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Worked solution

Compare an intersection event with the larger total-price event

Chicken and turkey masses in kg have independent normal distributions.

BirdMean massStandard deviation
Chicken2.20.5
Turkey10.52.1

Chickens cost 3 dollars per kg and turkeys 5 dollars per kg.

(i) Find the probability that a random chicken costs more than 7 dollars.

(ii) Find the probability that both a random chicken costs more than 7 dollars and a random turkey more than 55 dollars.

(iii) Find the probability that their total price exceeds 62 dollars.

(iv) Explain why(iii) is greater than (ii).

[2007/II/8]

CJC · Discussion · Question 11

Answer

(i)0.395; (ii)0.160; (iii)0.392 (3 s.f.); (iv)the joint threshold event is a proper subset of the total-price event.

Transform mass into selling price

Let C∼N(2.2,0.52)C\sim N(2.2,0.5^2) and T∼N(10.5,2.12)T\sim N(10.5,2.1^2) be independent masses. For prices X=3CX=3C and Y=5TY=5T, X∼N(2.2×3,0.52×32)=N(6.6,1.52)X\sim N(2.2\times3,0.5^2\times3^2)=N(6.6,1.5^2) and Y∼N(10.5×5,2.12×52)=N(52.5,10.52)Y\sim N(10.5\times5,2.1^2\times5^2)=N(52.5,10.5^2).

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(i) Method 1: price distribution

P(X>7)=0.395P(X>7)=0.395 to 3 significant figures.

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(i) Method 2: mass threshold

P(3C>7)=P(C>7/3)=0.39486≈0.395P(3C>7)=P(C>7/3)=0.39486\approx0.395 to 3 significant figures.

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(ii) Method 1: independent prices

P(X>7 and Y>55)=P(X>7)P(Y>55)=0.160P(X>7\text{ and }Y>55)=P(X>7)P(Y>55)=0.160 to 3 significant figures.

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(ii) Method 2: independent masses

P(3C>7 and 5T>55)=P(3C>7)P(5T>55)=P(C>7/3)P(T>11)=(0.39486)(0.405904)≈0.160P(3C>7\text{ and }5T>55)=P(3C>7)P(5T>55)=P(C>7/3)P(T>11)=(0.39486)(0.405904)\approx0.160 to 3 significant figures.

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(iii) Model the total selling price

X+Y∼N(6.6+52.5,1.52+10.52)=N(59.1,112.5)X+Y\sim N(6.6+52.5,1.5^2+10.5^2)=N(59.1,112.5). Thus P(X+Y>62)=0.392P(X+Y>62)=0.392 to 3 significant figures.

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(iv) Compare the two events

If X>7 and Y>55, then X+Y>62, so every outcome from (ii) is included in(iii). The source example X=2,Y=61 gives total 63, satisfying(iii) but not (ii). A whole neighbourhood of such price pairs also qualifies; independent nondegenerate normal densities give that region positive probability. Hence (iii) is strictly greater. A single exact pair alone has probability 0 in a continuous model.

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Common mistakes

Do not multiply probabilities for the total-price event. The source’s exact-pair example illustrates a region, not a positive point probability.

Solution notes

All four asks, complete mass table, all mass/price models, both source methods in(i) and (ii), total-price calculation and original example retained. Strict probability comparison justified by a positive-area region, qualifying the continuous exact-point illustration. No marks printed.

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