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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 

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distribution, normal or standard, probability 
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