The volumes, in millilitres, of large and small cups of tea are modelled by the distributions $$\(\mathrm{N}(200,30)\)$$ and $$\(\mathrm{N}(110,20)\)$$ respectively. Find the probability that the volume of a randomly chosen large cup of tea is more than twice the volume of a randomly chosen small cup of tea. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No：9709_m20_qp_62 Year：2020 Question No：6(b)
Answer:
\(\mathrm{P}(L-2 S>0)\)
\(\mathrm{E}(X)=200-2 \times 110\) or \(=-20\)
\(\operatorname{Var}=30+2^{2} \times 20\) or \(=110\)
\(\mathrm{N}(-20,110)\)
\(\frac{0-\left('-20^{\prime}\right)}{\sqrt{110^{\prime}}}(=1.907)\)
\(1-\Phi(' 1.907\) ')
\(=0.0283(3 \mathrm{sf})\)
\(\mathrm{E}(X)=200-2 \times 110\) or \(=-20\)
\(\operatorname{Var}=30+2^{2} \times 20\) or \(=110\)
\(\mathrm{N}(-20,110)\)
\(\frac{0-\left('-20^{\prime}\right)}{\sqrt{110^{\prime}}}(=1.907)\)
\(1-\Phi(' 1.907\) ')
\(=0.0283(3 \mathrm{sf})\)
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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