The answer
(a)(i) 55%
(ii) 31.5 min
(iii) outlier skews the mean
(iv) \(\approx 10.9\) min
(v) second group more consistent
(b)(ii) \(\dfrac{47}{120}\); \(\dfrac{33}{40}\)
O-Level E-Math 2016 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 2016 Paper 2. It tests reading a stem-and-leaf, in the Averages & spread / probability of combined events area. It is worth 11 marks: (a) 1 + 1 + 1 + 1 + 1, (b) 2 + 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) The times above 30 minutes are \(31, 32, 32, 32, 33, 37, 44, 46, 47, 48, 69\), that is 11 students. Percentage \(= \dfrac{11}{20} \times 100\% = 55\%\).
(ii) With 20 values, the median is the mean of the 10th and 11th values. Ordered, the 10th value is 31 and the 11th is 32, so median \(= \dfrac{31 + 32}{2} = 31.5\) minutes.
(iii) The value 69 minutes is an outlier, far above the rest of the data. It pulls the mean upwards so the mean would over-state a typical time; the median (or mode) is more representative here.
(iv) Mean \(= \dfrac{\sum x}{20} = \dfrac{691}{20} = 34.55\) min. Standard deviation \(= \sqrt{\dfrac{\sum x^2}{20} - \text{mean}^2}\). Here \(\sum x^2 = 26\,239\), so SD \(= \sqrt{\dfrac{26239}{20} - 34.55^2} = \sqrt{1311.95 - 1193.7025} = \sqrt{118.2475} = 10.9\) min (3 s.f.).
(v) The second group's standard deviation (8.64 min) is smaller than the first group's (10.9 min), so the second group's completion times are less spread out: they are more consistent / less varied than the first group's.
(b)(i) First branch P(W) \(= \dfrac{9}{16}\), P(R) \(= \dfrac{5}{16}\), P(B) \(= \dfrac{2}{16}\); second branches use one fewer ball of the chosen colour out of \(\dfrac{\cdot}{15}\) (e.g. after a white: P(W) \(= \dfrac{8}{15}\), P(R) \(= \dfrac{5}{15}\), P(B) \(= \dfrac{2}{15}\), and similarly for red and black first draws).
(ii) Same colour \(= P(WW) + P(RR) + P(BB) = \dfrac{9}{16}\cdot\dfrac{8}{15} + \dfrac{5}{16}\cdot\dfrac{4}{15} + \dfrac{2}{16}\cdot\dfrac{1}{15} = \dfrac{72 + 20 + 2}{240} = \dfrac{94}{240} = \dfrac{47}{120}\). At least one white \(= 1 - P(\text{no white})\). The 7 non-white balls are 5 red and 2 black, so \(P(\text{no white}) = \dfrac{7}{16}\cdot\dfrac{6}{15} = \dfrac{42}{240} = \dfrac{7}{40}\), giving \(1 - \dfrac{7}{40} = \dfrac{33}{40}\).
Answer: (a)(i) 55%
(ii) 31.5 min
(iii) outlier skews the mean
(iv) \(\approx 10.9\) min
(v) second group more consistent
(b)(ii) \(\dfrac{47}{120}\); \(\dfrac{33}{40}\)
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 reading a stem-and-leaf question from Averages & spread / probability of combined events, worth 11 marks: (a) 1 + 1 + 1 + 1 + 1, (b) 2 + 2 + 2.
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