Given \( X \sim N(13, 6^2) \), where \( \mu = 13 \) is the mean and \( \sigma^2 = 6^2 \) is the variance (with \( \sigma = 6 \) as the standard deviation), we are tasked with finding \( P(X > 11.5) \).

### Step 1: Standardize the Normal Distribution
To compute this, we convert \( X \) into the standard normal variable \( Z \) using the formula:

\[
Z = \frac{X - \mu}{\sigma}
\]

For \( X = 11.5 \):

\[
Z = \frac{11.5 - 13}{6} = \frac{-1.5}{6} = -0.25
\]

### Step 2: Find the Probability
We now want to find \( P(X > 11.5) = P(Z > -0.25) \).

Using the symmetry of the normal distribution and the standard normal tables, we know:

\[
P(Z > -0.25) = 1 - P(Z \leq -0.25)
\]

From standard normal tables or a calculator, \( P(Z \leq -0.25) \approx 0.4013 \).

Thus,

\[
P(X > 11.5) = 1 - 0.4013 = 0.5987
\]

### Final Answer:
\[
P(X > 11.5) \approx 0.5987
\]

Would you like further clarification on any part of this process or more details?

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Here are some related questions:
1. How do we interpret the standard normal table values?
2. What is the general process for finding probabilities using the normal distribution?
3. How does variance affect the shape of the normal distribution?
4. How would the probability change if \( \sigma \) were smaller or larger?
5. What is the relationship between the cumulative distribution function (CDF) and probability?

**Tip:** When working with normal distributions, always start by converting to the standard normal distribution to utilize tables or software efficiently.

Learn how to calculate the probability P(X > 11.5) for a normal distribution where X ~ N(13, 6^2). This solution uses the standard normal distribution and covers key concepts like Z-scores and the cumulative distribution function (CDF). Ideal for advanced high school or college students.

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