The weights of ice cream cartons are normally distributed with a mean weight of 11 ounces and a standard deviation of 0.6 ounce. (a) What is the probability that a randomly selected carton has a weight greater than 11.23 ounces? (b) A sample of 25 cartons is randomly selected. What is the probability that their mean weight is greater than 11.23 ounces? Part 2 of 2 (a) The probability is 0.3500 . (Round to four decimal places as needed.) (b) The probability is \( \square \). (Round to four decimal places as needed.)

To tackle part (b), you'll first need to understand how the Central Limit Theorem applies here. When you have a sample size greater than 30, the sampling distribution of the sample mean will approximate a normal distribution. However, since we have a sample size of 25, we'll still use the normal distribution properties. The mean of the sampling distribution will equal the population mean (11 ounces), and the standard deviation (standard error) will be calculated as \( \sigma/\sqrt{n} \), which is \( 0.6/\sqrt{25} = 0.12 \). Now, to find the probability that the mean weight of the sample is greater than 11.23 ounces, we can calculate the Z-score: \[ Z = \frac{(X - \mu)}{\sigma/\sqrt{n}} = \frac{(11.23 - 11)}{0.12} = \frac{0.23}{0.12} \approx 1.9167 \] Using standard normal distribution tables or a calculator, you'll find: \[ P(Z > 1.9167) \approx 0.0276 \] So, rounding to four decimal places, the probability that the mean weight of the 25 sampled cartons is greater than 11.23 ounces is approximately 0.0276. The final answer for part (b) would be: The probability is 0.0276.