Cumulative Frequency: WJEC GCSE Numeracy Higher November 2021 Unit 1 Q5

Numeracy Higher November 2021 Unit 1 Q5

WJECHigherCurrent spec12 marksCumulative Frequency

5. Bethan works as an office manager at a medical centre.

Last Monday, 60 patients each had an appointment with a doctor.
Bethan recorded how long each patient’s appointment lasted.
Her results are given in the table below.

Length of time, \(t\) (minutes)	\(0 \lt t \leqslant 4\)	\(4 \lt t \leqslant 8\)	\(8 \lt t \leqslant 12\)	\(12 \lt t \leqslant 16\)	\(16 \lt t \leqslant 20\)	\(20 \lt t \leqslant 24\)
Number of patients	4	24	18	6	2	6

(a) Complete the following cumulative frequency table. [1]

Time, \(t\) (minutes)	\(t \leqslant 0\)	\(t \leqslant 4\)	\(t \leqslant 8\)	\(t \leqslant 12\)	\(t \leqslant 16\)	\(t \leqslant 20\)	\(t \leqslant 24\)
Cumulative frequency	0	4	28				60

(b) On the graph paper below, draw a cumulative frequency diagram to show this information. [2]

Blank graph paper: Time, t (minutes) from 0 to 25 on the horizontal axis, Cumulative frequency from 0 to 60 on the vertical axis

(c) Each patient is given 10 minutes for their appointment.
Use your graph to give the best estimate for the number of appointments that lasted longer than 10 minutes.
Give your answer correct to the nearest whole number of appointments. [2]

(d) Of the patients seen last Monday, what percentage spent longer than 20 minutes with the doctor? [2]

(e) The median length of the appointments last Tuesday was 11·5 minutes.
How much shorter was the median length of the appointments on Monday?
Give your answer correct to the nearest minute. [2]

(f) Bethan is considering changing the time given for each patient’s appointment to 12 minutes.
She would set a target of 80% of patient appointments taking less than or equal to 12 minutes.
Would this target have been met last Monday?
You must show all your working. [3]

Mark scheme (b)

Scheme	Marks	Comments
Correct cumulative frequency diagram drawn, with points joined with a straight line or a curve	B2	FT only cumulative entries from (a) B1 for either correct plots (but not joined or with spurious or incorrect straight line or curve), or ‘their plots’ joined provided 5 or 6 plots are correct

Mark scheme (c)

Scheme	Marks	Comments
23 (patient appointments)	B2	STRICT FT ‘their cumulative graph’, i.e. correctly evaluated ‘60 – their 37’ B1 for sight of 60 – ‘their 37’ If ‘their graph’ is not cumulative or shows bars, FT for ‘60 – their 37’ provided a unique reading for 10 minutes (60 – ‘their reading at 10 minutes) but award B1 only (not B2) If no marks, award SC1 for 23 (patient appointments) calculated or unsupported, when not from ‘cumulative frequency graph’, e.g. (18/2 + 6 + 2 + 6 =) 23

Mark scheme (d)

Scheme	Marks	Comments
\(\dfrac{6}{60}\) ( × 100) or (100 ×) 1 - \(\dfrac{54}{60}\) ( × 100)	M1	FT ‘their 54’ (reading for t \(\leqslant\) 20) provided > 28
10 (%)	A1	Do not accept an answer with incorrect units, e.g. ‘10 people’

Mark scheme (e)

Scheme	Marks	Comments
Difference in medians 3 (minutes)	B2	Allow if calculated from the information in the table Must be correct to the nearest minute for ‘their cumulative frequency graph’ FT ‘their cumulative frequency graph’ Do not accept an answer from incorrect working, including without a cumulative graph seen B1 for either sight of Monday median 8.3 to 8.7 (minutes), or an answer for the difference in the medians in the range 2.8 to 3.2 (minutes) (from working with Monday median in the range 8.3 to 8.7 minutes) Apply the same tolerance of ±0.2 (minutes) when following through from ‘their graph’ On FT, if ‘their answer’ ≠ whole number of minutes, it must be rounded to the nearest minute

Mark scheme (f)

Scheme	Marks	Comments
Sight of any one of: \(\dfrac{4 + 24 + 18}{60}\) (× 100) \(\dfrac{46}{60}\) (× 100) 0.8 × 60 or equivalent 0.2 × 60 or equivalent	M1	FT ‘their 46’ from an incorrect cumulative total in (a) (i.e. ‘their value in the table for t \(\leqslant\) 12’)
Conclusion, ‘no (target not met)’ AND a correct evaluation: 76.666(%) rounded or truncated 48 (appointments) 12 and (60 – 12 =) 48 (appointments) OR 12 and (6 + 2 + 6 =) 14 (appointments)	A2	FT for correctly evaluated use of ‘their 46’ Allow for sight of 46 ÷ 60 = 0.76(…) or 100 × 46 ÷ 60 = 76(%) Allow, e.g. ‘over 0.7’ or ‘over 70(%) only provided the correct conclusion ‘no’ is given If 48 is not evaluated, accept the time from ‘their graph’ for 48 appointments instead A1 for a correct evaluation: 76.(66...%) rounded or truncated ‘over 70%’ or ‘over 0.7’ 0.76(...) or 0.77 48 (appointments) 12 (appointments)