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A steel rolling mill has two machines, A and B, for cutting steel bars. For each machine the length of a cut bar can be modelled by a normal distribution. Bars cut by machine A have a mean length of 1212 mm and a standard deviation of 5 mm.

i) Determine the probability that the length of a bar is greater than 1205 mm.

ii) Calculate the length exceeded, on average, by one bar in five hundred.

I have already gotten an answer for i) which is 0.919. Would appreciate help on ii) Thanks!

Hi Jordan,

It's a good thing that you've attempted the problem and, yes, the answer to part (i) is correct. For part (ii), what we need is to find from the normal distribution table the standard score where the probability to the right is one in five hundred i.e 1/500 = 0.002

From tables, this standard score z = 2.878 and we can calculate this length l from

(l - 1212)/5 = 2.878

l = 1226.39 mm

This means that the probability of a bar length exceeding 1226.39 mm = 1/500, as required.

Let me know if anything is unclear.

Regards

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