Question 769975: The shape of the distribution of the time required to get an oil change at a 20 min. Oil facility is unknown. However, records indicate that the mean is 21.8 mins., and the standard deviation is 4.8 mins. What is the probability that a random sample of n= 40 oil changes results in a sample mean time less than 20 mins.. The probability is approximately .? ( round to four decimal places as needed). I have tried many solutions on my TI 83 and come up with nothing cloce to four decimals. Help!
Answer by Edwin McCravy(20064)   (Show Source):
You can put this solution on YOUR website! The shape of the distribution of the time required to get an oil change at a 20 min. Oil facility is unknown. However, records indicate that the mean is 21.8 mins., and the standard deviation is 4.8 mins. What is the probability that a random sample of n= 40 oil changes results in a sample mean time less than 20 mins.. The probability is approximately .? ( round to four decimal places as needed). I have tried many solutions on my TI 83 and come up with nothing cloce to four decimals. Help!
Since 40 is a large sample, the central limit theorem tells us that
the distribution is approximately normal even though the shape of
the distribution is unknown. However this is a sample, not just a
single value, so we must divide the standard deviation by √n.
Press ON
Press CLEAR
Press 2ND
Press VARS
Press 2
You should see this---> normalcdf(
Make it read this way:
normalcdf(-1E99,20,21.8,4.8/√(40))
[To get the little E, press ALPHA then , (the comma key)]
Press ENTER
Read .00088530196
round to 0.0009
Edwin
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