Cumulative Frequency: Eduqas GCSE Higher June 2025 Paper 1 Q14

Higher June 2025 Paper 1 Q14

EduqasHigherCurrent spec13 marksCumulative Frequency Histograms

14. Cally, Deena and Juan are working on a GCSE project.
They are each in a different class.
Cally, Deena and Juan each record the journey times to school of pupils in their own class.

(a) The times for pupils in Cally’s class are shown in the table below.

Journey time, \(t\) (minutes)	Number of pupils
\(0 \lt t \leqslant 5\)	5
\(5 \lt t \leqslant 15\)	5
\(15 \lt t \leqslant 30\)	6
\(30 \lt t \leqslant 50\)	5
\(50 \lt t \leqslant 60\)	4

She decides to draw a histogram of her results.

Cally's histogram: frequency density 1 for 0-5, 2 for 5-15, 2.5 for 15-30, 4 for 30-50 and 2.5 for 50-60 minutes

Most of the frequency densities used in the histogram are incorrect.
What error did Cally make when she calculated the frequency densities? [1]

(b) Deena has correctly drawn the histogram for her results.
4 pupils take between 0 and 5 minutes to travel to school.

Deena's histogram of journey times from 0 to 30 minutes with class boundaries 0, 5, 10, 12.5, 17.5, 20, 25, 30; the frequency density axis has no scale

(i) Complete the scale on the frequency density axis. [2]

(ii) Calculate an estimate of the median journey time for the pupils in Deena’s class.
You must show all your working. [3]

Cumulative frequency graph through (0, 0), (16, 3), (24, 8), (30, 14), (35, 18), (45, 21) and (60, 24)

Use the cumulative frequency diagram to give the best estimates for the answers to each of the following questions.

(i) Calculate the percentage of pupils in Juan’s class that take 45 minutes or longer to travel to school. [3]

(ii) Complete the table for the journey time to school of the pupils in Juan’s class. [3]

Table to complete with columns Median, Lower quartile, Upper quartile, Inter-quartile range

(iii) Two pupils were absent when Juan collected his data.
The journey times of these two pupils are 6 minutes and 40 minutes.

What effect would including these two results have on the median journey time for the class? [1]

Mark scheme (a)

Scheme	Marks	Comments
An explanation relating to calculating incorrect frequency densities e.g. It should be frequency divided by width of bar for heights of bars not the other way around She has calculated width divided by frequency	E1	Allow e.g. ‘journey time’ rather than ‘width of bar’ Do not allow e.g. ‘Journey time x frequency density is not equal to the number of pupils’ The areas of the bars do not give the frequencies

Mark scheme (b)

Scheme	Marks	Comments
(b)(i) A correct uniform scale with intervals of 0·5 Minimum of fd labels 1 and 2, any other labels must be correct	B2	If not B2, award B1 for 1 label correct and none incorrect e.g. 0·8 marked correctly 0·5 or 1 marked correctly OR a correct uniform scale, with intervals of 0.5 to the top of the histogram with errors above OR 4/5 or 0·8 seen in calculations.
(b)(ii)		STRICT FT their labelling for a uniform scale only, provided uniform to at least the top of the histogram, for possible M1A1B1
\(5\times0{\cdot}8 + 5\times1 + 2{\cdot}5\times2 + 5\times1{\cdot}6 + 2{\cdot}5\times0{\cdot}8 + 5\times0{\cdot}6 + 5\times0{\cdot}2\) \((= 4+5+5+8+2+3+1)\)	M1	May be seen on the histogram M1 for all 7 area calculations with maximum of one slip using incorrect height
= 28	A1	CAO for their labelling May be implied by an answer of 12·5 provided M1 awarded
(median =) 12·5 (minutes)	B1	FT ‘their 28’ provided M1 awarded An unsupported 12·5 gains M0A0B1

Mark scheme (c)

Scheme

Marks

Comments

(c)(i)
(24 − 21 =) 3

May be implied in further working

\(3/24\ (\times100)\)

FT ‘their 3’ / 24, including use of 21

12·5(%)

FT
An answer of 87·5(%) is awarded B0M1A1

If no marks, award SC1 for 4 ÷ 25 (x 100) = 16(%)

(c)(ii)

Median	LQ	UQ	IQ range
28	21	35	14

Median = 28

LQ = 21 and UQ = 35

IQR = 14
FT ‘their UQ’ – ‘their LQ’, provided one is correct

(c)(iii)
The median will not change.

STRICT FT their table in c(ii) if ‘their median’ is outside the interval 6 - 40

(13)