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To determine the minimum score needed to receive the $500 award, which is given to the top 4.5% of students, we can use the properties of the normal distribution. Here are the steps:

1. **Identify the z-score that corresponds to the top 4.5%**:
   - The top 4.5% in a standard normal distribution corresponds to the 95.5th percentile (since 100% - 4.5% = 95.5%).
   - Using a z-table or calculator, the z-score for the 95.5th percentile is approximately $ z = 1.645 $.

2. **Convert the z-score to the actual score**:
   - Use the formula for the z-score: 
     \[
     z = \frac{X - \mu}{\sigma}
     \]
     where $ X $ is the score, $ \mu $ is the mean, and $ \sigma $ is the standard deviation.
   - Rearrange to solve for $ X $:
     \[
     X = \mu + z \cdot \sigma
     \]

Given:
- Mean ($ \mu $) = 350
- Standard deviation ($ \sigma $) = 80
- $ z $ = 1.645

Calculate $ X $:
\[
X = 350 + 1.645 \cdot 80 \approx 350 + 131.6 = 481.6
\]

Rounding to the nearest whole number, the minimum score needed to receive the award is approximately **482**.

Would you like more details or have any questions?

### Related Questions:
1. What is the z-score corresponding to the top 10% of a normally distributed dataset?
2. How would you calculate the score corresponding to the bottom 5% of students in this distribution?
3. If the mean were 400 and the standard deviation 70, what would be the new minimum score for the top 4.5%?
4. How does the standard deviation affect the spread of scores in a normal distribution?
5. Can you explain how to use a z-table to find percentiles in a normal distribution?

### Tip:
When dealing with normally distributed data, using z-scores and percentiles helps to translate between raw scores and their relative positions within the dataset.

A mandatory competency test for high school sophomores has a normal distribution with a mean of 350 and a standard deviation of 80. The top 4.5% of students receive $500. What is the minimum score needed to receive this award?

Find out the minimum score needed to be in the top 4.5% of students on a normal distribution competency test. Given a mean of 350 and a standard deviation of 80, learn how to calculate this using z-scores and percentiles.

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