The answer
(a)(i) \(120 < m \leqslant 140\)
(ii) \(117.5\) g
(iii) \(22.7\) g
(iv) added apples average \(98.75\) g, lighter than 110 g
(b)(i) extremes unknown
(ii) \(\dfrac{8}{15}\)
(iii) \(\dfrac{174}{5513}\)
O-Level E-Math 2024 Paper 2 Question 7 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 7 of the O-Level E-Math 2024 Paper 2. It tests grouped-data median & mean, in the Central tendency / Dispersion / Probability area. It is worth 9 marks: (a) 1 + 1 + 1 + 1, (b) 1 + 1 + 3. 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)(i) Median = average of the 60th and 61st values. Cumulative frequencies: 8, 29, 56, 102, 120. The 60th and 61st lie in \(120 < m \leqslant 140\).
(ii) Using midpoints 70, 90, 110, 130, 150: mean \(= \dfrac{8(70)+21(90)+27(110)+46(130)+18(150)}{120} = \dfrac{14100}{120} = 117.5\) g.
(iii) \(\text{s.d.} = \sqrt{\dfrac{\sum f x^2}{120} - \bar{x}^2} = \sqrt{\dfrac{1\,718\,400}{120} - 117.5^2} = \sqrt{14320 - 13806.25} = \sqrt{513.75} = 22.7\) g (3 s.f.).
(iv) Total mass before \(= 117.5 \times 120 = 14100\) g. After: \(110 \times 200 = 22000\) g, so the 80 added apples have total mass \(22000 - 14100 = 7900\) g, i.e. a mean of \(7900 \div 80 = 98.75\) g. The added apples are on average lighter (about 98.75 g, less than 110 g), which pulls the overall mean down.
(b)(i) The histogram only groups masses into intervals from 80 g to 180 g; the actual lightest tomato lies somewhere in \(80 < m \leqslant 100\) and the heaviest in \(160 < m \leqslant 180\), so the exact smallest and largest masses are not known. The range equals 100 g only if those extremes are exactly 80 g and 180 g, which is not guaranteed.
(ii) Counts: \(120\)-\(140\) is 32, \(140\)-\(160\) is 48. \(P = \dfrac{32+48}{150} = \dfrac{80}{150} = \dfrac{8}{15}\).
(iii) Less than 120 g: \(10 + 20 = 30\); greater than 160 g: \(40\). Choosing two of the 30 and one of the 40, without replacement, in any of the 3 orders: \(P = \dfrac{3 \times 30 \times 29 \times 40}{150 \times 149 \times 148} = \dfrac{174}{5513}\).
Answer: (a)(i) \(120 < m \leqslant 140\)
(ii) \(117.5\) g
(iii) \(22.7\) g
(iv) added apples average \(98.75\) g, lighter than 110 g
(b)(i) extremes unknown
(ii) \(\dfrac{8}{15}\)
(iii) \(\dfrac{174}{5513}\)
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
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.
It is a grouped-data median & mean question from Central tendency / Dispersion / Probability, worth 9 marks: (a) 1 + 1 + 1 + 1, (b) 1 + 1 + 3.
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