Mathematics • Year 9 • Unit 4 • Lesson 12
Data Displays — Mixed Challenge
Combine everything from Lessons 11 and 12 — five-number summaries, box plot shape, outliers, histograms and stem-and-leaf plots. Choose the right display, spot a classmate's mistake, and design your own box plot from scratch.
1. Mixed problems — choose the right display
Decide which display or measure to use before you start writing. Show working. 3 marks each
1.1 For 8, 10, 12, 15, 18, 20, 25, 30 find the five-number summary and the IQR.
1.2 A box plot has Min = 5, Q1 = 10, Median = 12, Q3 = 22, Max = 30. Without drawing it, describe the shape (symmetric / left-skewed / right-skewed) and justify in one sentence.
1.3 A back-to-back stem-and-leaf shows girls' and boys' Year 9 long-jump distances (cm). Girls (leaves read right-to-left from stem) on stem 3 = 5 2 1. Boys (leaves read left-to-right from stem) on stem 3 = 0 4 6. Write each group's raw data values.
1.4 For 6, 9, 11, 14, 16, 18, 22, 24, 80 use the 1.5 × IQR rule to determine whether 80 is an outlier. Where would the right whisker end on the box plot?
1.5 A data set is described as "right-skewed with one high outlier". Predict, without numbers, the relative order of mean, median and mode. Briefly justify.
1.6 For each scenario below, choose the most appropriate display (box plot, histogram or stem-and-leaf) and briefly explain why:
(a) Comparing the test scores of two whole classes (n ≈ 60 in each).
(b) Showing 15 individual student scores in a single class while keeping all original values visible.
(c) Showing the frequency distribution of 200 phone screen-time readings.
2. Find the mistake
Another student has constructed a box plot for the data 3, 5, 8, 10, 12, 14, 18, 22, 100. 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 — 3, 5, 8, 10, 12, 14, 18, 22, 100 (n = 9):
Line 1: Median = 5th value = 12.
Line 2: Lower half: 3, 5, 8, 10 → Q1 = (5 + 8) ÷ 2 = 6.5.
Line 3: Upper half: 14, 18, 22, 100 → Q3 = (18 + 22) ÷ 2 = 20.
Line 4: IQR = 20 − 6.5 = 13.5. Upper fence = 20 + 1.5(13.5) = 40.25.
Line 5: 100 > 40.25, so the right whisker extends all the way to 100.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected statement about the whisker and the outlier.
Stuck? Revisit lesson § "Misconceptions" — the whisker does NOT always extend to the maximum. Outliers are plotted separately, and the whisker stops at the last value inside the fence.3. Open-ended challenge — design two box plots
This question has more than one valid answer — there are many data sets that work. 4 marks
3.1 Design two data sets (call them Set X and Set Y), each of 8 positive whole numbers, that have the following properties when shown as box plots:
- Both sets have the same median.
- Set X is roughly symmetric (median in the middle of the box, whiskers about equal).
- Set Y is clearly right-skewed (longer right whisker than left).
- Set Y has a larger IQR than Set X.
For each set:
(i) Write the eight values in order.
(ii) Compute Min, Q1, Median, Q3, Max and IQR.
(iii) In one sentence, justify why your set has the required shape.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Five-number summary for 8, 10, 12, 15, 18, 20, 25, 30
n = 8. Min = 8. Max = 30. Median = (15 + 18) ÷ 2 = 16.5.
Lower half: 8, 10, 12, 15 → Q1 = (10 + 12) ÷ 2 = 11. Upper half: 18, 20, 25, 30 → Q3 = (20 + 25) ÷ 2 = 22.5.
Summary: 8, 11, 16.5, 22.5, 30. IQR = 22.5 − 11 = 11.5.
1.2 — Shape of Min 5, Q1 10, Med 12, Q3 22, Max 30
Median − Q1 = 12 − 10 = 2; Q3 − median = 22 − 12 = 10. The right side of the box is much wider, and the right whisker (30 − 22 = 8) is longer than the left (10 − 5 = 5). So the distribution is right-skewed (positive skew).
1.3 — Back-to-back stem-and-leaf
Girls (read right-to-left on stem 3, leaves 5 2 1 → reading from stem out gives 1, 2, 5): 31, 32, 35 cm.
Boys (read left-to-right on stem 3, leaves 0 4 6): 30, 34, 36 cm.
1.4 — Is 80 an outlier in 6, 9, 11, 14, 16, 18, 22, 24, 80?
n = 9, median = 5th = 16. Lower half: 6, 9, 11, 14 → Q1 = (9 + 11) ÷ 2 = 10. Upper half: 18, 22, 24, 80 → Q3 = (22 + 24) ÷ 2 = 23. IQR = 23 − 10 = 13.
Upper fence = 23 + 1.5(13) = 23 + 19.5 = 42.5. 80 > 42.5, so yes, 80 is an outlier. The right whisker ends at the largest value still inside the fence, 24; 80 is plotted as a separate dot.
1.5 — Right-skewed with one high outlier
For right-skewed data: mean > median > mode. The high outlier (and the long right tail) pulls the mean above the median; the mode sits at the clump of small values on the left.
1.6 — Choose the display
(a) Side-by-side box plots — best for comparing distributions of two groups.
(b) Stem-and-leaf plot — keeps every original value visible and works well for small data sets.
(c) Histogram — large data sets are easier to see as grouped frequencies than as individual values.
2 — Find the mistake
(a) The mistake is on Line 5.
(b) When an outlier is identified, the whisker does not extend to it. It stops at the largest data value still inside the fence (here, 22). The outlier 100 is plotted as a separate point beyond the whisker.
(c) Corrected statement: 100 > 40.25, so 100 is an outlier. The right whisker extends to 22 (the largest non-outlier), and 100 is plotted as an individual dot beyond the whisker.
3 — Open-ended challenge (sample solution)
Pick a common median of 10.
Set X (roughly symmetric): 4, 6, 8, 9, 11, 12, 14, 16.
Min = 4, Max = 16. Median = (9 + 11) ÷ 2 = 10. Lower half: 4, 6, 8, 9 → Q1 = (6 + 8) ÷ 2 = 7. Upper half: 11, 12, 14, 16 → Q3 = (12 + 14) ÷ 2 = 13. IQR = 6.
Median sits in the middle of the box (10 − 7 = 3; 13 − 10 = 3) and whiskers are similar (4 − 4 = 4 and 16 − 13 = 3) → symmetric ✓.
Set Y (right-skewed, larger IQR): 5, 6, 7, 9, 11, 15, 22, 30.
Min = 5, Max = 30. Median = (9 + 11) ÷ 2 = 10. Lower half: 5, 6, 7, 9 → Q1 = (6 + 7) ÷ 2 = 6.5. Upper half: 11, 15, 22, 30 → Q3 = (15 + 22) ÷ 2 = 18.5. IQR = 12.
Right whisker (30 − 18.5 = 11.5) is much longer than the left (6.5 − 5 = 1.5), and median is closer to Q1 → right-skewed ✓. IQR 12 > 6 ✓.
Other valid pairs are fine. Any pair that meets the four bullets earns full marks.
Marking: 1 mark for matching medians, 1 mark for Set X symmetric (with values and IQR shown), 1 mark for Set Y skewed, 1 mark for Set Y IQR > Set X IQR.