Frequency Polygons

A frequency polygon plots each point at the middle of its class.

The point for this class is plotted halfway between 20 and 30.

Each point is (midpoint of the class, frequency).

Find the highest point, then give the class it was plotted for.

Read the frequency of every point, then add them.

The point for a class goes at the middle of the class.

1. The point is plotted at the midpoint of the class.
2. Halfway between 160 and 180 is 170.
3. Answer: t=170.

Find the lowest point, then work out which class has that point at its middle.

1. The lowest point has a frequency of 4 and is plotted at m=140.
2. 140 is the midpoint of the class 130<m\le150.
3. Answer: 130<m\le150.

Which points are plotted for classes that start at 30 or more?

1. Times of more than 30 minutes are in the classes 30<t\le40 and 40<t\le50.
2. Their points are at t=35 (frequency 9) and t=45 (frequency 3).
3. 9+3=12
4. Answer: 12 people.

The point is (midpoint of the class, frequency of the class).

1. The midpoint of 5<w\le10 is \frac{5+10}{2}=7.5.
2. The frequency is 13.
3. Answer: (7.5,\ 13).

A frequency polygon is drawn with a ruler, and it is not a closed shape.

1. Each point is joined to the next one with a straight line, from left to right.
2. The first and last points are not joined to each other, and no curve is used.
3. Answer: with straight lines, in order, without joining the last point to the first.

Each point is (midpoint of the class, frequency).

Part (a)

1. Midpoint 50, frequency 10: the point is (50,\ 10).

Part (b)

1. Midpoint 70, frequency 15: the point is (70,\ 15).

Part (c)

1. Midpoint 110, frequency 3: the point is (110,\ 3).
2. Then join all five points in order with straight lines.

Use the midpoint of each class for the first coordinate.

1. The midpoints are 5, 15 and 25.
2. Each point is (midpoint, frequency): (5,\ 3), (15,\ 8), (25,\ 6).
3. Plotting at 10, 20, 30 uses the ends of the classes, which is a common mistake.

Use the key to tell the two polygons apart. Each point is at the middle of its class.

Part (a)

1. The points for 10<t\le20 are at t=15.
2. Girls: 13. Boys: 9.
3. 13-9=4 more girls.

Part (b)

1. More than 30 minutes is the class 30<t\le40, plotted at t=35.
2. Boys: 5. Girls: 3.
3. 5+3=8 children.

Height is along the horizontal axis. Which polygon has most of its points further along that axis?

1. Height goes along the horizontal axis.
2. Most Year 7 students are in the classes up to 150 cm, while most Year 11 students are in the classes from 150 cm upwards.
3. The Year 11 polygon is further to the right, so the Year 11 students tend to be taller.
4. The height of a point only tells you how many students are in that class.

Find the total number of runners first. Then find how many are in the classes up to 30 minutes.

1. The frequencies are 6,\ 10,\ 14,\ 12,\ 8, so there are 6+10+14+12+8=50 runners.
2. 30 minutes or less is the classes 10<t\le20 and 20<t\le30: 6+10=16 runners.
3. \frac{16}{50}=\frac{8}{25}
4. Answer: \frac{8}{25}

Check where along the time axis each point is plotted, then check each point's height against the table.

Part (a)

1. Leo's points are at t=10,\ 20,\ 30,\ 40,\ 50: the upper ends of the classes.
2. They should be at the midpoints: 5,\ 15,\ 25,\ 35,\ 45.

Part (b)

1. The heights of Leo's points are 6,\ 14,\ 18,\ 12,\ 4.
2. The table gives 6,\ 14,\ 18,\ 8,\ 4.
3. The fourth point, for 30<t\le40, should be at 8, not 12.

Check each point against the table, then compare her drawing with how a frequency polygon is joined: straight lines, in order, not closed.

1. Mia's points are at the midpoints 5,\ 15,\ 25,\ 35,\ 45 with the frequencies from the table, so they are correct.
2. Mistake 1: she joined them with a smooth curve. A frequency polygon uses straight lines.
3. Mistake 2: she joined the last point back to the first point. A frequency polygon is not closed.

The frequencies of all five classes must add up to 60.

Part (a)

1. The plotted frequencies are 8,\ 15,\ 18 and 5.
2. 8+15+18+5=46
3. 60-46=14

Part (b)

1. The midpoint of 40<a\le50 is 45 and the frequency is 14.
2. The point is (45,\ 14). It is joined to the points on each side with straight lines.

Find the total, work out which value in order is the middle one, then add up the frequencies class by class.

Part (a)

1. The frequencies are 7,\ 12,\ 9,\ 6,\ 5.
2. 7+12+9+6+5=39 students.

Part (b)

1. With 39 students the median is the \frac{39+1}{2}=20th value.
2. Running totals: 7, then 7+12=19, then 19+9=28.
3. The 20th value is after the first 19, so it is in the third class.
4. Answer: 20<t\le30. (The modal class, 10<t\le20, is not the same thing.)

Groups of different sizes are compared fairly with fractions or percentages, not with numbers of people.

Part (a)

1. 8+12+6+4=30 people.

Part (b)

1. 10+14+12+4=40 people.

Part (c)

1. 20 minutes or less is the first two classes.
2. Group A: 8+12=20 out of 30, and \frac{20}{30}\times100\approx66.7%.
3. Group B: 10+14=24 out of 40, and \frac{24}{40}\times100=60%.
4. Group B has more people, but a smaller proportion of them, so Ellie is wrong.