This set has ten original questions, ordered from easier to harder, covering all five lessons in data displays and cumulative reasoning. Questions 1 and 2 use histograms, 3 to 5 use cumulative frequency, 6 and 7 use scatter diagrams, and 8 to 10 ask you to compare and choose.
Attempt each question on paper, without a calculator unless you need one for a division, and write your working as you would in an exam.
Only then open the answer. Mark the ones you got wrong and use the routing list at the end. The non-calculator working trainer and the percentage-base explorer can help you check arithmetic, and the mistake log and retest queue is a good place to record errors.
Questions
1. A grouped table shows the ages, in years, of 65 people at a club.
| Age x | 0 < x ≤ 5 | 5 < x ≤ 15 | 15 < x ≤ 20 |
|---|---|---|---|
| Frequency | 10 | 35 | 20 |
Find the frequency density for each class, to draw a histogram.
Show answer
Class widths: 5, 10 and 5. Frequency density = frequency ÷ class width.
0 < x ≤ 5: 10 ÷ 5 = 2. 5 < x ≤ 15: 35 ÷ 10 = 3.5. 15 < x ≤ 20: 20 ÷ 5 = 4.
2. A histogram has three bars: 0 < x ≤ 20 with height 0.5, 20 < x ≤ 30 with height 2.2, and 30 < x ≤ 50 with height 1.4. (a) Find the total frequency. (b) Estimate the number of values with 25 < x ≤ 40.
Show answer
(a) Frequencies are height × width: 0.5 × 20 = 10, 2.2 × 10 = 22, 1.4 × 20 = 28. Total = 10 + 22 + 28 = 60.
(b) Part of the second bar from 25 to 30 is 5 wide: 2.2 × 5 = 11. Part of the third bar from 30 to 40 is 10 wide: 1.4 × 10 = 14. Estimate = 11 + 14 = 25. It assumes the values are spread evenly within each class.
3. The lengths of 60 phone calls (in minutes) give these cumulative frequencies: 3 by 10, 11 by 20, 29 by 30, 47 by 40, 56 by 50 and 60 by 60. Estimate the median length.
Show answer
n = 60, so the median is the 30th value. It lies between (30, 29) and (40, 47). The rise needed is 30 − 29 = 1 out of 18, so the length is 30 + 10 × 1/18 = 30 + 0.556 ≈ 30.6 minutes.
Check: the median should lie in the 30 to 40 class, where the cumulative frequency passes 30. It does.
4. Using the data in question 3, estimate the interquartile range.
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Lower quartile: position 60 ÷ 4 = 15, between (20, 11) and (30, 29): 20 + 10 × 4/18 ≈ 22.22.
Upper quartile: position 3 × 60 ÷ 4 = 45, between (30, 29) and (40, 47): 30 + 10 × 16/18 ≈ 38.89.
IQR ≈ 38.89 − 22.22 = 16.67, which is about 16.7 minutes.
5. Using the data in question 3, estimate how many calls lasted more than 35 minutes.
Show answer
At 35 minutes, halfway between 30 and 40, the cumulative frequency is 29 + 18 × 0.5 = 38. So 60 − 38 = 22 calls.
Note that the answer counts those above 35, not at or below it.
6. On a scatter diagram of daily temperature x (°C) against cold drinks sold y, the line of best fit passes through (28, 110) and (34, 140). (a) Find its equation. (b) Predict the sales at 31 °C. (c) Explain why a prediction for 40 °C is unreliable if the data covers 26 °C to 36 °C.
Show answer
(a) Gradient = (140 − 110) ÷ (34 − 28) = 30 ÷ 6 = 5. Then y = 5x + c. Using (28, 110): 110 = 140 + c, so c = −30. The line is y = 5x − 30. Check with (34, 140): 170 − 30 = 140. ✓
(b) y = 5 × 31 − 30 = 155 − 30 = 125 drinks.
(c) 40 °C is outside the data range, so this is extrapolation. The pattern may not continue, for example if very hot days keep people at home.
7. A scatter diagram shows that students who own more books tend to have higher scores. A classmate says, “So buying more books will raise your score.” Comment on this conclusion.
Show answer
The diagram shows association, not cause. A third factor, such as how much time students spend reading, could affect both the number of books owned and the score. The safe conclusion is that more books are associated with higher scores, and more evidence would be needed to claim that buying books causes the improvement.
8. Class P has a median of 64 and an IQR of 12. Class Q has a median of 64 and an IQR of 24. Write two comparisons in context (test marks).
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The medians are the same (64), so the typical mark is the same in both classes. Class P has a smaller IQR (12 against 24), so its marks are more consistent, while Class Q’s marks are more spread out.
9. In Form A, 21 of 50 students use the library at lunch. In Form B, 36 of 120 do. Which form has the greater proportion using the library?
Show answer
Form A: 21 ÷ 50 = 0.42 = 42%. Form B: 36 ÷ 120 = 0.30 = 30%. Form A has the greater proportion, even though Form B has more students overall and its count of 36 is larger than 21.
10. A student spends RM45 of a RM300 allowance on transport. (a) What angle does transport take in a pie chart? (b) Which display would you choose to find the median of 200 grouped exam marks, and which would you choose to compare two groups of different sizes?
Show answer
(a) 45 ÷ 300 × 360 = 0.15 × 360 = 54°.
(b) For the median of grouped marks, use a cumulative frequency curve and read at the 100th value (200 ÷ 2). To compare groups of different sizes, use percentages or relative frequencies, for example in a bar chart, so that the group sizes do not distort the picture.
If you got these wrong
- Questions 1 and 2 (heights read as frequencies, or width forgotten): go back to reading a histogram using frequency density.
- Questions 3 to 5 (wrong position, wrong axis, plotting at the wrong point): revisit estimating a median from cumulative frequency.
- Questions 6 and 7 (extrapolating, or claiming a cause): see using a scatter diagram without inferring causation.
- Questions 8 and 9 (comparing counts, or one measure only): work through comparing two distributions with compatible scales.
- Question 10 (choosing a display): review choosing a display that preserves the data meaning.
Fix the lesson first, then retry a fresh question a few days later. If the same slip returns more than once, it is worth having someone look at your working, which is what our teachers do in online one-to-one Mathematics tuition.