Mathematics • Year 8 • Unit 4 • Lesson 10
Range and Outliers — Mixed Challenge
Pull together everything from Lesson 10: calculating range, identifying outliers, comparing the impact of outliers on mean / median / range, and reasoning about when to include or exclude them.
1. Mixed problems — choose the right move
Each question uses a different idea from Lesson 10. Show your working. 3 marks each
1.1 Calculate the range for each set:
(a) 14, 22, 8, 19, 11, 27, 16.
(b) 1.4, 1.6, 1.3, 1.8, 1.5, 1.7.
(c) −5, 3, 12, 0, −2, 8.
1.2 A data set has range 25, minimum 12. What is the maximum?
1.3 Identify the outlier in each list:
(a) 12, 13, 14, 15, 16, 17, 50.
(b) 1, 102, 103, 104, 105, 106.
(c) 21, 22, 24, 25, 23, 21, 22 (none / which one?).
1.4 For the data set 30, 32, 34, 36, 38, 40, 80, find the mean, median, and range. Then find each WITHOUT the outlier. Build a small table comparing the three statistics with and without the outlier.
1.5 Match each statement to the correct statistic (mean / median / range):
(a) Most affected by an outlier.
(b) Equal to maximum minus minimum.
(c) Resistant to outliers.
1.6 Two data sets have the same range of 10 but different shapes. (a) Give an example of each. (b) Explain why "same range" does not mean "same spread overall".
2. Find the mistake
Another student attempted this range/outlier problem. 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
Problem: Data: 45, 48, 50, 52, 55, 58, 200. (a) State the range with and without the outlier. (b) State the mean with and without.
Line 1: Outlier = 200 (well above cluster 45–58).
Line 2: Range with: 200 − 45 = 155.
Line 3: Range without: 58 − 45 = 13.
Line 4: Mean with: Σ = 45+48+50+52+55+58+200 = 508. Mean = 508 ÷ 7 ≈ 72.6.
Line 5: Mean without: Σ = 308. n = 6. Mean = 308 ÷ 6 ≈ 51.3.
Line 6: The mean is barely affected by the outlier — only changes by 1.3.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write the correct conclusion sentence.
Stuck? The mean WITH outlier is 72.6 and WITHOUT is 51.3 — that's a difference of over 21, not 1.3. The conclusion is wrong.3. Open-ended challenge — quality control report
This question has many valid answers. 4 marks
3.1 Your job: you are a quality-control analyst at a chocolate factory. Each chocolate bar should weigh 100 g, with a tolerance of ±2 g. Invent measurements for 10 chocolate bars that include at least one obvious outlier, then write a one-paragraph QC report.
Write up your report with:
(i) Your 10 chocolate-bar weights (g), listed in original order.
(ii) Identify the outlier(s) and explain how you spotted them.
(iii) Calculate range, mean, and median WITH the outlier(s).
(iv) Calculate range, mean, and median WITHOUT the outlier(s).
(v) Write a 2–3 sentence QC report explaining whether production is meeting target weight, whether you would include or exclude the outlier(s) in the public report, and why.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Ranges
(a) 27 − 8 = 19. (b) 1.8 − 1.3 = 0.5. (c) 12 − (−5) = 17.
1.2 — Maximum from range
Max = min + range = 12 + 25 = 37.
1.3 — Outliers
(a) Outlier = 50 (well above cluster 12–17). (b) Outlier = 1 (well below cluster 102–106). (c) No clear outlier — all values cluster between 21 and 25.
1.4 — Compare statistics with and without outlier
With outlier (80): Σx = 290; mean = 290 ÷ 7 ≈ 41.4. Median = 4th value = 36. Range = 80 − 30 = 50.
Without 80: Σx = 210; mean = 210 ÷ 6 = 35. Median = (34 + 36)/2 = 35. Range = 40 − 30 = 10.
Comparison: mean drops by 6.4; median drops by 1; range drops by 40. Range and mean are heavily affected; median is resistant.
1.5 — Match statistic
(a) Mean (and range, both). (b) Range. (c) Median.
1.6 — Same range, different spread
(a) Set A: 10, 11, 12, 13, 14, 15, 16, 17, 18, 20 (range 10, tightly clustered with one outlier). Set B: 10, 12, 14, 15, 16, 16, 17, 18, 19, 20 (range 10, more evenly spread).
(b) Range uses only two values (max and min). Two data sets with the same range can have very different distributions of values in between — same range ≠ same shape.
2 — Find the mistake
(a) The mistake is on Line 6.
(b) The mean changed from 72.6 to 51.3 — that's a difference of over 21, not 1.3. The student made an arithmetic / interpretation error in writing the conclusion; the mean is clearly heavily affected by this outlier.
(c) Correct conclusion: "The mean is heavily affected by the outlier — it changes by about 21 (from 72.6 down to 51.3), confirming that the mean is NOT resistant to outliers."
3 — Open-ended (sample QC report)
(i) Weights (g): 99, 100, 101, 100, 102, 99, 101, 100, 102, 75.
(ii) Outlier = 75 g — isolated 24 g below the cluster of 99–102 g. Spotted because it's far below all other values and outside the ±2 g tolerance.
(iii) With outlier: Σx = 979. Mean = 97.9 g. Ordered: 75, 99, 99, 100, 100, 100, 101, 101, 102, 102. Median = (100 + 100)/2 = 100 g. Range = 102 − 75 = 27 g.
(iv) Without outlier: Σx = 904. Mean = 904 ÷ 9 ≈ 100.4 g. Median = 5th value = 100 g. Range = 102 − 99 = 3 g.
(v) QC report: Aside from one bar at 75 g (likely a defective bar or a measurement error), production is meeting target — the other 9 bars all fall within the ±2 g tolerance, mean ≈ 100.4 g, range only 3 g. I would FLAG the 75 g bar for investigation but EXCLUDE it from the public QC report on typical performance, because including it inflates the range from 3 g to 27 g and misrepresents typical production quality. The bar should be retested or marked as a non-conforming unit and the cause investigated.
Marking: 1 mark for plausible 10-value data with at least one outlier; 1 mark for correctly identifying the outlier with reasoning; 1 mark for accurate statistics with and without the outlier; 1 mark for a coherent QC report that explains the include/exclude decision with reasoning.