Handling Data 6
4 marks · NormalThe histogram below shows the time taken, in minutes, for a group of students to complete a test.
The class widths for all intervals are equal except for the last bar, which covers a wider interval.
Use the histogram to write down the class width for the interval 50 < t ≤ 60.
Work out the frequency (number of students) for the interval 40 < t ≤ 50.
Write down the total number of students in the group.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) 10 minutes
(b) 10
(c) 55
Common mistakes: Mixing up frequency and frequency density; Adding heights rather than area (width × density)
Show mark scheme and worked solution
Marking points
B1 (a) · 1 mark · Identifies the class width for 50 < t ≤ 60 as 10 minutes.
M1 (b) · 1 mark · Calculates the frequency for 40 < t ≤ 50 using frequency = width × density (10 × 1.0 = 10).
M2 (c) · 1 mark · Uses the correct method: adds total frequencies from all intervals.
A1 (c) · 1 mark · Correct total of 55 students. (follow-through)
Worked solution
a) The class width for the interval 50 < t ≤ 60 is 10 minutes.
b) Frequency for 40 < t ≤ 50 = width × frequency density = 10 × 1.0 = 10.
c) Find frequencies for each interval:
- 20 < t ≤ 30: width = 10, density = 1.8 → freq = 18 - 30 < t ≤ 40: width = 10, density = 2.2 → freq = 22 - 40 < t ≤ 50: width = 10, density = 1.0 → freq = 10 - 50 < t ≤ 60: width = 5, density = 0.6 → freq = 5 × 0.6 = 3 (this is incorrect if widths are not specified for this bar: if all widths are 10, then freq = 10 × 0.6 = 6) - Total frequency = 18 + 22 + 10 + 6 = 56.
*Please ensure bar widths; if last bar width is actually 10, final total is 56. If 5, then 53.