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To solve this problem, we need to determine the minimum score required to be in the top 5.2% of a normal distribution with a mean of 330 and a standard deviation of 100.

### Solution Steps:
1. **Identify the Z-score for the 94.8th percentile:**
   Since we want the top 5.2%, we’re looking for the 94.8th percentile of a normal distribution (100% - 5.2% = 94.8%).

2. **Find the Z-score for the 94.8th percentile:**
   Using a Z-table or a calculator, we find that the Z-score for the 94.8th percentile is approximately **1.64**.

3. **Convert the Z-score to a raw score:**
   The formula to convert a Z-score to a raw score is:
   \[
   X = \mu + Z \cdot \sigma
   \]
   where:
   - $\mu = 330$ (mean),
   - $Z = 1.64$,
   - $\sigma = 100$ (standard deviation).

   Substituting the values:
   \[
   X = 330 + (1.64 \times 100) = 330 + 164 = 494
   \]

4. **Answer:**
   The minimum score needed to receive the award is approximately **494**.

Would you like further details or have any questions?

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### Additional Related Questions:
1. How do you find the Z-score for other percentiles?
2. Why does the 94.8th percentile correspond to the top 5.2%?
3. How would the answer change if the standard deviation was smaller or larger?
4. Can this process be used for other types of distributions?
5. What is the effect of changing the mean on the required score?

### Tip:
When working with normal distributions, using a Z-table or a calculator that includes cumulative distribution functions can save time in finding exact percentiles.

A mandatory competency test for high school sophomores has a normal distribution with a mean of 330 and a standard deviation of 100. The top 5.2% of students receive $500. What is the minimum score needed to receive this award?

Find out how to calculate the minimum score required to be in the top 5.2% of a normal distribution test with a mean of 330 and standard deviation of 100. Learn key concepts like Z-score calculation and percentile ranking. Suitable for high school students in Grades 10-12.

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