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Statistics - Mean, Median, Mode, and Range

Grade 10IGCSE

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

🔑Concepts

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Mean: The arithmetic average of a data set.

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Median: The middle value when data is arranged in ascending or descending order.

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Mode: The value that appears most frequently in a data set.

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Range: The difference between the highest and lowest values, indicating spread.

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Estimated Mean: Used for grouped frequency data where midpoints of intervals are used as representative values.

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Modal Class: The class interval with the highest frequency in grouped data.

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Median Class: The class interval that contains the middle value (position (n+1)/2(n+1)/2).

📐Formulae

Mean for discrete data(xˉ)=∑xn\text{Mean for discrete data} (\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}

Estimated Mean (Grouped Data)=∑(f×midpoint)∑f\text{Estimated Mean (Grouped Data)} = \frac{\sum (f \times \text{midpoint})}{\sum f}

Range=Maximum Value−Minimum Value\text{Range} = \text{Maximum Value} - \text{Minimum Value}

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

💡Examples

Problem 1:

Find the mean, median, mode, and range for the following data set: 5, 8, 3, 5, 10, 4, 7.

Solution:

Sorted: 3, 4, 5, 5, 7, 8, 10. Mean: 6. Median: 5. Mode: 5. Range: 7.

Explanation:

  1. Mean: (3+4+5+5+7+8+10)/7=42/7=6(3+4+5+5+7+8+10)/7 = 42/7 = 6. 2. Median: The 4th term in the sorted list is 5. 3. Mode: 5 appears twice. 4. Range: 10−3=710 - 3 = 7.

Problem 2:

A frequency table shows scores: Score 1 (freq 2), Score 2 (freq 5), Score 3 (freq 3). Calculate the mean score.

Solution:

Mean = 2.1

Explanation:

Sum of f×x=(1×2)+(2×5)+(3×3)=2+10+9=21f \times x = (1 \times 2) + (2 \times 5) + (3 \times 3) = 2 + 10 + 9 = 21. Total frequency ∑f=2+5+3=10\sum f = 2 + 5 + 3 = 10. Mean =21/10=2.1= 21 / 10 = 2.1.

Problem 3:

In a grouped frequency table, the class 10<x≤2010 < x \le 20 has a frequency of 5 and 20<x≤3020 < x \le 30 has a frequency of 8. Find the midpoints for these classes.

Solution:

Midpoints are 15 and 25.

Explanation:

Midpoint is calculated as (Lower Bound+Upper Bound)/2(\text{Lower Bound} + \text{Upper Bound}) / 2. For the first class: (10+20)/2=15(10+20)/2 = 15. For the second: (20+30)/2=25(20+30)/2 = 25.