Answer to Question #319625 in Statistics and Probability for gel

Question

A manufacturer of flashlight batteries took a sample of 13 batteries from a day’s production and used them continuously until they failed to work. The life of the batteries in hours until failure is:  342, 426, 317, 545, 264, 451, 1049, 631, 512, 266, 492, 562, 298 . What would you advise if the manufacturer wanted to say in advertisement that these batteries last more than 400 hours?

Answer 1

$\mu=(342+426+317+545+264+451+1049+631+512+266+482+562+297)/13=473$

$\sigma=\sqrt{\frac{\sum(x-\mu)^2}{n-1}}=\sqrt{\frac{17161+2209+24336+5184+43681+484+331776+24964+1521+42849+81+7921+30976}{12}}=206$

$\mu \plusmn z \frac{\sigma}{\sqrt{n}}=473 \plusmn 1.96\times206/3.6=473 \plusmn 112$

So 400 falls within this range.