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Statistics - Mean, Median, Mode, and Range

Grade 9IGCSE

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

🔑Concepts

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Measures of Central Tendency: Mean, Median, and Mode are used to find the 'center' or typical value of a dataset.

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Measure of Spread: The Range indicates the dispersion or how spread out the data is.

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Ordered Data: To find the Median, data must be arranged in ascending or descending order first.

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Discrete vs. Grouped Data: Mean calculation differs slightly when data is presented in a simple list versus a frequency table or grouped intervals (where midpoints are used).

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Outliers: Extremely high or low values can significantly affect the Mean but usually have little impact on the Median or Mode.

📐Formulae

Mean (xˉ)=∑xn (Sum of values ÷ Number of values)\text{Mean } (\bar{x}) = \frac{\sum x}{n} \text{ (Sum of values } \div \text{ Number of values)}

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

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

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

💡Examples

Problem 1:

The heights (in cm) of 6 students are: 150, 162, 155, 162, 170, 155, 162. Find the Mean, Median, Mode, and Range.

Solution:

Mean: 159.4, Median: 162, Mode: 162, Range: 20

Explanation:

  1. Order the data: 150, 155, 155, 162, 162, 162, 170. 2. Mean: (150+155+155+162+162+162+170)/7=1116/7≈159.4(150+155+155+162+162+162+170) / 7 = 1116 / 7 \approx 159.4. 3. Median: The middle (4th) value is 162. 4. Mode: 162 occurs most frequently (3 times). 5. Range: 170−150=20170 - 150 = 20.

Problem 2:

A survey of the number of pets owned by 20 families is shown in a frequency table: Pets(x): [0, 1, 2, 3], Frequency(f): [4, 8, 5, 3]. Calculate the mean number of pets.

Solution:

Mean = 1.35 pets

Explanation:

  1. Multiply each value by its frequency (f×xf \times x): (0×4)=0,(1×8)=8,(2×5)=10,(3×3)=9(0 \times 4)=0, (1 \times 8)=8, (2 \times 5)=10, (3 \times 3)=9. 2. Sum of fx=0+8+10+9=27fx = 0 + 8 + 10 + 9 = 27. 3. Sum of ff (Total families) = 4+8+5+3=204 + 8 + 5 + 3 = 20. 4. Mean = 27/20=1.3527 / 20 = 1.35.

Problem 3:

Estimate the mean for the following grouped data: Weight(w) 0 < w ≤ 10 (Freq: 2), 10 < w ≤ 20 (Freq: 8).

Solution:

14

Explanation:

  1. Find midpoints (xx) of intervals: (0+10)/2=5(0+10)/2 = 5 and (10+20)/2=15(10+20)/2 = 15. 2. Multiply midpoints by frequencies: (5×2)=10(5 \times 2) = 10 and (15×8)=120(15 \times 8) = 120. 3. Sum of fx=130fx = 130. 4. Sum of f=10f = 10. 5. Estimated Mean = 130/10=13130 / 10 = 13.