The answer
(a) \(\approx 13.5\) hours
(b)(i) \(n \approx 14.7\)
(b)(ii) \(\dfrac{12}{35} \approx 0.343\)
O-Level E-Math 2020 Paper 1 Question 23 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 23 of the O-Level E-Math 2020 Paper 1. It tests estimated mean of grouped data, in the Data handling area. It is worth 5 marks: (a) 1, (b)(i) 2, (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.
(a) Using mid-interval values \(2.5, 7.5, 11.5, 14.5, 18, 22.5\): mean \(= \dfrac{6(2.5) + 13(7.5) + 32(11.5) + 62(14.5) + 22(18) + 5(22.5)}{140} = \dfrac{1888}{140} = 13.49 \approx 13.5\) hours.
(b)(i) "More than \(n\)" for \(40\%\) means \(60\%\) watch \(\leqslant n\). Read the curve at cumulative frequency \(0.6 \times 140 = 84\): this gives \(n \approx 14.7\) hours *(graph read; the answer key accepts \(\approx 14.6\)-\(14.75\))*.
(ii) Read the curve at \(8\) h and \(13.5\) h: cumulative frequencies \(\approx 12\) and \(\approx 60\), so about \(60 - 12 = 48\) adults lie between. Probability \(= \dfrac{48}{140} = \dfrac{12}{35} \approx 0.343\) (graph read).
Answer: (a) \(\approx 13.5\) hours
(b)(i) \(n \approx 14.7\)
(b)(ii) \(\dfrac{12}{35} \approx 0.343\)
Same structure, different numbers
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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Genius Plus Academy · O-Level & IP Mathematics
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It is a estimated mean of grouped data question from Data handling, worth 5 marks: (a) 1, (b)(i) 2, (b)(ii) 2.
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