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

Students are highly encouraged to discuss freely for alternative solutions too.

I typed this, so pardon me if got errors. There should be a handful of errors as I just type and solve as I go. Just refresh as you go along as I’m typing with wifi in a cafe. AND I SKIP A LOT OF STEPS, including denoting my random variables. I’ll get to them probably tomorrow morning.

Question 1

Discriminant
= b^2 - 4ac
= (k-4)^2 -4(-2)(2k)
= k^2 -8k + 16 + 16k
= k^2 + 8k + 16
= (k+4)^2 \ge 0 for all real values of k.

Thus, there will be at least one real root for all real values of k.

\Rightarrow there exist no real values of k for 2k + (k-4)x - 2x^2 to be always negative.


 

    Question 2

(i)
\frac{d}{dx} \frac{3}{(2x-1)^4} = -\frac{24}{(2x-1)^5}

(ii)
\int_{\frac{1}{2}}^1 (x+ \frac{2}{x})^2 dx

= \int_{\frac{1}{2}}^1 (x^2 + 4 + \frac{4}{x^2})dx

= \frac{x^3}{3} + 4x - \frac{4}{x} \biggl|_\frac{1}{2}^1

= \frac{151}{24}


 

    Question 3

(i)
\frac{dy}{dx} = 12 - 16e^{-2x}

Let \frac{dy}{dx}=0 \Rightarrow, x = \frac{1}{2}\mathrm{ln}\frac{4}{3}

(ii)
\int_0^a 12x + 8e^{-2x} dx

= 6x^2 - 4 e^{-2x} \biggl|_0^a

= 6a^2 - 4e^{-2a} + 4


 

    Question 4

Let H denote the height of big triangle with side y and h denote the height of small triangle with side x.

By Pythagoras’ Theorem, y^2 = (\frac{1}{2}y)^2 + H^2

\Rightarrow H = \frac{\sqrt{3}}{2} y

Area of big triangle = 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2}y \times \frac{y}{2} = \frac{\sqrt{3}}{4}y^2

Perimeter = 3x + 3(y-2x) = 30

\Rightarrow y = 10 + x

By Pytheagoras’ Theorem, x^2 = (\frac{1}{2}x)^2 + h^2

\Rightarrow h = \frac{\sqrt{3}}{2}x

Area of small triangle = 2\times \frac{1}{2} \times \frac{\sqrt{3}{2}}x \times \frac{x}{2} = \frac{\sqrt{3}}{4}x^2

Required Area = \frac{\sqrt{3}}{4}y^2 - 3 \times \frac{\sqrt{3}}{4}x^2

= \frac{\sqrt{3}}{4}(10+x)^2 - 3 \times \frac{\sqrt{3}}{4}x^2

= \frac{\sqrt{3}}{2} [\frac{1}{2}(10+x)^2 - \frac{3x^2}{2}]

= \frac{\sqrt{3}}{2} [\frac{1}{2}(100 + 20x + x^2) - \frac{3x^2}{2}]

= \frac{\sqrt{3}}{2} (50 + 10x - x^2)

\frac{dA}{dx} = \frac{\sqrt{3}}{2}(10-2x) = 0

\Rightarrow x = 5

\frac{d^2A}{dx^2} = -\sqrt{3} < 0

Therefore, A is maximum when x = 5

A = \frac{\sqrt{3}}{2}(50 + 50 -25) = \frac{75 \sqrt{3}}{2}


 

    Question 5
Graph 5i
Graph 5i

(ii)
Using GC, \frac{dy}{dx}\biggl|_{x=0.5} = -1.1567 \approx -1.157

(iii)
y - 0.3016416731 = \frac{1}{1.1567}(x-0.5)

When x = 0, y =    When latex y = 0, x =

By Pythagoras’ Theorem, Length = \sqrt{}


 

    Question 6

Let X denote the mass of peaches sold by a shop, in grams.

X \sim N(\mu, \sigma^2)

\mathrm{P}(X < 40) = 0.2

\frac{40-\mu}{\sigma} = -0.841621 \rightarrow (1)

\mathrm{P}(X \le 60) = 0.75

\frac{60- \mu}{\sigma} = 0.674489 \rightarrow (2)

Using GC, \mu = 51.10237 \approx 51.1 and \sigma = 13.19165 \Rightarrow \sigma^2 = 174.01 \approx 174


 

    Question 7

(i)
Using the ordered list, we pick a random number from 1 to \frac{1200}{100} = 12 and then proceed to survey every 12^{th} in the list until we have collected 100 names.

(ii)
Due to the periodic and cyclical nature, we might not have a representative sample.

(ii)
Stratified Sampling


 

    Question 8

(i)
0.42 = p + 2p - 0.03

p = 0.15

(ii)
\mathrm{P}(A \cup B')

= \mathrm{P}(A) + \mathrm{P}(B') - \mathrm{P}(A \cap B')

= 0.15 + 0.7 - [\mathrm{P} (A \cup B) - \mathrm{P}(B)]

= 0.85 - 0.42 + 0.3

= 0.73

(iii)
\mathrm{P}(A) \times \mathrm{P}(B') = 0.15 \times 0.7 \ne \mathrm{P}(A \cap B')

Thus, A and B' are not independent.


 

    Question 9

(i)
Let X denote the number of six Kai obtain out of 8 throws.

X \sim \mathrm{Bin}(8, \frac{1}{6})

\mathrm{P} (X=3) \approx 0.104

(ii)
\mathrm{P} (X \le 3) \approx 0.969343

(iii)
Let Y denote the number of six Lam obtain out of 600 throws.
Y \sim \mathrm{Bin} (600, \frac{1}{6})

Since n = 600 is large, np = 100 > 5 and n(1-p) = 500 > 5,

\mathbb{E}(Y)=100 and \mathrm{Var}(Y) = \frac{500}{6}

Y \sim \mathrm{N}(100, \frac{500}{6}) approximately

\mathrm{P}(90 \le Y \le 100)

= \mathrm{P} (89.5 \le Y \le 100.5) by continuity correction

\approx 0.397


 

    Question 10

(i)
Graph to be inserted

(ii)
r = 0.922441 \approx 0.922
The r values shows a strong positive linear correlation between the height and weight of the rowing club members.

(iii)
Using GC, w = 58.0325h -6.3004 \Rightarrow w = 58.0h - 6.30 (3 SF)

(iv)
Using GC, w = 90.033 \approx 90kg

The estimate is reliable since its within data range and the r-value is also close to 1.


 

Question 11
(i)
Let X and Y denote the masses of Men and Women respective, in kg

X \sim \mathrm{N}(77, 9.8^2)

Y \sim \mathrm{N}(62, 10.6^2)

\mathrm{P}(-2 < X - 77 < 2)

\mathrm{P}(75 < X < 79)

\approx 0.162
(ii)
Let Q = X_1 + X_2 + X_3 - (Y_1 + \ldots + Y_4)

\mathbb{E}(Q) = -17

\mathrm{Var}(Q) = 737.56

Q \sim \mathrm{N}(-17, 737.56)

\mathrm{P}(Q > 0) \approx 0.266

(iii)
Let T = X_1 + X_2 + X_3 + (Y_1 + \ldots + Y_4)

\mathbb{E}(T) = 479

\mathrm{Var}(T) = 737.56

T \sim \mathrm{N}(479, 737.56)

\mathrm{P}(T \le 460) \approx 0.242


 

    Question 12

(i)
To be inserted

(ii)
Required Probability = \frac{3}{4} \times \frac{2}{5} + \frac{1}{4} \times \frac{3}{4} \times \frac{2}{5} = \frac{3}{8}

(iii)
Required Probability
= \mathrm{P}(\mathrm{Student~succeeds~in~qualification~} | \mathrm{~student~fails~Part~I})

= \frac{\mathrm{P}(\mathrm{Student~succeeds~in~qualification} \cap \mathrm{student~fails~Part~I})}{\mathrm{P}(\mathrm{student~fails~Part~I})}

= \frac{0.25 \times 0.75 \times 0.4}{0.25}

= 0.3

(iv)
Let X denote the number of students out of 5 who passes the qualification.

X \sim \mathrm{Bin}(5, \frac{3}{8})

Required Probability = 1 - \mathrm{P}(X \le 1) \approx 0.61853


Question 13
(i)
H_0: \mu = 15.2

H_1: \mu \ne 15.2

Under H_0, \bar{X} \sim \mathrm{N}(\mu, \frac{2.1^2}{30})

Test Statistic, Z = \frac{\bar{X}-\mu}{\frac{\sigma}{n}} \sim \mathrm{N}(0,1)

We performa a two tailed Z-Test at 5% level significance level.

Using GC, p =0.06788 > 0.05

Thus, we do not reject H_0 and conclude with insufficient evidence…

(ii)
Unbiased estimate of population mean = \frac{-32}{40} +18 = 17.2

s^2 = \frac{1}{39}(325 - \frac{32^2}{40}) = \frac{499}{65}

(iii)
An unbiased estimate means the expectations of the estimate is its true population parameter..

(iv)
H_0: \mu = 18

$latex H_1: \mu < 18$ Under $latex H_0, \bar{X} \sim \mathrm{N}(\mu, \frac{s^2}{30})$ Test Statistic, $latex Z = \frac{\bar{X}-\mu}{\frac{s}{n}} \sim \mathrm{N}(0,1)$ To reject $latex H_0, p \le \alpha$ From GC, $latex p = 0.03391$ $latex \therefore \{\alpha \in \mathbb{R}| 3.40 < \alpha < 100\}$   Check out Christine's Comments on the recent GP Essays here too.

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MF15

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