Mathematics • Year 10 • Unit 4 • Lesson 10
Standard Deviation, Skill Drill
Build fluency with Lesson 10's two key calculations: the sample standard deviation s = √[Σ(x − x̄)² / (n − 1)] and the population standard deviation σ = √[Σ(x − μ)² / n]. Practise the four-step process (mean → deviations → squared deviations → square root) and connect the value back to the lesson's interpretation: small SD = clustered, large SD = spread out.
1. I do, fully worked example
Read every step. The reason on the right explains why.
Problem. Find the sample standard deviation (s) of 4, 6, 8, 10, 12.
Step 1, Find the mean x̄.
x̄ = (4 + 6 + 8 + 10 + 12) ÷ 5 = 40 ÷ 5 = 8.
Step 2, Find each deviation (x − x̄).
−4, −2, 0, 2, 4.
Reason: Lesson 10 Key Terms, deviation = x − x̄. Some are negative, some positive.
Step 3, Square each deviation.
16, 4, 0, 4, 16.
Reason: Lesson 10 Learning Intentions, squaring makes all deviations positive so they don't cancel out.
Step 4, Sum of squared deviations.
Σ(x − x̄)² = 16 + 4 + 0 + 4 + 16 = 40.
Step 5, Divide by (n − 1) and take square root.
Variance s² = 40 ÷ (5 − 1) = 40 ÷ 4 = 10.
s = √10 ≈ 3.16 (2 d.p.).
Reason: divide by n − 1 for SAMPLE SD (Lesson 10 Key Terms); divide by n for population.
Answer: s ≈ 3.16. The typical distance of values from the mean (8) is about 3 units.
2. We do, fill in the missing steps
Fill in each blank. 4 marks
Problem. Find the sample standard deviation of 70, 72, 74, 76, 78.
Step 1, Mean.
x̄ = (70 + 72 + 74 + 76 + 78) ÷ 5 = ________ ÷ 5 = ________.
Step 2, Deviations.
x − x̄: ________, ________, ________, ________, ________
Step 3, Squared deviations.
(x − x̄)²: ________, ________, ________, ________, ________
Step 4, Sum.
Σ(x − x̄)² = ________
Step 5, Variance and SD.
s² = ________ ÷ (5 − 1) = ________. s = √________ ≈ ________ (2 d.p.).
3. You do, independent practice
Eight graduated questions. Show all five steps for SD calculations; for interpretation, write a clear sentence.
Foundation, interpret without computing
3.1 Without calculating, which has the LARGER standard deviation?
Set A: 5, 5, 5, 5, 5. Set B: 1, 3, 5, 7, 9. Justify in one sentence using "spread". 1 mark
3.2 Without calculating, which has the SMALLER standard deviation?
Set A: 10, 12, 14, 16, 18. Set B: 10, 11, 12, 13, 14. 1 mark
3.3 If every value in a data set is the SAME, what is the standard deviation? Explain in one sentence using the Lesson 10 misconception card. 1 mark
Standard, calculate sample SD
3.4 Find s for the data 2, 4, 6, 8, 10 (mean = 6). Show all five steps. 2 marks
3.5 Find s for the test scores 65, 70, 75, 80, 85 (mean = 75). 2 marks
3.6 A class's test results were 60, 64, 68, 72, 76, 80 (mean = 70). Find s correct to 2 d.p. 2 marks
Extension, interpret SD in context, compare two sets
3.7 Class A has 25 test scores with mean 70 and s = 4. Class B has 25 test scores with mean 70 and s = 12. Compare the two classes using BOTH centre and spread (Lesson 10 Remember callout). 2 marks
3.8 A student's six test scores are 78, 82, 76, 80, 84, 80 (mean = 80). (a) Find s correct to 2 d.p. (b) Using Lesson 10's interpretation, describe what s tells the student about the consistency of her performance. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2, We do (70-78 in steps of 2)
Step 1: Σx = 370, x̄ = 74.
Step 2: deviations = −4, −2, 0, 2, 4.
Step 3: squared = 16, 4, 0, 4, 16.
Step 4: Σ(x − x̄)² = 40.
Step 5: s² = 40 ÷ 4 = 10. s = √10 ≈ 3.16 (2 d.p.).
3.1, Compare A and B (no calc)
Set B has the larger SD. Set A has all values equal (zero spread → SD = 0), while Set B's values are scattered around the mean, every deviation is non-zero.
3.2, Smaller SD?
Set B has the smaller SD: its values are closer together (steps of 1) than Set A's (steps of 2), so they cluster more tightly around the mean.
3.3, All values the same
Every deviation is 0, so Σ(x − x̄)² = 0 and SD = 0. The Lesson 10 Key Term is reinforced: standard deviation measures spread, so no spread → SD = 0.
3.4, s for 2, 4, 6, 8, 10
x̄ = 6. Deviations: −4, −2, 0, 2, 4. Squared: 16, 4, 0, 4, 16. Sum = 40. s² = 40 ÷ 4 = 10. s = √10 ≈ 3.16.
3.5, s for 65, 70, 75, 80, 85
x̄ = 75. Deviations: −10, −5, 0, 5, 10. Squared: 100, 25, 0, 25, 100. Sum = 250. s² = 250 ÷ 4 = 62.5. s = √62.5 ≈ 7.91 (2 d.p.).
3.6, s for 60, 64, 68, 72, 76, 80
x̄ = 70. Deviations: −10, −6, −2, 2, 6, 10. Squared: 100, 36, 4, 4, 36, 100. Sum = 280. s² = 280 ÷ 5 = 56. s = √56 ≈ 7.48 (2 d.p.).
3.7, Compare Class A and B
Same centre (mean 70), so the typical performance is identical. Class A's s = 4 is much smaller than Class B's s = 12, so Class A's scores cluster tightly around 70 (more consistent results) while Class B's scores are widely scattered around 70 (some students well above, some well below). Lesson 10 Remember callout: always report both centre AND spread.
3.8, Student's six tests
(a) Deviations from 80: −2, 2, −4, 0, 4, 0. Squared: 4, 4, 16, 0, 16, 0. Sum = 40. s² = 40 ÷ 5 = 8. s = √8 ≈ 2.83 (2 d.p.).
(b) s ≈ 2.83 means the typical score sits within about 3 marks of her mean (80). Her performance is very consistent small SD relative to the marks scale (out of ~100).