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.