S3: Grouped Data
Interpret and construct diagrams for grouped data: histograms with equal/unequal class intervals, cumulative frequency graphs
S3 Grouped Data
๐ Key Concepts
Definition: Grouped data organises values into class intervals. When class intervals have different widths, we need special techniques to represent them fairly.
Key Terms
| Term | Definition |
|---|---|
| Class Interval | A range of values grouped together (e.g. 0-10) |
| Frequency | Number of items in each class |
| Frequency Density | Frequency รท Class Width |
| Cumulative Frequency | Running total of frequencies |
| Upper Class Boundary | The highest value in a class interval |
Higher Tier Only: Histograms with unequal class intervals and cumulative frequency graphs are Higher tier topics. You must understand frequency density.
๐ Histograms with Equal Class Intervals
When class widths are equal: The height of each bar represents the frequency directly. Bars touch each other (no gaps).
Example 1
Draw a histogram for this data (equal class widths):
| Height (cm) | Frequency |
|---|---|
| 150-160 | 5 |
| 160-170 | 12 |
| 170-180 | 18 |
| 180-190 | 10 |
Solution:
Class width = 10 cm (same for all)
Draw bars with heights: 5, 12, 18, 10
Bars touch each other, x-axis labelled with boundaries (150, 160, 170, 180, 190)
๐ Histograms with Unequal Class Intervals
Key Formula: Frequency Density = Frequency รท Class Width
The area of each bar represents the frequency, not the height!
The area of each bar represents the frequency, not the height!
Why use frequency density? With unequal intervals, a larger class would unfairly look bigger if we used height. Frequency density makes comparisons fair.
Example 2
Draw a histogram for this data:
| Time (seconds) | Frequency |
|---|---|
| 0-20 | 30 |
| 20-40 | 45 |
| 40-80 | 60 |
| 80-120 | 20 |
Solution:
Calculate frequency density for each class:
| Time (s) | Width | Frequency | Freq Density |
|---|---|---|---|
| 0-20 | 20 | 30 | 30รท20 = 1.5 |
| 20-40 | 20 | 45 | 45รท20 = 2.25 |
| 40-80 | 40 | 60 | 60รท40 = 1.5 |
| 80-120 | 40 | 20 | 20รท40 = 0.5 |
Draw histogram with frequency density on y-axis. The width of each bar corresponds to class width.
Example 3
A histogram has a bar from 0-10 with frequency density 4. What is the frequency?
Solution:
Frequency = Frequency Density ร Class Width
Frequency = 4 ร 10 = 40
Remember: Area of bar = frequency!
Example 4
A histogram bar has width 5 cm and represents a frequency of 30. Find the frequency density.
Solution:
Frequency Density = Frequency รท Width = 30 รท 5 = 6
๐ Cumulative Frequency Graphs
Cumulative Frequency: A running total of frequencies. We plot this against the upper class boundary of each interval.
How to construct:
- Find the cumulative frequency for each class (add frequencies)
- Plot points at upper class boundaries
- Join points with a smooth curve
- Start from (lower boundary of first class, 0)
Example 5
Construct a cumulative frequency table and graph for:
| Marks | Frequency |
|---|---|
| 0-20 | 8 |
| 20-40 | 15 |
| 40-60 | 22 |
| 60-80 | 12 |
| 80-100 | 3 |
Solution:
| Upper Boundary | Cumulative Frequency |
|---|---|
| 20 | 8 |
| 40 | 8+15 = 23 |
| 60 | 23+22 = 45 |
| 80 | 45+12 = 57 |
| 100 | 57+3 = 60 |
Plot points: (20,8), (40,23), (60,45), (80,57), (100,60)
Start from (0, 0) and draw a smooth curve through all points.
๐ Finding Median and Quartiles
From a Cumulative Frequency Graph:
- Median (Q2): Go to half the total frequency on y-axis, read across to curve, then down to x-axis
- Lower Quartile (Q1): Go to quarter of total frequency
- Upper Quartile (Q3): Go to three-quarters of total frequency
Example 6
A cumulative frequency graph has total frequency 80. Find the median and interquartile range.
Solution:
Median position: 80 รท 2 = 40
Lower quartile position: 80 รท 4 = 20
Upper quartile position: (3ร80) รท 4 = 60
Read values from graph at these cumulative frequencies.
If median = 45, Q1 = 32, Q3 = 58:
Interquartile Range = Q3 - Q1 = 58 - 32 = 26
๐ Estimating from Histograms
You can estimate values between classes by reading the frequency density and calculating area.
Example 7
A histogram shows ages 20-40 with frequency density 2.5. Estimate how many people are aged 20-30.
Solution:
Width from 20-30 = 10 years
Estimated frequency = 2.5 ร 10 = 25 people
Assumes uniform distribution within the class.
โ Practice Questions
Q1: A histogram bar has width 25 and frequency density 3.2. Find the frequency.
Q2: Calculate the frequency density for a class with width 15 and frequency 45.
Q3: A cumulative frequency graph has total 200. What cumulative frequency gives the median?
Q4: In a histogram, one bar from 0-20 has frequency 50. What is the frequency density?
Q5: The cumulative frequencies are: at 40 = 25, at 60 = 70, at 80 = 120. How many values are between 60 and 80?
โ Answers
- Frequency = 3.2 ร 25 = 80
- Frequency density = 45 รท 15 = 3
- Median position = 200 รท 2 = 100
- Frequency density = 50 รท 20 = 2.5
- 120 - 70 = 50 values between 60 and 80
๐ฏ Exam Tips
- Histograms have NO gaps between bars (continuous data)
- For unequal class intervals, always calculate frequency density
- Area of bar = frequency (not height!)
- Cumulative frequency graphs start at (0, 0) or the lowest boundary
- Use the graph to find median, quartiles, and percentiles
- Read scales carefully - cumulative frequency graphs have different scales
๐ง Problem-Solving Strategies
Problem-Solving
For grouped data problems: (1) For histograms, always calculate frequency density = frequency รท class width, (2) Remember: area of bar = frequency, NOT height, (3) For cumulative frequency, plot at UPPER class boundaries, (4) To find median from cumulative frequency graph, go to half the total on y-axis, read across to curve, then down to x-axis, (5) For estimated mean from grouped data, use midpoints: mean = ฮฃ(f ร midpoint) รท ฮฃf.
Multi-Step Problem
The table shows heights of 60 plants:
| Height (cm) | Frequency |
|---|---|
| 0โ10 | 8 |
| 10โ20 | 15 |
| 20โ40 | 25 |
| 40โ60 | 12 |
(a) Calculate frequency densities. (b) Estimate the mean height.
Solution: (a) 0โ10: 8/10=0.8, 10โ20: 15/10=1.5, 20โ40: 25/20=1.25, 40โ60: 12/20=0.6. (b) Midpoints: 5, 15, 30, 50. Mean = (8ร5+15ร15+25ร30+12ร50)/60 = (40+225+750+600)/60 = 1615/60 โ 26.9 cm.
โ ๏ธ Common Errors
Watch Out!
1. Wrong: Using frequency as the bar height in a histogram with unequal class intervals Correct: Use frequency density on the y-axis โ area of bar = frequency, so height = frequency รท class width
2. Wrong: Plotting cumulative frequency at the midpoint of each class interval Correct: Plot cumulative frequency at the UPPER class boundary โ cumulative frequency has reached that value by that point
3. Wrong: Using class width of 10 for the interval 10โ20 (getting 10 instead of the correct 10, but wrong for intervals like 10โ20 where boundaries matter) Correct: Check class boundaries carefully โ "10โ20" usually means 10 โค x < 20, giving width 10, but always verify
โ๏ธ 6-Mark Exam Question
Extended Answer
6 marks: A histogram shows the times taken for 200 runners to complete a race. One bar covers 20โ30 minutes with frequency density 4. Another bar covers 30โ50 minutes with frequency density 3. (a) Find the frequency for each class. (b) A third class 50โ60 minutes has frequency 20. Find its frequency density. (c) Explain why using frequency (not frequency density) on the y-axis would give a misleading histogram.
(a) 20โ30: Frequency = 4 ร 10 = 40. 30โ50: Frequency = 3 ร 20 = 60.
(b) 50โ60: Width = 10. Frequency density = 20/10 = 2.
(c) If frequency were used on the y-axis, the 30โ50 class (frequency 60) would have a bar of height 60, and the 20โ30 class (frequency 40) would have height 40. But the 30โ50 bar would be twice as wide, making its area 1200 vs 400 for 20โ30. This visually over-represents the 30โ50 group because area is what the eye perceives. Frequency density ensures area = frequency, making visual comparison fair.
Mark scheme: M1 for frequency = density ร width, A1 for 40 and 60, M1 for density = frequency/width, A1 for 2, M1 for explaining area interpretation, A1 for clear explanation of why frequency alone misleads
๐ AO3: Reason & Interpret
Reasoning and Interpretation
A cumulative frequency graph for 120 exam scores shows: the curve passes through (40, 0), (50, 15), (60, 48), (70, 85), (80, 108), (90, 118), (100, 120).
(a) Use the graph to estimate the median and interquartile range.
(b) A score of 65 is needed to pass. Estimate how many students passed.
(c) The teacher says "Most students scored between 60 and 80." Use the data to evaluate this claim.
Answers: (a) Median at CF=60: read from curve โ 64. Q1 at CF=30: โ 55. Q3 at CF=90: โ 73. IQR = 73โ55 = 18. (b) At score 65, read CF โ 50. Students passing = 120โ50 = 70. (c) Between 60 and 80: CF at 80 = 108, CF at 60 = 48. Students in range = 108โ48 = 60 out of 120 = 50%. "Most" means more than 50%, so the claim is borderline โ exactly half, not "most".
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