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).