Answer:
Probability that a selected cup will be overfilled is 0.15866 .
Step-by-step explanation:
We are given that the amount of hot chocolate dispensed by a hot chocolate machine is normally distributed with a mean of 16.0 oz. and a standard deviation of 2 oz.
Let X = Amount of hot chocolate dispensed by a hot chocolate machine, so X ~ N([tex]\mu = 16 oz, \sigma^{2} = 2^{2} oz[/tex])
The standard normal z distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
We are also given that the cups hold 18.0 oz.
Now, probability that a selected cup will be overfilled = P(X > 18.0 oz)
P(X > 18.0) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{18-16}{2}[/tex] ) = P(Z > 1) = 1 - P(Z <= 1)
= 1 - 0.84134= 0.15866
Therefore, the probability that a selected cup will be overfilled is 0.15866.
