The continuous random variable 𝑋𝑋 has probability density function 𝑓𝑓 given by 𝑓𝑓(𝑥𝑥) = � |𝑎𝑎𝑎𝑎| −3 ⩽ 𝑥𝑥 ⩽ 3 0 otherwise where 𝑎𝑎 is a constant. The value of 𝑎𝑎 is:

To find the value of \( a \), we use the fact that the total probability for any probability density function (PDF) must be 1, i.e.,

\[
\int_{-\infty}^{\infty} f(x) \, dx = 1
\]

Given the probability density function \( f(x) \):

\[
f(x) = 
\begin{cases} 
a & -3 \leq x \leq 3 \\
0 & \text{otherwise} 
\end{cases}
\]

We focus on the range where \( f(x) = a \) (i.e., between -3 and 3), and integrate over this range:

\[
\int_{-3}^{3} a \, dx = 1
\]

Since \( a \) is constant, we can factor it out of the integral:

\[
a \int_{-3}^{3} 1 \, dx = 1
\]

The integral of 1 from -3 to 3 is simply the length of the interval:

\[
a \cdot (3 - (-3)) = 1
\]

\[
a \cdot 6 = 1
\]

Solving for \( a \):

\[
a = \frac{1}{6}
\]

Thus, the value of \( a \) is \( \frac{1}{6} \).

---

Would you like further details or have any questions?

Here are 5 related questions you could explore:
1. How would you compute the expected value \( E(X) \) for this PDF?
2. What is the variance \( \text{Var}(X) \) for this distribution?
3. How would you calculate the median of the distribution \( X \)?
4. How does the value of \( a \) affect the shape of the distribution?
5. Can you extend this to find the cumulative distribution function (CDF)?

**Tip**: When working with probability density functions, always remember that the integral of the PDF over the entire possible range of \( X \) must equal 1!

The continuous random variable X has probability density function f given by f(x) = a | a a | -3 ≤ x ≤ 3 0 otherwise where a is a constant. The value of a is:

Learn how to find the constant 'a' for the probability density function f(x) defined over the interval -3 to 3. This problem involves the normalization of the PDF using integration. Suitable for undergraduate students studying statistics.

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