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Worked solution

Scale fruit masses before combining independent prices

Durians cost 7 dollars per kg; a durian’s mass is normal with mean 1.5 kg and standard deviation 0.2 kg. Mangoes are sold in bags of 5 at 2 dollars per kg. One mango’s mass is normal with mean 0.4 kg and standard deviation 0.05 kg. Fruit masses are independent; empty bag mass is negligible.

(i) Find the probability that a random bag of mangoes costs more than 4.20 dollars. marks

(ii) Find the probability that both a random durian costs more than 10 dollars and a random mango bag more than 4.20 dollars. marks

(iii) Find the probability that 2 durians and 3 mango bags cost less than 33.50 dollars in total. marks

[2009/IJC/II/9]

CJC · Review problems · Question 2

Answer

(i)0.186; (ii)0.119; (iii)0.598 (3 s.f.).

(i) Model a bag of five mangoes

Let D∼N(1.5,0.22)D\sim N(1.5,0.2^2) and M∼N(0.4,0.052)M\sim N(0.4,0.05^2) be masses in kg. For bag price C=2(M1+M2+M3+M4+M5)C=2(M_1+M_2+M_3+M_4+M_5), E(C)=2×5×0.4=4E(C)=2\times5\times0.4=4 and Var⁡(C)=22×5×0.052=0.05\operatorname{Var}(C)=2^2\times5\times0.05^2=0.05. Thus C∼N(4,0.05)C\sim N(4,0.05) and P(C>4.20)=0.186P(C>4.20)=0.186 to 3 significant figures.

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(ii) Use independent selling prices

Durian price X=7D∼N(7×1.5,72×0.22)=N(10.5,1.42)X=7D\sim N(7\times1.5,7^2\times0.2^2)=N(10.5,1.4^2). Hence P(X>10 and C>4.20)=P(X>10)P(C>4.20)=0.119P(X>10\text{ and }C>4.20)=P(X>10)P(C>4.20)=0.119 to 3 significant figures.

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(iii) Add separate fruit and bag prices

T=X1+X2+C1+C2+C3∼N(10.5+10.5+4+4+4,1.42+1.42+0.05+0.05+0.05)=N(33,4.07)T=X_1+X_2+C_1+C_2+C_3\sim N(10.5+10.5+4+4+4,1.4^2+1.4^2+0.05+0.05+0.05)=N(33,4.07). Therefore P(T<33.50)=0.598P(T<33.50)=0.598 to 3 significant figures.

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Common mistakes

Three bags are three independent bag prices, not 3 times one realised bag price.

Solution notes

All asks/marks, negligible-bag qualification, every source mass/price model and expanded mean/variance stage retained.

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