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How many pages should the manufacturer advertise for each cartridge if it wants to be correct 99% of the time

Sagot :

The manufacturer should advertise 13811 pages for each cartridge if it wants to be correct 99% of the time

z-score is the number of standard deviations from the mean value of the reference population

99%=0.99

The z-value associate with 0.99 is 2.33:

2.33 = (X - 12425) / 595:

X - 12425 = 2.33 * 595

X = (2.33 * 595) + 12425

X = 138111

Therefore, The manufacturer should advertise 13811 pages for each cartridge if it wants to be correct 99% of the time

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