The manufacturer of a laser printer reports the mean number of pages a cartridge will print before it needs replacing is 12,450. The distribution of pages printed per cartridge closely follows the normal probability distribution and the standard deviation is 570 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 90 percent of the time? (Round z value to 2 decimal places. Round your answer to the nearest whole number.)

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 12450, standard deviation of 570:

This means that [tex]\mu = 12450, \sigma = 570[/tex]

How many pages should the manufacturer advertise for each cartridge if it wants to be correct 90 percent of the time?

They should advertise the 10th percentile, which is X when Z has a p-value of 0.1, so X when Z = -1.28. Then

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-1.28 = \frac{X - 12450}{570}[/tex]

[tex]X - 12450 = -1.28*570[/tex]

[tex]X = 11720[/tex]

The manufacturer should advertise 11720 pages.