S3Histograms

Edexcel GCSE Maths (1MA1), Higher tier · Statistics

Practise Histograms. 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.

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Sample questions

Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.

Question 1Easy4 marks

The table gives information about the heights, in cm, of some tomato plants. One of the frequencies is missing.

[object Object]

(a) Work out the frequency density for the class \(10 \lt h \le 20\).[2]

(b) The frequency density for the class \(20 \lt h \le 25\) is 2.
Work out the frequency for this class.[2]

Show the answer and mark scheme

(a) Answer: 1.6

M1 for \(16 \div 10\)
A1 for 1.6

Worked solution: Frequency density = frequency ÷ class width \(= 16 \div 10 = 1.6\)

(b) Answer: 10

M1 for \(2 \times 5\)
A1 for 10

Worked solution: Frequency = frequency density × class width \(= 2 \times 5 = 10\)

Question 2Medium4 marks

The table shows information about the wingspans, in mm, of a sample of moths caught in a trap.

[object Object]

(a) On the grid, draw a histogram for this information.[3]

[object Object]

(b) Explain why the heights of the bars are not the same as the frequencies.[1]

Show the answer and mark scheme

(a) Answer: Frequency densities 15, 24, 18, 9, 3. Bars: 20 to 24 with height 15; 24 to 28 with height 24; 28 to 32 with height 18; 32 to 40 with height 9; 40 to 52 with height 3.

M1 for frequency ÷ class width for at least 3 classes (frequency densities 15, 24, 18, 9, 3)
A1 for at least 3 bars of different widths drawn at the correct heights
A1 for a fully correct histogram: all 5 bars with the correct class boundaries and heights, each within ½ small square

Worked solution: Frequency density = frequency ÷ class width: \(60 \div 4 = 15\), \(96 \div 4 = 24\), \(72 \div 4 = 18\), \(72 \div 8 = 9\), \(36 \div 12 = 3\).
Draw each bar across its class, with height equal to its frequency density. There are no gaps between the bars.

(b) Answer: The class widths are not all equal, so the area of each bar (not its height) must represent the frequency: height = frequency ÷ class width.

C1 for a correct explanation, e.g. the classes have different widths, so the area of each bar shows the frequency

Worked solution: In a histogram the area of a bar represents the frequency. The classes have different widths, so each height is the frequency density, frequency ÷ class width.

Question 3Hard5 marks

The histogram gives information about the times, \(t\) minutes, that 100 customers spent in a shop.

[object Object]

(a) Jake says, 'The tallest bar is for \(15 \lt t \le 20\), so more customers spent between 15 and 20 minutes in the shop than in any other class.'
Explain why Jake is wrong.[2]

(b) Work out an estimate for the number of customers who spent between 12 minutes and 30 minutes in the shop.[3]

Show the answer and mark scheme

(a) Answer: The height is frequency density, not frequency. \(15 \lt t \le 20\) has \(5 \times 5 = 25\) customers but \(20 \lt t \le 40\) has \(2 \times 20 = 40\).

M1 for frequency = frequency density × class width for at least one class, e.g. \(5 \times 5 = 25\) or \(2 \times 20 = 40\)
C1 for explaining that \(20 \lt t \le 40\) has more customers (40 > 25), because the bar height shows frequency density, not frequency

Worked solution: Frequencies: \(1.5 \times 10 = 15\), \(4 \times 5 = 20\), \(5 \times 5 = 25\), \(2 \times 20 = 40\).
The class \(20 \lt t \le 40\) has the most customers; its bar is lower because it is wider.

(b) Answer: 57

M1 for \(3 \times 4\) (= 12) for 12 to 15 minutes
M1 for \(10 \times 2\) (= 20) for 20 to 30 minutes
A1 for 57

Worked solution: 12 to 15: \(3 \times 4 = 12\). 15 to 20: 25. 20 to 30: \(10 \times 2 = 20\).
Estimate \(= 12 + 25 + 20 = 57\) customers.

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