A machine fills containers with a particular product. The standard deviation of filling weights is known from past data to be .6 ounce. If only 2% of the containers hold less than 18 ounces, what is the mean filling weight for the machine? That is, what must μ equal? Assume the filling weights have a normal distribution

In this problem, you are asked to determine the mean filling weight, denoted as \( \mu \), given that:

- The filling weights follow a normal distribution.
- The standard deviation (\( \sigma \)) is 0.6 ounces.
- 2% of the containers hold less than 18 ounces.

We can solve this using the properties of the normal distribution. Let's follow the steps:

### Step 1: Understanding the problem
Given the information:
- \( \sigma = 0.6 \) ounces (standard deviation).
- 2% of the containers hold less than 18 ounces.

In a normal distribution, the percentage of data below a certain value corresponds to the cumulative probability up to that value. The 2% threshold corresponds to a z-score, which we can find from the standard normal distribution table.

### Step 2: Find the z-score for 2%
The cumulative probability of 2% (or 0.02) corresponds to a z-score. Looking up this value in a z-score table (or using a calculator), we find that:

\[
z \approx -2.05
\]

This means that 18 ounces is 2.05 standard deviations below the mean.

### Step 3: Use the z-score formula
The z-score formula is:

\[
z = \frac{x - \mu}{\sigma}
\]

Where:
- \( z \) is the z-score,
- \( x = 18 \) is the observed value (the threshold of 18 ounces),
- \( \mu \) is the mean (what we are solving for),
- \( \sigma = 0.6 \) is the standard deviation.

Substitute the known values:

\[
-2.05 = \frac{18 - \mu}{0.6}
\]

### Step 4: Solve for \( \mu \)

\[
18 - \mu = -2.05 \times 0.6
\]

\[
18 - \mu = -1.23
\]

\[
\mu = 18 + 1.23
\]

\[
\mu = 19.23
\]

### Conclusion
The mean filling weight for the machine is approximately **19.23 ounces**.

Would you like more details or have any questions?

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Here are 5 related questions that could expand this topic:
1. How would the mean change if the standard deviation was larger or smaller?
2. What percentage of containers would hold more than 20 ounces?
3. How can you calculate the probability of a container holding between two specific weights?
4. What would the z-score be for a container that weighs 18.5 ounces?
5. How does changing the percentage (e.g., 5% instead of 2%) affect the mean?

**Tip**: When working with the normal distribution, always check the cumulative probability values to correctly find the z-score!

A machine fills containers with a particular product. The standard deviation of filling weights is known from past data to be .6 ounce. If only 2% of the containers hold less than 18 ounces, what is the mean filling weight for the machine? That is, what must μ equal? Assume the filling weights have a normal distribution.

Solve a normal distribution problem where 2% of containers hold less than 18 ounces, and the standard deviation is 0.6 ounces. Learn how to use the z-score formula to calculate the mean filling weight, approximately 19.23 ounces, for the machine. Suitable for students in Grades 10-12.

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