SOLUTION: The manufacturer of a laser printer reports the mean number of pages a cartridge will print before it needs replacing is 12,350. The distribution of pages printed per cartridge clo

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Question 1034322: The manufacturer of a laser printer reports the mean number of pages a cartridge will print before it needs replacing is 12,350. The distribution of pages printed per cartridge closely follows the normal probability distribution and the standard deviation is 670 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 99 percent of the time? (Round z value to 2 decimal places. Round your answer to the nearest whole number.)

****I really only want to know the formulas. I am unable to find the X value to make a Z value to come up with the final answer. ***
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
The manufacturer of a laser printer reports the mean number of pages a cartridge will print before it needs replacing is 12,350. The distribution of pages printed per cartridge closely follows the normal probability distribution and the standard deviation is 670 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 99 percent of the time? (Round z value to 2 decimal places. Round your answer to the nearest whole number.)
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Find the z-value with a left tail of 0.99
invNorm(0.99) = 2.3263
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Find the corresponding x value using x = z*s + u
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Cheers,
Stan H
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