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The salary of college professors from more than 25,000 universities and colleges in USA is normally distributed with a mean, $85,900 and its standard deviation, $11,000. If you collect a random sample from 35 professors, find the probability that the sample mean is between $70,000 and $100,000?

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Answer:

The answer is below

Step-by-step explanation:

The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by the formula:

[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mean,\ \sigma=standard\ deviation.\\\\For \ a\ sample\ size(n):\\\\z=\frac{x-\mu}{\sigma/\sqrt{n} }[/tex]

Therefore given that μ = $85900, σ = $11000, n = 35.

For x > 70000:

[tex]z=\frac{x-\mu}{\sigma/\sqrt{n} } =\frac{70000-85900}{11000/\sqrt{35} } =-8.55\\[/tex]

For x < 100000

[tex]z=\frac{x-\mu}{\sigma/\sqrt{n} } =\frac{100000-85900}{11000/\sqrt{35} } =7.58[/tex]

From the normal distribution table, P(70000< x < 100000) = P(-8.55 < z < 7.58) = P(z < 7.58) - P(z < -8.55) = 1 - 0 = 1 = 100%

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