SOLUTION: The mean salary offered to students who are graduating from Coastal State University this year is $24,215, with a standard deviation of $3678. A random sample of 85 Coastal State s

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Question 1118388: The mean salary offered to students who are graduating from Coastal State University this year is $24,215, with a standard deviation of $3678. A random sample of 85 Coastal State students graduating this year has been selected. What is the probability that the mean salary offer for these 85 students is $24,500 or less?
Found 2 solutions by Theo, stanbon:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
the population mean is 24215
the population standard deviation is 3678.

the sample size is 85.
the standard deviation of the distribution of sample means, called the standard error, from samples whose size is 85 is equal to 3678 / sqrt(85) = 398.9351119.

the z-score for a sample mean of 24500 or less from this population, with a sample size of 85, is z = (24500 - 24215) / 398.9351119 = .7143732313.

the area under the normal distribution curve to the left of this z-score is equal to .7625107243.

that means the probability that you would get a sample of size 85 whose mean is 24500 or less is .7625107243, or approximately 76.25%.

this can be seen visually in the following graph.

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Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The mean salary offered to students who are graduating from Coastal State University this year is $24,215, with a standard deviation of $3678. A random sample of 85 Coastal State students graduating this year has been selected. What is the probability that the mean salary offer for these 85 students is $24,500 or less?
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z(24500) = (24500-24215)/[3678/sqrt(85)] = 0.7144
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P(x-bar <= 24500) = P(z <= 0.7144) = normalcdf(-100,0.7144) = 0.7625
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Cheers,
Stan H.
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