Interpreting Histograms

Use the area of the relevant bar.

1. In a histogram the area of a bar gives the frequency: frequency = class width \times frequency density.
2. The bar for 10<t\le20 has width 10 minutes and frequency density 3.
3. Number of journeys =10\times3=30.

Find the frequency represented by each bar before adding.

1. Frequency = class width \times frequency density for each bar.
2. 0<t\le10: 10\times2=20. 10<t\le20: 10\times3=30. 20<t\le40: 20\times1=20.
3. Total =20+30+20=70 journeys.

For unequal-width histogram classes, compare frequency densities.

1. The modal class is the class with the greatest frequency density.
2. The tallest bar is for 10<t\le20 minutes.

Class frequencies are represented by areas, even when bar heights differ.

1. The three areas are approximately 5\times4=20, 15\times2=30 and 5\times5=25.
2. The middle class has the greatest frequency, despite the last bar being taller.

Include every class up to the stated boundary.

1. The first two classes cover 0<t\le20.
2. Their frequencies are about 10\times2=20 and 10\times3=30.
3. The estimate is 20+30=50 journeys.

Compare both the widths and the heights of the bars.

1. The first bar has area 5\times4=20. The second has area 10\times2=20.
2. The first two classes have the same frequency, even though their heights differ.

Use the part of the bar over the requested interval.

1. The interval 25<t\le40 is part of the class 20<t\le40, whose bar has frequency density 1.
2. Assume the times are spread evenly across the class, so use the area of the bar above 25 to 40 only.
3. Use the highlighted bar areas, including only the needed part of each class.

4. Width =40-25=15 minutes, so the estimate is 15\times1=15 journeys.

Split the requested interval at any class boundary it crosses.

1. The interval 5<t\le15 crosses the class boundary at 10, so split it into 5 to 10 and 10 to 15.
2. Assume the times are spread evenly within each class and use the area of each part.
3. Use the highlighted bar areas, including only the needed part of each class.

4. From 5 to 10: 5\times2=10. From 10 to 15: 5\times3=15.
5. Estimate =10+15=25 journeys.

Find the relevant bar area and divide by the known total.

1. The journeys longer than 20 minutes are in the bar for 20<t\le40.
2. Frequency = class width \times frequency density =20\times1=20 journeys.
3. Probability =\frac{20}{70}=\frac27.

Find each class frequency, then multiply it by the midpoint of the class.

1. First find each frequency (class width \times frequency density): 10\times2=20, 10\times3=30 and 20\times1=20.
2. We don't know the exact times, so use the midpoint of each class: 5, 15 and 30 minutes.
3. Estimated total time =20\times5+30\times15+20\times30=100+450+600=1150 minutes.
4. Estimated mean =1150\div70=16.428\ldots, which is 16.4 minutes to 1 decimal place.

Find the class containing the 50th score, then work out how far into that class you need to go.

1. Frequencies (class width \times frequency density): 0 to 20: 20\times0.5=10; 20 to 40: 20\times1.5=30; 40 to 50: 10\times2=20.
2. The median is the 50th score. 10+30=40 scores are 40 or less, so the median is in the class 40<s\le50.
3. We need 50-40=10 more scores. That class has 2 scores per mark, so 10 scores take up 10\div2=5 marks (assuming the scores are spread evenly).
4. Use the highlighted bar areas, including only the needed part of each class.

5. Estimated median =40+5=45.

Think about what grouping hides about the individual times.

1. The interval from 10 to 15 occupies half the class width.
2. Allocating half the frequency assumes times are spread uniformly throughout that class.

A class boundary is not necessarily one of the observed values.

1. The last occupied class is 20<t\le40, so the greatest observed time lies in that class.
2. The plot does not show its exact position within the class.

Only half of the bar for 30<a\le50 is over 40.

1. Members aged over 40 are in part of the 30<a\le50 class and all of the 50<a\le80 class.
2. Assume ages are spread evenly in each class. From 40 to 50: 10\times1.8=18 members.
3. From 50 to 80: 30\times0.8=24 members. So about 18+24=42 members are over 40.
4. Use the highlighted bar areas, including only the needed part of each class.

5. One third of them go to the class: 42\div3=14 members.

Use only the stated eligible group as the probability denominator.

1. The times more than 20 minutes are in the bar for 20<t\le25: frequency =5\times5=25.
2. The time is chosen from the 55 times that are more than 5 minutes, so 55 is the total for this probability, not all the times.
3. Use the highlighted bar areas, including only the needed part of each class.

4. Probability =\frac{25}{55}=\frac5{11}.

Count back from the right-hand end of the histogram until you have 30 students.

1. 30\% of 100 students is 30 students, so we need the score with 30 students above it.
2. The last bar, 80<s\le100, has 20\times0.5=10 students. We need 30-10=20 more from the class 50<s\le80.
3. That class has frequency density 1, so 20 students take up 20\div1=20 marks (assuming the scores are spread evenly).
4. Use the highlighted bar areas, including only the needed part of each class.

5. Counting back 20 marks from 80: the estimate is 80-20=60.

Use the known class area to calibrate the frequencies of the other bars.

1. Frequency is proportional to the area of each bar, so compare the areas with the first bar.
2. Count the small squares up each bar: the bars are 8, 4 and 12 squares tall. The first bar is 10 minutes wide, so its area is 10\times8=80 units, which represents 20 people. So 4 units represent 1 person.
3. Second bar: 20\times4=80 units, so 20 people. Third bar: 10\times12=120 units, so 120\div4=30 people.
4. Total =20+20+30=70 people.

Find the relevant bar areas for each group before combining their frequencies.

1. For A, the part from 15 to 20 represents about 5\times2=10, and the last bar about 5\times5=25, giving 35.
2. Count the times above 15 minutes in Group A.

3. For B, the last bar represents about 5\times6=30.
4. Count the times above 15 minutes in Group B.

5. The combined count is about 35+30=65 out of 145.
6. The estimated percentage is \frac{65}{145}\times100=44.827\ldots\%, or 44.8\%.

Find the areas of the bars in grid squares, then compare the two groups.

1. Frequency is proportional to the area of each bar, so measure each area in grid squares (width \times height).
2. 3 km or less: 2\times3=6 and 1\times8=8, so 14 squares.
3. More than 3 km: 3\times2=6 and 4\times1=4, so 10 squares, which represents n members.
4. So 1 square represents \frac{n}{10} members, and 14 squares represent \frac{14n}{10}=\frac{7n}{5} members.

Find the 25th and 75th scores, working through the classes in order.

1. Frequencies (class width \times frequency density): 10, 30, 20, 30 and 10, from left to right.
2. Lower quartile (25th score): 10 scores are 20 or less, so we need 15 more from 20<s\le40, which has 1.5 per mark. 15\div1.5=10 marks, so the lower quartile is about 20+10=30.
3. Accumulate frequency as area from the left until the required quartile count is reached.

4. Upper quartile (75th score): 10+30+20=60 scores are 50 or less, so we need 15 more from 50<s\le80, which has 1 per mark. So the upper quartile is about 50+15=65.
5. Accumulate frequency as area from the left until the required quartile count is reached.

6. Interquartile range =65-30=35.