Mathematics • Year 9 • Unit 4 • Lesson 11
Measures of Centre and Spread
Build fluency with the five summary statistics — mean, median, mode, range and IQR — by walking from a fully worked example through a guided fill-in and into eight graduated practice problems.
1. I do — fully worked example
Read every step. The reason is on the right so you can see why, not just what.
Problem. For the data set 2, 5, 7, 8, 10, 12, 20 find the mean, median, mode, range and IQR. Decide whether 20 is an outlier.
Step 1 — Order the data and count.
Already ordered: 2, 5, 7, 8, 10, 12, 20. n = 7.
Reason: every measure of centre and spread assumes the data is in order.
Step 2 — Mean = sum ÷ n.
Sum = 2+5+7+8+10+12+20 = 64. Mean = 64 ÷ 7 ≈ 9.14.
Reason: x̄ = Σx ÷ n.
Step 3 — Median = middle value.
Position = (n+1) ÷ 2 = 4. The 4th value is 8. Median = 8.
Reason: with an odd n, the single middle value is the median.
Step 4 — Mode = most frequent.
Every value appears once → no mode.
Reason: a data set with no repeats has no modal value — write "no mode", not 0.
Step 5 — Range = max − min, IQR = Q3 − Q1.
Range = 20 − 2 = 18.
Lower half (excluding median): 2, 5, 7 → Q1 = 5.
Upper half: 10, 12, 20 → Q3 = 12.
IQR = 12 − 5 = 7.
Step 6 — Outlier check.
Upper fence = Q3 + 1.5 × IQR = 12 + 1.5(7) = 22.5.
20 < 22.5, so 20 is not an outlier (only just).
Answer: Mean ≈ 9.14, Median = 8, No mode, Range = 18, IQR = 7. No outliers.
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank line. 5 marks
Problem. For 4, 6, 6, 8, 10, 12, find mean, median, mode, range and IQR.
Step 1 — Order and count: already ordered. n = ______ .
Step 2 — Mean:
Sum = 4 + 6 + 6 + 8 + 10 + 12 = ______ . Mean = ______ ÷ ______ = ______ .
Step 3 — Median: n is even, so average the two middle values.
Middle pair = ______ and ______ . Median = (___ + ___) ÷ 2 = ______ .
Step 4 — Mode: the value that appears most often = ______ .
Step 5 — Spread:
Range = ___ − ___ = ______ .
Lower half (3 values) = ___, ___, ___ → Q1 = ______ .
Upper half (3 values) = ___, ___, ___ → Q3 = ______ .
IQR = ___ − ___ = ______ .
3. You do — independent practice
Show working under each problem. Foundation = single measure; Standard = two or three measures; Extension = compare or judge.
Foundation — single measure
3.1 Find the mean of 3, 5, 7, 9. 1 mark
3.2 Find the median of 6, 2, 9, 4, 11, 7, 8. 1 mark
3.3 Find the mode of 2, 4, 4, 5, 7, 9, 4, 11. 1 mark
3.4 Find the range of 14, 22, 9, 31, 18, 27. 1 mark
Standard — combine measures
3.5 For 5, 8, 8, 10, 12, 15, 20 find the mean, median and mode. 3 marks
3.6 For 2, 4, 6, 8, 10, 12, 14, 16 find the range and IQR. 3 marks
Extension — judge or check
3.7 For 10, 12, 15, 18, 20, 22, 50, calculate the IQR and use the 1.5 × IQR rule to test whether 50 is an outlier. 3 marks
3.8 A data set of 6 values has mean 12. A new value of 26 is added. What is the new mean? 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (4, 6, 6, 8, 10, 12)
n = 6. Sum = 46. Mean = 46 ÷ 6 = ≈ 7.67.
Middle pair: 6 and 8. Median = (6 + 8) ÷ 2 = 7.
Mode = 6 (appears twice).
Range = 12 − 4 = 8.
Lower half: 4, 6, 6 → Q1 = 6. Upper half: 8, 10, 12 → Q3 = 10. IQR = 10 − 6 = 4.
3.1 — Mean of 3, 5, 7, 9
Sum = 24, n = 4. Mean = 24 ÷ 4 = 6.
3.2 — Median of 6, 2, 9, 4, 11, 7, 8
Order: 2, 4, 6, 7, 8, 9, 11. n = 7, middle position = 4. Median = 7.
3.3 — Mode of 2, 4, 4, 5, 7, 9, 4, 11
4 appears three times — more often than any other. Mode = 4.
3.4 — Range of 14, 22, 9, 31, 18, 27
Max = 31, min = 9. Range = 31 − 9 = 22.
3.5 — Mean, median, mode of 5, 8, 8, 10, 12, 15, 20
Sum = 78, n = 7. Mean = 78 ÷ 7 ≈ 11.14.
Middle (4th) = 10.
8 appears twice. Mode = 8.
3.6 — Range and IQR of 2, 4, 6, 8, 10, 12, 14, 16
Range = 16 − 2 = 14.
Lower half: 2, 4, 6, 8 → Q1 = (4 + 6) ÷ 2 = 5.
Upper half: 10, 12, 14, 16 → Q3 = (12 + 14) ÷ 2 = 13.
IQR = 13 − 5 = 8.
3.7 — Is 50 an outlier in 10, 12, 15, 18, 20, 22, 50?
n = 7, median = 18 (4th value).
Lower half: 10, 12, 15 → Q1 = 12. Upper half: 20, 22, 50 → Q3 = 22.
IQR = 22 − 12 = 10.
Upper fence = Q3 + 1.5 × IQR = 22 + 15 = 37. 50 > 37, so yes, 50 is an outlier.
3.8 — New mean after adding 26
Original sum = 6 × 12 = 72. New sum = 72 + 26 = 98. New n = 7. New mean = 98 ÷ 7 = 14.
Adding a value larger than the original mean pulled the mean up.