Question

business weekly conducted a survey of graduates from 30 top mba programs. on the basis of the survey, assume the mean annual salary for graduates 10 years after graduation is 161000 dollars. assume the standard deviation is 43000 dollars. suppose you take a simple random sample of 88 graduates.

find the probability that a single randomly selected graduate has a salary between 155499.4 and 174293.1 dollars.
\\(p(155499.4 < x < 174293.1) =\\) (enter your answers as numbers accurate to 4 decimal places.)

find the probability that a random sample of size \\(n = 88\\) has a mean value between 155499.4 and 174293.1 dollars.
\\(p(155499.4 < m < 174293.1) =\\) (enter your answers as numbers accurate to 4 decimal places.)

Explanation:

Calculate the standard error of the mean

$$ \sigma_M = \frac{\sigma}{\sqrt{n}} = \frac{43000}{\sqrt{88}} \approx 4583.7915 $$

Compute the z-scores for the sample mean boundaries

$$
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$$

Find the probability using the standard normal distribution

$$
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$$

Answer:

Find the probability that a random sample of size \(n = 88\) has a mean value between 155499.4 and 174293.1 dollars.
\(P(155499.4 < M < 174293.1) =\) <blank>0.8830</blank>