Year 7 Mathematics · AC9M7ST02

Numerical Data Displays and Distribution Comparison

create different types of numerical data displays including stem-and-leaf plots using software where appropriate; describe and compare the distribution of data…

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Learning goalsSay it simply

We are learning to construct and compare numerical data displays, including stem-and-leaf plots, and interpret distributions in context.

A numerical data display should make the distribution visible without changing the values it represents. A stem-and-leaf plot separates each value into a leading stem and a final leaf, preserves individual observations, and requires ordered leaves and a key such as 4 | 7 = 47 so the scale cannot be misread.

Centre describes where a distribution is located, spread describes how widely its values vary, and shape includes clusters, gaps, symmetry or a longer tail. Reading only a median or only the tallest stack can hide important differences between distributions.

A comparison must use the same variable, units and scale and must connect numerical evidence to context. Two groups can have the same median but very different ranges or clustering, so a strong conclusion cites at least one measure of centre and one feature of spread or shape.

Success criteria

  • I can construct a stem-and-leaf plot with ordered leaves, a title and an unambiguous key.
  • I can recover original values and calculate centre and spread from a numerical display.
  • I can compare two distributions using evidence about centre, spread, clusters, gaps and extreme values.
Key conceptTeach from the board

Use these models to connect the mathematical idea to values, diagrams, coordinates, graphs or structure before moving to symbolic calculation.

Follow each example from representation and setup through calculation/reasoning, interpretation and an independent check.

Example 1

Reading centre and spread from a stem-and-leaf plot.

  1. Use the key 4 | 7 = 47 and read the ordered rows 3 | 2 5 8, 4 | 1 3 3 7 9, 5 | 0 4 6 and 6 | 2 as twelve values.
  2. The sixth and seventh values are 43 and 47, so the median is their mean, 45.
  3. The minimum is 32 and the maximum is 62, so the range is 62 minus 32, which equals 30.
  4. The value 43 occurs twice while every other value occurs once, and most observations lie in the 40s and 50s.

Final answer: The distribution has median 45, range 30, mode 43 and a concentration of values in the 40s and 50s.

Check: Reconstructing all twelve values in order confirms the middle pair, endpoints and repeated value.

Example 2

Comparing groups with equal medians but different spread.

  1. Group A has values 32, 40, 45, 47, 47, 50 and 62, while Group B has 40, 44, 46, 47, 48, 50 and 54.
  2. Both groups have seven values and the fourth value is 47, so both medians are 47.
  3. Group A has range 62 minus 32 equals 30, while Group B has range 54 minus 40 equals 14.
  4. Conclude that the groups share the same centre by median, but Group B is more tightly clustered and Group A is much more variable.

Final answer: Both medians are 47, but Group A has range 30 and greater spread, whereas Group B has range 14 and stronger clustering around the high 40s.

Check: The comparison uses the complete ordered values and reports both a centre feature and a spread feature in the same units.

Example 3

Application problem 1

Problem: Two classes both have median score 68. Class A ranges from 52 to 84; Class B ranges from 64 to 73. Compare the distributions using centre and spread without claiming they are 'the same'.

  1. Plan: Represent the information first, then calculate, interpret and independently check the result.
  2. Work: The centres are equal by median, but Class A has much greater spread. Class B is more tightly clustered, so equal medians do not mean equal distributions.
  3. Interpret: This synthesis problem combines the chapter's core representation, calculation and reasoning skills.

Final answer: The centres are equal by median, but Class A has much greater spread. Class B is more tightly clustered, so equal medians do not mean equal distributions.

Check: This synthesis problem combines the chapter's core representation, calculation and reasoning skills.

Example 4

Application problem 2

Problem: Two stem-and-leaf plots use identical rows, but Plot X has key 2 | 5 = 25 centimetres and Plot Y has key 2 | 5 = 2.5 metres; explain why their numerical patterns cannot be compared responsibly without conversion.

  1. Plan: Read one mark from each key as a full measurement and express both measurements in the same unit.
  2. Work: The keys use different scales and units, so every value must be converted to a common unit before centre, spread or contextual magnitude is compared.
  3. Interpret: Similar-looking marks can represent different quantities when keys and units differ.

Final answer: The keys use different scales and units, so every value must be converted to a common unit before centre, spread or contextual magnitude is compared.

Check: Similar-looking marks can represent different quantities when keys and units differ.

Clean visual examplesOne-page board

Clean one-page examples

AC9M7ST02 - Numerical Data Displays and Distribution Comparison
Example 1

Example 1 Reading centre and spread from a stem-and-leaf plot. Use the key 4 | 7 = 47 and read the ordered rows 3 | 2 5 8, 4 | 1 3 3 7 9, 5 | 0 4 6 and 6 | 2 as twelve values. The sixth and seventh values are 43 and 47, so the median is their mean, 45. The minimum is 32 and the maximum is 62, so the range is 62 minus 32, which equals 30. The value 43 occurs twice while every other value occurs once, and most observations lie in the 40s and 50s. Final answer: The distribution has median 45, range 30, mode 43 and a concentration of values in the 40s and 50s. Check: Reconstructing all twelve values in order confirms the middle pair, endpoints and repeated value.

Example 2

Example 2 Comparing groups with equal medians but different spread. Group A has values 32, 40, 45, 47, 47, 50 and 62, while Group B has 40, 44, 46, 47, 48, 50 and 54. Both groups have seven values and the fourth value is 47, so both medians are 47. Group A has range 62 minus 32 equals 30, while Group B has range 54 minus 40 equals 14. Conclude that the groups share the same centre by median, but Group B is more tightly clustered and Group A is much more variable. Final answer: Both medians are 47, but Group A has range 30 and greater spread, whereas Group B has range 14 and stronger clustering around the high 40s. Check: The comparison uses the complete ordered values and reports both a centre feature and a spread feature in the same units.

Example 3

Example 3 Application problem 1 Problem: Two classes both have median score 68. Class A ranges from 52 to 84; Class B ranges from 64 to 73. Compare the distributions using centre and spread without claiming they are 'the same'. Plan: Represent the information first, then calculate, interpret and independently check the result. Work: The centres are equal by median, but Class A has much greater spread. Class B is more tightly clustered, so equal medians do not mean equal distributions. Interpret: This synthesis problem combines the chapter's core representation, calculation and reasoning skills. Final answer: The centres are equal by median, but Class A has much greater spread. Class B is more tightly clustered, so equal medians do not mean equal distributions. Check: This synthesis problem combines the chapter's core representation, calculation and reasoning skills.

Example 4

Example 4 Application problem 2 Problem: Two stem-and-leaf plots use identical rows, but Plot X has key 2 | 5 = 25 centimetres and Plot Y has key 2 | 5 = 2.5 metres; explain why their numerical patterns cannot be compared responsibly without conversion. Plan: Read one mark from each key as a full measurement and express both measurements in the same unit. Work: The keys use different scales and units, so every value must be converted to a common unit before centre, spread or contextual magnitude is compared. Interpret: Similar-looking marks can represent different quantities when keys and units differ. Final answer: The keys use different scales and units, so every value must be converted to a common unit before centre, spread or contextual magnitude is compared. Check: Similar-looking marks can represent different quantities when keys and units differ.

Curriculum examplesCopied content

Content description: create different types of numerical data displays including stem-and-leaf plots using software where appropriate; describe and compare the distribution of data, commenting on the shape, centre and spread including outliers and determining the range, median, mean and mode.

Questions and answersWith answers

Attempt each problem before opening Check answer. The set moves from core understanding to application and synthesis.

  1. 1. In a stem-and-leaf plot with key 4 | 7 = 47, what value is represented by 5 | 3?

    Check answer

    Answer: The entry 5 | 3 represents 53.

    Hint: Use the example key to combine stem 5 with leaf 3 in the same place-value pattern.

    Why: The stem gives the tens digit and the leaf gives the ones digit under the stated key.

  2. 2. Write the values represented by the row 3 | 2 5 8 when the key is 3 | 2 = 32.

    Check answer

    Answer: The row represents 32, 35 and 38.

    Hint: Keep the stem digit 3 and attach each leaf in turn.

    Why: Each leaf is paired with the common stem to reconstruct one original value.

  3. 3. A student leaves each stem's leaves in collection order. Explain why the leaves should instead be ordered.

    Check answer

    Answer: Ordered leaves allow the data to be read in numerical order so the median, clusters, gaps and extremes can be identified accurately.

    Hint: Think about which distribution features depend on knowing the order of values.

    Why: Ordering improves interpretation without changing any observation.

  4. 4. Construct a stem-and-leaf plot for 21, 24, 24, 29, 31, 35 and 38 using tens as stems, and include a key.

    Check answer

    Answer: The plot has rows 2 | 1 4 4 9 and 3 | 1 5 8, with a valid key such as 2 | 1 = 21.

    Hint: Group values by tens digit, order each set of ones digits and then state how one mark is read.

    Why: Every value appears once with ordered leaves beside its tens stem and an unambiguous key.

  5. 5. For the ordered stem-and-leaf values 12, 15, 18, 21, 23, 23, 27 and 30, calculate the median, mode and range.

    Check answer

    Answer: The median is 22, the mode is 23 and the range is 18.

    Hint: There are eight values, so use the middle pair before checking repetitions and endpoints.

    Why: The median averages 21 and 23, the repeated value is 23, and 30 minus 12 equals 18.

  6. 6. Class A has median 47 and range 30, while Class B has median 47 and range 14 for the same test; write a supported comparison.

    Check answer

    Answer: The classes have the same centre by median, but Class B's scores are less spread out because its range is 16 points smaller.

    Hint: Use one sentence for the equal medians and another clause for the ranges.

    Why: A sound comparison distinguishes the shared centre from the different spread.

  7. 7. A student creates the row 4 | 9 1 7 3 with key 4 | 1 = 41; identify the presentation error and correct the row.

    Check answer

    Answer: The leaves are unordered, so the corrected row is 4 | 1 3 7 9.

    Hint: Arrange the ones digits in numerical order while keeping the same stem.

    Why: Leaves must increase from left to right within a stem.

  8. 8. Compare Group A values 32, 40, 45, 47, 47, 50, 62 with Group B values 40, 44, 46, 47, 48, 50, 54 using centre, spread and clustering.

    Check answer

    Answer: Both groups have median 47, but Group A has range 30 and widely separated endpoints, whereas Group B has range 14 and is more tightly clustered from 40 to 54.

    Hint: Find the fourth value and endpoint difference for each group before describing where most values sit.

    Why: The groups share a median but differ substantially in variability and concentration.

  9. 9. Two stem-and-leaf plots use identical rows, but Plot X has key 2 | 5 = 25 centimetres and Plot Y has key 2 | 5 = 2.5 metres; explain why their numerical patterns cannot be compared responsibly without conversion.

    Check answer

    Answer: The keys use different scales and units, so every value must be converted to a common unit before centre, spread or contextual magnitude is compared.

    Hint: Read one mark from each key as a full measurement and express both measurements in the same unit.

    Why: Similar-looking marks can represent different quantities when keys and units differ.

  10. 10. Two classes both have median score 68. Class A ranges from 52 to 84; Class B ranges from 64 to 73. Compare the distributions using centre and spread without claiming they are 'the same'.

    Check answer

    Answer: The centres are equal by median, but Class A has much greater spread. Class B is more tightly clustered, so equal medians do not mean equal distributions.

    Hint: Represent the information first, then calculate, interpret and independently check the result.

    Why: This synthesis problem combines the chapter's core representation, calculation and reasoning skills.

Practice and reviewReady for practice

Common mistake: Leaves can be written in the order in which the data were collected.

Correction: Leaves should be ordered within each stem so that the median, gaps and shape can be read reliably.

Common mistake: A stem-and-leaf plot does not need a key because the digits are visible.

Correction: A key is essential because the same marks could represent ones, tenths or another place-value scale.

Common mistake: Two distributions with the same median must be practically the same.

Correction: Equal medians can occur with very different ranges, clusters, gaps and extreme values.

Curriculum alignmentStart here

We are learning to construct and compare numerical data displays, including stem-and-leaf plots, and interpret distributions in context.

A numerical data display should make the distribution visible without changing the values it represents. A stem-and-leaf plot separates each value into a leading stem and a final leaf, preserves individual observations, and requires ordered leaves and a key such as 4 | 7 = 47 so the scale cannot be misread.

Centre describes where a distribution is located, spread describes how widely its values vary, and shape includes clusters, gaps, symmetry or a longer tail. Reading only a median or only the tallest stack can hide important differences between distributions.

A comparison must use the same variable, units and scale and must connect numerical evidence to context. Two groups can have the same median but very different ranges or clustering, so a strong conclusion cites at least one measure of centre and one feature of spread or shape.

Success criteria

  • I can construct a stem-and-leaf plot with ordered leaves, a title and an unambiguous key.
  • I can recover original values and calculate centre and spread from a numerical display.
  • I can compare two distributions using evidence about centre, spread, clusters, gaps and extreme values.
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