Normal Distribution (OCR S2)

I have two questions from the OCR textbook that I am struggling with:

a) The time taken by a garage to replace worn-out break pads follows a normal distribution with the mean time 90 minutes and the standard deviation 5.8.

The garage claims to complete the replacements in 'a to b minutes' If this claim is correct for 90% of the repairs, find a and b correct to 2 significant figures, based on a symmetrical interval centred on the mean.

b) Fluorescent light tubes have lifetimes which are normally distributed with a mean of 2010 hours and a standard deviation of 20 hours.

If the company agrees to replace light tubes which fail to meet a given lifetime, calculate the value of this minimum lifetime if the company wishes to replace only 3% of tubes sold.

My workings so far:

a) X ~ N(90,33.64)

P[(a-90)/5.8<z<(b-90)/5.8] = 0.9

b) X ~ N(2010,400)

P[(z-2010)/20<t] = 0.03

Are these workings right so far? In terms of proceeding from here, I understand that you need to look in the normal distribution tables for the above values (0.9 and 0.03) and replace them. But then what, and how can this help me form and solve equations for the unknown values?

You've got the right working.

For a) it may help to use symmetry to think and think of what P(X<b) would be, as you'd be able to read that off from the table.

Symmetry in b may help. The aim for these types of questions is to somehow get to P(Z<z) using symmetry and other things you know.

Thanks for this.

Continuing on:

a) Becomes: P[z<(b-90)/5.8] - P[z<(a-90)/5.8] = 0.9

Yet a may be negative, in which case P[z<(b-90)/5.8] - {1-P[(a-90)/5.8]} = 0.9 ??

b) I am still quite unsure of how to continue for this question. Given my previous working, which you say is correct, I have two unknowns: z and t (the guaranteed minimum life). Yet I only need to find t. Must I find z first or is this not necessary?

While your statement for a is true, it'll still leave you with too many unknowns to do anything. The corresponding z value for a will be negative, yes.

I seem to have assumed that 5% of repairs take less than a minutes and 5% take more than b minutes, but using that, think of what % of values are less than or equal to b. That will be a nice probability that you can find which z value has that value of P(Z<z) , and work backwards to find what b is equal to.

For b, I was a bit confused by your notation but I didn't manage to point it out. It should be something like P(T<t) = P(Z<(t-2010)/20) = 0.03, where t is the minimum lifetime, so you have to look for the z value. The problem is that you can't find what z value satisfies P(Z<z) = 0.03 immediately. But you know that P(Z<-z) = .... so z is ....