A government specification calls for a drug to have a therap

A government specification calls for a drug to have a therapeutic effectiveness for a mean period of 30 hours with the standard deviation of the random variable X that characterizes the period of effectiveness known to be 15 hours. Suppose that the actual mean period of effectiveness of the drug in a given shipment is 28 hours. Before purchasing the drug, we are allowed to select a sample and test for effectiveness. Shipments of the drug are to be accepted if the mean period of the sample lies between 24 and 32 hours. In a random sample of 16 packages, determine the probability that a given shipment will be accepted according to the government specification?

Solution

Mean ( u ) =28
Standard Deviation ( sd )=15
Number ( n ) = 16
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 24) = (24-28)/15/ Sqrt ( 16 )
= -4/3.75
= -1.0667
= P ( Z <-1.0667) From Standard Normal Table
= 0.14306
P(X < 32) = (32-28)/15/ Sqrt ( 16 )
= 4/3.75 = 1.0667
= P ( Z <1.0667) From Standard Normal Table
= 0.85694
P(24 < X < 32) = 0.85694-0.14306 = 0.7139                  

probability that a given shipment will be accepted according to the government specification is 0.7139