Questions: The lengths of lumber a machine cuts are normally distributed with a mean of 101 inches and a standard deviation of 0.4 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 45 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.)

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 using the formula:

\[ z = \frac{X - \mu}{\sigma} = \frac{101.15 - 101}{0.4} = 0.375 \]

Thus, the Z-score for \( 101.15 \) inches is \( z = 0.375 \).

Next, we determine the probability that a board is longer than \( 101.15 \) inches. This is given by:

\[ P(X > 101.15) = P(Z > 0.375) = \Phi(\infty) - \Phi(0.375) \]

Using the standard normal distribution, we find:

\[ P(X > 101.15) = 0.3538 \]

Now, we calculate the probability that the mean length of a sample of \( 45 \) boards is greater than \( 101.15 \) inches. The Z-score for the sample mean is calculated as follows:

\[ z = \frac{X - \mu}{\sigma / \sqrt{n}} = \frac{101.15 - 101}{0.4 / \sqrt{45}} \approx 2.5156 \]

Then, we find the probability:

\[ P(\bar{X} > 101.15) = P(Z > 2.5156) = \Phi(\infty) - \Phi(2.5156) \]

This results in:

\[ P(\bar{X} > 101.15) = 0.0059 \]

Final Answer