The masses, in kilograms, of female and male animals of a certain species have the distributions $$\(\mathrm{N}\left(102,27^{2}\right)\)$$ and $$\(\mathrm{N}\left(170,55^{2}\right)\)$$ respectively. Find the probability that a randomly chosen female has a mass that is less than half the mass of a randomly chosen male. ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................
Exam No：9709_w20_qp_63 Year：2020 Question No：3
Answer:
\( \mathrm{F}-0.5 \mathrm{M} \)
\( \begin{aligned} & \\ \sim \mathrm{N}\left(17,27^{2}+0.25 \times 55^{2}\right) \\ \frac{0-{ }^{\prime} 17^{\prime}}{\sqrt{1485.25^{\prime}}}(=-0.4411) \end{aligned} \)
\(\mathrm{P}(\mathrm{F}-0.5 \mathrm{M}<0)=\phi\left({ }^{\prime}-0.4411^{\prime}\right)=1-\phi\left({ }^{\prime} 0.4411^{\prime}\right)\)
\(=0.330(3 \mathrm{sf})\)
Alternative method for question 3
\(2 F-M\)
\(\sim \mathrm{N}\left(34, \quad 2^{2} \times 27^{2}+55^{2}\right)\)
\(\frac{0-{ }^{-34^{\prime}}}{\sqrt{15941^{\prime}}}(=-0.4411)\)
\(\mathrm{P}(2 \mathrm{~F}-\mathrm{M}<0)=\phi\left({ }^{\prime}-0.4411^{\prime}\right)=1-\phi\left(\left(^{\prime} 0.4411^{\prime}\right)\right.\)
\(=0.330(3 \mathrm{sf})\)
\( \begin{aligned} & \\ \sim \mathrm{N}\left(17,27^{2}+0.25 \times 55^{2}\right) \\ \frac{0-{ }^{\prime} 17^{\prime}}{\sqrt{1485.25^{\prime}}}(=-0.4411) \end{aligned} \)
\(\mathrm{P}(\mathrm{F}-0.5 \mathrm{M}<0)=\phi\left({ }^{\prime}-0.4411^{\prime}\right)=1-\phi\left({ }^{\prime} 0.4411^{\prime}\right)\)
\(=0.330(3 \mathrm{sf})\)
Alternative method for question 3
\(2 F-M\)
\(\sim \mathrm{N}\left(34, \quad 2^{2} \times 27^{2}+55^{2}\right)\)
\(\frac{0-{ }^{-34^{\prime}}}{\sqrt{15941^{\prime}}}(=-0.4411)\)
\(\mathrm{P}(2 \mathrm{~F}-\mathrm{M}<0)=\phi\left({ }^{\prime}-0.4411^{\prime}\right)=1-\phi\left(\left(^{\prime} 0.4411^{\prime}\right)\right.\)
\(=0.330(3 \mathrm{sf})\)
Knowledge points:
6.2.1.5 if X and Y have independent normal distributions then aX + bY has a normal distribution
Solution:
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