Statistics I
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: Calculate and interpret measures of central tendency and spread for a data set.
Imagine a see-saw balanced perfectly on a fulcrum. Each data point is a sack of maize placed at a specific mark along the beam. The beam balances in perfect equilibrium at exactly the mean (\(\bar{x}\)). If you move a sack further away from the center, the balance shifts to compensate. This is the essence of central tendency and spread: finding where the center of the data sits and measuring how widely the data is distributed around that center.
Interactive Data Fulcrum & Spread Visualizer
Click anywhere on the number line to place or remove a data point (up to 7 points). Observe how the balance point (Mean, green) and middle point (Median, purple) respond!
(a) Concrete Scenario
A farmer in Nakuru records the daily milk yield (in litres) from her cow over five days: 4, 7, 9, 10, 12. To describe the overall daily performance, we sum the values (\(42\)) and divide by the number of days (\(5\)), yielding a mean of 8.4 litres. Notice that 8.4 is not one of the recorded yields — it is the single balanced summary value.
(b) Geometric Insight
On a number line, the mean is the physical center of mass (the fulcrum where total leftward deviations equal total rightward deviations: \(\sum (x_i - \bar{x}) = 0\)). The median is the middle rank when ordered, unaffected by extreme values (outliers). The range and standard deviation measure how far values wander from the central balance.
(c) Algebraic Structure
Let a dataset consist of \(n\) values \(x_1, x_2, \dots, x_n\):
- Mean: \(\bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_i\)
- Deviation of \(x_i\): \(d_i = (x_i - \bar{x})\)
- Variance: \(\sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \bar{x})^2\)
Key Formulas
1. Arithmetic Mean (\(\bar{x}\))
\[\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}\]
Sum of all observations divided by the total number of observations \(n\).
2. Median (\(\tilde{x}\))
\[\text{Position of Median} = \frac{n + 1}{2}\text{th value in an ordered dataset}\]
- If \(n\) is odd: the exact middle value.
- If \(n\) is even: the arithmetic average of the two middle values.
3. Mode
\[\text{Mode} = \text{The data value(s) with the highest frequency}\]
A data set can have no mode, one mode (unimodal), or multiple modes (bimodal/multimodal).
4. Range
\[\text{Range} = x_{\text{max}} - x_{\text{min}}\]
The simplest measure of spread, measuring the absolute span of the data.
5. Population Variance (\(\sigma^2\)) & Standard Deviation (\(\sigma\))
\[\sigma^2 = \frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n} = \frac{\sum x_i^2}{n} - (\bar{x})^2\]
\[\sigma = \sqrt{\sigma^2} = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n}}\]
\(\sigma\) is in the original measurement units and quantifies the typical distance of data points from the mean.
Worked Examples
Example 1 (Easy): Basic Mean and Range
Problem: A duka owner in Machakos recorded the sales of bags of sugar across five days: 14, 18, 12, 16, 20. Find the mean and range of the sugar sales.
Step-by-step Solution:
- Calculate the Sum:
\[\sum x = 14 + 18 + 12 + 16 + 20 = 80\] - Compute the Mean (\(\bar{x}\)):
\[\bar{x} = \frac{80}{5} = 16\text{ bags}\] - Compute the Range:
\[\text{Range} = x_{\text{max}} - x_{\text{min}} = 20 - 12 = 8\text{ bags}\]
Answer: Mean = 16 bags, Range = 8 bags
Example 2 (Medium): Median with Even \(n\) & Finding an Unknown
Problem: Six students scored the following marks in a test: 55, 68, 72, 85, 90, \(k\). The mean mark is 74. Find the value of \(k\) and hence calculate the median mark.
Step-by-step Solution:
- Use the Mean Formula to solve for \(k\):
\[\bar{x} = \frac{55 + 68 + 72 + 85 + 90 + k}{6} = 74\]\[370 + k = 74 \times 6 = 444\]\[k = 444 - 370 = 74\] - Arrange all 6 marks in ascending order:
\(55, 68, 72, 74, 85, 90\) - Locate the median position:
Since \(n = 6\) (even), the median is the average of the 3rd and 4th values:
\[\text{Median} = \frac{72 + 74}{2} = \frac{146}{2} = 73\]
Answer: \(k = 74\), Median = 73
Example 3 (Hard): Variance and Standard Deviation Computation
Problem: Four tea pickers in Kericho picked the following kilograms of tea in an hour: 6, 8, 11, 15. Calculate the population variance and standard deviation.
Step-by-step Solution:
- Compute the Mean (\(\bar{x}\)):
\[\bar{x} = \frac{6 + 8 + 11 + 15}{4} = \frac{40}{4} = 10\text{ kg}\] - Compute Deviations and Squared Deviations:
\(x_i\) \(x_i - \bar{x}\) \((x_i - \bar{x})^2\) 6 \(6 - 10 = -4\) 16 8 \(8 - 10 = -2\) 4 11 \(11 - 10 = +1\) 1 15 \(15 - 10 = +5\) 25 Total 0 \(\sum (x_i - \bar{x})^2 = 46\) - Calculate Variance (\(\sigma^2\)):
\[\sigma^2 = \frac{46}{4} = 11.5\] - Calculate Standard Deviation (\(\sigma\)):
\[\sigma = \sqrt{11.5} \approx 3.39\text{ kg}\]
Answer: Variance = 11.5, Standard Deviation \(\approx\) 3.39 kg
Common Mistakes
Misconception 1: Finding the Median Without Ordering the Data First
Mistake: Picking the value in the middle position of an unsorted list (e.g., in \(12, 4, 18\), claiming the median is 4).
Why it happens: Students remember the rule "median is the middle number" but forget that rank and order are fundamental to position.
Correction: Always sort the numbers in ascending order first: \(4, 12, 18 \implies \text{Median} = 12\).
Misconception 2: Believing the Mean Must Be a Value in the Dataset
Mistake: Expecting the mean family size or mean score to equal one of the recorded integers.
Why it happens: Concrete reasoning assumes an "average" person or score must physically exist in the sample.
Correction: The mean is a mathematical equilibrium point (center of mass), and frequently results in a decimal fraction (e.g., average children per household = 2.4).
Misconception 3: Forgetting that Outliers Distort the Mean but not the Median
Mistake: Relying exclusively on the mean to describe salaries or crop yields when extreme outliers exist.
Why it happens: The mean is familiar and simple to calculate.
Correction: If 4 workers earn KES 15,000 each and the director earns KES 300,000, the mean is KES 72,000, which misrepresents typical worker pay. The median (KES 15,000) provides a far more representative central measure in skewed data.
Real World
Mean = 12.0%, Standard Deviation = 0.11%
Mean = 12.0%, Standard Deviation = 2.83%
Practice