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Worked solution

Use the lower-tail quantile to set the new filling mean

A manufacturer produces margarine packets with normally distributed mass, mean 502 g and standard deviation 0.8 g. Find the proportion containing less than 500 g.

The mean is increased to μg and the standard deviation is unchanged. Only 1 packet in 1000, on average, now contains less than 500 g. Find μ correct to 1 decimal place.

[2007/HCI/H1/6]

CJC · Discussion · Question 3

Answer

Original proportion 0.00621 (3 s.f.); new mean 502.5 g (1 d.p.).

Original mass distribution and lower tail

Let X be the original mass in grams: X∼N(502,0.82)X\sim N(502,0.8^2). Then P(X<500)=0.00621P(X<500)=0.00621 to 3 significant figures.

How did this part go?

Set the new model and quantile equation

Let Y∼N(μ,0.82)Y\sim N(\mu,0.8^2). The target is P(Y<500)=0.001P(Y<500)=0.001, so P(Z<(500−μ)/0.8)=0.001P(Z<(500-\mu)/0.8)=0.001, with Z∼N(0,1)Z\sim N(0,1). The source lower-tail quantile gives (500−μ)/0.8=−3.090232308(500-\mu)/0.8=-3.090232308.

How did this part go?

Solve and round at the end

Thus μ=500+0.8(3.090232308)≈502.472\mu=500+0.8(3.090232308)\approx502.472, giving μ=502.5 g to 1 decimal place. The rounded value is the reported mean; the quantile equation uses the unrounded value.

How did this part go?

Common mistakes

Keep the standard deviation 0.8 in the denominator;0.8² is the variance.

Solution notes

Both asks, original/new models, source probability, complete standardisation, printed quantile and final rounding retained. No marks printed.

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