Questions: Annual income: The mean annual income for people in a certain city (in thousands of dollars) is 41, with a standard deviation of 35. A pollster draws a sample of 91 people to interview. Part 1 of 5 (a) What is the probability that the sample mean income is less than 37? Round the answer to at least four decimal places. The probability that the sample mean income is less than 37 is . Part 2 of 5 (b) What is the probability that the sample mean income is between 40 and 45? Round the answer to at least four decimal places. The probability that the sample mean income is between 40 and 45 is

Solution Steps

To find the probability that the sample mean income is less than 37, we first calculate the Z-score for \( x = 37 \):

\[ Z_{end} = \frac{37 - \mu}{\sigma / \sqrt{n}} = \frac{37 - 41}{35 / \sqrt{91}} \approx -1.0902 \]

Using the Z-score calculated, we find the probability:

\[ P(X < 37) = \Phi(Z_{end}) - \Phi(-\infty) = \Phi(-1.0902) \approx 0.1378 \]

Thus, the probability that the sample mean income is less than 37 is:

\[ \boxed{0.1378} \]

Next, we calculate the Z-scores for the range \( 40 \) to \( 45 \):

For \( x = 40 \):

\[ Z_{start} = \frac{40 - \mu}{\sigma / \sqrt{n}} = \frac{40 - 41}{35 / \sqrt{91}} \approx -0.2726 \]

For \( x = 45 \):

\[ Z_{end} = \frac{45 - \mu}{\sigma / \sqrt{n}} = \frac{45 - 41}{35 / \sqrt{91}} \approx 1.0902 \]

Now, we find the probability that the sample mean income is between 40 and 45:

\[ P(40 < X < 45) = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(1.0902) - \Phi(-0.2726) \approx 0.4696 \]

Thus, the probability that the sample mean income is between 40 and 45 is:

\[ \boxed{0.4696} \]

Final Answer