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O-Level E-Math · 2018 · P2 Q9 Cumulative frequency, averages & spread / probability · Frequencies from a CF curve 11 marks: (a) 1 + 1 + 1 + 1 + 2, (b) 1 + 2 + 2 · statistics & probability (cumulative frequency, mean & sd, probability without replacement) difficulty 4 of 5

O-Level E-Math 2018 Paper 2, Question 9: Frequencies from a CF curve

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.

Step-by-step solution

(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)\(2.0 \leqslant m < 2.5\)\(2.5 \leqslant m < 3.0\)\(3.0 \leqslant m < 3.5\)\(3.5 \leqslant m < 4.0\)\(4.0 \leqslant m < 4.5\)
Frequency41118134

(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

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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Questions students ask

What does O-Level E-Math 2018 Paper 2 Question 9 test?

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.

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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