The random variable $X$ is defined by its probability density function (PDF) $f(x)$ as follows: $f(x) = \begin{cases} 0, & x < -\frac{\pi}{2} \\ a \cdot \cos(x), & -\frac{\pi}{2} \le x < \frac{\pi}{2} \\ 0, & x \ge \frac{\pi}{2} \end{cases}$ The problem asks to find the value of the parameter $a$ and the cumulative distribution function (CDF) $F(x)$.

1. Problem Description

The random variable $X$ is defined by its probability density function (PDF) $f(x)$ as follows:

$f(x) = \begin{cases}

0, & x < -\frac{\pi}{2} \\

a \cdot \cos(x), & -\frac{\pi}{2} \le x < \frac{\pi}{2} \\

0, & x \ge \frac{\pi}{2}

\end{cases}$

The problem asks to find the value of the parameter $a$ and the cumulative distribution function (CDF) $F(x)$.

2. Solution Steps

First, we need to find the value of $a$. The integral of the probability density function over the entire real line must be equal to 1:

$\int_{-\infty}^{\infty} f(x) dx = 1$

Since $f(x)$ is non-zero only between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, we have:

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} a \cdot \cos(x) dx = 1$

$a \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(x) dx = 1$

$a [\sin(x)]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = 1$

$a [\sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2})] = 1$

$a [1 - (-1)] = 1$

$2a = 1$

$a = \frac{1}{2}$

Now, we need to find the CDF $F(x)$. The CDF is defined as:

$F(x) = \int_{-\infty}^{x} f(t) dt$

We have three cases:

1. $x < -\frac{\pi}{2}$:

$F(x) = \int_{-\infty}^{x} 0 dt = 0$

2. $-\frac{\pi}{2} \le x < \frac{\pi}{2}$:

$F(x) = \int_{-\infty}^{-\frac{\pi}{2}} 0 dt + \int_{-\frac{\pi}{2}}^{x} \frac{1}{2} \cos(t) dt = 0 + \frac{1}{2} [\sin(t)]_{-\frac{\pi}{2}}^{x} = \frac{1}{2} [\sin(x) - \sin(-\frac{\pi}{2})] = \frac{1}{2} [\sin(x) + 1]$

3. $x \ge \frac{\pi}{2}$:

$F(x) = \int_{-\infty}^{-\frac{\pi}{2}} 0 dt + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{2} \cos(t) dt + \int_{\frac{\pi}{2}}^{x} 0 dt = 0 + \frac{1}{2} [\sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2})] + 0 = \frac{1}{2} [1 - (-1)] = \frac{1}{2} (2) = 1$

Therefore, the CDF $F(x)$ is:

$F(x) = \begin{cases}

0, & x < -\frac{\pi}{2} \\

\frac{1}{2} (\sin(x) + 1), & -\frac{\pi}{2} \le x < \frac{\pi}{2} \\

1, & x \ge \frac{\pi}{2}

\end{cases}$

3. Final Answer

$a = \frac{1}{2}$

$F(x) = \begin{cases}

0, & x < -\frac{\pi}{2} \\

\frac{1}{2} (\sin(x) + 1), & -\frac{\pi}{2} \le x < \frac{\pi}{2} \\

1, & x \ge \frac{\pi}{2}

\end{cases}$