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26 January, 10:51

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.

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  1. 26 January, 10:55
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    Answer: 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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