The problem in the image asks for the number of peaches expected to have weights of at least 16 ounces, assuming the peach weights are normally distributed with a mean of 16 ounces and a standard deviation of 0.2 ounces.

Step-by-Step Solution:

1. Set up the problem:

   • Mean ($\mu$) = 16 ounces
   • Standard deviation ($\sigma$) = 0.2 ounces
   • We need to find the number of peaches that weigh at least 16 ounces.

2. Convert the weight to a Z-score: Since 16 ounces is exactly the mean, the Z-score for this value is:

   $Z = \frac{X - \mu}{\sigma} = \frac{16 - 16}{0.2} = 0$

3. Look up the Z-score in the standard normal distribution table: The Z-score of 0 corresponds to a cumulative area of 0.5, which means that 50% of the peaches are expected to weigh less than 16 ounces, and 50% are expected to weigh more.

4. Determine the number of peaches: If there are 1000 peaches, then 50% of 1000 peaches are expected to weigh 16 ounces or more.

   $\text{Number of peaches} = 1000 \times 0.5 = 500$

Conclusion:

About 500 peaches are expected to have weights of at least 16 ounces.

Would you like more details or clarifications? Here are some related questions you might find useful:

1. What if the standard deviation were different? How would that change the answer?
2. How do we calculate the probability for peaches weighing more than a specific value, like 16.5 ounces?
3. What is the Z-score formula and how can it be applied to other similar problems?
4. How does the concept of normal distribution help in quality control in fruit production?
5. Could we calculate how many peaches fall between two weights, like between 15 and 17 ounces?

Tip: The Z-score tells you how many standard deviations a value is from the mean, which is useful in many real-world applications like grading on a curve.