Histograms with unequal widths
3 exam-style questions. Answers and the mark for each step are on the last page.
- Question 1
(a) A histogram class covers 15 < t 25 minutes and contains 18 journeys, where t is the journey time. Work out the frequency density for this class.
- Question 2
A histogram has class widths 10, 15 and 5. Its corresponding drawn bar heights are 2, 4 and 6 cm. The total frequency is 330.
(a) Find the frequency in the second class.
(b) Explain why the frequencies cannot be read directly from the bar heights.
- Question 3
A table of waiting times for 65 patients has some gaps. : frequency 8. : frequency density 1.4. : frequency 18. : frequency density 0.8. : frequency 9.
(a) Show that the frequencies are consistent with a total of 65.
(b) Work out an estimate for the median waiting time. Give your answer to 1 decimal place.
Answers and marks
Question 1
(a)
- P1 Establishing or an equivalent valid method.
- A1 Correct answer:
Question 2
(a)
- P1 Calculate each rectangle’s drawn area, which is proportional to frequency.
- P1 Allocate the total using the middle area’s share.
- A1 Correct answer:
(b) The widths differ, so frequency is proportional to bar area. Height alone represents frequency density up to the vertical scale.
- C1 Correct conclusion with supporting reasoning: The widths differ, so frequency is proportional to bar area. Height alone represents frequency density up to the vertical scale.
Question 3
(a)
- P1 Both missing frequencies from density times width.
- A1 The total shown as 65.
(b) 25.8 minutes
- P1 Locating the median class by cumulative frequency.
- P1 Interpolating within the class.
- A1 Correct answer: 25.8 minutes.