The answer
(a)(i) median \(= 71.5\)
(ii) range \(= 43\)
(iii) \(58.3\%\)
(iv) see below
(b)(i) he used replacement
(iii) \(\dfrac{12}{25}\)
O-Level E-Math 2019 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 2019 Paper 2. It tests median & range from a stem-and-leaf diagram, in the Data handling & probability area. It is worth 10 marks: 1 + 1 + 1 + 2 + 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) Group A scores (read each leaf back to the stem): \(53, 58, 59\); \(60, 62, 64, 65, 66, 67, 68\); \(71, 71, 72, 75, 76, 77, 77, 79\); \(82, 87, 88, 88\); \(90, 95\), that is 24 scores in all. Group B scores: \(50, 51, 55, 57\); \(62, 62, 66\); \(71, 73, 74, 75, 75, 77, 78, 79\); \(80, 86, 87\); \(91, 93\), which is 20 scores.
(i) With 24 values, the median is the average of the 12th and 13th in order. These are \(71\) and \(72\), so median \(= \dfrac{71 + 72}{2} = 71.5\).
(ii) Group B range \(= 93 - 50 = 43\).
(iii) Group A with score above 70: from stem 7 (eight scores \(> 70\)), stem 8 (four), stem 9 (two) \(= 14\) students. Percentage \(= \dfrac{14}{24} \times 100 = 58.3\%\) (3 s.f.).
(iv) Two valid comparisons: (1) a larger proportion of group B earned a merit, about \(\dfrac{13}{20} = 65\%\) scored above 70, compared with \(\dfrac{14}{24} \approx 58\%\) of group A; (2) group A produced more distinctions, with 5 students scoring above 85 (87, 88, 88, 90, 95) against 4 in group B (86, 87, 91, 93). So group B did slightly better at merit level overall while group A had more top performers.
(b)(i) Shen multiplied \(\dfrac{9}{25} \times \dfrac{9}{25} = \dfrac{81}{625}\), treating the two draws as if the first counter were replaced. The counters are taken without replacement, so after one blue is removed there are 8 blue out of 24 left; the correct probability is \(\dfrac{9}{25} \times \dfrac{8}{24} = \dfrac{3}{25}\).
(ii) Tree diagram (total 25 counters). First draw: Red \(\dfrac{16}{25}\) or Blue \(\dfrac{9}{25}\). Second draw (no replacement, 24 left): after Red, Red \(\dfrac{15}{24}\) or Blue \(\dfrac{9}{24}\); after Blue, Red \(\dfrac{16}{24}\) or Blue \(\dfrac{8}{24}\).
(iii) "Only one red" means Red then Blue, or Blue then Red: \[P = \frac{16}{25}\cdot\frac{9}{24} + \frac{9}{25}\cdot\frac{16}{24} = \frac{144}{600} + \frac{144}{600} = \frac{288}{600} = \frac{12}{25}.\]
Answer: (a)(i) median \(= 71.5\)
(ii) range \(= 43\)
(iii) \(58.3\%\)
(iv) see below
(b)(i) he used replacement
(iii) \(\dfrac{12}{25}\)
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 median & range from a stem-and-leaf diagram question from Data handling & probability, worth 10 marks: 1 + 1 + 1 + 2 + 1 + 2 + 2.
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