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March 4th, 2016, 01:37 PM   #1
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Normal/Standard Probability Distribution

How would I solve this with a TI 83 Calculator?

A certain printer cartridge has demonstrated a mean time usage of 9 hours and a standard deviation of 22 minutes. What time usage guarantee (in hours) should the manufacturer advertise in order to ensure that only 4% of the cartridges fail to meet the guaranteed time usage?
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March 4th, 2016, 02:06 PM   #2
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Mean = 9*60 = 540 mins
S.D = 22 mins
CI = Mean ± 1.75*S.D
= 540 ± 1.75*22
= 540 ± 38.5
CI = (501.5,578.5)
8 hrs 31.5 mins to 9hrs 38.5 mins
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