Question

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.

Answer 1

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