Final answer:
The percentage of steel beams that will be rejected based on a normal distribution with a mean of 8 feet and a standard deviation of 4 inches, with rejection criteria of being more than 10 inches from the mean, is 1.24%.
Step-by-step explanation:
The question deals with the normal distribution of the lengths of steel beams. With a mean of 8 feet and a standard deviation of 4 inches, we need to calculate the percentage of beams that will be rejected if they are more than 10 inches different from the mean. To do this, we first convert the rejection criteria from inches to feet (10 inches = 10/12 feet = 5/6 feet). Beams longer than 8 + 5/6 feet or shorter than 8 - 5/6 feet will be rejected. Then we convert the standard deviation to feet (4 inches = 4/12 feet = 1/3 feet).
Next, we calculate the z-scores for the rejection points. For the upper limit (8 + 5/6 = 8.8333 feet), the z-score is (8.8333 - 8) / (1/3) = 2.5. For the lower limit (8 - 5/6 = 7.1667 feet), the z-score is (7.1667 - 8) / (1/3) = -2.5. Looking these values up in the standard normal distribution table, or using a calculator, gives us the tail probabilities for each end. The total percentage of rejection is the sum of the two tail probabilities, which need to be converted from the z-score probabilities to the actual percentages.
For example, a z-score of 2.5 gives a tail probability of approximately 0.9938, and by symmetry, the same value will apply to a z-score of -2.5. Hence, the acceptance area is 0.9938 + 0.9938 = 1.9876, and the rejection area is 1 - 1.9876 = 0.0124, or 1.24%. Therefore, 1.24% of the beams will be rejected due to quality requirements.