SOLUTION: A phone maker claims that its phones’ batteries last for 13.7 hours on average with standard usage with a standard deviation of 0.9 hours. The distribution of battery lives is n

Question 1189989: A phone maker claims that its phones’ batteries last for 13.7 hours on average with standard usage with a standard deviation of 0.9 hours. The distribution of battery lives is normal.
(a) If we accept that claim, what is the probability that a battery lasts between 11 and 12 hours?
(b) Find the 85th percentile for the battery lives- that is, a time for which 85% of the
batteries last less than that time.
(c) The company has just tested 19 devices. Find the probability that the mean battery life of the 19 phones is more than 14 hours.

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
z=(x-mean)/sd
so z (11 hr)=(11-13.7)/0.9 and z (12 hr)=(12-13.7)/0.9
on a calculator, use 2ndVARS 2normalcdf(11,12,13.7,0.9) for probability of 0.0281
Another way has z between -3 and -1.89 for prob. 0.00135 and 0.0294 for the same answer, subtracting the first from the second.
-
85th percentile is z=1.036=(x-13.7)/0.9, and x=14.63 hours.
-
this is z>(x-mean)/sigma/sqrt(n)
so z>(14-13.7)/0.9/sqrt(19)=1.453
That has probability of 0.0731

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