Porisma

Worked solution

Keep both tails of a call-duration difference and all handling fees

Independent call durations to City A are normal with mean 8 minutes and standard deviation 2.5 minutes; to City B, mean 10 and standard deviation 3.2.

(i) Find the probability that two random City A call durations in total differ from three times a third City A duration by more than 3 minutes. marks

Calls cost 22 cents per minute to A and 30 cents per minute to B, plus a 10 cent handling fee per call. Ann makes two calls to A and one to B.

(ii) Find her expected total cost. mark

(iii) Show that the probability her cost exceeds 8.50 dollars is 0.0870 to 3 significant figures. marks

[2009/AJC/II/13(part of)]

CJC · Review problems · Question 4

Answer

(i)0.819; (ii)6.82 dollars; (iii)0.0870 (3 s.f.).

(i) Model the duration difference

Let X∼N(8,2.52)X\sim N(8,2.5^2) and Y∼N(10,3.22)Y\sim N(10,3.2^2) be call durations. Put D=X1+X2−3X3D=X_1+X_2-3X_3. Then E(D)=E(X)+E(X)−3E(X)=−8E(D)=E(X)+E(X)-3E(X)=-8 and Var⁡(D)=Var⁡(X)+Var⁡(X)+32Var⁡(X)=68.75\operatorname{Var}(D)=\operatorname{Var}(X)+\operatorname{Var}(X)+3^2\operatorname{Var}(X)=68.75, so D∼N(−8,68.75)D\sim N(-8,68.75).

How did this part go?

(i) Use the absolute difference

P(∣D∣>3)=P(D>3 or D<−3)=P(D<−3)+P(D>3)=0.819P(|D|>3)=P(D>3\text{ or }D<-3)=P(D<-3)+P(D>3)=0.819 to 3 significant figures. The bars are present in the rendered source; a one-sided tail would answer a different question.

How did this part go?

(ii) Model each call price in dollars

For A, F=0.22X+0.1F=0.22X+0.1, E(F)=0.22(8)+0.1=1.86E(F)=0.22(8)+0.1=1.86 and Var⁡(F)=0.222(2.52)=0.552\operatorname{Var}(F)=0.22^2(2.5^2)=0.55^2, so F∼N(1.86,0.552)F\sim N(1.86,0.55^2). For B, G=0.30Y+0.1G=0.30Y+0.1, E(G)=0.30(10)+0.1=3.1E(G)=0.30(10)+0.1=3.1 and Var⁡(G)=0.302(3.22)=0.962\operatorname{Var}(G)=0.30^2(3.2^2)=0.96^2, so G∼N(3.1,0.962)G\sim N(3.1,0.96^2).

How did this part go?

(ii) Total expected cost

C=F1+F2+GC=F_1+F_2+G, so E(C)=1.86+1.86+3.1=6.82E(C)=1.86+1.86+3.1=6.82 dollars. All three handling fees are included.

How did this part go?

(iii) Add independent price variances

Var⁡(C)=0.552+0.552+0.962=1.5266\operatorname{Var}(C)=0.55^2+0.55^2+0.96^2=1.5266. Hence C∼N(6.82,1.5266)C\sim N(6.82,1.5266) and P(C>8.50)=0.0870P(C>8.50)=0.0870 to 3 significant figures, as required.

How did this part go?

Common mistakes

The fixed handling fees change means but not variances; independent calls contribute separate variances.

Solution notes

All asks/marks, every duration and cost transformation/moment, both duration tails and all fixed fees retained. Printed absolute-value bars restored from source image.

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