Questions: The lengths of lumber a machine cuts are normally distributed with a mean of 98 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 98.15 inches? (b) A sample of 40 boards is randomly selected. What is the probability that their mean length is greater than 98.15 inches?

Solution Steps

To find the probability that a randomly selected board cut by the machine has a length greater than 98.15 inches, we first calculate the Z-score using the formula:

\[ z = \frac{X - \mu}{\sigma} = \frac{98.15 - 98}{0.4} = 0.375 \]

Next, we determine the probability that a single board is greater than 98.15 inches. This is given by:

\[ P(X > 98.15) = P(Z > 0.375) = \Phi(\infty) - \Phi(0.375) \approx 0.3538 \]

For a sample of 40 boards, we need to calculate the probability that their mean length is greater than 98.15 inches. The Z-score for the sample mean is calculated as follows:

\[ z = \frac{X - \mu}{\sigma / \sqrt{n}} = \frac{98.15 - 98}{0.4 / \sqrt{40}} \approx 2.3717 \]

Then, we find the probability:

\[ P(\bar{X} > 98.15) = P(Z > 2.3717) = \Phi(\infty) - \Phi(2.3717) \approx 0.0089 \]

Final Answer