The answer
(a)(i) \(4, 11, 18, 13, 4\)
(ii) \(\approx 3.27\) kg
(iii) \(\approx 0.529\) kg
(iv) grouped data, so midpoints assumed
(v) \(\approx 84\%\), claim not supported
(b)(i) \(\dfrac{2}{5}\)
(ii)(a) \(\dfrac{13}{50}\)
(ii)(b) \(\dfrac{9}{50}\)
O-Level E-Math 2018 Paper 2 Question 9 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 9 of the O-Level E-Math 2018 Paper 2. It tests frequencies from a cf curve, in the Cumulative frequency, averages & spread / probability area. It is worth 11 marks: (a) 1 + 1 + 1 + 1 + 2, (b) 1 + 2 + 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)(i) Reading the cumulative frequencies at the class boundaries gives about \(4, 15, 33, 46, 50\) at \(m = 2.5, 3.0, 3.5, 4.0, 4.5\). Differencing gives the frequencies:
| Mass (\(m\) kg) | \(2.0 \leqslant m < 2.5\) | \(2.5 \leqslant m < 3.0\) | \(3.0 \leqslant m < 3.5\) | \(3.5 \leqslant m < 4.0\) | \(4.0 \leqslant m < 4.5\) |
|---|---|---|---|---|---|
| Frequency | 4 | 11 | 18 | 13 | 4 |
(They total \(50\).) (Read from the curve, so the last three frequencies are estimates.)
(ii) Using midpoints \(2.25, 2.75, 3.25, 3.75, 4.25\): \[\bar{m} = \frac{4(2.25) + 11(2.75) + 18(3.25) + 13(3.75) + 4(4.25)}{50} = \frac{163.5}{50} = 3.27 \text{ kg.}\]
(iii) Standard deviation \(= \sqrt{\dfrac{\sum f x^2}{50} - \bar{m}^2}\). With \(\sum f x^2 = 4(2.25^2) + 11(2.75^2) + 18(3.25^2) + 13(3.75^2) + 4(4.25^2) = 548.625\), this is \(\sqrt{\dfrac{548.625}{50} - 3.27^2} = \sqrt{10.9725 - 10.6929} = \sqrt{0.2796} = 0.529\) kg (3 s.f.).
(iv) The babies are grouped into classes, so the exact mass of each baby is unknown. Using the class midpoints as representative values makes the mean and standard deviation estimates rather than exact figures.
(v) From the curve the cumulative frequency at \(2.8\) kg is about \(8\), so the number of babies with mass greater than \(2.8\) kg is about \(50 - 8 = 42\), i.e. about \(\dfrac{42}{50} = 84\%\). Since \(84\%\) is below \(90\%\), the hospital data does not support the article's claim. (Read from graph.)
(b)(i) Mothers under 30 \(= 3 + 7 = 10\) out of \(25\), so \(P = \dfrac{10}{25} = \dfrac{2}{5}\).
(ii) Two mothers are chosen without replacement from 25. (a) Girls' mothers number \(7 + 6 = 13\): \(P(\text{both girl}) = \dfrac{13}{25} \times \dfrac{12}{24} = \dfrac{13}{25} \times \dfrac12 = \dfrac{13}{50}\). (b) Among the \(30\)-or-over mothers there are \(9\) boy-mothers and \(6\) girl-mothers. "Only one has a baby boy" means one of each, in either order: \[P = 2 \times \frac{9}{25} \times \frac{6}{24} = 2 \times \frac{9}{25} \times \frac14 = \frac{18}{100} = \frac{9}{50}.\]
Answer: (a)(i) \(4, 11, 18, 13, 4\)
(ii) \(\approx 3.27\) kg
(iii) \(\approx 0.529\) kg
(iv) grouped data, so midpoints assumed
(v) \(\approx 84\%\), claim not supported
(b)(i) \(\dfrac{2}{5}\)
(ii)(a) \(\dfrac{13}{50}\)
(ii)(b) \(\dfrac{9}{50}\)
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 frequencies from a cf curve question from Cumulative frequency, averages & spread / probability, worth 11 marks: (a) 1 + 1 + 1 + 1 + 2, (b) 1 + 2 + 2.
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