The problem in the image involves a normal distribution with the following given values:

• Mean ($\mu$) = 9.12 ounces
• Standard deviation ($\sigma$) = 0.05 ounces

We need to use the empirical rule (68-95-99.7 rule) to answer two questions.

(a) What percent of bags weigh less than 9.07 ounces?

To solve this:

1. First, find the z-score for 9.07 ounces: $z = \frac{x - \mu}{\sigma} = \frac{9.07 - 9.12}{0.05}$

2. Then, we will use the z-score to determine the percentage.

(b) What proportion of bags weigh between 8.97 and 9.17 ounces?

For this:

1. Calculate the z-scores for 8.97 ounces and 9.17 ounces: $z_1 = \frac{8.97 - 9.12}{0.05}$ $z_2 = \frac{9.17 - 9.12}{0.05}$

2. Use the z-scores to find the proportion of bags within this range.

Let me calculate these values.Here are the calculated z-scores:

• For part (a), the z-score for 9.07 ounces is approximately $-1.00$.
• For part (b), the z-scores are:
  • For 8.97 ounces: $-3.00$
  • For 9.17 ounces: $1.00$

Now, I'll use these z-scores to find the respective percentages and proportions using the standard normal distribution. Let's proceed.Here are the results:

• (a) Approximately 15.87% of the bags weigh less than 9.07 ounces.
• (b) The proportion of bags that weigh between 8.97 and 9.17 ounces is approximately 0.840.

Would you like any additional details or clarifications?

Here are 5 related questions for further practice:

1. What percentage of bags weigh more than 9.17 ounces?
2. What is the probability of a bag weighing between 9.07 and 9.22 ounces?
3. What percentage of bags weigh less than 9.00 ounces?
4. What is the z-score for a bag weighing exactly 9.12 ounces?
5. If the standard deviation changed to 0.10 ounces, how would that affect the proportion between 8.97 and 9.17 ounces?

Tip: The z-score tells you how many standard deviations a value is from the mean. This is essential for converting between raw scores and probabilities in a normal distribution.