The manufacturer of a laser printer reports the mean number of pages a cartridge

Question

The manufacturer of a laser printer reports the mean number of pages a cartridge will print before it needs replacing is 12,250. The distribution of pages printed per cartridge closely follows the normal probability distribution and the standard deviation is 770 pages. The manufacturer wants to provide guidelines to potential customers as to how long they can expect a cartridge to last. How many pages should the manufacturer advertise for each cartridge if it wants to be correct 95 percent of the time?

Explanation / Answer

If X denote the number of pages the cartridge lasts, you are given X follows a Normal distribution with mean 12250 and sd=770

So Y=(X-12250)/770 follows a standard Normal distibution (i.e., mean=0, sd=1)

If you want to find a value a such that P(X>a)=0.95
this is same as saying P(Y>(a-12250)/770)=0.95
i.e., b=(a-12250)/770 is such that the area to the right of b is 0.95 under the standard nromal curve

Clearly the value b should be a negative number.
Since most normal probability tables give values of the areas to the left of z for various values of z>0,
we use the symmetry of the normal curve.

P(Y>b)=0.95, is same as saying P(Y<(-b)) =0.95

So we look in normal probability table for a value z such that the area left of z is 0.95
This value is 1.645 in my table
So (-b)=1.645 or b=(-1.645)

Thus (a-12250)/770=(-1.645)
Solving this we get a=10983

So in your notation x is 10983

Do check the calculations!

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