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O-Level E-Math · 2019 · P1 Q24 Averages & spread (grouped data) · Median class 5 marks: (a) 1 + 1, (b) 1 + 1, (c) 1 · statistics & probability difficulty 4 of 5

O-Level E-Math 2019 Paper 1, Question 24: Median class

The answer

(a)(i) \(180 < h \leqslant 200\)
(ii) \(\approx 179\) cm
(b)(i) \(\approx 23\)
(ii) \(\approx 50\) cm
(c) the second set's heights are less spread out (more consistent)

O-Level E-Math 2019 Paper 1 Question 24 · Verified worked solution by the Genius Plus Academy teaching team

What this question tests

This is Question 24 of the O-Level E-Math 2019 Paper 1. It tests median class, in the Averages & spread (grouped data) area. It is worth 5 marks: (a) 1 + 1, (b) 1 + 1, (c) 1. 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

The histogram gives frequencies 4, 5, 8, 11, 14, 12, 6 for the intervals 100 to 120, 120 to 140, 140 to 160, 160 to 180, 180 to 200, 200 to 220, 220 to 240 (total 60).

(a)(i) Median = average of the 30th and 31st values. Cumulative frequencies: 4, 9, 17, 28, 42, 54, 60. The 30th and 31st lie in \(180 < h \leqslant 200\).

(ii) Using midpoints 110, 130, 150, 170, 190, 210, 230: mean \(= \dfrac{110(4)+130(5)+150(8)+170(11)+190(14)+210(12)+230(6)}{60} = \dfrac{10720}{60} = 178.7\) cm (3 s.f.).

(b)(i) \(1.92 \text{ m} = 192\) cm. From the curve the cumulative frequency at 192 cm is about 37, so the number with height at least 192 cm is about \(60 - 37 = 23\). (Read from graph; accept roughly 22 to 25.)

(ii) From the curve, \(Q_1\) (15th value) \(\approx 155\) cm and \(Q_3\) (45th value) \(\approx 205\) cm, so IQR \(\approx 205 - 155 = 50\) cm. (Read from graph.)

(c) An interquartile range of 36 cm is smaller than about 50 cm, so the middle half of the second set's heights is less spread out: the second set of trees has more consistent (less varied) heights.

Answer: (a)(i) \(180 < h \leqslant 200\)
(ii) \(\approx 179\) cm
(b)(i) \(\approx 23\)
(ii) \(\approx 50\) cm
(c) the second set's heights are less spread out (more consistent)

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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Our O-Level E-Math tuition trains the same recognise-the-structure method these worked solutions show, taught by a team that has marked these papers for years. It runs within our weekly Secondary Math programme, Sec 1 to 4 and IP.

Questions students ask

What does O-Level E-Math 2019 Paper 1 Question 24 test?

It is a median class question from Averages & spread (grouped data), worth 5 marks: (a) 1 + 1, (b) 1 + 1, (c) 1.

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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