To solve this problem, we'll use properties of the normal distribution, specifically the z-scores corresponding to the percentiles in question.

Given:

• Mean $\mu = 17.1$ pounds
• Standard deviation $\sigma = 4$ pounds

(a) 10th Percentile Calculation:

The 10th percentile means that we are looking for the value $x$ such that 10% of the data falls below this point. We need to find the z-score corresponding to the 10th percentile, then use the formula for a normal distribution:

$x = \mu + z \sigma$

From z-tables or a calculator, the z-score for the 10th percentile is approximately $z = -1.28$.

Now, we can substitute the values into the formula:

$x = 17.1 + (-1.28)(4) = 17.1 - 5.12 = 11.98$

So, the annual per capita utilization of oranges that represents the 10th percentile is approximately 11.98 pounds.

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(b) Third Quartile (75th Percentile) Calculation:

For the third quartile, we need the z-score corresponding to the 75th percentile. The z-score for the 75th percentile is approximately $z = 0.675$.

Using the same formula:

$x = \mu + z \sigma$

Substitute the values:

$x = 17.1 + (0.675)(4) = 17.1 + 2.7 = 19.8$

So, the annual per capita utilization of oranges that represents the third quartile is approximately 19.8 pounds.

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Let me know if you'd like more details on any step! Below are some follow-up questions related to this problem:

1. How would the calculation change if the mean utilization was 20 pounds instead of 17.1?
2. What does it mean if a person consumes oranges above the third quartile?
3. How do z-scores relate to standard deviations in normal distributions?
4. What would be the 90th percentile utilization of oranges?
5. How would increasing the standard deviation affect the percentile calculations?

Tip: When working with normal distributions, remember that z-scores give you a way to standardize values so you can compare them across different normal distributions.