2014 A-level H2 Mathematics (9740) Paper 2 Question 11 Suggested Solutions

JC Mathematics

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Let X and Y be the number of originals and prints sold in a week, respectively.
X \sim \mathrm{Po}(2)

Y \sim \mathrm{Po}(11)

(a)\mathrm{P}(Y > 8) = 1 - \mathrm{P}(Y \le 8) = 0.768

(b) X + Y \sim \mathrm{Po}(13)

\mathrm{P}(X + Y < 15) = 1 - \mathrm{P}(X + Y \le 14) = 0.675

(ii) Let W be the number of originals sold in n weeks.

W \sim \mathrm{Po}(2n)

\mathrm{P}(W < 3)<0.01

\mathrm{P}(W = 2) + \mathrm{P}(W = 1) + \mathrm{P}(W = 0)<0.01

e^{-2n} + e^{-2n}(2n) + e^{-2n}\frac{(2n)^{2}}{2}<0.01

e^{-2n}(1 + 2n + 2n^{2})<0.01

Using GC, smallest possible n = 5.

(iii) Since n=52 is large, by Central Limit Theorem,

B_1 + ... + B_52 \sim N(572, 572) approximately.

\mathrm{P}(B_1 + ... + B_52 > 550) = 0.821

(iv) Firstly the mean number of paintings sold per week may not be constant due to seasonal factors. Secondly, the sales might not be independent as a buyer might have preferential taste for particular artists.

Personal Comments

For (ii), students are expected to draw out the table to illustrate the choice of n. For (iii), students can either use CLT or approximate by normal distribution but do remember to do continuity correction then. I did CLT here since I haven’t done CLT for the entire paper while I did approximation in Question 7.If students do by approximation of normal distribution, they should expect the answer to be 0.816.

2014 A-level H2 Mathematics (9740) Paper 2 Question 9 Suggested Solutions

JC Mathematics

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Let \mu be the population mean number of minutes that the bus is late.
H_0: \mu = 4.3
H_1: \mu \textless 4.3

(ii)
s^{2} = \frac{10}{9}(3.2) = \frac{32}{9}

Under H_0. \frac{\bar{T}-4.3}{\sqrt{\frac{32/9}{10}}}~\sim~ t(9)

Since H_0 is not rejected, t_{calc}~>~t_{crit}

\frac{\bar{t}-4.3}{\sqrt{\frac{32/9}{10}}} > -1.3830287

\bar{t} > 3.48

Required set =~\{ \bar{t} \in \mathbb{R} ~|~ \bar{t} > 3.48\}

(iii)
For H_0 to be rejected, t_{calc}~\le~t_{crit}

\frac{4.0-4.3}{\sqrt{\frac{k^{2}}{9}}} \le -1.3830287

k^{2} \le 0.423

Required set =~\{ k^{2} \in \mathbb{R} ~|~ 0 \textless k^{2} \le 0.423\}

Personal Comments:
To find the t_{crit} value in (ii), students are required to know how to use InvT(0.1,9) or read the T-table. The last part wants range of k^{2} and not k, do read carefully and note that variance cannot be negative so it is necessary to bound it by zero. Also the question asked for set of values, so students need to give answers in set.

2014 A-level H2 Mathematics (9740) Paper 2 Question 8 Suggested Solutions

JC Mathematics

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(a)
(i)

Graph of 8ai
Graph of 8ai

(ii)

Graph of 8aii
Graph of 8aii

(b)
(i)
Model A: r = -0.9470
Model B: r= -0.9749

(ii) Since the |r| is close to 1 for Model B, Model B is the better model.
P = -33659.72 \mathrm{ln}m + 195693.559
P \approx -33700 \mathrm{ln}m + 196000 (3 S.F.)

(iii) Required estimated price = -33659.72 \mathrm{ln}(50) + 195693.559 = \$64000

Personal comments:
(a) is rather straight forward so long as you follow the given instructions, bear in mind the graph is quadratic in nature. Since no scatter was drawn in part (b), we can simply justify the choice of regression model using the r-value found.

2014 A-level H2 Mathematics (9740) Paper 2 Question 7 Suggested Solutions

JC Mathematics

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Let X = ~\mathrm{number ~ of ~ sixes ~ out ~ of  ~ 10 ~ rolls}.
X \sim~ \mathrm{B} (10, \frac{1}{6})
\mathrm{P}(X~=~3) = 0.155

(ii)
Let Y = ~\mathrm{number ~ of ~ sixes ~ out ~ of  ~ 60 ~ rolls}.
Y \sim~ \mathrm{B} (60, \frac{1}{6})
Since n=60 is large, np=10 > 5, nq=50 > 5, Y \sim~ \mathrm{N} (10, \frac{50}{6}) approximately.
\mathrm{P}(5 \le Y \le 8) = \mathrm{P}(4.5 \le Y \le 8.5) by continuity correction.
\therefore, \mathrm{Required ~ Probability}=0.273

(iii)
Let W = ~\mathrm{number ~ of ~ sixes ~ out ~ of  ~ 60 ~ rolls ~ of ~ the ~ biased ~ die}.
W \sim~ \mathrm{B} (60, \frac{1}{15})
Since n=60 is large, np=4 \textless 5,~  W \sim~ \mathrm{Po} (4) approximately.
\mathrm{P}(5 \le W \le 8) = \mathrm{P}(W \le 8) - \mathrm{P}(W \le 4)
\therefore, \mathrm{Required ~ Probability}=0.350

Personal Comments:
Quite a standard binomial and approximation question. Some students still forget to do continuity correction or do wrong. Please go learn the correct technique to do. At the same time, you need to state the conditions to approximate the distribution here.

2014 A-level H2 Mathematics (9740) Paper 2 Question 6 Suggested Solutions

JC Mathematics

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Required number of ways = {3 \choose 1} \times {8 \choose 4} \times {5 \choose 2} \times {6 \choose 4} = 31500

(ii)
Number of ways for both brother are in team = {3 \choose 1} \times {8 \choose 4} \times {4 \choose 1} \times {5 \choose 3} = 8400
Number of ways for both brother are not in team = {3 \choose 1} \times {8 \choose 4} \times {4 \choose 2} \times {5 \choose 4} = 6300
Required number of ways = 31500 - 8400 - 6300 = 16800 ways.

(iii)
We consider three cases here.
1. Remaining midfield plays as midfielder
2. particular midfield plays defender
3. particular midfield not playing

Required number of ways = {3 \choose 1} [{8 \choose 4} \times {3 \choose 1} + {8 \choose 3} \times {3 \choose 2} + {8 \choose 4} \times {3 \choose 2}] {5 \choose 4} = 8820 ways.

Personal Comments:
Students can solve (ii) by consider two cases, i.e., midfield bother is in team while attacker brother is not and midfield bother is not team while attacker brother i. (iii), I was a bit lazy to type it out to many times so I simply factorise everything, knowing that the number of ways to choose goalkeeper and attackers are the same.

2014 A-level H2 Mathematics (9740) Paper 2 Question 5 Suggested Solutions

JC Mathematics

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)

interval = \frac{10000}{500} = 20
Order the list of 10000 regular customers, then randomly select a number from 1 to 20, e.g. 4. Then select the 4th, 24th, 44th, 64th, … customers on the list until all 500 customers are surveyed.

(ii)
Advantage:
1. Sample is evenly spread out over the list of customers.

Disadvantage:
1. Not representative as the list could exhibit periodic or cyclic pattern.
2. Biased as the list could exhibit periodic or cyclic pattern.

Personal Comments:
Students, please go memorise how to describe these methods and know how to write in context. We MUST write in context of the question.