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 March 4th, 2016, 01:37 PM #1 Newbie   Joined: Feb 2016 From: Illinois Posts: 8 Thanks: 0 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?
 March 4th, 2016, 02:06 PM #2 Senior Member   Joined: Jul 2014 From: भारत Posts: 1,178 Thanks: 230 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 Thanks from Ore0Kidd

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### a certain printer ribbon has demonstrated a mean average time usage of 9 hours and a standard deviation of 22 minutes.what time usage guarantee in hrs should the manufacturer advertise to ensure only 4% of the ribbons fail to meet the guaranteed Time usag

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