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O-Level E-Math · 2015 · P2 Q10 Cumulative frequency / probability · Median & quartiles from a CF curve 10 marks: (a)(i) 1 + 2 + 2, (a)(ii) 1, (b)(i) 1 + 1, (b)(ii) 2 · statistics & probability difficulty 4 of 5

O-Level E-Math 2015 Paper 2, Question 10: Median & quartiles from a CF curve

The answer

(a)(i)(a) \(\approx 40\) min
(b) \(\approx 14\) min
(c) \(\approx 10\%\)
(a)(ii) the women's curve is shifted to the right
(b)(i)(a) \(\dfrac{1}{16}\)
(b) \(\dfrac{31}{120}\)
(b)(ii) \(0.0638\)

O-Level E-Math 2015 Paper 2 Question 10 · Verified worked solution by the Genius Plus Academy teaching team

What this question tests

This is Question 10 of the O-Level E-Math 2015 Paper 2. It tests median & quartiles from a cf curve, in the Cumulative frequency / probability area. It is worth 10 marks: (a)(i) 1 + 2 + 2, (a)(ii) 1, (b)(i) 1 + 1, (b)(ii) 2. It is a worded / diagram-based question, so open your Ten-Year Series (TYS) or the official paper at this question, then follow our full worked solution below.

Step-by-step solution

(a)(i)(a) Median \(=\) time at cumulative frequency \(\dfrac{120}{2} = 60\): about \(40\) minutes. (Read from graph.)

(b) Lower quartile (CF \(= 30\)) \(\approx 37\) min; upper quartile (CF \(= 90\)) \(\approx 51\) min; IQR \(\approx 51 - 37 = 14\) minutes. (Read from graph.)

(c) At least one hour \(=\) time \(\geqslant 60\) min. The cumulative frequency at 60 min is about 108, so about \(120 - 108 = 12\) men took at least an hour, i.e. \(\dfrac{12}{120} = 10\%\). (Read from graph; accept roughly 8 to 12%.)

(ii) The same interquartile range means the curve has the same overall steepness in the middle, and a higher median means it is shifted to the right (towards longer times). So the women's curve has the same shape as the men's but is translated horizontally to the right.

(b) Men total \(35+46+24+15 = 120\); women total \(27+34+41+18 = 120\); overall 240. (i)(a) Men aged 50 or more \(= 15\), so \(P = \dfrac{15}{240} = \dfrac{1}{16}\).

(b) Aged under 30 \(= 35 + 27 = 62\), so \(P = \dfrac{62}{240} = \dfrac{31}{120}\).

(ii) Women aged under 40 \(= 27 + 34 = 61\). Choosing two without replacement: \(P = \dfrac{61}{240} \times \dfrac{60}{239} = \dfrac{3660}{57\,360} = 0.0638\) (3 s.f.).

Answer: (a)(i)(a) \(\approx 40\) min
(b) \(\approx 14\) min
(c) \(\approx 10\%\)
(a)(ii) the women's curve is shifted to the right
(b)(i)(a) \(\dfrac{1}{16}\)
(b) \(\dfrac{31}{120}\)
(b)(ii) \(0.0638\)

Same structure, different numbers

A question is hard because of its structure, not its surface.

Swap the constants, dress a quadratic as a length, hide a derivative inside an integral, and a student sees a brand new problem. The structure underneath is the same, and so is the method. Once a student can name the structure, a whole row of questions that look different start to open the same way.

That is where marks really leak: in choosing the method, not in the algebra that follows. We call it Lock and Key, name the lock, then the key follows.

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All O-Level E-Math 2015 worked solutions →

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What does O-Level E-Math 2015 Paper 2 Question 10 test?

It is a median & quartiles from a cf curve question from Cumulative frequency / probability, worth 10 marks: (a)(i) 1 + 2 + 2, (a)(ii) 1, (b)(i) 1 + 1, (b)(ii) 2.

Is this the same as IP Math?

Yes. IP (Integrated Programme) schools teach the same O-Level Mathematics content; they just sequence it differently and set their own internal exams, so these worked solutions apply to IP students too.

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