Let X be normally distributed with mean -3.8 and variance 12.96. Find **a** (correct up to at least 3 decimal places) such that 

**P(X > 2a) = P( X < a)**.

Given that \( X \) is normally distributed with mean \( \mu = -3.8 \) and variance \( \sigma^2 = 12.96 \), we need to find the value of \( a \) such that:

\[
P(X > 2a) = P(X < a)
\]

### Step 1: Standardize the normal variable
Let \( Z \) be the standard normal variable, where:

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

Given \( \mu = -3.8 \) and \( \sigma = \sqrt{12.96} = 3.6 \), the standardization for the two probabilities is:

\[
P\left(\frac{X - \mu}{\sigma} > \frac{2a - \mu}{\sigma}\right) = P\left(\frac{X - \mu}{\sigma} < \frac{a - \mu}{\sigma}\right)
\]

This simplifies to:

\[
P\left(Z > \frac{2a + 3.8}{3.6}\right) = P\left(Z < \frac{a + 3.8}{3.6}\right)
\]

### Step 2: Use the symmetry property of the normal distribution
The standard normal distribution is symmetric about zero. The equality \( P(Z > z_1) = P(Z < z_2) \) holds when \( z_2 = -z_1 \). Hence:

\[
\frac{2a + 3.8}{3.6} = -\frac{a + 3.8}{3.6}
\]

### Step 3: Solve for \( a \)
Now, equate and solve:

\[
2a + 3.8 = -a - 3.8
\]

Add \( a \) to both sides:

\[
3a + 3.8 = -3.8
\]

Subtract 3.8 from both sides:

\[
3a = -7.6
\]

Finally, divide by 3:

\[
a = \frac{-7.6}{3} = -2.5333\ldots
\]

### Final Answer:
Thus, the value of \( a \) that satisfies the condition is \( \boxed{-2.533} \).

Would you like more details or have any questions? Here are 5 related questions you might find interesting:

1. What is the probability \( P(X > a) \) for the calculated value of \( a \)?
2. How would the solution change if the variance were different?
3. What is the significance of symmetry in normal distribution problems?
4. How do you interpret the z-scores obtained during standardization?
5. How would you compute \( P(X > b) \) for any value \( b \) using the given mean and variance?

**Tip:** When working with normal distributions, it's helpful to familiarize yourself with standard normal tables or use software to compute probabilities directly from z-scores.

Discover how to find the value of 'a' in a normal distribution problem where X is normally distributed with mean -3.8 and variance 12.96. Solve for a such that P(X > 2a) = P(X < a). Learn about normal distribution standardization and the symmetry property of normal distributions.

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