Answer to Question #350224 in Statistics and Probability for frwan

Question

The time usage of a certain printer cartridge follows a normal probability distribution with mean 58 hours and standard deviation of 183 minutes. Around what time usage guarantee (in hours) should the manufacturer advertise in order to ensure that only 0.1306% of the cartridges fail to meet the guaranteed time usage?

Answer 1

183 min=3.05 hrs.

$P(X<x)=P(Z<\frac{x-58}{3.05})=0.1306.$

$\frac{x-58}{3.05}=-1.12.$

$x=58-3.05*1.12=54.58.$

The manufacturer should advertise the time usage of 54.58 hours.