Porisma

Worked solution

Apply prices after adding independent masses

Melons cost $1.50 per kilogram and their masses are normal with mean 0.8 kg and standard deviation 0.1 kg. Pumpkins cost $0.50 per kilogram and their masses are normal with mean 1.2 kg and standard deviation 0.2 kg. Find the probability that the total price of 5 randomly chosen melons and 3 randomly chosen pumpkins exceeds $8.

[N2002/II/28]

CJC · Lecture · Example 9

Define separate fruit masses

Let M∼N(0.8,0.12)M\sim N(0.8,0.1^2) and P∼N(1.2,0.22)P\sim N(1.2,0.2^2) be masses in kg. The source sum models assume independently selected fruit masses; state this sampling assumption before adding variances.

How did this part go?

Add five melon masses, then scale by the price

M1+M2+M3+M4+M5∼N(5(0.8),5(0.12))=N(4,0.05)M_1+M_2+M_3+M_4+M_5\sim N(5(0.8),5(0.1^2))=N(4,0.05). Their price is 1.5(M1+⋯+M5)∼N(1.5(4),1.52(0.05))=N(6,0.1125)1.5(M_1+\cdots+M_5)\sim N(1.5(4),1.5^2(0.05))=N(6,0.1125) in dollars.

How did this part go?

Add three pumpkin masses, then scale by the price

P1+P2+P3∼N(3(1.2),3(0.22))=N(3.6,0.12)P_1+P_2+P_3\sim N(3(1.2),3(0.2^2))=N(3.6,0.12). Their price is 0.5(P1+P2+P3)∼N(0.5(3.6),0.52(0.12))=N(1.8,0.03)0.5(P_1+P_2+P_3)\sim N(0.5(3.6),0.5^2(0.12))=N(1.8,0.03).

How did this part go?

Combine both prices and evaluate the upper tail

Let T=1.5(M1+⋯+M5)+0.5(P1+P2+P3)T=1.5(M_1+\cdots+M_5)+0.5(P_1+P_2+P_3). Then T∼N(6+1.8,0.1125+0.03)=N(7.8,0.1425)T\sim N(6+1.8,0.1125+0.03)=N(7.8,0.1425). Thus P(T>8)=0.298P(T>8)=0.298 to three significant figures. The independent numerical value is 0.2981208153…; convert mass uncertainty to price uncertainty before evaluating the event.

How did this part go?

Common mistakes

Five separately chosen melons give 5Var(M), not 25Var(M); the separate price factor 1.5 then contributes its square.

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