Interpreting and drawing frequency polygons
Interpreting and drawing frequency polygons
- Unit 1: Probability
- Unit 2: Data Collection
- Unit 3: Interpreting and discussing results
🎯 In this topic you will
- Draw and interpret frequency polygons for discrete and continuous data
🧠 Key Words
- frequency polygon
- midpoint
Show Definitions
- frequency polygon: A graph made by joining the midpoints of class intervals in a frequency distribution with straight lines.
- midpoint: The middle value of a class interval, used to represent the entire group when plotting graphs.
You already know how to draw frequency diagrams for discrete and continuous data. You can also draw a frequency polygon for continuous data.
Drawing a frequency polygon is a useful way to show patterns, or trends, in the data. When you draw a frequency polygon, you plot the frequency against the midpoint of the class interval.
❓ EXERCISE
1. The table shows the heights of the students in class 9R.
a) Copy and complete the table.
b) Copy and complete the frequency polygon.
| Height, $h$ (cm) | Frequency | Midpoint |
|---|---|---|
| $140 \le h < 150$ | 7 | 145 |
| $150 \le h < 160$ | 13 | 155 |
| $160 \le h < 170$ | 6 | 165 |
| $170 \le h < 180$ | 2 | 175 |
👀 Show answer
a) Midpoints are the class centres: 145, 155, 165, 175 (already filled above).
b) Plot points (145, 7), (155, 13), (165, 6), (175, 2) and join with straight lines. Optionally add boundary points at 135 and 185 with frequency 0 to close the polygon.
2. The table shows the masses of the students in class 9T.
a) Copy and complete the table.
b) Draw a frequency polygon for this data.
c) How many students are there in class 9T?
d) What fraction of the students have a mass less than 60 kg?
e) Arun says: “The frequency polygon shows that the heaviest student has a mass of 65 kg.” Is Arun correct? Explain your answer.
| Mass, $m$ (kg) | Frequency | Midpoint |
|---|---|---|
| $40 \le m < 50$ | 4 | 45 |
| $50 \le m < 60$ | 12 | 55 |
| $60 \le m < 70$ | 8 | 65 |
👀 Show answer
a) Midpoints: 45, 55, 65 (filled above).
b) Plot (45, 4), (55, 12), (65, 8) and join with straight lines. Optionally add boundary points at 35 and 75 with frequency 0.
c) Total students = 4 + 12 + 8 = 24.
d) Less than 60 kg = first two classes → 4 + 12 = 16 students → fraction $16/24 = \mathbf{2/3}$.
e) Not correct. A frequency polygon uses class midpoints; 65 kg is the midpoint of the 60–70 class, not the heaviest student. The heaviest student is somewhere in $[60,70)$, at most just below 70 kg, but we don’t know the exact value.
🧠 Think like a Mathematician
Data (ages of cycling-club members, 28 values):
25, 30, 44, 18, 33, 13, 43, 32, 54, 49, 35, 29, 60, 72, 61, 10, 75, 69, 52, 32, 27, 16, 36, 22, 47, 58, 41, 21
- a) Record this information in a frequency table with equal class intervals.
- b) Draw a frequency polygon to show the data.
- c) Compare your choices with others. Which classes are best? Explain why.
a) Frequency table (class width = 10 years)
| Age class (years) | Frequency | Midpoint |
| 10 ≤ a < 20 | 4 | 14.5 |
| 20 ≤ a < 30 | 5 | 24.5 |
| 30 ≤ a < 40 | 6 | 34.5 |
| 40 ≤ a < 50 | 5 | 44.5 |
| 50 ≤ a < 60 | 3 | 54.5 |
| 60 ≤ a < 70 | 3 | 64.5 |
| 70 ≤ a < 80 | 2 | 74.5 |
| Total | 28 | — |
b) How to draw the frequency polygon
- Horizontal axis: age (years). Mark the class boundaries 10, 20, 30, …, 80.
- Plot points at each class midpoint with height equal to the class frequency: (14.5,4), (24.5,5), (34.5,6), (44.5,5), (54.5,3), (64.5,3), (74.5,2).
- Join the points with straight lines and return to the baseline at the ends (at 10 and 80) for a closed polygon.
c) Which classes are best?
- The range is 10–75. Width 10 gives 7 clear classes—good balance of detail and clarity.
- Width 5 would produce many small classes with several tiny frequencies; width 15 would hide useful shape (e.g., the peak around 30–39).
- Recommendation: use equal width 10-year classes with inclusive–exclusive notation (e.g., 30 ≤ a < 40) to avoid overlaps.
👀 quick check
Counts by class: 4, 5, 6, 5, 3, 3, 2 (sum 28). Midpoints: 14.5, 24.5, 34.5, 44.5, 54.5, 64.5, 74.5.
❓ EXERCISES
4. Here are the times (minutes) it took 24 people to complete a puzzle:
| 17 | 21 | 28 | 27 | 13 | 28 | 14 | 33 | 37 | 22 | 44 | 38 |
| 35 | 42 | 30 | 32 | 25 | 34 | 36 | 22 | 25 | 39 | 17 | 48 |
a) Record this information in a frequency table. Choose your own suitable classes.
b) Draw a frequency polygon to show the data.
| Time, $t$ (min) | Frequency | Midpoint |
|---|---|---|
| $10 \le t < 20$ | 4 | 14.5 |
| $20 \le t < 30$ | 8 | 24.5 |
| $30 \le t < 40$ | 9 | 34.5 |
| $40 \le t < 50$ | 3 | 44.5 |
| Total | 24 |
👀 Show answer
The four equal-width classes $10$–$19$, $20$–$29$, $30$–$39$, $40$–$49$ give frequencies $4, 8, 9, 3$ respectively (midpoints $14.5, 24.5, 34.5, 44.5$). For the frequency polygon, plot these midpoint–frequency pairs and join with straight lines; add boundary points at $t=9.5$ and $t=49.5$ with frequency $0$ to close the polygon.
5. Ahmad surveyed waiting times (minutes) to see a doctor at two surgeries.
| Oaklands Surgery Time, $t$ (min) |
Frequency | Midpoint |
|---|---|---|
| $0 \le t < 10$ | 25 | 5 |
| $10 \le t < 20$ | 10 | 15 |
| $20 \le t < 30$ | 12 | 25 |
| $30 \le t < 40$ | 3 | 35 |
| Birchfields Surgery Time, $t$ (min) |
Frequency | Midpoint |
|---|---|---|
| $0 \le t < 10$ | 8 | 5 |
| $10 \le t < 20$ | 14 | 15 |
| $20 \le t < 30$ | 17 | 25 |
| $30 \le t < 40$ | 11 | 35 |
a) How many people were surveyed at each surgery?
b) Copy and complete the tables (midpoints are shown above).
c) On the same grid, draw a frequency polygon for each set of data. Make sure you label which polygon is which surgery.
d) Compare the two frequency polygons. What can you say about the waiting times at the two surgeries?
👀 Show answer
a) Oaklands total = $25+10+12+3=\mathbf{50}$; Birchfields total = $8+14+17+11=\mathbf{50}$.
c) Plot Oaklands at $(5,25)$, $(15,10)$, $(25,12)$, $(35,3)$ and Birchfields at $(5,8)$, $(15,14)$, $(25,17)$, $(35,11)$; join each set with straight lines (add 0-frequency boundary points if you wish).
d) Birchfields’ polygon lies higher for the larger midpoints ($25,35$), showing longer waits on average. Oaklands has many more very short waits ($0$–$10$), so its overall waiting times are shorter and less variable.
🧠 Think like a Mathematician
Grouped data
| Time, t (hours) | Frequency | Midpoint |
| 0 ≤ t < 2 | 6 | 1 |
| 2 ≤ t < 4 | 8 | 3 |
| 4 ≤ t < 6 | 2 | 5 |
- a) Critique Sofia’s and Zara’s methods.
- b) Whose method do you prefer? Why?
👀 show answer
- Correct recipe for a frequency polygon: 1) Find each class midpoint. 2) Plot points (midpoint, frequency). Here: (1,6), (3,8), (5,2). 3) Join the points with straight lines. 4) Optionally add end points on the baseline at the outer class boundaries (0,0) and (6,0) to “close” the polygon.
- Sofia: Draws a histogram then joins the midpoints of the bar tops. This gives the same polygon if the bars are correct and she really uses the midpoints. It’s fine but longer (two graphs for one result) and easy to slip into joining bar corners instead of midpoints.
- Zara: Computes midpoints first (1, 3, 5), plots (1,6), (3,8), (5,2), then joins them. This is the direct, standard method and avoids histogram-related mistakes.
- Prefer: Zara’s method — quicker, cleaner, and it emphasises that polygons are based on class midpoints, not bar edges.
- Common pitfalls to watch: • plotting points at class boundaries rather than midpoints; • unequal class widths (not here) — then use density, not raw frequency; • forgetting to label axes (Time t in hours; Frequency) or omitting baseline points when a closed polygon is required.
❓ EXERCISE
7. Jeff grew 40 plants. He grew 20 in a greenhouse and 20 outdoors.
a) The heights of the 20 plants in the greenhouse are shown in the table. Draw a frequency polygon for the data.
| Height, $h$ (cm) | Frequency | Midpoint |
|---|---|---|
| $0 \le h < 10$ | 2 | 5 |
| $10 \le h < 20$ | 4 | 15 |
| $20 \le h < 30$ | 8 | 25 |
| $30 \le h < 40$ | 6 | 35 |
b) The diagram shows the frequency polygon for 20 plants grown outdoors. Compare it with your polygon from part a). What can you say about the heights of the two sets of plants?
Heights of plants grown outdoors
👀 Show answer
a) Plot the greenhouse polygon at midpoints with their frequencies: (5,2), (15,4), (25,8), (35,6). Join with straight lines; optionally add points at (0,0) and (40,0).
b) Greenhouse plants: most are in the 20–30 cm and 30–40 cm ranges, so they are generally taller. Outdoor plants: peak is in the 10–20 cm and 20–30 cm ranges, so they are generally shorter. Overall, greenhouse plants tend to grow taller than outdoor plants.
8. Liza carried out a survey on the number of hours that some students spent doing homework each week. The frequency diagrams show the results of her survey.
a) On the same grid, draw a frequency polygon for each set of data.
b) Compare the two frequency polygons. What can you say about the amount of time that boys and girls spend doing homework?
c) How many boys and how many girls were surveyed?
d) Do you think it is fair to make a comparison using these sets of data? Explain your answer.
👀 Show answer
a) Read the heights of the bars and plot midpoints at $2,6,10,14,18$ hours for each class interval (0–4, 4–8, 8–12, 12–16, 16–20). Join each set with straight lines.
b) From the polygons/histograms: boys peak around 8–12 hours, while girls peak around 12–16 hours. Overall, girls tend to spend more time on homework each week (their graph is shifted to the right).
c) Approximating the bar heights gives:
• Boys: $6+10+15+8+3=\mathbf{42}$ surveyed.
• Girls: $7+8+12+18+5=\mathbf{50}$ surveyed.
d) A direct comparison of raw frequencies is only partly fair because the sample sizes are different (42 vs 50). A better comparison is with relative frequencies/percentages for each class, which removes the effect of different totals. Even then, the class widths and centres are the same, so comparing shapes and centres is reasonable.
9. The table shows the lengths of 40 sea turtles.
| Length, $l$ (cm) | Frequency |
|---|---|
| $190 \le l < 210$ | 5 |
| $210 \le l < 230$ | 8 |
| $230 \le l < 250$ | 11 |
| $250 \le l < 270$ | 7 |
| $270 \le l < 290$ | 5 |
| $290 \le l < 310$ | 4 |
a) Draw a frequency polygon for the data in the table.
b) Marcus regroups the data into fewer classes:
| Length, $l$ (cm) | Frequency |
|---|---|
| $190 \le l < 230$ | ? |
| $230 \le l < 270$ | ? |
| $270 \le l < 310$ | ? |
i) Copy and complete Marcus’s frequency table.
ii) Draw a frequency polygon for Marcus’s table.
c) Compare your frequency polygons in parts a and bi. Which gives better information on the lengths of the sea turtles? Explain.
d) Arun wants to draw a table with more groups of width $10$ cm (e.g. $190\le l<200$, $200\le l<210$, …).
i) How many groups will there be from $190$ to $310$?
ii) Can Arun fill in the correct frequencies using only the original table? Explain.
👀 Show answer
a) Plot midpoints and join with straight lines. Midpoints: $200,220,240,260,280,300$ with frequencies $5,8,11,7,5,4$.
bi) Combine the original classes:
- $190 \le l < 230$: $5+8=\mathbf{13}$
- $230 \le l < 270$: $11+7=\mathbf{18}$
- $270 \le l < 310$: $5+4=\mathbf{9}$
Use midpoints $210,250,290$ for the polygon in bii.
c) The polygon from a (six narrower classes) gives more detail about where lengths cluster (e.g., a clear peak at 230–250). Marcus’s regrouped polygon is smoother but loses information. So the fine-class polygon is better for describing the distribution.
d)
i) Range $[190,310)$ split into $10$-cm groups ⇒ $[190,200),[200,210),[210,220),[220,230),[230,240),[240,250),[250,260),[260,270),[270,280),[280,290),[290,300),[300,310)$ → 12 groups.
ii) No. The original table uses $20$-cm classes, so we don’t know how many turtles fall into each $10$-cm sub-interval (e.g., $230$–$240$ vs $240$–$250$). We’d need the raw data or an assumption (like uniform spread) to split them.
⚠️ Be careful!
- Plot at midpoints only: each point is (class midpoint, frequency). Don’t plot at class boundaries or bar tops.
- Equal class widths: frequency polygons assume equal-width classes; if widths differ, use frequency density (advanced) rather than raw frequency.
- Join in order with straight lines: connect adjacent midpoint points left→right; no curves.
- Label axes clearly: horizontal = variable (with units), vertical = frequency (start at 0).
- Close the ends sensibly: many courses return the line to the baseline at the two end class boundaries (or via zero-frequency midpoints one class-width beyond) to “close” the polygon—follow your course convention.
- One polygon ≠ cumulative: don’t mix up with an ogive (cumulative frequency curve). A frequency polygon uses midpoints and straight segments.
- Multiple sets need a key: if you plot two polygons on the same axes, use a clear legend and identical class boundaries.
- Read height, not area: the y-value is the frequency at that midpoint; don’t interpret area under the line as a total (unless using densities/histograms).
- Totals check: the sum of class frequencies should match the sample size shown by your table.
- Discrete data: if using single-value classes (e.g., exact ages), midpoints equal the values; plot those and join as usual.