Questions: The lengths of lumber a machine cuts are normally distributed with a mean of 101 inches and a standard deviation of 0.5 inch. (a) What is the probability that a randomly selected board cut by the machine has a length greater than 101.15 inches? (b) A sample of 41 boards is randomly selected. What is the probability that their mean length is greater than 101.15 inches? (a) The probability is . (Round to four decimal places as needed.) (b) The probability is (Round to four decimal places as needed.)

Solution Steps

To find the probability that a randomly selected board cut by the machine has a length greater than \( 101.15 \) inches, we first calculate the Z-score for \( 101.15 \):

\[ Z = \frac{X - \mu}{\sigma} = \frac{101.15 - 101}{0.5} = 0.3 \]

Next, we find the probability using the cumulative distribution function \( \Phi \):

\[ P(X > 101.15) = 1 - \Phi(Z) = 1 - \Phi(0.3) \]

From the output, we have:

\[ P(X > 101.15) = 0.6179 \]

For a sample of \( n = 41 \) boards, we need to find the probability that their mean length is greater than \( 101.15 \) inches. The standard error of the mean is calculated as:

\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{0.5}{\sqrt{41}} \approx 0.0781 \]

Now, we calculate the Z-score for the sample mean:

\[ Z = \frac{\bar{X} - \mu}{SE} = \frac{101.15 - 101}{0.0781} \approx 1.9209 \]

We then find the probability:

\[ P(\bar{X} > 101.15) = 1 - \Phi(Z) = 1 - \Phi(1.9209) \]

From the output, we have:

\[ P(\bar{X} > 101.15) = 0.9726 \]

Final Answer