Question 1100421: The manufacturer of a laser printer reports the mean number of pages a cartridge will print before it needs replacing is 12,225. The distribution of pages printed per cartridge closely follows the normal probability distribution and the standard deviation is 795 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% of the time? (Round z value to 2 decimal places and your final answer to the nearest whole number.)
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 12,225. The distribution of pages printed per cartridge closely follows the normal probability distribution and the standard deviation is 795 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% of the time? (Round z value to 2 decimal places and your final answer to the nearest whole number.)
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mean = 12,225
std = 795
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Find the z-value with a left-tail of 10%
invNorm(0.10) = -1.2816
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Find the corresponding # of pages:: -1.2816*795+12,225 = 11206
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Ans: 11206 pages
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Cheers,
Stan H.
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