A soft drink machine outputs a mean of 28 ounces per cup. The machine's output is normally distributed with a standard deviation of 2 ounces. What is the probability of filling a cup between 30 and 31 ounces? Round your answer to four decimal places.

Step-by-step answer:

Normally distributed = > get ready the normal probability table, or equivalent.

mu = mean = 28 oz

sigma = standard deviation = 2 oz

Need to find P (30 to 31), the probability of filling between 30 and 31 oz.

Solution:

With normal probabilities,

P (30 to 31) = P (X<31) - P (X<30), i. e. the difference of the right tails for X=30 and X=31.

Calculate the Z scores

Z (X) = (X-mu) / sigma

Z (31) = (31-28) / 2 = 1.5

P (X<31) = P (Z<1.5) = 0.9331928

Z (30) = (30-28) / 2 = 1.0

P (X<30) = P (Z<1.0) = 0.8413447

Therefore,

probability of filling between 30 and 31 oz

= P (X<31) - P (X<30)

= 0.9331928 - 0.8413447

= 0.09184805

= 0.0918 (to 4 decimal places)