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13 November, 15:53

The average hourly wage of workers at a fast food restaurant is $6.50/hr. assume the wages are normally distributed with a standard deviation of $0.45. if a worker at this fast food restaurant is selected at random, what is the probability that the worker earns more than $6.75

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  1. 13 November, 17:26
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    Salaries are normally distributed

    Average salary per hour: $6.50 ⇒ μ=$6.50

    σ=$0.45 (Standard deviation)

    Probability that the selected worker will earn $6.75:

    P (x>6.75)

    Now you must standardize the normal random variables:

    If x~N (μσ²) = > Z = (x-μ) / σ

    Z = (6.75-6.50) / 0.45

    Z=0.556 = > P (Z>0.556) = ? (To find this probability, you must use the tables for Standard normal distribution).

    Then:

    P (Z>0.556) = 0.2877

    The probability of selecting a worker who earns more than $ 6.75 is 28.7%
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