Question #101770
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?
1
Expert's answer
2020-01-29T05:56:01-0500
mean=960 min=540 minmean=9\cdot60\ min=540\ minS.D.=22minS.D.=22 minα/2=0.04,zα/2=1.7507\alpha/2=0.04, z_{\alpha/2}=1.7507

CI=mean±zα/2S.DCI=mean\pm z_{\alpha/2}\cdot S.D

CI=540±1.750722CI=540\pm 1.7507\cdot 22

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

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


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS