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29 November, 07:06

The mean annual salary for intermediate level executives is about $74000 per year with a standard deviation of $2500. A random sample of 50 intermediate level executives is selected. What is the probability that the mean annual salary of the sample is between $71000 and $73500? A 0.079B. 0.500C. 0.487D. 0.306

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  1. 29 November, 10:40
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    Answer: D. 0.306

    Step-by-step explanation:

    Assuming a normal distribution for the annual salary for intermediate level executives, the formula for normal distribution is expressed as

    z = (x - u) / s

    Where

    x = annual salary for intermediate level executives

    u = mean annual salary

    s = standard deviation

    From the information given,

    u = $74000

    s = $2500

    We want to find the probability that the mean annual salary of the sample is between $71000 and $73500. It is expressed as

    P (71000 lesser than or equal to x lesser than or equal to 73500)

    For x = 71000,

    z = (71000 - 74000) / 2500 = - 1.2

    Looking at the normal distribution table, the probability corresponding to the z score is 0.1151

    For x = 73500,

    z = (73500 - 74000) / 2500 = - 0.2

    Looking at the normal distribution table, the probability corresponding to the z score is 0.4207

    P (71000 lesser than or equal to x lesser than or equal to 73500) is

    0.4207 - 0.1151 = 0.306
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