Mathematics • Year 10 • Unit 4 • Lesson 10
Standard Deviation — Mixed Challenge
Pull together every idea from Lesson 10: sample vs population SD, comparing data sets using BOTH centre and spread, the "SD is never negative" rule, what happens when all values are equal, and how transformations affect SD. Then spot a Year 10 mistake and design a data set to a strict SD brief.
1. Mixed problems
Each question uses a different idea from Lesson 10. 3 marks each
1.1 Find the sample standard deviation of 5, 7, 9, 11, 13 (mean = 9). Show all five steps.
1.2 Find the POPULATION standard deviation (σ) of 4, 6, 8, 10, 12 (mean = 8). Then compare your answer to the sample SD (≈ 3.16) and explain in one sentence why they differ.
1.3 For the data 2, 4, 6, 8, 10 the mean is 6. Verify (without computing variance) that the sum of (x − x̄) values is 0. Why is this ALWAYS true for any data set?
1.4 A test set of 5 student marks has mean 70 and s = 8. If the teacher adds 5 to every student's mark, what happens to (i) the mean and (ii) the standard deviation? Justify each in one sentence. (Hint: SD measures spread, not centre.)
1.5 Two data sets each contain five values. Set A: 60, 60, 60, 60, 60. Set B: 30, 50, 60, 70, 90. (a) State each SD without computing. (b) State each mean. (c) Explain in one sentence what this comparison shows about the Lesson 10 misconception "same mean implies same data".
1.6 A student claims the standard deviation of {3, 3, 3, 3} is −2. Use the Lesson 10 Key Term and the formula s = √[Σ(x − x̄)² / (n − 1)] to show why this must be wrong in two sentences.
2. Find the mistake
A Year 10 student is computing the sample SD of 2, 4, 6, 8, 10 (mean = 6). Their working is below. Exactly one line contains the error. Spot it, explain, re-do. 3 marks
Student's working:
Line 1: Mean x̄ = 6.
Line 2: Deviations: −4, −2, 0, 2, 4.
Line 3: Sum of deviations = (−4) + (−2) + 0 + 2 + 4 = 0. So s = 0.
Line 4: Final answer: s = 0.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong (use Lesson 10's formula).
(c) Re-do the calculation correctly, showing the squared deviations step.
Stuck? The formula uses Σ(x − x̄)² — you must SQUARE each deviation before summing. The sum of raw deviations is always 0 (which is exactly why we square them).3. Open-ended challenge — design a data set
This question has many valid answers. Follow every rule. 4 marks
3.1 Design TWO data sets of exactly 5 values each that satisfy ALL of the following Lesson 10 properties:
- both sets have the same mean of 50,
- Set X has sample SD between 1 and 3 (consistent),
- Set Y has sample SD greater than 15 (highly variable),
- no negative values, no values above 100.
Show:
(i) the two data sets,
(ii) verification that each mean = 50,
(iii) full SD calculation (all 5 steps) for each, correct to 2 d.p.,
(iv) one sentence using the Lesson 10 Key Term "spread" to describe the contrast between the two sets.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Sample SD of 5, 7, 9, 11, 13
x̄ = 9. Deviations: −4, −2, 0, 2, 4. Squared: 16, 4, 0, 4, 16. Sum = 40. s² = 40 ÷ 4 = 10. s = √10 ≈ 3.16 (2 d.p.).
1.2 — Population SD of 4, 6, 8, 10, 12
Same deviations as Section 1 (squared sum 40), but divide by n = 5 instead of n − 1 = 4. σ² = 40 ÷ 5 = 8. σ = √8 ≈ 2.83. The sample SD (3.16) is slightly larger because dividing by (n − 1) instead of n produces a bigger variance — this is Lesson 10's Bessel correction for samples.
1.3 — Σ(x − x̄) = 0 always
Deviations: −4, −2, 0, 2, 4. Sum = 0 ✓. This is always true because the mean is defined as the value that balances the data — the negative deviations (values below the mean) exactly cancel the positive deviations (values above). This is exactly WHY we square deviations in the SD formula: otherwise the sum would always be 0.
1.4 — Add 5 to every value
(i) Mean increases by 5: new mean = 75 (the "add k" rule from Lesson 6 applied to the mean).
(ii) SD is unchanged (s = 8). Each deviation x − x̄ shifts by (+5) − (+5) = 0; squared deviations are identical → variance and SD are identical. SD measures spread, not centre — translation doesn't change spread.
1.5 — Compare A and B
(a) Set A: all 60s → SD = 0. Set B: spread out around 60, so SD > 0 (computation: deviations −30, −10, 0, 10, 30; squared sum 2000; s² = 500; s ≈ 22.36).
(b) Both means = 60.
(c) Two data sets with identical means can have wildly different spreads (SD 0 vs ~22). Lesson 10 misconception: same mean does NOT imply same data — you always need a spread measure too.
1.6 — Why SD = −2 is impossible
SD is defined as the square root of the (non-negative) variance, so by definition s ≥ 0 — a square root can never be negative. (Also, for {3, 3, 3, 3}, all deviations are 0 → Σ(x − x̄)² = 0 → s = 0 ≠ −2.) Lesson 10 Key Term: standard deviation is a distance from the mean, and a distance cannot be negative.
2 — Find the mistake
(a) The mistake is on Line 3.
(b) The formula is s = √[Σ(x − x̄)² ÷ (n − 1)], i.e. square each deviation first, then sum, then divide, then root. The student forgot to square — and as Lesson 10 explains, the sum of raw deviations is always 0 (which is exactly why we must square).
(c) Squared deviations: 16, 4, 0, 4, 16. Sum = 40. s² = 40 ÷ 4 = 10. s = √10 ≈ 3.16.
3 — Open-ended challenge (sample solution)
Set X (consistent): 48, 49, 50, 51, 52. Mean = 250 ÷ 5 = 50 ✓. Deviations −2, −1, 0, 1, 2. Squared: 4, 1, 0, 1, 4. Sum = 10. s² = 10 ÷ 4 = 2.5. s ≈ 1.58 (within 1-3 ✓).
Set Y (highly variable): 20, 35, 50, 65, 80. Mean = 250 ÷ 5 = 50 ✓. Deviations −30, −15, 0, 15, 30. Squared: 900, 225, 0, 225, 900. Sum = 2250. s² = 2250 ÷ 4 = 562.5. s ≈ 23.72 (> 15 ✓).
(iv) Both sets share the same centre (mean 50), but Set Y's spread is roughly 15 times larger than Set X's — Set X clusters tightly within 2 of the mean, while Set Y stretches over 60 units around the mean.
Marking: 1 mark for valid Set X with SD 1-3, 1 for valid Set Y with SD > 15, 1 for both means = 50 verified, 1 for full SD working and sentence using "spread".