0.4 = Sampling Cuestion 7,5.4.37-T \( 1 \in \) Part 1 of 2 The lengths of lumber a machine cuts are normally distributed with a mean of 87 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 87.11 inches? \( 60.42 \%, 4.83 \) of 8 points (b) A sample of 44 boards is randomly selected. What is the probability that their mean length is greater than 87.11 inches? (a) The probability is \( \square \). (Round to four decimal places as needed.)

To solve part (a), we need to find the probability that a randomly selected board cut by the machine has a length greater than 87.11 inches using the properties of the normal distribution. First, we need to standardize the value using the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] Where: - \( X = 87.11 \) inches (the value we are comparing) - \( \mu = 87 \) inches (the mean) - \( \sigma = 0.4 \) inches (the standard deviation) Plugging in the values: \[ Z = \frac{87.11 - 87}{0.4} = \frac{0.11}{0.4} = 0.275 \] Next, we need to find the probability corresponding to a Z-score of 0.275. Using the standard normal distribution table or calculator, we find: P(Z > 0.275) = 1 - P(Z < 0.275) Looking up the Z-score of 0.275, we find P(Z < 0.275) ≈ 0.6103. Hence, P(Z > 0.275) ≈ 1 - 0.6103 = 0.3897. So, rounding to four decimal places, the probability is **0.3897**. For part (b), we want to determine the probability that the mean length of a sample of 44 boards is greater than 87.11 inches. We will use the Central Limit Theorem here, which states that the sampling distribution of the sample mean will be normally distributed with: Mean \( \mu_{\bar{X}} = \mu = 87 \) inches Standard deviation \( \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{0.4}{\sqrt{44}} \approx 0.0603 \) inches Now, we use the Z-score formula again. \[ Z = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}} = \frac{87.11 - 87}{0.0603} \] Calculating this: \[ Z = \frac{0.11}{0.0603} \approx 1.826 \] Then we can find the probability: P(Z > 1.826) = 1 - P(Z < 1.826) Looking up the Z-score of 1.826, we find P(Z < 1.826) ≈ 0.9651. Thus, P(Z > 1.826) ≈ 1 - 0.9651 = 0.0349. So for part (b), the probability that the mean length of the sample of boards is greater than 87.11 inches is approximately **0.0349** (rounded to four decimal places). In summary: (a) The probability is **0.3897**. (b) The probability is **0.0349**.