Zima, a clear malt beverage from the 90s produced by the Coors Brewing Company, came in 12-ounce bottles. In the manufacturing process, the actual amount of Zima that made it into a bottle had an expected value of 12 ounces with a standard deviation of 0.1 ounces.
What is the probability that the sample mean of 35 randomly chosen bottles of Zima from the production line is greater than 11.99?
Note: Round to the nearest thousandth.
To find the probability that the sample mean of 35 randomly chosen bottles of Zima is greater than 11.99, we can use the Central Limit Theorem (CLT) and the concept of standard error.
Step 1: Calculate the standard error of the sample mean.
The standard error (SE) is the standard deviation of the population (0.1 ounces) divided by the square root of the sample size (35). So, SE = 0.1 / √35.
Step 2: Calculate the z-score.
The z-score is calculated by subtracting the sample mean (11.99) from the population mean (12) and dividing it by the standard error. So, z = (11.99 - 12) / SE.
Step 3: Determine the probability using the z-table.
Once we have the z-score, we can look it up in the z-table to find the probability associated with that value. Specifically, we want to find the probability of getting a z-score greater than the calculated z-value.
Note: The z-table provides the area under the standard normal distribution curve, so we may need to subtract the obtained probability from 1 to get the desired probability.
Let's calculate the probability step by step:
Step 1: Calculate the standard error:
SE = 0.1 / √35 ≈ 0.016944
Step 2: Calculate the z-score:
z = (11.99 - 12) / 0.016944 ≈ -2.960
Step 3: Lookup probability from the z-table:
Looking up -2.96 in the z-table gives us a cumulative probability of 0.0015.
Step 4: Calculate the probability of the sample mean being greater than 11.99:
Prob(mean > 11.99) = 1 - 0.0015 ≈ 0.9985
Therefore, the probability that the sample mean of 35 randomly chosen bottles of Zima from the production line is greater than 11.99 is approximately 0.9985 (or 99.85%).