Assume the machine shifts and is filling the vials with a mean amount of 9.96 milligrams and a standard deviation of 0.05 milligrams you select 5 vials and find the mean amount compounded added.
What is the probability that you select a sample of five vial that has a mean that is within the acceptable range?
What formula is being used for this?
I have the answer(0.3264)
I don't understand how they are getting this answer and how they are solving it.
mean amount = 9.96 milligrams
standard deviation =.05 milligrams
n = 5
Considering the answer I can only conclude that the acceptable range of the sample is within +/-1 of the standard deviation.
P( mean - sd < X < mean + sd ) = P( ( ( mean - sd ) - mean )/(sd/sqrt(n)) < ( X - mean )/(sd/sqrt(n)) < ( ( mean + sd ) - mean )/(sd/sqrt(n)) )
P( 9.96 - 0.05 < X < 9.96 + 0.05) = P( ( ( 9.96 - 0.05 ) - 9.96 )/(0.05/sqrt(5)) < ( X - 9.96 )/(0.05/sqrt(5)) < ( ( 9.96 + 0.05 ) - 9.96 )/(0.05/sqrt(5)) )
P( 9.995 < X < 9.965 ) = P( (- 0.05)/(0.05/sqrt(5)) < ( X - 9.96 )/(0.05/sqrt(5)) < ( (0.05 )/(0.05/sqrt(5)) )
P( 9.995 < X < 9.965 ) = P( -0.4472 < t < 0.4472 )
df = n-1 = 5-1 = 4
P( 9.995 < X < 9.965 ) = 2*P( 0 < t < 0.4472 )
P( 9.995 < X < 9.965 ) = 2*0.1632
P( 9.995 < X < 9.965 ) = 0.3264
Assume the machine shifts and is filling the vials with a mean amount of 9.96 milligrams...
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