Mathematics Advanced • Year 12 • Module 5 • Lesson 6

Measures of Centre and Spread

Build procedural fluency in calculating mean, median, mode, range, IQR and standard deviation — and in spotting outliers with the 1.5 × IQR rule.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the formulas:

Sample mean: x̄ = ____________

Interquartile range: IQR = ____________

Sample standard deviation: s = ____________

Q1.2 State the lower and upper fences used in the 1.5 × IQR outlier test.

Lower fence = ____________

Upper fence = ____________

Q1.3 Which measures of centre/spread are robust (little affected by outliers)? Circle: Mean · Median · Mode · Range · IQR · Standard deviation

Stuck? Revisit lesson § Formula Reference and § Outliers.

2. Worked example — five-number summary & SD for 12, 15, 18, 20, 22, 25, 28, 100

Follow each step. The work is laid out exactly as it should appear in your book.

Problem. For the data 12, 15, 18, 20, 22, 25, 28, 100, find the mean, median, range, IQR and sample standard deviation, and decide whether 100 is an outlier.

Step 1 — Order the data and count.

Already in order; n = 8.

Step 2 — Mean.

Σx = 12 + 15 + 18 + 20 + 22 + 25 + 28 + 100 = 240
x̄ = 240 / 8 = 30

Step 3 — Median (average of 4th and 5th values).

Median = (20 + 22) / 2 = 21

Step 4 — Range & quartiles.

Range = 100 − 12 = 88
Q₁ = median of {12, 15, 18, 20} = (15 + 18)/2 = 16.5
Q₃ = median of {22, 25, 28, 100} = (25 + 28)/2 = 26.5
IQR = 26.5 − 16.5 = 10

Step 5 — Sample standard deviation (computational form).

Σx² = 144 + 225 + 324 + 400 + 484 + 625 + 784 + 10000 = 12986
Σ(x − x̄)² = Σx² − (Σx)²/n = 12986 − 240²/8 = 12986 − 7200 = 5786
s = √(5786 / 7) ≈ √826.6 ≈ 28.75

Reason: use n − 1 because this is sample data.

Step 6 — Outlier test.

Lower fence = 16.5 − 1.5(10) = 1.5
Upper fence = 26.5 + 1.5(10) = 41.5
100 > 41.5, so 100 is an outlier.

Conclusion. Mean 30, median 21, range 88, IQR 10, s ≈ 28.75. The mean and SD are inflated by the outlier 100; the median and IQR are largely unaffected.

3. Faded example — fill in the missing steps

Find the mean, median, IQR and sample SD of 4, 6, 7, 8, 10, 12, 13. Fill in each blank line. 4 marks

Step 1 — Count and order. n = ____ (data is already ordered).

Step 2 — Mean. Σx = ____________ ⇒ x̄ = ________ / 7 = ____________

Step 3 — Median. Position = (7 + 1)/2 = 4th value ⇒ median = ____________

Step 4 — Quartiles. Lower half = {4, 6, 7}, Q₁ = ____________ · Upper half = {10, 12, 13}, Q₃ = ____________ · IQR = ____________

Step 5 — SD using computational form.

Σx² = 16 + 36 + 49 + 64 + 100 + 144 + 169 = ____________

Σ(x − x̄)² = Σx² − (Σx)²/n = __________ − __________ = ____________

s = √( ________ / 6 ) ≈ ____________

Conclusion. Mean ____, Median ____, IQR ____, s ≈ ____.

Stuck? Revisit lesson § Worked Example.

4. Graduated practice — calculate the summary statistic asked for

Show your working. Quote answers to 2 decimal places where appropriate. Use sample SD (divide by n − 1) unless told otherwise.

Foundation — single-step calculations (4 questions)

Q	Data & question	Answer
4.1 1	4, 7, 8, 8, 9, 12, 15 — find the mean.
4.2 1	4, 7, 8, 8, 9, 12, 15 — find the median and mode.
4.3 1	2, 5, 6, 8, 10, 12, 15 — find the range.
4.4 1	2, 5, 6, 8, 10, 12, 15 — find Q₁, Q₃ and the IQR.

Standard — typical HSC difficulty (6 questions)

Show your working in the space below each part.

4.5 Find the sample standard deviation of 2, 5, 6, 8, 10, 12, 15. 2 marks

4.6 Add 3 to every value in 2, 5, 6, 8, 10, 12, 15. Without recalculating from scratch, state the new mean, median, range and sample SD. Briefly justify. 2 marks

4.7 The data set is 10, 12, 14, 15, 16, 18, 50. Use the 1.5 × IQR rule to test whether 50 is an outlier. 2 marks

4.8 Estimate the mean from this grouped data (use mid-interval values):
0–5 (f = 4), 5–10 (f = 8), 10–15 (f = 6), 15–20 (f = 2). 2 marks

4.9 For the population data 8, 9, 10, 11, 12, find the population standard deviation σ (divide by n). 2 marks

4.10 A data set has mean 50 and sample SD 6. Every value is multiplied by 4. State the new mean and new SD, with a one-line justification. 2 marks

Extension — combine concepts (2 questions)

4.11 A class of 10 students has mean test mark 70. A new student joins, scoring 92. Without listing the individual marks, find the new mean of the 11 students. 3 marks

4.12 A data set has Q₁ = 22 and Q₃ = 34. A new value of 60 is added. (i) Show 60 is an outlier under the 1.5 × IQR rule (assume IQR unchanged). (ii) Explain in one sentence why this single point will increase the mean and SD significantly but leave the median almost unchanged. 3 marks

Stuck on 4.11? Use Σx = n × x̄.

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 I divided the total (63) by 7 to get mean = 9 — I did not just "pick the middle".

For 4.2 I named 8 as both the median (middle value when ordered) and the mode (it appears twice).

For 4.3 my range used max − min = 15 − 2 = 13, not (Q₃ − Q₁).

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Formulas

x̄ = Σx / n · IQR = Q₃ − Q₁ · s = √[ Σ(x − x̄)² / (n − 1) ].

Q1.2 — Outlier fences

Lower fence = Q₁ − 1.5 × IQR. Upper fence = Q₃ + 1.5 × IQR.

Q1.3 — Robust measures

Robust: Median, Mode, IQR. Not robust: Mean, Range, Standard deviation (these are pulled by extreme values).

Q3 — Faded example for 4, 6, 7, 8, 10, 12, 13

Step 1: n = 7.
Step 2: Σx = 60; x̄ = 60/7 ≈ 8.57.
Step 3: 4th value = 8 → median = 8.
Step 4: Q₁ = 6, Q₃ = 12, IQR = 6.
Step 5: Σx² = 578; Σ(x − x̄)² = 578 − 60²/7 = 578 − 514.286 ≈ 63.71; s = √(63.71/6) ≈ 3.26.
Conclusion: mean ≈ 8.57, median 8, IQR 6, s ≈ 3.26.

Q4.1 — Mean of 4, 7, 8, 8, 9, 12, 15

Σx = 63, n = 7, x̄ = 63/7 = 9.

Q4.2 — Median and mode

Ordered already; middle value (4th of 7) = median 8. Most frequent value = mode 8.

Q4.3 — Range of 2, 5, 6, 8, 10, 12, 15

Range = max − min = 15 − 2 = 13.

Q4.4 — Quartiles and IQR

Lower half (excl. median 8): {2, 5, 6} → Q₁ = 5. Upper half: {10, 12, 15} → Q₃ = 12. IQR = 12 − 5 = 7.

Q4.5 — Sample SD of 2, 5, 6, 8, 10, 12, 15

x̄ = 58/7 ≈ 8.286. Σx² = 4 + 25 + 36 + 64 + 100 + 144 + 225 = 598. Σ(x − x̄)² = 598 − 58²/7 = 598 − 480.571 ≈ 117.43. s = √(117.43 / 6) ≈ 4.42.

Q4.6 — Add 3 to every value

Adding a constant shifts every value by 3, so the mean increases by 3 (≈ 11.29), the median increases by 3 (= 11), but distances between points are unchanged, so the range and SD are unchanged (range = 13, s ≈ 4.42).

Q4.7 — Is 50 an outlier in 10, 12, 14, 15, 16, 18, 50?

Median = 15 (4th value). Lower half {10, 12, 14}: Q₁ = 12. Upper half {16, 18, 50}: Q₃ = 18. IQR = 6. Upper fence = 18 + 1.5(6) = 27. Since 50 > 27, 50 is an outlier.

Q4.8 — Grouped-data mean

Midpoints m: 2.5, 7.5, 12.5, 17.5. Σf = 20. Σ(f·m) = 4(2.5) + 8(7.5) + 6(12.5) + 2(17.5) = 10 + 60 + 75 + 35 = 180. x̄ ≈ 180 / 20 = 9.

Q4.9 — Population SD of 8, 9, 10, 11, 12

μ = 50/5 = 10. (x − μ)² values: 4, 1, 0, 1, 4 → sum 10. σ = √(10/5) = √2 ≈ 1.41.

Q4.10 — Multiply every value by 4

Multiplying every value by 4 multiplies both the centre and the spread by 4. New mean = 4 × 50 = 200; new SD = 4 × 6 = 24.

Q4.11 — New mean after adding a student

Old total Σx = 10 × 70 = 700. New total = 700 + 92 = 792. New mean = 792 / 11 = 72.

Q4.12 — Outlier reasoning

(i) IQR = 34 − 22 = 12. Upper fence = 34 + 1.5(12) = 52. Since 60 > 52, 60 is an outlier.
(ii) The mean uses every data point and SD squares the deviations, so a single extreme value of 60 substantially increases both. The median only depends on rank order, so adding one new value at the top shifts the central rank by at most half a position — the median barely moves.