Answer to Question #101770 in Statistics and Probability for Awais Ashraf

Question

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?

Answer 1

$$mean=9\cdot60\ min=540\ min$$ $$S.D.=22 min$$ $$\alpha/2=0.04, z_{\alpha/2}=1.7507$$

$$CI=mean\pm z_{\alpha/2}\cdot S.D$$

$$CI=540\pm 1.7507\cdot 22$$

$$CI=(501.5\ min; 578.5\ min )$$

$$CI=(8\ h \ 21.5\ min; 9\ h\ 38.5\ min )$$