Mathematics • Year 10 • Unit 4 • Lesson 8
Range and IQR, Skill Drill
Build fluency with Lesson 8's two measures of spread. Practise the calculations: range = max − min; quartiles Q1, Q2 (median) and Q3 from ordered data; IQR = Q3 − Q1. Then apply the 1.5 × IQR outlier rule: any value below Q1 − 1.5×IQR or above Q3 + 1.5×IQR is an outlier.
1. I do, fully worked example
Read every step. Each one has a short reason on the right.
Problem. For the data 8, 12, 15, 18, 21, 24, 28, 32, 36, find the range, the IQR, and check for outliers using the 1.5×IQR rule.
Step 1, Confirm ordered. Count n.
Already ordered. n = 9.
Step 2, Range = max − min.
Range = 36 − 8 = 28.
Reason: Lesson 8 Key Terms, "Range: difference between max and min".
Step 3, Median (Q2) splits data into halves.
Median position = (9+1)/2 = 5th value. Q2 = 21. Lower half: 8, 12, 15, 18. Upper half: 24, 28, 32, 36.
Step 4, Q1 = median of lower half, Q3 = median of upper half.
Q1 = (12 + 15)/2 = 13.5. Q3 = (28 + 32)/2 = 30.
Reason: Lesson 8 misconception, "Q1 is a single number" (the value below which 25% sits), not a set.
Step 5, IQR = Q3 − Q1.
IQR = 30 − 13.5 = 16.5.
Step 6, Outlier rule.
Lower fence = Q1 − 1.5×IQR = 13.5 − 24.75 = −11.25.
Upper fence = Q3 + 1.5×IQR = 30 + 24.75 = 54.75.
All values lie between −11.25 and 54.75, so there are NO outliers.
Answer: range = 28, IQR = 16.5, no outliers.
2. We do, fill in the missing steps
Same structure as Section 1. Fill in each blank. 4 marks
Problem. For the data set 5, 7, 9, 11, 13, 15, 17, find the range, IQR, and check for outliers.
Step 1, n = ____ (odd).
Step 2, Range.
Range = ____ − ____ = ________
Step 3, Median (Q2).
Median position = (7+1)/2 = ____. Q2 = ________
Step 4, Lower half = ________; Upper half = ________.
Step 5, Q1 = median of lower half = ________; Q3 = median of upper half = ________.
Step 6, IQR = Q3 − Q1 = ________.
Step 7, Outlier fences.
Lower = Q1 − 1.5×IQR = ________. Upper = Q3 + 1.5×IQR = ________. Outliers? ________
3. You do, independent practice
Eight graduated questions. Show ordering and Q1/Q3 working. Foundation (range only), Standard (IQR), Extension (1.5×IQR rule, why IQR is robust).
Foundation, range only
3.1 Find the range of 12, 5, 18, 9, 23, 7. 1 mark
3.2 Find the range of −3, 4, 2, −7, 5, 1, −1. 1 mark
3.3 A class's test marks were: 64, 71, 58, 82, 49, 73, 90, 67. Find the range. 1 mark
Standard, IQR
3.4 For the ordered data 2, 4, 6, 8, 10, 12, 14, 16, find Q1, Q3 and the IQR. (n = 8, so split into halves of 4 and 4.) 2 marks
3.5 Find Q1, Q3, range and IQR for 14, 16, 18, 20, 22, 24, 26, 28, 30. (n = 9, odd.) 2 marks
3.6 The daily rainfall (mm) at a school over 10 days was 0, 0, 2, 3, 5, 6, 8, 10, 12, 18. Calculate Q1, Q3 and the IQR. 2 marks
Extension, 1.5 × IQR outlier rule, robustness
3.7 For the data 5, 8, 9, 10, 11, 12, 13, 14, 15, 40: find Q1, Q3 and IQR. Then apply the 1.5×IQR rule: is 40 an outlier? Show the upper fence calculation. 3 marks
3.8 Compare these two data sets using the Lesson 8 misconception card.
Set A: 10, 20, 30, 40, 50. Set B: 10, 20, 30, 40, 200.
(a) Find the range of each.
(b) Find the IQR of each.
(c) In one sentence, explain why the range "exaggerates" the difference between A and B but the IQR does not. (This is exactly what the Lesson 8 misconception warns about.) 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2, We do (5, 7, 9, 11, 13, 15, 17)
Step 1: n = 7.
Step 2: range = 17 − 5 = 12.
Step 3: median position = 4. Q2 = 11.
Step 4: lower half = 5, 7, 9; upper half = 13, 15, 17.
Step 5: Q1 = 7; Q3 = 15.
Step 6: IQR = 15 − 7 = 8.
Step 7: lower fence = 7 − 12 = −5; upper fence = 15 + 12 = 27. All values lie between −5 and 27. No outliers.
3.1, Range of 12, 5, 18, 9, 23, 7
Max = 23, min = 5. Range = 23 − 5 = 18.
3.2, Range of −3, 4, 2, −7, 5, 1, −1
Max = 5, min = −7. Range = 5 − (−7) = 12.
3.3, Test marks range
Max = 90, min = 49. Range = 41.
3.4, Q1, Q3, IQR for 2,4,6,8,10,12,14,16
n = 8. Lower half: 2, 4, 6, 8. Upper half: 10, 12, 14, 16. Q1 = (4+6)/2 = 5. Q3 = (12+14)/2 = 13. IQR = 13 − 5 = 8.
3.5, Quartiles for 14,16...,30
n = 9, median (Q2) = 22 (5th value). Lower half (excluding median): 14, 16, 18, 20. Upper half: 24, 26, 28, 30. Q1 = (16+18)/2 = 17. Q3 = (26+28)/2 = 27. Range = 30 − 14 = 16. IQR = 27 − 17 = 10.
3.6, Rainfall data
n = 10. Lower half: 0, 0, 2, 3, 5. Upper half: 6, 8, 10, 12, 18. Q1 = 2 (median of 5 values, 3rd). Q3 = 10. IQR = 10 − 2 = 8 mm.
3.7, Outlier check for ...,15,40
n = 10. Lower half: 5, 8, 9, 10, 11. Upper half: 12, 13, 14, 15, 40. Q1 = 9, Q3 = 14, IQR = 5.
Upper fence = Q3 + 1.5×IQR = 14 + 7.5 = 21.5. Since 40 > 21.5, 40 IS an outlier.
3.8, Range vs IQR sensitivity
(a) Range A = 50 − 10 = 40. Range B = 200 − 10 = 190.
(b) Set A (n = 5): lower half {10, 20} → Q1 = 15; upper half {40, 50} → Q3 = 45; IQR = 30. Set B (n = 5): lower half {10, 20} → Q1 = 15; upper half {40, 200} → Q3 = (40 + 200)/2 = 120; IQR = 120 − 15 = 105.
(c) Replacing 50 with 200 caused the range to explode from 40 to 190 (a 375% increase) because the range depends entirely on the extreme values. The IQR also increased (from 30 to 105) because the outlier 200 entered the upper half for this tiny data set, but unlike the range, the IQR's job is to describe the middle 50%, so for larger data sets an outlier would not even affect Q3. Lesson 8 misconception: the range IS heavily affected by outliers because it uses max and min directly; the IQR is the more robust alternative.