Mathematics • Year 8 • Unit 4 • Lesson 10
Range and Outliers
Build fluency calculating the range, identifying outliers, and explaining how outliers affect mean, median, and range differently.
1. I do — fully worked example
Read every line. Each step has a short reason so you can see why we do it, not just what we do.
Problem. Sprint times (seconds) for 7 athletes: 11.2, 11.4, 11.5, 11.6, 11.8, 12.0, 15.3. (a) Find the range with and without the outlier. (b) Compare the effect on mean, median, and range.
Step 1 — Identify the outlier.
15.3 is well above the cluster of 11.2–12.0. Outlier = 15.3 s.
Reason: an outlier is isolated from the main group.
Step 2 — Range with outlier.
Range = max − min = 15.3 − 11.2 = 4.1 s.
Step 3 — Range without outlier.
Without 15.3: range = 12.0 − 11.2 = 0.8 s.
Step 4 — Compare effects.
Mean with: ≈ 11.97 s. Without: ≈ 11.58 s. Change ≈ 0.39 s (large).
Median with: 11.6 s (4th of 7). Without: 11.5 s (3rd of 6, average with 4th). Change ≈ 0.1 s (small).
Range change: 4.1 → 0.8 (huge).
Answer: The outlier heavily inflates the range and significantly pulls the mean, but barely moves the median.
2. We do — fill in the missing steps
Same shape as Section 1, but with the working faded. Fill in each blank. 4 marks
Problem. Data: 50, 52, 54, 56, 58 (and then add an outlier of 2 to make a new set). Find the range of the original set and the range of the new set.
Step 1 — Original range:
Max = ______ . Min = ______ . Range = ______ − ______ = ______ .
Step 2 — Add the outlier 2 to the data:
New data set: 2, 50, 52, 54, 56, 58.
Step 3 — New range:
New max = ______ . New min = ______ . New range = ______ − ______ = ______ .
Step 4 — One-sentence comment:
A single outlier changed the range from ______ to ______ — a change of ______ . This shows the range is _____________________.
3. You do — independent practice
Show your working in the space under each problem. The first four are foundation. The middle two are standard. The last two are extension.
Foundation — quick recall
3.1 Calculate the range of: 25, 13, 48, 31, 22, 40, 17. 1 mark
3.2 Identify the outlier in: 47, 50, 2, 53, 51, 48, 55. 1 mark
3.3 Which statement about the range is correct?
(a) The range tells us the average value.
(b) A small range indicates consistent, closely clustered data.
(c) The range is not affected by outliers.
(d) A larger range always means a larger mean. 1 mark
3.4 A data set has an outlier that is the maximum value. When removed, which statistic changes the MOST? (mean / median / mode / range) 1 mark
Standard — compare two data sets
3.5 Set A: 45, 60, 70, 80, 95. Set B: 60, 63, 66, 68, 70. (a) Find the range of each. (b) Which set is more consistent? Justify in one sentence. 2 marks
3.6 Two athletes' jumps (m): A: 6.1, 6.2, 6.2, 6.3, 6.3, 6.4, 6.4, 6.5. B: 5.2, 5.8, 6.1, 6.4, 6.5, 6.7, 7.0, 7.3. (a) Find each athlete's range. (b) Who is more consistent? Justify. 2 marks
Extension — outlier analysis
3.7 Data: 52, 55, 57, 58, 60, 62, 98. (a) Identify the outlier. (b) Calculate mean and median with the outlier. (c) Calculate mean and median without the outlier. (d) Which statistic changed more? Why? 2 marks
3.8 A coach records sprint times where one athlete tripped, getting an outlier time of 18.5 s vs the others all around 11.5 s. The coach asks: "Should the 18.5 s be included in the analysis?" Give a justified Yes/No answer in one or two sentences. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (effect of an outlier on range)
Step 1: Max = 58, min = 50, range = 8.
Step 3: New max = 58, new min = 2, new range = 56.
Step 4: Range changed from 8 to 56 — a change of 48. The range is heavily affected by outliers.
3.1 — Range
Max = 48, min = 13. Range = 48 − 13 = 35.
3.2 — Outlier
2 — isolated far below the cluster of 47–55.
3.3 — Range statement
(b) A small range indicates consistent, closely clustered data.
3.4 — Statistic most changed
Range — the max value is part of the range calculation, so removing it directly changes the range. (Mean would also change noticeably, but the median is resistant.)
3.5 — Consistency comparison
(a) Set A range = 95 − 45 = 50. Set B range = 70 − 60 = 10.
(b) Set B is more consistent — its range of 10 is much smaller, showing all values clustered closely together.
3.6 — Athletes' jumps
(a) A range = 6.5 − 6.1 = 0.4 m. B range = 7.3 − 5.2 = 2.1 m.
(b) Athlete A is more consistent — only 0.4 m of variation across 8 jumps, whereas B varies by 2.1 m.
3.7 — Outlier analysis
(a) Outlier = 98 (isolated above the cluster 52–62).
(b) Σx = 442. Mean = 442 ÷ 7 ≈ 63.1. Median = 4th value = 58.
(c) Without 98: Σx = 344. n = 6. Mean = 344 ÷ 6 ≈ 57.3. Median = (57 + 58) ÷ 2 = 57.5.
(d) The mean changed by ~5.8 (large); the median only changed by 0.5 (small). The mean uses all values, so the outlier pulls it; the median is resistant.
3.8 — Tripped athlete time
Exclude — the 18.5 s was caused by an external event (the athlete tripping), not their actual sprint ability. Including it would heavily inflate the mean and range and misrepresent typical performance. The time should still be recorded in the raw data but labelled as an anomaly.