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Statistics - Mean, Median, Mode, and Range

Grade 11IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Mean: The arithmetic average calculated by dividing the sum of all values by the total count.

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Median: The middle value in a data set when the values are arranged in ascending or descending order.

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Mode: The value that appears most frequently in a data set. A set can be bimodal or have no mode.

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Range: The difference between the highest and lowest values, representing the spread of the data.

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Grouped Data: For frequency tables, the mean is estimated using midpoints of class intervals.

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Modal Class: In grouped data, the interval with the highest frequency.

📐Formulae

Mean(xˉ)=∑xn\text{Mean} (\bar{x}) = \frac{\sum x}{n}

Mean from Frequency Table=∑(f×x)∑f\text{Mean from Frequency Table} = \frac{\sum (f \times x)}{\sum f}

Median Position=n+12\text{Median Position} = \frac{n + 1}{2}

Range=xmax−xmin\text{Range} = x_{max} - x_{min}

Estimated Mean (Grouped)=∑(f×mid−value)∑f\text{Estimated Mean (Grouped)} = \frac{\sum (f \times mid-value)}{\sum f}

💡Examples

Problem 1:

Find the mean, median, mode, and range for the following data set: 5, 8, 3, 8, 10, 2.

Solution:

  1. Ordered set: 2, 3, 5, 8, 8, 10.
  2. Mean: (2+3+5+8+8+10)/6=36/6=6(2+3+5+8+8+10)/6 = 36/6 = 6.
  3. Median: Average of 3rd and 4th terms (5+8)/2=6.5(5+8)/2 = 6.5.
  4. Mode: 8.
  5. Range: 10−2=810 - 2 = 8.

Explanation:

To find the median, the data must first be ordered. Since there is an even number of values (n=6), the median is the midpoint between the two central values.

Problem 2:

A frequency table shows the number of goals scored in matches: 0 goals (f=3), 1 goal (f=5), 2 goals (f=2). Calculate the mean number of goals.

Solution:

Sum of (goals ×\times frequency) = (0×3)+(1×5)+(2×2)=0+5+4=9(0 \times 3) + (1 \times 5) + (2 \times 2) = 0 + 5 + 4 = 9. Total frequency = 3+5+2=103 + 5 + 2 = 10. Mean = 9/10=0.99 / 10 = 0.9.

Explanation:

Multiply each value by its frequency to find the total sum, then divide by the total number of matches (sum of frequencies).

Problem 3:

Estimate the mean for the following grouped data: 0<x≤100 < x \leq 10 (f=4), 10<x≤2010 < x \leq 20 (f=6).

Solution:

  1. Find midpoints: 5 and 15.
  2. ∑(f×x)=(4×5)+(6×15)=20+90=110\sum (f \times x) = (4 \times 5) + (6 \times 15) = 20 + 90 = 110.
  3. Total frequency = 4+6=104 + 6 = 10.
  4. Estimated Mean = 110/10=11110 / 10 = 11.

Explanation:

Since the exact values within intervals are unknown, we use the midpoint of each class interval as a representative value to estimate the mean.