density is frequency divided by class width: 12\div4=3.

Find the gap between the two class boundaries.

1. The bar extends from 10 to 25 on the horizontal scale.
2. Its class width is 25-10=15.

Use the area of the bar.

1. Frequency equals class width multiplied by frequency density.
2. The frequency is 8\times2.5=20.

The frequencies are equal, but the class widths differ.

1. The class widths are 10 and 20.
2. The densities are 20\div10=2 and 20\div20=1.
3. The wider bar has half the height so that the areas are equal.

Work out each class width and compare the resulting bar heights.

1. The class widths are 5,10,5.
2. The frequency densities are 10\div5=2, 10\div10=1 and 15\div5=3.
3. Only plot B has these three heights and the correct widths.

Use the boundaries of the missing class to find its width.

1. The missing class has width 30-10=20.
2. Its required height is frequency density 30\div20=1.5.
3. The completed bar spans 10 to 30 and reaches density 1.5.

The tallest bar has the greatest frequency density, which is not always the class with the greatest frequency.

1. The height of each bar is its frequency density: frequency density = frequency \div class width.
2. The class widths are 2, 2, 6 and 10 kg.
3. The frequency densities are 8\div2=4, 18\div2=9, 24\div6=4 and 12\div10=1.2.
4. The tallest bar is for 2<m\le4, with frequency density 9. The class 4<m\le10 has the most parcels, but it is wider, so its bar is lower.

Consider what changes when a bar has the same height but a greater width.

1. The second bar has width 10, while the first has width 5.
2. With the same height, its area is twice as large. It represents twice the frequency.

First find the density, then apply the stated drawing scale.

1. The frequency density is 18\div4=4.5.
2. Each cm represents 3 density units, so the bar height is 4.5\div3=1.5 cm.

Find the frequencies already shown before using the total.

1. In a histogram, frequency = class width \times frequency density (the area of the bar).
2. The 0<t\le10 bar represents 10\times2=20 times. The 20<t\le40 bar represents 20\times2=40 times.
3. So the missing class has frequency 90-20-40=30.
4. Its class width is 10, so its frequency density is 30\div10=3.

Divide each frequency by its class width, even when the width is less than 1.

Part (a)

1. Frequency density = frequency \div class width.
2. The class width is 0.5 km, so the frequency density is 18\div0.5=36.
3. Dividing by 0.5 doubles the frequency, so this bar is taller than the frequency.

Part (b)

1. The class width is 4-2=2 km.
2. The frequency density is 12\div2=6.

Check which feature must be proportional to frequency in a histogram.

1. For a histogram, each bar area must represent its class frequency.
2. When widths are unequal, using frequency as height gives an extra multiplication by width.
3. Dividing each frequency by its class width gives the required frequency density.

Frequency is the area of a bar: class width multiplied by frequency density.

Part (a)

1. Frequency = class width \times frequency density.
2. The bar for 10<t\le15 has width 5 minutes and reaches frequency density 4.4 (each small square is 0.2).
3. Number of calls =5\times4.4=22.

Part (b)

1. Frequency density = frequency \div class width.
2. The class 25<t\le40 has width 15 minutes and frequency 9.
3. Frequency density =9\div15=0.6, so the bar should reach 0.6 on the vertical scale.

Add the frequencies and use the full width of the new class.

1. Frequency density = frequency \div class width.
2. The merged class has frequency 10+10=20 and width 15-0=15.
3. Frequency density =\frac{20}{15}=\frac43.

Share the frequency in proportion to the widths of the two new classes.

Part (a)

1. If the values are spread evenly, each part of the class gets a share of the frequency in proportion to its width.
2. The new class is 5 wide out of 20, which is \frac{5}{20}=\frac14 of the original class.
3. Estimated frequency =\frac14\times60=15.

Part (b)

1. The class 5<t\le20 gets the rest of the frequency: 60-15=45.
2. Its width is 20-5=15.
3. Frequency density =45\div15=3. This is the same as the original bar's height, 60\div20=3, because the values were assumed to be spread evenly.

The bars that are drawn show the other 85\% of the people.

Part (a)

1. Frequency = class width \times frequency density for each bar that is drawn.
2. 10<a\le15: 5\times4=20. 15<a\le20: 5\times6=30. 20<a\le30: 10\times3.5=35.
3. These 20+30+35=85 people are the 100\%-15\%=85\% aged 30 or under.
4. So 1\% is 85\div85=1 person, and the total is 100 people.

Part (b)

1. 15\% of 100 people is 15 people in the class 30<a\le50.
2. This class has width 50-30=20 years.
3. Frequency density =15\div20=0.75.

Calibrate the vertical scale using the first class, then use the density of the second.

1. The first class has density 40\div10=4. Thus 2 cm represents density 4, or 2 density units per cm.
2. The second class has density 30\div5=6.
3. Its bar should be 6\div2=3 cm tall.

Find how many customers the two drawn bars represent, then share the rest in the ratio 1:2.

Part (a)

1. Frequency = class width \times frequency density. The drawn bars represent 10\times0.5=5 and 10\times2.5=25 customers.
2. That leaves 75-5-25=45 customers in the two missing classes.
3. The ratio 10<t\le20 : 30<t\le60 is 1:2, so 45\div3=15 customers have 10<t\le20 and 30 have 30<t\le60.
4. Frequency density for 10<t\le20: 15\div10=1.5.

Part (b)

1. From part (a), 30 customers have 30<t\le60.
2. This class has width 60-30=30 minutes.
3. Frequency density =30\div30=1. The class has twice as many customers as 10<t\le20 but is three times as wide, so its bar is lower.

Use the area ratio to share the total frequency before finding the height.

1. The first bar represents \frac38\times64=24 observations.
2. Its frequency density is 24\div6=4.

Write each bar area in terms of the unknown boundary and use their total.

1. The class widths are a and 20-a. Their frequencies are 2a and 5(20-a).
2. The total gives 2a+5(20-a)=70.
3. Simplifying gives 100-3a=70, so a=10.
4. The frequencies are then 20 and 50, which sum to 70.