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The problem asks for the bag weight that corresponds to the 84th percentile of a normal distribution, with a mean \( \mu = 9.12 \) ounces and a standard deviation \( \sigma = 0.05 \) ounces. 

### Using the Empirical Rule:
The empirical rule tells us that:
- 68% of the data falls within 1 standard deviation of the mean,
- 95% falls within 2 standard deviations, and
- 99.7% falls within 3 standard deviations.

The 84th percentile is approximately one standard deviation above the mean, as 68% of the data is within one standard deviation, and half of the remaining 32% is distributed evenly between below and above. So:

\[
\text{Percentile of } 84\% \approx 1 \text{ standard deviation above the mean}.
\]

### Calculation:
The weight corresponding to the 84th percentile is:

\[
\text{Weight} = \mu + \sigma = 9.12 + 0.05 = 9.17 \text{ ounces}.
\]

Therefore, the bag weight at the 84th percentile is **9.17 ounces**.

Let me know if you'd like any further details or have any questions!

Here are five related questions:
1. How would you find the 97.5th percentile using the empirical rule?
2. What would the bag weight be at the 50th percentile (the median)?
3. How would the weight at the 16th percentile compare to the 84th percentile?
4. How does the standard deviation affect the distribution of weights?
5. How would you calculate the exact percentile if you didn’t use the empirical rule?

**Tip:** The empirical rule is useful for quick estimates, but for exact percentiles, use a Z-table or normal distribution calculator.

What bag weight is at the 84th percentile of the distribution if the weights of 9-ounce bags of a particular brand of potato chips can be modeled by a normal distribution with a mean of 9.12 ounces and standard deviation of 0.05 ounce?

Learn how to use the empirical rule to find the bag weight at the 84th percentile of a normal distribution with a mean of 9.12 ounces and a standard deviation of 0.05 ounces. Perfect for high school students studying statistics.

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