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Worked solution

Interpret a probability inequality as a range of mass thresholds

Chain-link mass is normal with mean 800 g and standard deviation 20 g.

(i) Find the probability one link exceeds 805 g. mark

A locking sleeve increases the link’s mass by 10%.

(ii) State the sleeved-link mass distribution and find the probability it lies between 865.35 and 895.5 g. marks

Hook mass is normal with mean 750 g;15% weigh below 735.6 g.

(iii) Find its standard deviation. marks

Five independent sleeved links and one hook are packed in a fixed 1 kg wooden box. The probability that the box and contents exceed k grams is more than 0.013.

(iv) Find k to 1 decimal place, stating all distribution parameters used. marks

[2018/NYJC/Prelim/8(modified)]

CJC · Review problems · Question 10

Answer

(i)0.401; (ii)S∼N(880,222)S\sim N(880,22^2), probability 0.507; (iii)σ≈13.9 g; (iv)k<q with q≈6263.8 g, not one uniquely determined k.

(i) Original link model

Let L∼N(800,202)L\sim N(800,20^2). Then P(L>805)=0.40129≈0.401P(L>805)=0.40129\approx0.401 to 3 significant figures.

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(ii) Scale the mass and its standard deviation

S=1.1L∼N(880,222)S=1.1L\sim N(880,22^2). Thus P(865.35<S<895.5)=0.50672≈0.507P(865.35<S<895.5)=0.50672\approx0.507 to 3 significant figures.

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(iii) Infer the hook standard deviation

For H∼N(750,σ2)H\sim N(750,\sigma^2), P(H<735.6)=0.15P(H<735.6)=0.15 gives (735.6−750)/σ=−1.0364(735.6-750)/\sigma=-1.0364. Thus σ≈13.894 g, or 13.9 g to 3 significant figures.

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(iv) Include independent items and the fixed box mass

Let C=S1+S2+S3+S4+S5+H+1000C=S_1+S_2+S_3+S_4+S_5+H+1000. The source treats the hook and all five links as independent, and uses rounded hook σ=13.9 g. Thus C∼N(6150,2613.21)C\sim N(6150,2613.21), with variance 5(222)+13.925(22^2)+13.9^2. Using the unrounded hook quantile gives variance≈2613.0377 instead; both round the threshold to 6263.8 g at 1 decimal place.

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(iv) Keep the full inequality and identify the boundary

P(C>k)>0.013P(C>k)>0.013 means P(C≤k)<0.987P(C\le k)<0.987, so k<qk<q, where q is the 98.7 th percentile of C, approximately 6263.8 g. This matches the source’s inequality k<6263.8 with its boundary rounded to 1 decimal place. The printed condition does not determine a unique k: it gives all thresholds below the unrounded q. Rounding q does not turn the inequality into an equality or an exact rounded cutoff.

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Common mistakes

A probability greater than 0.013 gives a strict range of k values. The source’s 6263.8 is the rounded boundary, not an equality.

Solution notes

All asks/marks, original/scaled/hook/total distributions, each decimal stage and source threshold inequality retained. Independent hook assumption and source rounded-hook variance 2613.21 explicit; unrounded variance and strict/underdetermined threshold qualification added. No unique k invented.

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