A factory produces two types of nut and bolt with independent, normally distributed masses, each with standard deviation 0.2 g. Type A bolts have mean 20.5 g and Type A nuts have mean 5 g; Type B bolts have mean 20 g and Type B nuts have mean 4.7 g. Each bolt is fitted with two nuts. What is the probability that the total mass of a Type A bolt-and-nuts unit is greater than the total mass of a Type B bolt-and-nuts unit?

0.988

0.994

0.997

0.999

Explanation

Let

D

= (Type A total)

-

(Type B total). Mean

(D) = 30.5-29.4=1.1

g. Since each total sums 3 independent components each with variance

0.2^2=0.04

\text{Var}(D) = 6(0.04) = 0.24

, so

\text{sd}(D)=\sqrt{0.24}\approx0.490

P(D>0) = P\left(Z > \dfrac{-1.1}{0.490}\right) = P(Z>-2.245) \approx 0.988

. Using only the two nut variances and forgetting the bolt variance gives

\text{sd}=\sqrt{0.16}=0.4

and

z=2.75

, i.e.

0.997

. Halving the variance by forgetting to add both sides' contributions gives

\text{sd}=\sqrt{0.12}\approx0.346

and

z\approx3.18

, i.e.

0.999

. Rounding

\sqrt{0.24}

0.44

gives

z=2.5

, i.e.

0.994

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

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