Mathematics • Year 9 • Unit 4 • Lesson 11
Statistics — Mixed Challenge
Combine every idea from Lesson 11 — mean, median, mode, range, IQR and outliers. You'll choose the right tool for each problem, fix a classmate's error, and tackle an open-ended challenge.
1. Mixed problems — choose the right tool
Decide which measure to use before you start writing. Show working. 3 marks each
1.1 Find the mean, median and mode of 4, 7, 7, 9, 11, 14, 18, 18, 25.
1.2 For 6, 9, 12, 14, 17, 20, 23, 25 find the range and the IQR. State which is larger and why.
1.3 A data set of 5 values has mean 20. The values are 12, 15, 22, 25 and one missing value. Find the missing value.
1.4 Use the 1.5 × IQR rule to test whether 2 is an outlier in 2, 14, 16, 18, 20, 22, 25.
1.5 A data set has mean 30 and median 22. Sketch (in words) what its distribution most likely looks like, and name the type of skew.
1.6 Data: 8, 10, 11, 12, 13, 14, 15, 16, 90. Compute the mean, median and IQR. Then state which measure of centre and which measure of spread you would use to summarise this data, and explain in one sentence why.
2. Find the mistake
Another student has tried to find the median and IQR of the data set 5, 8, 11, 13, 17, 20, 22, 28. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — 5, 8, 11, 13, 17, 20, 22, 28 (n = 8):
Line 1: n = 8 is even, so median = (13 + 17) ÷ 2 = 15.
Line 2: Lower half = 5, 8, 11, 13. Q1 = middle of lower half = 13.
Line 3: Upper half = 17, 20, 22, 28. Q3 = (20 + 22) ÷ 2 = 21.
Line 4: IQR = 21 − 13 = 8.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected IQR.
Stuck? The lower half has 4 values — like a mini even-n data set. Q1 should be the average of its two middle values, not one of them.3. Open-ended challenge — design a data set
This question has more than one valid answer — there are many data sets that work. 4 marks
3.1 Design a data set of seven positive whole numbers that has all of the following properties at once:
- Median = 10
- Mode = 8
- Range = 20
- Mean is greater than the median (so the distribution is positively skewed).
For your data set:
(i) Write the seven values in order.
(ii) Show the working that confirms each of the four properties above.
(iii) Briefly explain in one sentence which value forced the mean to be larger than the median.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Mean, median, mode of 4, 7, 7, 9, 11, 14, 18, 18, 25
Sum = 113, n = 9. Mean = 113 ÷ 9 ≈ 12.56. Median = 5th value = 11. Modes = 7 and 18 (each appears twice — bimodal).
1.2 — Range and IQR of 6, 9, 12, 14, 17, 20, 23, 25
Range = 25 − 6 = 19. Lower half: 6, 9, 12, 14 → Q1 = (9 + 12) ÷ 2 = 10.5. Upper half: 17, 20, 23, 25 → Q3 = (20 + 23) ÷ 2 = 21.5. IQR = 21.5 − 10.5 = 11. Range > IQR because range uses all values (including extremes) while IQR uses only the middle 50%.
1.3 — Missing value, mean of 5 numbers = 20
Sum of all five must be 5 × 20 = 100. Sum of the four known values = 12 + 15 + 22 + 25 = 74. Missing value = 100 − 74 = 26.
1.4 — Is 2 an outlier in 2, 14, 16, 18, 20, 22, 25?
n = 7, median = 18. Lower half: 2, 14, 16 → Q1 = 14. Upper half: 20, 22, 25 → Q3 = 22. IQR = 22 − 14 = 8.
Lower fence = Q1 − 1.5 × IQR = 14 − 12 = 2. 2 is not less than 2, so 2 is right on the fence — not an outlier by the strict 1.5 × IQR rule (we need a value below 2).
1.5 — Mean 30, median 22
Mean > median, so the distribution has a positive skew (right skew): most values are clumped at the lower end, with a tail of large values stretching to the right that drags the mean above the median.
1.6 — 8, 10, 11, 12, 13, 14, 15, 16, 90
Sum = 189, n = 9. Mean = 189 ÷ 9 = 21. Median = 5th value = 13.
Lower half: 8, 10, 11, 12 → Q1 = (10 + 11) ÷ 2 = 10.5. Upper half: 14, 15, 16, 90 → Q3 = (15 + 16) ÷ 2 = 15.5. IQR = 15.5 − 10.5 = 5.
Use the median (13) and IQR (5) — both are robust to the outlier 90, which has badly inflated the mean.
2 — Find the mistake
(a) The mistake is on Line 2.
(b) The lower half (5, 8, 11, 13) has 4 values, so Q1 must be the average of its two middle values, not one of them. The correct Q1 = (8 + 11) ÷ 2 = 9.5.
(c) Corrected working:
Median = (13 + 17) ÷ 2 = 15. ✓
Lower half = 5, 8, 11, 13. Q1 = (8 + 11) ÷ 2 = 9.5.
Upper half = 17, 20, 22, 28. Q3 = (20 + 22) ÷ 2 = 21.
IQR = 21 − 9.5 = 11.5.
3 — Open-ended challenge (sample solution)
One valid data set: 5, 8, 8, 10, 12, 15, 25.
- Median = 4th value = 10 ✓.
- Mode = 8 (appears twice; no other value repeats) ✓.
- Range = 25 − 5 = 20 ✓.
- Sum = 83. Mean = 83 ÷ 7 ≈ 11.86, which is greater than the median 10 ✓.
The 25 at the top end is what dragged the mean above the median.
Other valid sets: e.g. 5, 7, 8, 10, 12, 14, 25 (mean ≈ 11.57) or 6, 8, 8, 10, 11, 13, 26 (mean ≈ 11.71). Any set that satisfies the four bullets is fine.
Marking: 1 mark for each verified property (median, mode, range, mean > median) plus partial credit for attempts that satisfy three out of four. Cap at 4 marks.