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Worked solution

Convert both rejection rates into quantile equations

A grading machine rejects apples as too large if their mass exceeds 150 g, and too small if below 110 g. Over a long period 10% are too large and 15% too small. Apple mass in grams may be modelled by N(μ,σ2)N(\mu,\sigma^2). Find μ and σ.

[2005/Specimen Paper/II/9(modified)]

CJC · Discussion · Question 4

Answer

μ≈128\mu\approx128g, σ≈17.3\sigma\approx17.3g (3 s.f.).

Upper rejection rate to cumulative probability

For X∼N(μ,σ2)X\sim N(\mu,\sigma^2), P(X>150)=0.10P(X>150)=0.10 implies P(X<150)=1−0.10=0.90P(X<150)=1-0.10=0.90. Thus P(Z<(150−μ)/σ)=0.90P(Z<(150-\mu)/\sigma)=0.90, giving (150−μ)/σ=1.28155(150-\mu)/\sigma=1.28155. The first equation is μ+1.28155σ=150\mu+1.28155\sigma=150.

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Lower rejection rate

P(X<110)=0.15P(X<110)=0.15 gives P(Z<(110−μ)/σ)=0.15P(Z<(110-\mu)/\sigma)=0.15. Hence (110−μ)/σ=−1.03643(110-\mu)/\sigma=-1.03643, so μ−1.03643σ=110\mu-1.03643\sigma=110.

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Solve both equations

Subtract the second equation from the first: (1.28155+1.03643)σ=40(1.28155+1.03643)\sigma=40. Solving gives σ≈17.3 g and μ≈128 g, both to 3 significant figures, as in the source.

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Common mistakes

Use one positive and one negative standard-normal quantile; σ must be positive.

Solution notes

Complete ask, both rejection probabilities/complements, all standardisations and both printed quantile equations/results retained. Extra subtraction step explains the source simultaneous solve. No marks printed.

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