Question

business weekly conducted a survey of graduates from 30 top mba programs. on the basis of the survey, the mean annual salary for graduates 10 years after graduation is 161000 dollars and the standard deviation is 31000 dollars. suppose you take a simple random sample of 75 graduates. assume that the population of all salaries for mba graduates is normally distributed. enter your answers as numbers accurate to 4 decimal places. find the probability that a single randomly selected salary is at most 164000 dollars. answer = find the probability that a sample of size n = 75 is randomly selected with a mean that is at most 164000 dollars. answer =

Explanation:

Step1: Calculate the z - score for a single salary

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 164000\), \(\mu=161000\), and \(\sigma = 31000\).

$$z=\frac{164000 - 161000}{31000}=\frac{3000}{31000}\approx0.0968$$

Step2: Find the probability for a single salary

Using the standard normal distribution table or a calculator with a normal - distribution function (e.g., \(\text{normcdf}\) on a TI - 84: \(\text{normcdf}(-\infty,0.0968)\)), the probability \(P(X\leq164000)\) is \(P(Z\leq0.0968)\).

$$P(Z\leq0.0968)\approx0.5383$$

Step3: Calculate the standard error for the sample mean

The formula for the standard error is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\), where \(n = 75\) and \(\sigma=31000\).

$$\sigma_{\bar{x}}=\frac{31000}{\sqrt{75}}\approx\frac{31000}{8.6603}\approx3579.5373$$

Step4: Calculate the z - score for the sample mean

The formula for the z - score of the sample mean is \(z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}\), where \(\bar{x}=164000\), \(\mu = 161000\), and \(\sigma_{\bar{x}}\approx3579.5373\)

$$z=\frac{164000 - 161000}{3579.5373}=\frac{3000}{3579.5373}\approx0.8381$$

Step5: Find the probability for the sample mean

Using the standard normal distribution table or a calculator with a normal - distribution function (e.g., \(\text{normcdf}\) on a TI - 84: \(\text{normcdf}(-\infty, 0.8381)\)), the probability \(P(\bar{X}\leq164000)\) is \(P(Z\leq0.8381)\)

$$P(Z\leq0.8381)\approx0.7980$$

Answer:

The probability that a single randomly selected salary is at most \(164000\) dollars is \(0.5383\).
The probability that a sample of size \(n = 75\) is randomly selected with a mean that is at most \(164000\) dollars is \(0.7980\).