Mathematics Standard • Year 12 • Module 8 • Lesson 4

Displaying Data — Skill Drill

Build fluency in constructing and reading frequency tables, histograms, stem-and-leaf plots and dot plots — and naming the distribution shape.

Build · Skill Drill

1. Quick recall

Answer each in the space provided. 1 mark each

Q1.1 A histogram bar runs from 150 to 160 cm. The class width is ____________ cm and the midpoint is ____________ cm.

Q1.2 Match each display to the data type it suits best.
Stem-and-leaf → ____________    Histogram → ____________    Dot plot → ____________    Pie chart → ____________

Options: small numerical data sets, large grouped continuous data, small discrete data with low repeats, categorical (proportion of a whole).

Q1.3 Name the three common distribution shapes: ____________, ____________ (tail on right), ____________ (tail on left).

Stuck? Revisit lesson § Key Ideas — Histogram, Stem-and-leaf, Class width.

2. Worked example — frequency table → histogram → shape

Heights (cm) of 20 students: 152, 155, 158, 160, 162, 163, 165, 166, 168, 170, 171, 172, 175, 178, 180, 182, 185, 188, 190, 195.

Step 1 — Group with class width 10 cm.

Class (cm)Frequency
150 – 1593
160 – 1696
170 – 1796
180 – 1894
190 – 1991

Step 2 — Histogram (sketch). Bars of equal width, no gaps between them, height = frequency.

▮▮▮ ▮▮▮▮▮▮ ▮▮▮▮▮▮ ▮▮▮▮ ▮

Step 3 — Stem-and-leaf (stem = tens, leaves = units).

15 | 2 5 8
16 | 0 2 3 5 6 8
17 | 0 1 2 5 8
18 | 0 2 5 8
19 | 0 5

Conclusion. Roughly symmetric, unimodal, centred around 170 cm. Most students fall in the 160-179 cm range.

3. Faded example — group the raw data

The number of texts sent yesterday by 18 students: 0, 1, 2, 4, 5, 6, 8, 10, 12, 14, 17, 20, 23, 28, 35, 42, 48, 60. Use class width 10. 4 marks

Class (texts)TallyFrequency
0 – 9________________________________
10 – 19________________________________
20 – 29________________________________
30 – 39________________________________
40 – 49________________________________
50 – 59________________________________
60 – 69________________________________

Step — Describe the shape: The distribution is ____________-skewed (most students texted few times; a few texted very often).

Stuck? Revisit lesson § Histogram — equal-width classes, frequency = number in each class.

4. Graduated practice — display and describe

Foundation — one-step reading (4 questions)

QProblemAnswer
4.1 1A class has width 10, lower boundary 30. The midpoint is ____.
4.2 1A histogram has bars at 0-9, 10-19, 20-29 with heights 5, 12, 8. How many total observations?
4.3 1A distribution has its tail on the right. Name the shape.
4.4 1If a stem-and-leaf row reads "4 | 0 3 7", list the data values.

Standard — typical HSC difficulty (6 questions)

4.5 Group these test scores into classes of width 10 (starting 50-59): 52, 55, 58, 61, 63, 65, 68, 71, 74, 76, 78, 81, 85, 88, 92, 95. Make a frequency table. 2 marks

4.6 Draw a back-to-back stem-and-leaf for two classes' marks. Class A: 62, 65, 67, 70, 71, 74, 75, 78, 82, 85. Class B: 55, 58, 60, 62, 65, 68, 70, 71, 74, 78. (Use stems 5, 6, 7, 8.) 2 marks

4.7 A histogram shows the bars (left to right): 4, 8, 12, 15, 9, 3. Describe the shape (symmetric / left-skew / right-skew). 1 mark

4.8 A pie chart shows favourite drinks at a cafe: coffee 45%, tea 25%, juice 15%, water 10%, other 5%. What angle (in degrees) represents tea? 2 marks

4.9 Draw a dot plot for: 3, 5, 5, 6, 6, 6, 7, 7, 8, 8, 9. Then state the mode and describe the shape. 2 marks

4.10 A stem-and-leaf has rows 3|2 4 9, 4|0 1 5 7 8, 5|2 3 6, 6|1. Find n (total values) and the median. 2 marks

Extension — read and interpret (2 questions)

4.11 A histogram of exam marks (class width 10) has frequencies: 50-59 (5), 60-69 (12), 70-79 (15), 80-89 (10), 90-99 (3). (a) How many students took the exam? (b) Estimate the median by finding the class containing the (n+1)/2-th value. 3 marks

4.12 The same data is shown as a histogram and as a stem-and-leaf plot. State one advantage of the histogram and one advantage of the stem-and-leaf for this data set. 3 marks

Stuck on 4.12? Histograms compress big data sets; stem-and-leaf preserves every original value.

5. Self-check the easy 3

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Class width and midpoint

Width = 160 − 150 = 10 cm. Midpoint = (150 + 160) / 2 = 155 cm.

Q1.2 — Match displays

Stem-and-leaf → small numerical data sets. Histogram → large grouped continuous data. Dot plot → small discrete data with low repeats. Pie chart → categorical (proportion of a whole).

Q1.3 — Distribution shapes

Symmetric, right-skewed (positive skew, tail on right), left-skewed (negative skew, tail on left).

Q3 — Faded example (texts)

Frequencies: 0-9 (7), 10-19 (4), 20-29 (3), 30-39 (1), 40-49 (2), 50-59 (0), 60-69 (1). Total = 18. Distribution is right-skewed (positive skew) — most students sent few texts; a few students sent many.

Q4.1–4.4 — Foundation

4.1: midpoint = (30 + 40)/2 = 35.   4.2: total = 5 + 12 + 8 = 25.   4.3: right-skewed (positive skew).   4.4: data = 40, 43, 47.

Q4.5 — Frequency table (test scores)

50-59: 3 (52, 55, 58).   60-69: 4 (61, 63, 65, 68).   70-79: 4 (71, 74, 76, 78).   80-89: 3 (81, 85, 88).   90-99: 2 (92, 95). Total = 16.

Q4.6 — Back-to-back stem-and-leaf

Class A | Stem | Class B
               | 5 | 5 8
    7 5 2 | 6 | 0 2 5 8
 8 5 4 1 0 | 7 | 0 1 4 8
    5 2 | 8 |

(Class A leaves read right-to-left from the stem; Class B leaves read left-to-right.)

Q4.7 — Shape

Bar heights 4, 8, 12, 15, 9, 3 rise then fall, peak slightly right of centre, slightly longer left tail (4, 8, 12) than right (9, 3). Best description: roughly symmetric (very slight left-skew is acceptable).

Q4.8 — Pie slice for tea

Tea = 25% of 360° = 0.25 × 360 = 90°.

Q4.9 — Dot plot

3: •
5: ••
6: •••
7: ••
8: ••
9: •

Mode = 6 (3 dots, the tallest column). Shape: roughly symmetric, unimodal.

Q4.10 — n and median from stem-and-leaf

Data: 32, 34, 39, 40, 41, 45, 47, 48, 52, 53, 56, 61. n = 12. Median = (6th + 7th)/2 = (45 + 47)/2 = 46.

Q4.11 — Histogram exam marks

(a) Total = 5 + 12 + 15 + 10 + 3 = 45 students.
(b) Median position = (45+1)/2 = 23rd value. Cumulative: 5, 17, 32, 42, 45. The 23rd value falls in the 70-79 class. So estimated median ≈ 74-75 marks (i.e. in the 70-79 group).

Q4.12 — Histogram vs stem-and-leaf

Histogram advantage: shows the shape and frequency of large data sets at a glance — easy to compare bars visually. Stem-and-leaf advantage: retains every original value, so you can still find the exact median, mode and individual data points after plotting.