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Numerical Answers (click the questions for workings/explanation)
Question 1:
Question 2:
Question 3:
Question 4:
Question 5:
Question 6:
Question 7:
Question 8:
Question 9: ;
latex 0.0336; \bar{y}=0.64, s^2 = 0.0400
latex \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; 0.362
latex X
latex 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 + \mu
latex \mu = 1.69241, \sigma = 0.1098079154
latex = 1.69
latex = 0.0121
latex X
latex X \sim \text{B}(8, 0.7)latex \text{P}(X =5) = 0.254
latex Y
latex 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 W
latex W \sim \text{B} (6, 0.19410435)latex \text{P} ( W \le 2)latex = 0.9082304639 \approx 0.908
latex {{6}\choose{3}} \times 3! \times {{8}\choose{3}} \times 3! = 40320
latex \bigg[ {{5}\choose{1}} \times \frac{3!}{2!} + {{5}\choose{2}} \times 3! \bigg] \times {{7}\choose{1}} \times \frac{3!}{2!} = 1575
latex = \frac{1575}{6^3 \times 8^3} = 0.0142
latex \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}=0.978
latex a=0.182, b=2.56
latex y = 0.1821185456 x + 2.564419724
latex x = 2, y = 0.1821185456 (2) + 2.564419724latex \Rightarrow y = 2.9287
293.
(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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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.
it’s ok!! Thank u for doing this 🙂