Final answer:
To find the percentage of items weighing at most 11.7 ounces, calculate the z-score and find the cumulative probability from the standard normal distribution. The z-score for 11.7 ounces, with a mean of 8 ounces and standard deviation of 2 ounces, is 1.85.
Step-by-step explanation:
The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces. To determine the percentage of items that will weigh at most 11.7 ounces, we can use the properties of the normal distribution. First, we need to calculate the z-score for 11.7 ounces, which is given by (X - μ) / σ, where X is the value in question, μ is the mean, and σ is the standard deviation.
The calculation would be (11.7 - 8) / 2 = 1.85. We would then use the z-score to find the cumulative probability from the standard normal distribution table or use a calculator with normal distribution functions. The value would represent the percentage of items expected to weigh at the most 11.7 ounces.