Histograms & Cumulative Freq
Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Objective: Master the construction, interpretation, and analysis of grouped continuous data using Histograms (via Frequency Density) and Cumulative Frequency Curves (Ogives).
Context: The Kericho Tea Harvest
Imagine a tea cooperative in Kericho weighing daily harvests from hundreds of smallholder farmers. Because weights vary continuously (e.g., \(12.4\text{ kg}, 18.7\text{ kg}, 35.2\text{ kg}\)), we group them into class intervals. But when classes have different widths, drawing simple bar heights is misleading! We must draw Frequency Density so that the area of each rectangular bar represents the true number of farmers.
Interactive Ogive & Histogram Explorer
Adjust class frequencies to see how the Cumulative Frequency Curve (Ogive) develops and locate the median position (\(n/2\)).
Class Frequencies (Marks)
Median Position (\(n/2\)): 15
Target CF: 15 | Estimated Mark: —
Core Concepts & Principles
- Continuous Data & Histograms: In a histogram, intervals touch without gaps. The area of each rectangle is directly proportional to the class frequency: \[\text{Area} = \text{Class Width} \times \text{Frequency Density} = \text{Frequency}\]
- The Cumulative Frequency Curve (Ogive): An ogive represents the running total of frequencies. Points are strictly plotted at the upper class boundaries against cumulative frequency, starting from \((\text{lower boundary of first class}, 0)\).
- Finding Statistical Measures:
- Median (\(Q_2\)): Read at cumulative frequency \(\frac{N}{2}\).
- Lower Quartile (\(Q_1\)): Read at cumulative frequency \(\frac{N}{4}\).
- Upper Quartile (\(Q_3\)): Read at cumulative frequency \(\frac{3N}{4}\).
- Interquartile Range (IQR): \(\text{IQR} = Q_3 - Q_1\).
Key Formulas
1. Frequency Density
\[\text{Frequency Density} = \frac{\text{Class Frequency}}{\text{Class Width}}\]Used as the vertical axis of a histogram to ensure areas remain strictly proportional to frequencies even when intervals are unequal.
2. Cumulative Frequency (\(CF\))
\[CF_k = \sum_{i=1}^{k} f_i = f_1 + f_2 + \dots + f_k\]The total number of observations scoring less than or equal to the upper boundary of class \(k\).
3. Ogive Quartile Positions
\[\text{Lower Quartile Position } (Q_1) = \frac{N}{4}\] \[\text{Median Position } (Q_2) = \frac{N}{2}\] \[\text{Upper Quartile Position } (Q_3) = \frac{3N}{4}\] \[\text{Interquartile Range (IQR)} = Q_3 - Q_1\] \[\text{Semi-Interquartile Range} = \frac{Q_3 - Q_1}{2}\]4. Relative Frequency & Proportion
\[\text{Relative Frequency} = \frac{\text{Class Frequency}}{\text{Total Observations } (N)}\]Worked Examples
Example 1 (Easy): Calculating Frequency Density
Problem: In a test distribution, the class interval \(20 - 30\) contains \(16\) students. Find the frequency density (height of the histogram bar) for this interval.
Solution:
- Identify class limits: Lower boundary = \(20\), Upper boundary = \(30\).
- Calculate class width \(w\): \[w = 30 - 20 = 10\]
- Compute Frequency Density: \[\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}} = \frac{16}{10} = 1.6\]
Answer: The bar is drawn to a height of \(1.6\) units.
Example 2 (Medium): Cumulative Frequency Table
Problem: The table below shows the distribution of weights (in kg) of 40 bags of maize harvested in Nakuru:
| Weight (kg) | \(10-19\) | \(20-29\) | \(30-39\) | \(40-49\) |
|---|---|---|---|---|
| Frequency | 6 | 14 | 12 | 8 |
Determine the upper class boundaries and the cumulative frequency for each interval.
Solution:
- Because data is continuous, upper class boundaries are: \(19.5, 29.5, 39.5, 49.5\).
- Calculate running totals:
- Class 1 (\(\le 19.5\)): \(CF = 6\)
- Class 2 (\(\le 29.5\)): \(CF = 6 + 14 = 20\)
- Class 3 (\(\le 39.5\)): \(CF = 20 + 12 = 32\)
- Class 4 (\(\le 49.5\)): \(CF = 32 + 8 = 40\)
- The ogive is plotted through points: \((9.5, 0), (19.5, 6), (29.5, 20), (39.5, 32), (49.5, 40)\).
Example 3 (Hard): Interpreting an Unequal Interval Histogram
Problem: A histogram consists of three bars with the following dimensions:
- Class A: Interval \(0 - 10\), Frequency Density = \(1.8\)
- Class B: Interval \(10 - 30\), Frequency Density = \(1.2\)
- Class C: Interval \(30 - 40\), Frequency Density = \(0.8\)
Find the total number of items \(N\) and the percentage of items scoring greater than 10.
Solution:
- Calculate frequency (area) of each interval: \[f_A = \text{Width} \times \text{Density} = (10 - 0) \times 1.8 = 10 \times 1.8 = 18\] \[f_B = (30 - 10) \times 1.2 = 20 \times 1.2 = 24\] \[f_C = (40 - 30) \times 0.8 = 10 \times 0.8 = 8\]
- Total observations \(N\): \[N = 18 + 24 + 8 = 50\]
- Items scoring greater than 10: \[f_B + f_C = 24 + 8 = 32\]
- Calculate percentage: \[\text{Percentage} = \frac{32}{50} \times 100\% = 64\%\]
Common Mistakes
Misconception 1: Plotting cumulative frequency at class midpoints
Misconception 2: Using bar height as frequency in unequal class widths
Misconception 3: Reading the median at \(50\) instead of \(N/2\)
Real World
Practice