a manufacturer of ice pops flls each plastic container with

a manufacturer of ice pops flls each plastic container with a fruity liquid,

leaving enough room so that consumers can freeze the product. A flling machine is set so

that the amount of liquid in each ice pop is normally distributed with mean 8.00 ounces and

standard deviation 0.25 ounces. Suppose 16 ice pops are randomly selected.

a. Find the probability that the sample mean is more than 8.15 ounces.

b. Suppose the filling machine operator can fine-tune the process by controlling the stan-

dard deviation of the fill. Find a value of such that the probability that the sample

mean is more than 8.05 ounces is 0.05

Solution

Mean ( u ) =8
Standard Deviation ( sd )=0.25
Number ( n ) = 16
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
a)
P(X > 8.15) = (8.15-8)/0.25/ Sqrt ( 16 )
= 0.15/0.063= 2.4
= P ( Z >2.4) From Standard Normal Table
= 0.0082                  
              

b)
P ( Z < X ) = 0.05
Value of z to the cumulative probability of 0.05 from normal table is -1.645
P( X-u/s.d < X - U /0.25 ) = 0.05
That is, ( 8.05 - U /0.25 ) = -1.6449
--> ( 8.05 - U ) = -1.6449*0.25
--> ( 8.05 - U /0.25 ) = -2.2414
--> U = 0.4112 + 8.05 = 8.46124