Porisma

Worked solution

Recover normal parameters and combine independent masses

(i) Factory A’s nut mass is normal with mean μ\mu grams and standard deviation σ\sigma grams. For a random sample of 50, P(Xˉ<247)=0.018079P(\bar X<247)=0.018079 and P(∑Xj>12600)=0.78397P(\sum X_j>12600)=0.78397, each correct to 5 s.f. Find μ,σ\mu,\sigma to the nearest gram. marks

(ii) State clearly the parameters of every normal distribution used. Factory B’s independent mass models are:

ProductMean (g)Standard deviation (g)
Bolt7457.3
Nut2505

(a) Find the probability that a randomly chosen bolt’s mass differs from three times one randomly chosen nut’s mass by at least 40 g. marks

(b) A new process reduces each nut’s mass by 10%. Find the probability that the total mass of 10 randomly chosen new nuts is below 2.24 kg. marks

[2017/MJC/II/10]

CJC · Review problems · Question 8

Answer

(i) μ=255\mu=255 g, σ=27\sigma=27 g (nearest gram). (ii)(a) 0.02140.0214; (b) 0.2410.241.

(i) Two quantile equations

Xˉ∼N(μ,σ2/50)\bar X\sim N(\mu,\sigma^2/50) and ∑Xj∼N(50μ,50σ2)\sum X_j\sim N(50\mu,50\sigma^2). Let a=Φ−1(0.018079)≈−2.095146a=\Phi^{-1}(0.018079)\approx-2.095146 and b=Φ−1(1−0.78397)≈−0.7856714b=\Phi^{-1}(1-0.78397)\approx-0.7856714. Then (247−μ)/(σ/50)=a(247-\mu)/(\sigma/\sqrt{50})=a and (12600−50μ)/(50σ)=b(12600-50\mu)/(\sqrt{50}\sigma)=b. Equivalently μ−0.2962984σ=247\mu-0.2962984\sigma=247 and 50μ−0.785671450σ=1260050\mu-0.7856714\sqrt{50}\sigma=12600. Solving gives μ≈254.99995\mu\approx254.99995 g and σ≈26.99963\sigma\approx26.99963 g, hence 255 g and 27 g to the nearest gram.

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(ii)(a) A two-tailed difference event

For Factory B let Y∼N(250,52)Y\sim N(250,5^2) and W∼N(745,7.32)W\sim N(745,7.3^2) independently. Then D=W−3Y∼N(−5,7.32+9(52))=N(−5,278.29)D=W-3Y\sim N(-5,7.3^2+9(5^2))=N(-5,278.29). The absolute-difference event is ∣D∣≥40|D|\ge40, so its probability is P(D≤−40)+P(D≥40)≈0.0214P(D\le-40)+P(D\ge40)\approx0.0214.

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(ii)(b) Scale before summing

The new total is T=0.9∑j=110Yj∼N(10(0.9)(250),10(0.92)(52))=N(2250,202.5)T=0.9\sum_{j=1}^{10}Y_j\sim N(10(0.9)(250),10(0.9^2)(5^2))=N(2250,202.5). Convert 2.24 kg to 2240 g. Then P(T<2240)≈0.241P(T<2240)\approx0.241.

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Common mistakes

“At least 40 g difference” includes both tails; a 10% mass reduction multiplies variance by 0.9².

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