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Numerical Answers (click the questions for workings/explanation)

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Question 2:

Question 3:

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Question 8:

Question 9: ; latex 0.0336; \bar{y}=0.64, s^2 = 0.0400latex \frac{48+x}{80+x}, \frac{32+x}{80+x}; x= 16; \frac{25}{32}; \frac{7}{16}; \frac{341}{8930}latex 0.773; 0.0514; 0.866; 0.362latex Xlatex X \sim \text{N} (\mu, \sigma^2) latex \text{P}( X \textless 1.6) = 0.2latex \text{P}( Z \textless \frac{1.6 – \mu}{\sigma}) = 0.2latex \frac{1.6 – \mu}{\sigma} = -0.8416212335 latex 1.6 = -0.8416212335 \sigma + \mu latex \text{P}( X \textgreater 1.75 ) = 0.3latex \text{P}( X \textless 1.75 ) = 0.7latex \text{P}( Z \textless \frac{1.75 – \mu}{\sigma} ) = 0.7latex \frac{1.75 – \mu}{\sigma} = 0.5244005101latex 1.75 = 0.5244005101 \sigma + \mulatex \mu = 1.69241, \sigma = 0.1098079154latex = 1.69latex = 0.0121latex Xlatex X \sim \text{B}(8, 0.7)latex \text{P}(X =5) = 0.254latex Ylatex Y \sim \text{B}(8, 0.3)latex \text{P}(Y \ge 4)latex = 1 – \text{P}(Y le 3)latex = 0.19410435 \approx 0.194 latex Wlatex W \sim \text{B} (6, 0.19410435)latex \text{P} ( W \le 2)latex = 0.9082304639 \approx 0.908latex {{6}\choose{3}} \times 3! \times {{8}\choose{3}} \times 3! = 40320latex \bigg[ {{5}\choose{1}} \times \frac{3!}{2!} + {{5}\choose{2}} \times 3! \bigg] \times {{7}\choose{1}} \times \frac{3!}{2!} = 1575latex = \frac{1575}{6^3 \times 8^3} = 0.0142latex \text{P}(\text{has~}2\text{~as~its~first~character}) = \frac{1}{6}latex \text{P}(\text{has~H~as~its~sixth~character}) = \frac{1}{8}latex \text{P}(\text{has~}2\text{~as~its~first~character~and~has~H~as~its~sixth~character}) = \frac{1}{6 \times 8} = \frac{1}{48}latex = \frac{1}{6} + \frac{1}{8} – 2(\frac{1}{48}) = \frac{1}{4}latex \text{r}=0.978latex a=0.182, b=2.56 latex y = 0.1821185456 x + 2.564419724latex x = 2,ย y = 0.1821185456 (2) + 2.564419724latex \Rightarrow y = 2.9287293.

(v)

Firstly, the linear model is not appropriate here. Since it suggests as the number of years increases, the weekly earnings will increase proportionately, which is not realistic.

Secondly, is out of data range and this is extrapolation, which is a bad practice since our trend might not continue out of data range.

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(i)

Using GC,

There is a probability of 0.0366 of observing a result equal to or more extreme than that is actually observed when the mean volume is indeed 0.6.

(ii)

It is not necessary since the sample size of 50 is sufficiently large for us to approximate the volume of juices to a normal distribution by Central Limit Theorem.

(iii)

Unbiased estimate of population mean

Unbiased estimate of population variance

(iv)

(i)

(ii)

and are independent

(iii)

(iv)

It is the probability that a student takes Finance given that he takes Marketing.

(v)

Number of students that take exactly two of these three subjects

Required probability

(i)

Let denote the journey times, in minutes by bus and by train, respectively.

(ii)

(iii)

The cases found by (ii) is a subset of the cases found by . Thus, will be greater.

(iv)

(v)

It is the probability that 3 times the cost of one bus journey is less than 2 times the cost of one train journey by not more than $3.

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### KS Comments

Hello!! Will you be uploading the answers for this paper? Thank you so much!!!!

yes, sorry was having a H2 math class for J2 just now.

itโs ok!! Thank u for doing this ๐