A filling machine in a food processing plant was found to pr

A filling machine in a food processing plant was found to produce pint juice cartons with a mean ontent of 16.25 fluid ounces, with a standard deviation of 0.12 fluid ounces. Assume tht the quantity of juice is normally distrubuted.(please show work-that way I can understand)

A) What is the probability of getting (in the long run average) less than the official amount (16 ounces) that you oay for/?

B) What is the probability of getting between 16 and 16.45 fluid ounces of juice?

Solution

Normal Distribution
Mean ( u ) =16.25
Standard Deviation ( sd )=0.12
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X < 16) = (16-16.25)/0.12
= -0.25/0.12= -2.0833
= P ( Z <-2.0833) From Standard Normal Table
= 0.0186                  

b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 16) = (16-16.25)/0.12
= -0.25/0.12 = -2.0833
= P ( Z <-2.0833) From Standard Normal Table
= 0.01861
P(X < 16.45) = (16.45-16.25)/0.12
= 0.2/0.12 = 1.6667
= P ( Z <1.6667) From Standard Normal Table
= 0.95221
P(16 < X < 16.45) = 0.95221-0.01861 = 0.9336