Curriculum

Course: Statistics — Third Secondary

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Curriculum

Statistics — Third Secondary

Unit 1: Correlation and Regression

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Unit 2: Advanced Statistical Measures

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Unit 3: Probability

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Unit 4: Random Variables

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Unit 5: The Normal Distribution

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Text lesson

Advanced Measures of Dispersion

Statistics — Third Secondary

Advanced Measures of Dispersion

Worked examples

Worked examples

Example 1→Find the range, the variance and the standard deviation of the values: 4, 6, 8, 10, 12.

Comparison aspect	Main idea	Keywords
•	Range = 12 − 4 = 8	Range
Mean	x̄ = (4 + 6 + 8 + 10 + 12) ÷ 5 = 40 ÷ 5 = 8
Deviations	4. Their squares: 16 with a sum of 40	squares
Variance	σ² = 40 ÷ 5 = 8
Standard deviation	σ = √8 = 2√2 ≈ 2.83
Example 2	The exam scores of two classes are compared. Class A has a mean of 60 and a standard deviation of 6. Class B has a mean of 80 and a standard deviation of 12. Which class is relatively more consistent?	scoresclassescomparedClass

Coefficient of variation for Class A = (6 ÷ 60) × 100% = 10%.
Coefficient of variation for Class B = (12 ÷ 80) × 100% = 15%.
Class A is relatively more consistent. Its standard deviation is also smaller in absolute value, but the means are different, so the coefficient of variation gives the fairer comparison.
Example 3 Find the standard deviation of the grouped data with midpoints 15, 25, 35, 45 and frequencies 3, 5, 8, 4.

Comparison aspect	Main idea	Keywords
•	Total frequency Σf = 3 + 5 + 8 + 4 = 20	Totalfrequency
•	Σ(f × x) = 3 × 15 + 5 × 25 + 8 × 35 + 4 × 45 = 45 + 125 + 280 + 180 = 630
Mean	x̄ = 630 ÷ 20 = 31.5
•	Σ(f × (x − x̄)²) = 3 × 272.25 + 5 × 42.25 + 8 × 12.25 + 4 × 182.25 = 816.75 + 211.25 + 98 + 729 = 1855	1855
Variance	σ² = 1855 ÷ 20 = 92.75	1855
Check with the shortcut	Σ(f × x²) = 3 × 225 + 5 × 625 + 8 × 1225 + 4 × 2025 = 21700 so σ² = 21700 ÷ 20 − 31.5² = 1085 − 992.25 = 92.75 ✓	12252025217001085
Standard deviation	σ = √92.75 ≈ 9.63
Example 4	A set of 8 values has Σx = 64 and Σx² = 560. Find its standard deviation. Then find the mean and the standard deviation of the new set y = 2x + 5	valuesFindstandarddeviation

Comparison aspect	Main idea	Keywords
Mean	x̄ = 64 ÷ 8 = 8
Variance	σ² = 560 ÷ 8 − 8² = 70 − 64 = 6
Standard deviation	σ = √6 ≈ 2.45
Mean of the new set	ȳ = 2 × 8 + 5 = 21
Variance of the new set	σ²y = 2² × 6 = 24 so σy = √24 ≈ 4.90
•	Notice that adding 5 did not change the spread while multiplying by 2 doubled the standard deviation	Noticeaddingchangespread

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