Porisma

Worked solution

Use the given variances and the central eighty percent

Oil volumes in Small, Regular and Large bottles are normal.

BottleMean (ml)Variance (ml²)
Small61025
Regular1050σ2\sigma^2
Large120050

(i) Explain why Small volume can be modelled as normal although oil volume cannot be negative. mark

(ii) Three Small bottles are randomly chosen. Find the probability that one contains more than 615 ml and the other two less than 602 ml each. marks

(iii) Find σ if 80% of Regular bottles contain between 1040 and 1060 ml. marks

(iv) Find the probability that two random Large bottles contain more than four times the volume in one random Small bottle. marks

The source also prints “The mass of an empty bag is assumed to be negligible”; this unrelated bag note has no role in the bottle-volume model.

[2009/TJC/II/11]

CJC · Review problems · Question 3

Answer

(i)negative tail negligible; (ii)0.00143; (iii)σ≈7.80 ml; (iv)0.0368 (3 s.f.).

Define the three volume distributions

S∼N(610,25)S\sim N(610,25), R∼N(1050,σ2)R\sim N(1050,\sigma^2) and L∼N(1200,50)L\sim N(1200,50) in ml. Distinct selected bottles are treated as independent in the source calculations.

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(i) Check the impossible negative tail

P(S<0)≈0P(S<0)\approx0, so impossible negative volumes have negligible probability. This addresses the objection to the approximation; it does not by itself prove the actual factory distribution is normal.

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(ii) Count the possible positions of the larger bottle

P(S1>615)P(S2<602)P(S3<602)(3!/2!)=0.00143P(S_1>615)P(S_2<602)P(S_3<602)(3!/2!)=0.00143 to 3 significant figures. The factor 3 chooses which bottle exceeds 615 ml.

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(iii) Standardise the central interval boundary

Symmetry leaves 0.10 in each tail. Thus P(R<1040)=0.1P(R<1040)=0.1, so P((R−1050)/σ<(1040−1050)/σ)=P(Z<−10/σ)=0.1P((R-1050)/\sigma<(1040-1050)/\sigma)=P(Z<-10/\sigma)=0.1. Hence −10/σ=−1.281551567-10/\sigma=-1.281551567 and σ≈7.80 ml to 3 significant figures. The source retains R∼N(1050,7.8030414542)R\sim N(1050,7.803041454^2) before final rounding.

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(iv) Form the correct difference

E(L1+L2−4S)=2E(L)−4E(S)=−40E(L_1+L_2-4S)=2E(L)-4E(S)=-40 and Var⁡(L1+L2−4S)=2Var⁡(L)+42Var⁡(S)=500\operatorname{Var}(L_1+L_2-4S)=2\operatorname{Var}(L)+4^2\operatorname{Var}(S)=500. Therefore L1+L2−4S∼N(−40,500)L_1+L_2-4S\sim N(-40,500) and P(L1+L2−4S>0)=0.0368P(L_1+L_2-4S>0)=0.0368 to 3 significant figures. The source distribution line prints −2 S; this is a typo, since its prompt, moments and final event all use −4 S.

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Compare your sketch

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Normal oil-volume curve centred at 1050 ml. The shaded central 80% lies between 1040 and 1060 ml; each remaining tail has probability 0.10.
Normal oil-volume curve centred at 1050 ml. The shaded central 80% lies between 1040 and 1060 ml; each remaining tail has probability 0.10.
Normal oil-volume curve centred at 1050 ml. The shaded central 80% lies between 1040 and 1060 ml; each remaining tail has probability 0.10.

Common mistakes

In part (iv), the single Small volume is multiplied by 4, contributing 16 times its variance.

Solution notes

All asks/marks, full table, unrelated printed bag note, source independent-bottle model, order factor, complete quantile stages/unrounded model and every difference moment retained. Printed −2 S distribution typo qualified against −4 S working. Central 80% source sketch reconstructed accessibly.

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