Answer:
Step-by-step explanation:
Given that a  cola-dispensing machine is set to dispense a mean of 2.02 liters into a bottle labeled 2 liters.
Std deviation =0.015 litres
X- litres contained in a bottle is N(2.02, 0.15)
Z score is obtained as [tex]z=\frac{x-2.02}{0.015}[/tex]
a) probability a bottle will contain between 2.00 and 2.03 liters
=P(2<x<2.03) = P(-1.33<Z<2)
= 0.4082+0.4772
=0.8854
b) P(X<2) = P(Z<-1.33) =0.5-0.4082 = 0.0918
c) 2% of containers
|z|<0.11
X lies between 0.6883 and 3.352 l
