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7 September, 14:19

The average starting salary for this year's graduates at a large university (LU) is $50,000 with a standard deviation of $8,000. Furthermore, it is known that the starting salaries are normally distributed. What is the probability that a randomly selected LU graduate will have a starting salary of at least $55,200?

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  1. 7 September, 18:07
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    0.74215

    Step-by-step explanation:

    The Z score formula = z = (x-μ) / σ, where

    x is the raw score,

    μ is the population mean,

    and σ is the population standard deviation.

    In the question, we are given

    x is the raw score = Salary of a randomly selected LU graduate will be a starting salary of at least = 55,200

    μ is the population mean = average starting salary for this year's graduates at a large university (LU) = 50,000

    σ is the population standard deviation = 8,000

    z = (x-μ) / σ,

    z = (55,200 - 50,000) : 8000

    z = 5,200 : 8000

    z = 0.65

    Therefore,

    P (z < 0.65)

    Using the z score table for probability,

    z < 0.65 = 0.74215

    Therefore, the probability that a randomly selected LU graduate will have a starting salary of at least $55,200 in decimal form is 0.74215
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