A soft drink machine outputs a mean of 27 ounces per cup. The machine's output is normally distributed with a standard deviation of 3 ounces. What is the probability of filling a cup between 22 and 29 ounces? Round your answer to four decimal places.

P(22 ≤ x ≤ 29) = 0.703

Step-by-step explanation:

Since the machine's output is normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = output of the machine in ounces per cup.

µ = mean output

σ = standard deviation

From the information given,

µ = 27

σ = 3

The probability of filling a cup between 22 and 29 ounces is expressed as

P(22 ≤ x ≤ 29)

For x = 22,

z = (22 - 27)/3 = - 1.67

Looking at the normal distribution table, the probability corresponding to the z score is 0.047

For x = 29,

z = (29 - 27)/3 = 0.67

Looking at the normal distribution table, the probability corresponding to the z score is 0.75

Therefore,

P(22 ≤ x ≤ 29) = 0.75 - 0.047 = 0.703

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