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The problem involves determining a test score that represents the 14th percentile of a normally distributed set of scores. To solve this, we can use the following information:

- **Mean** (\(\mu\)): 69.2
- **Standard deviation** (\(\sigma\)): 6.5
- **Percentile**: 14th percentile (which corresponds to a z-score for \(P = 0.14\))

### Step 1: Find the Z-score corresponding to the 14th percentile
We use standard normal distribution tables or a calculator to find the z-score for \(P = 0.14\). The z-score corresponding to the 14th percentile is approximately \(-1.08\).

### Step 2: Use the Z-score formula to find the test score
The formula to convert a z-score to a raw score \(X\) in a normal distribution is:

\[
X = \mu + Z \cdot \sigma
\]

Substitute the values:

\[
X = 69.2 + (-1.08) \cdot 6.5
\]

\[
X = 69.2 - 7.02
\]

\[
X \approx 62.2
\]

Thus, the test score that determines the boundary of the F grades is approximately **62.2**.

Would you like more details or further explanation?

Here are 5 related questions:
1. How do you calculate percentiles in a normal distribution?
2. What does a z-score represent in a standard normal distribution?
3. How would you determine the score for the 85th percentile?
4. What happens to the boundary score if the standard deviation increases?
5. How is the 50th percentile related to the mean in a normal distribution?

**Tip:** Remember that in a normal distribution, about 68% of the data falls within 1 standard deviation from the mean, 95% within 2 standard deviations, and 99.7% within 3 standard deviations.

A set of test scores are normally distributed with a mean of 69.2 and a standard deviation of 6.5 points. If the scores in the fourteenth percentile and below will receive an F, what test score will determine the boundary of the F grades? Round your answer to the nearest tenth.

Learn how to calculate the boundary score for an F grade in a normally distributed set of test scores, where the 14th percentile and below receive an F. This problem involves using the Z-score formula and understanding normal distribution properties.

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