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.

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)

Step-by-step explanation:

question is incomplete.

a = 11 : 4

b = 38

c = 1 for every 6