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The manufacturer of a laser printer reports the mean number of pages a cartridge will print before it needs replacing is 12,200. The distribution of pages printed per cartridge closely follows the normal probability distribution and the standard deviation is 820 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% of the time

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Answer:

14110

Step-by-step explanation:

Let x = number of pages ;

μ = 12200 ; σ = 820

Zscore = (x - μ) / s

Zscore = (x - 12200) / 820

Zcritical at 99% confidence, for a 1 - tail test is 2.33

Hence, Z at 99% = 2.33

Put Zscore = 2.33

Zscore = (x - 12200) / 820

2.33 = (x - 12200) / 820

2.33 * 820 = x - 12200

1910.6 = x - 12200

1910.6 + 12200 = x

x = 14110.6

x = 14111

The number of pages the manufacturer should advertise is 14,110

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