Questions: The incomes in a certain large population of college teachers have a Normal distribution, with mean 75,000 and standard deviation 10,000. Sixteen teachers are selected at random from this population to serve on a committee. What is the probability that their average salary is more than 77,500?

Solution Steps

We are given that the incomes of college teachers follow a Normal distribution with a mean (\( \mu \)) of \$75,000 and a standard deviation (\( \sigma \)) of \$10,000. We need to find the probability that the average salary of a sample of 16 teachers exceeds \$77,500.

To find the probability, we first calculate the Z-score for the sample mean of \$77,500. The formula for the Z-score is given by:

\[ Z = \frac{X - \mu}{\sigma / \sqrt{n}} \]

Where:

• \( X = 77500 \)
• \( \mu = 75000 \)
• \( \sigma = 10000 \)
• \( n = 16 \)

Substituting the values, we have:

\[ Z = \frac{77500 - 75000}{10000 / \sqrt{16}} = \frac{2500}{2500} = 1.0 \]

Next, we need to find the probability that the sample mean is greater than \$77,500, which can be expressed as:

\[ P(\bar{X} > 77500) = 1 - P(\bar{X} \leq 77500) = 1 - \Phi(Z) \]

Using the Z-score calculated, we find:

\[ P(\bar{X} > 77500) = 1 - \Phi(1.0) \]

From standard normal distribution tables or calculators, we know:

\[ \Phi(1.0) \approx 0.8413 \]

Thus, the probability becomes:

\[ P(\bar{X} > 77500) = 1 - 0.8413 = 0.1587 \]

Final Answer