SOLUTION: Hi! Please help with the following question! Thank you so much! A coffee machine dispenses normally distributed amounts of coffee with a mean of 12 ounces and a standard deviat

Question 1179208: Hi! Please help with the following question! Thank you so much!
A coffee machine dispenses normally distributed amounts of coffee with a mean of 12 ounces and a standard deviation of 0.2 ounce. If a sample of 9 cups is selected, find the probability that the mean of the sample will be greater than 12.1 ounces.
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
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.

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