Mathematics • Year 9 • Unit 4 • Lesson 12

Box Plots and Data Visualisation

Build fluency with the five-number summary, box plots, histograms and stem-and-leaf displays — from a fully worked example through guided practice to independent problems.

Build · I Do / We Do / You Do

1. I do — fully worked example

Read every step. The reason on the right tells you why, not just what.

Problem. For the data 5, 7, 9, 11, 13, 15, 17 construct the five-number summary and sketch a box plot.

Step 1 — Order the data and count.

Already ordered. n = 7.

Reason: every box plot starts from ordered data.

Step 2 — Identify Min and Max.

Min = 5, Max = 17.

Step 3 — Median.

Middle position = (7 + 1) ÷ 2 = 4. Median = 4th value = 11.

Step 4 — Q1 and Q3.

Lower half (below median): 5, 7, 9 → Q1 = 7.

Upper half (above median): 13, 15, 17 → Q3 = 15.

Step 5 — Five-number summary.

Min = 5, Q1 = 7, Median = 11, Q3 = 15, Max = 17. IQR = 15 − 7 = 8.

Step 6 — Sketch the box plot.

   |---|====|========|====|---|
   5   7    11       15  17
  min  Q1  median   Q3  max

Reason: a box plot is one-dimensional — the box stretches from Q1 to Q3 with the median marked inside; whiskers reach to Min and Max (no outliers here).

Answer: Box from 7 to 15, median line at 11, whiskers to 5 and 17.

Stuck? Revisit lesson § "Box Plots" — the box ALWAYS spans Q1 to Q3.

2. We do — fill in the missing steps

Same structure as Section 1, but with the working faded. Fill in each blank line. 5 marks

Problem. Construct the five-number summary for 6, 8, 10, 12, 14, 16, 18, 20 and describe the box plot.

Step 1 — Order and count: n = ______ .

Step 2 — Min and Max: Min = ______ . Max = ______ .

Step 3 — Median (even n, average the two middle values):

Middle pair = ______ and ______ . Median = (___ + ___) ÷ 2 = ______ .

Step 4 — Q1 (median of lower half) and Q3 (median of upper half):

Lower half: 6, 8, 10, 12 → Q1 = (___ + ___) ÷ 2 = ______ .

Upper half: 14, 16, 18, 20 → Q3 = (___ + ___) ÷ 2 = ______ .

Step 5 — Five-number summary and IQR:

Min = ___, Q1 = ___, Median = ___, Q3 = ___, Max = ___, IQR = ___ − ___ = ______ .

Step 6 — Shape: describe whether the box plot is symmetric, left-skewed or right-skewed.

Stuck? Revisit lesson § "Interpreting Box Plots" — symmetric means the median sits roughly in the centre of the box.

3. You do — independent practice

Show working under each problem. Foundation = read a summary; Standard = construct a five-number summary; Extension = include outliers or compare displays.

Foundation — read a box plot

3.1 A box plot has Min = 4, Q1 = 8, Median = 12, Q3 = 16, Max = 20. Find the range. 1 mark

3.2 Using the same summary as 3.1, find the IQR. 1 mark

3.3 A stem-and-leaf plot has stems 1, 2, 3 with leaves shown. Stem 2 has leaves 1 3 3 5. Write out the actual data values for stem 2. 1 mark

3.4 A histogram shows two bars: 0-10 has frequency 5, 10-20 has frequency 8. How many data values in total? 1 mark

Standard — construct a summary

3.5 Find the five-number summary for 4, 8, 10, 12, 15, 18, 22, 25, 30. 3 marks

3.6 Draw a stem-and-leaf plot for 12, 15, 21, 23, 23, 30, 31, 35, 42. Use the tens digit as the stem. 3 marks

Extension — include outliers / compare

3.7 For 2, 3, 5, 7, 8, 10, 12, 15, 50 use the 1.5 × IQR rule to decide whether 50 should be plotted as an outlier on a box plot, and state where the right whisker should extend to. 3 marks

3.8 Two classes sat the same test. Class A: Min 40, Q1 55, Med 70, Q3 80, Max 95. Class B: Min 50, Q1 60, Med 65, Q3 75, Max 90. Which class had the higher median? Which had the larger IQR? Which would you describe as more consistent? 3 marks

Stuck on 3.8? Higher median = better central performance. Smaller IQR = more consistent (less spread in the middle 50%).

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Section 2 — We do (6, 8, 10, 12, 14, 16, 18, 20)

n = 8. Min = 6. Max = 20.
Middle pair: 12 and 14. Median = (12 + 14) ÷ 2 = 13.
Lower half: 6, 8, 10, 12 → Q1 = (8 + 10) ÷ 2 = 9.
Upper half: 14, 16, 18, 20 → Q3 = (16 + 18) ÷ 2 = 17.
Summary: 6, 9, 13, 17, 20. IQR = 17 − 9 = 8.
Shape: symmetric — the median sits roughly in the centre of the box and the whiskers are equal in length.

3.1 — Range

Range = Max − Min = 20 − 4 = 16.

3.2 — IQR

IQR = Q3 − Q1 = 16 − 8 = 8.

3.3 — Stem-and-leaf values

Stem 2, leaves 1 3 3 5 → data values 21, 23, 23, 25.

3.4 — Histogram total

Total = 5 + 8 = 13 data values.

3.5 — Five-number summary for 4, 8, 10, 12, 15, 18, 22, 25, 30

n = 9. Min = 4. Max = 30. Median = 5th value = 15.
Lower half: 4, 8, 10, 12 → Q1 = (8 + 10) ÷ 2 = 9.
Upper half: 18, 22, 25, 30 → Q3 = (22 + 25) ÷ 2 = 23.5.
Summary: 4, 9, 15, 23.5, 30.

3.6 — Stem-and-leaf for 12, 15, 21, 23, 23, 30, 31, 35, 42

Stem | Leaf
  1  | 2 5
  2  | 1 3 3
  3  | 0 1 5
  4  | 2

3.7 — Is 50 an outlier?

n = 9, median = 8 (5th value). Lower half: 2, 3, 5, 7 → Q1 = (3 + 5) ÷ 2 = 4. Upper half: 10, 12, 15, 50 → Q3 = (12 + 15) ÷ 2 = 13.5. IQR = 13.5 − 4 = 9.5.
Upper fence = 13.5 + 1.5(9.5) = 13.5 + 14.25 = 27.75. 50 > 27.75, so plot 50 as an outlier (separate point). The right whisker extends to the largest value still inside the fence — i.e. to 15.

3.8 — Compare Class A and Class B

Class A median = 70, Class B median = 65 → Class A higher median.
Class A IQR = 80 − 55 = 25; Class B IQR = 75 − 60 = 15 → Class A larger IQR.
Class B is more consistent because its IQR (and overall range) are smaller, even though its typical mark is a bit lower.