Questions: The lengths of lumber a machine cuts are normally distributed with a mean of 86 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 86.14 inches? (b) A sample of 40 boards is randomly selected. What is the probability that their mean length is greater than 86.14 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 \( 86.14 \) inches, we first calculate the Z-score using the formula:

\[ z = \frac{X - \mu}{\sigma} = \frac{86.14 - 86}{0.4} = 0.35 \]

Next, we determine the probability corresponding to this Z-score. The probability that a board is longer than \( 86.14 \) inches is given by:

\[ P(X > 86.14) = P(Z > 0.35) = \Phi(\infty) - \Phi(0.35) = 0.3632 \]

For a sample of \( 40 \) boards, we need to find the probability that their mean length is greater than \( 86.14 \) inches. First, we calculate the Z-score for the sample mean:

\[ z = \frac{X - \mu}{\sigma / \sqrt{n}} = \frac{86.14 - 86}{0.4 / \sqrt{40}} \approx 2.2136 \]

Now, we calculate the probability:

\[ P(\bar{X} > 86.14) = P(Z > 2.2136) = \Phi(\infty) - \Phi(2.2136) = 0.0134 \]

Final Answer