Step 1 :Given that the weights of ice cream cartons are normally distributed with a mean weight of 8 ounces and a standard deviation of 0.4 ounce.
Step 2 :For part (a), we need to calculate the z-score for the weight 8.14 ounces. The z-score is a measure of how many standard deviations an element is from the mean. We can calculate it using the formula: \(z = \frac{X - \mu}{\sigma}\) where X is the value we are interested in, \(\mu\) is the mean and \(\sigma\) is the standard deviation. Substituting the given values, we get \(z = \frac{8.14 - 8}{0.4} = 0.35\).
Step 3 :Once we have the z-score, we can use a z-table or a statistical function to find the probability that a randomly selected carton has a weight greater than 8.14 ounces. The probability is approximately 0.3632.
Step 4 :For part (b), we need to calculate the probability that the mean weight of a sample of 36 cartons is greater than 8.14 ounces. The standard deviation of the sample mean is given by \(\frac{\sigma}{\sqrt{n}}\), where n is the sample size. Substituting the given values, we get \(\frac{0.4}{\sqrt{36}} = 0.0667\).
Step 5 :We can then calculate the z-score for the sample mean using the formula: \(z = \frac{X - \mu}{\sigma}\). Substituting the given values, we get \(z = \frac{8.14 - 8}{0.0667} = 2.1\).
Step 6 :Using a z-table or a statistical function, we find the probability that the mean weight of a sample of 36 cartons is greater than 8.14 ounces. The probability is approximately 0.0179.
Step 7 :Final Answer: (a) The probability that a randomly selected carton has a weight greater than 8.14 ounces is approximately \(\boxed{0.3632}\). (b) The probability that the mean weight of a sample of 36 cartons is greater than 8.14 ounces is approximately \(\boxed{0.0179}\).