The annual per capita utilization of oranges (in pounds) in the United States can be approximated by a normal distribution with a mean of 17.1 and a standard deviation of 4                                                        a) What annual per capita utilization of apples represents the 10th percentile?
b) What annual per capita utilization of apples represents the third quartile?  


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.

The annual per capita utilization of oranges (in pounds) in the United States can be approximated by a normal distribution with a mean of 17.1 and a standard deviation of 4. a) What annual per capita utilization of oranges represents the 10th percentile? b) What annual per capita utilization of oranges represents the third quartile?

Learn how to calculate the 10th and 75th percentiles of annual per capita orange utilization using a normal distribution with a mean of 17.1 pounds and a standard deviation of 4 pounds. This problem involves understanding z-scores and applying the normal distribution formula, ideal for high school students studying probability and statistics.

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