Question 1179208
Here's how to solve this problem:

**1. Understand the Problem:**

* We're dealing with a normally distributed population (coffee dispensed) with a known mean and standard deviation.
* We're interested in the probability of a sample mean being greater than a certain value.
* This requires using the Central Limit Theorem and z-scores.

**2. Define the Parameters:**

* Population mean (μ): 12 ounces
* Population standard deviation (σ): 0.2 ounces
* Sample size (n): 9
* Sample mean (x̄): 12.1 ounces

**3. Calculate the Standard Error of the Mean:**

* The standard error (SE) is the standard deviation of the sampling distribution of the mean.
* SE = σ / √n = 0.2 / √9 = 0.2 / 3 ≈ 0.0667 ounces

**4. Calculate the Z-Score:**

* The z-score tells us how many standard errors the sample mean is away from the population mean.
* z = (x̄ - μ) / SE = (12.1 - 12) / 0.0667 = 0.1 / 0.0667 ≈ 1.5

**5. Find the Probability:**

* We want to find P(x̄ > 12.1), which is the same as P(z > 1.5).
* Using a z-table or calculator:
    * P(z < 1.5) ≈ 0.9332
    * P(z > 1.5) = 1 - P(z < 1.5) = 1 - 0.9332 ≈ 0.0668

**Answer:**

The probability that the mean of the sample will be greater than 12.1 ounces is approximately 0.0668.