S2.1Histograms, box plots and cumulative frequency
Edexcel A level Maths (9MA0) · Statistics › Data presentation and interpretation
Practise Histograms, box plots and cumulative frequency. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
The table shows the times, in minutes, taken by 80 people to complete a puzzle.
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(a) A student draws a histogram of these data in which the height of each bar is equal to the frequency. Explain why the student's histogram is misleading.[1]
(b) Calculate the frequency density for each class.[2]
(c) Estimate the number of people who took between 12 and 25 minutes.[2]
(d) Use linear interpolation to estimate the median time.[2]
Show the answer and mark scheme
(a)Answer: The classes have different widths; in a histogram the area of a bar, not its height, represents frequency, so wide classes (such as 30–50) look far too large.
B1 for the class widths are unequal, so heights should be frequency densities (area proportional to frequency); e.g. the 30–50 bar is 20 minutes wide and so looks much bigger than 6 people
Worked solution: The classes have unequal widths. In a histogram the area of each bar represents frequency, so heights must be frequency densities. Using frequency as height makes wide classes such as \(30 \le t \lt 50\) look far more important than they are.
(b)Answer: 1.2, 3.6, 4.8, 2, 0.3
M1 for frequency ÷ class width for at least two classes
Worked solution: The median is the 40th value. 30 people took less than 15 minutes, and the next class has 24 people, so the median \(\approx 15 + \frac{40 - 30}{24} \times 5 = 17.1\) minutes.
Question 3Hard7 marks
The table shows the ages for a random sample of 204 members of a gym. A histogram is drawn to represent these data. On the histogram, the bar for the class 15 ≤ a < 20 is 1 cm wide and 4 cm tall.
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(a) Find the width and the height of the bar for the class 30 ≤ a < 40.[3]
(b) Estimate the probability that a member chosen at random from this sample has \(a\) between 23 and 44.[3]
(c) State, giving a reason, whether a normal distribution is likely to be a suitable model for these ages.[1]
Show the answer and mark scheme
(a)Answer: Width 2 cm, height 5.4 cm
B1 for width 2 cm (5 years is represented by 1 cm)
M1 for using area proportional to frequency, e.g. \(1 \times 4 = 4 \text{cm}^2\) represents 20, so \(1 \text{cm}^2\) represents 5 members
A1 for height 5.4 cm
Worked solution: Horizontal scale: 5 years is 1 cm, so the width is \(10 \times 0.2 = 2\) cm. Area: \(4 \text{cm}^2\) represents 20, so the bar for 54 members has area \(54 \div 5 = 10.8 \text{cm}^2\) and height \(10.8 \div 2 = 5.4\) cm.
M1 for a correct method for a part class, e.g. \(\dfrac{2}{5} \times 40\)
M1 for adding the frequencies of the whole classes and both part classes, and dividing by 204
A1 for awrt 0.584
Worked solution: \(8.0 \times 2 + 7.6 \times 5 + 5.4 \times 10 + 2.8 \times 4 = 119.2\), so the probability is \(\dfrac{119.2}{204} = 0.5843\).
(c)Answer: Not suitable: the distribution is not symmetrical (it has positive skew, with a longer tail to the right), whereas a normal distribution is symmetrical about its mean.
B1 for not suitable, with a reason based on the lack of symmetry (positive skew) of the data