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Paper 4 (9164, Section A) · June 2007 · Normal Distribution

A factory produces two types of nut and bolt with normally distributed masses. Type A bolts have mean mass 20.5 g and Type A nuts have mean mass 5 g; Type B bolts have mean mass 20 g and Type B nuts have mean mass 4.7 g. Each bolt is fitted with two nuts. What is the mean, in grams, of (total mass of a Type A bolt-and-nuts unit) minus (total mass of a Type B bolt-and-nuts unit)?

Model answer

1.1

Also accepted: 1.1 g

Explanation

Mean total mass of Type A =20.5+2(5)=30.5= 20.5 + 2(5) = 30.5 g. Mean total mass of Type B =20+2(4.7)=29.4= 20 + 2(4.7) = 29.4 g. The mean of the difference is 30.5−29.4=1.130.5 - 29.4 = 1.1 g.

Derived from ZIMSEC Maths Paper 1, June 2007, Q2 (Statistics)

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