Question 722361: Swift Auto Service has taken advantage of some of the results
you provided (HW #4) and has improved their process. Their oil &
lube “service time per car” still follows a normal distribution, but
now with a mean of 30 minutes per car, and a standard deviation
of 4.0 minutes.
b) if you select a random sample of 16 oil & lube jobs, what is
the probability the sample mean “service time” from that group
will be shorter than 28 minutes ?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! follows a normal distribution, but
now with a mean of 30 minutes per car, and a standard deviation
of 4.0 minutes.
b) if you select a random sample of 16 oil & lube jobs, what is
the probability the sample mean “service time” from that group
will be shorter than 28 minutes ?
-----
x-bar = 28 min
--------------------
z(28) = (28-30)/[4/sqrt(16)) = -2
----
P(x-bar < 28) = P(z < -2) = 0.0228
==============
Cheers,
Stan H.
==============
|
|
|
