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: Master the core measures of central tendency—mean, median, and mode—for ungrouped data sets through balance, order, and frequency.
Interactive Fulcrum: Center of Balance
Enter comma-separated values (e.g., daily harvest weights in kg) to see how the Mean (balance point) and Median (ordered middle) behave:
The Story of the Three Representatives: Suppose five farmers bring sacks of maize to a cooperative weighing \(40\text{ kg}\), \(45\text{ kg}\), \(45\text{ kg}\), \(50\text{ kg}\), and \(90\text{ kg}\).
• The Mean is the fair share if all maize were pooled together and divided equally.
• The Median is the exact middle sack when arranged from lightest to heaviest (protects against the extreme \(90\text{ kg}\) sack).
• The Mode is the most common weight encountered (\(45\text{ kg}\)).
1. The Mean (Centre of Gravity)
Algebraically, the mean is the sum of all numerical values divided by the total number of observations: \[ \bar{x} = \frac{\sum_{i=1}^n x_i}{n} \] Because every single value contributes to the sum, the mean is sensitive to outliers (extreme low or high values).
2. The Median (Positional Middle)
The median is positional. After ordering \(n\) values in ascending sequence:
- If \(n\) is odd: The median is at position \(\frac{n+1}{2}\).
- If \(n\) is even: The median is the arithmetic mean of the two middle values at positions \(\frac{n}{2}\) and \(\frac{n}{2} + 1\).
3. The Mode (Frequency Champion)
The mode is the value that occurs with the highest frequency. A data set may have no mode (all unique), one mode (unimodal), or multiple modes (bimodal/multimodal).
Key Formulas
Arithmetic Mean (Ungrouped Data):
\[ \bar{x} = \frac{\sum x}{n} = \frac{x_1 + x_2 + \dots + x_n}{n} \]Where \(\sum x\) is the sum of all scores and \(n\) is the total number of items.
Median (Ungrouped Data):
First, sort data in ascending order: \(x_{(1)} \le x_{(2)} \le \dots \le x_{(n)}\).
\[ \text{Median} = \begin{cases} x_{\left(\frac{n+1}{2}\right)}, & \text{if } n \text{ is odd} \\[10pt] \dfrac{x_{\left(\frac{n}{2}\right)} + x_{\left(\frac{n}{2} + 1\right)}}{2}, & \text{if } n \text{ is even} \end{cases} \]Mode:
\[ \text{Mode} = \arg\max_x (\text{Frequency of } x) \]The observation that appears with the highest frequency count.
Worked Examples
A boda-boda rider in Kisumu recorded the number of trips made per day over 5 days: \(12, 18, 15, 12, 13\). Find the mean, median, and mode.
- Mean: Add all values and divide by \(n = 5\): \[ \bar{x} = \frac{12 + 18 + 15 + 12 + 13}{5} = \frac{70}{5} = 14 \text{ trips} \]
- Median: Sort in ascending order: \(12, 12, 13, 15, 18\). With \(n = 5\) (odd), the median position is \(\frac{5+1}{2} = 3\text{rd}\) item. \[ \text{Median} = 13 \]
- Mode: The number \(12\) appears twice; all others appear once. \[ \text{Mode} = 12 \]
Answer: Mean = 14, Median = 13, Mode = 12.
Six tea pluckers in Kericho picked the following masses (in kg) on a Tuesday: \(24, 31, 19, 28, 56, 26\). Calculate the mean and median.
- Mean: \[ \bar{x} = \frac{24 + 31 + 19 + 28 + 56 + 26}{6} = \frac{184}{6} \approx 30.67\text{ kg} \]
- Median: First arrange in ascending order: \[ 19, 24, 26, 28, 31, 56 \] With \(n = 6\) (even), take the average of the \(3\text{rd}\) (26) and \(4\text{th}\) (28) values: \[ \text{Median} = \frac{26 + 28}{2} = \frac{54}{2} = 27\text{ kg} \]
Pedagogical Insight: Notice how the outlier (\(56\text{ kg}\)) pulls the mean up to \(30.67\text{ kg}\), while the median (\(27\text{ kg}\)) remains a more representative central value.
The mean score of 5 test scores is \(72\). Four of the scores are \(65, 80, 70,\) and \(60\). Find the 5th score, and then determine the median of all 5 scores.
- Step 1: Use the mean formula to find the sum: \[ \bar{x} = \frac{\sum x}{n} \implies 72 = \frac{65 + 80 + 70 + 60 + x_5}{5} \] \[ 72 \times 5 = 275 + x_5 \implies 360 = 275 + x_5 \] \[ x_5 = 360 - 275 = 85 \]
- Step 2: Determine the median of all 5 scores: Arrange all 5 scores in ascending order: \(60, 65, 70, 80, 85\). The middle value (3rd score) is \(70\).
Answer: The 5th score is 85, and the median is 70.
Common Mistakes
Real World
Practice