https://file.aiquickdraw.com/mathbot/file/1726366583576zod1m2w4.jpg

Let's solve the given problem step by step.

### Part (a) - Finding the value of \( c \):

The probability density function (PDF) must satisfy the condition that the total probability over the entire space is 1. In other words, the integral of the PDF \( f(x) \) over the range of \( X \) must be equal to 1:

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

Since \( f(x) = 0 \) for \( x \notin (-1, 1) \), we only need to integrate over \( x \in (-1, 1) \):

\[
\int_{-1}^{1} c(1 - x^2) \, dx = 1
\]

Let's solve this integral to find the value of \( c \).

\[
\int_{-1}^{1} c(1 - x^2) \, dx = c \int_{-1}^{1} (1 - x^2) \, dx
\]

Now, break the integral into two parts:

\[
c \left( \int_{-1}^{1} 1 \, dx - \int_{-1}^{1} x^2 \, dx \right)
\]

1. First integral: \( \int_{-1}^{1} 1 \, dx = 2 \)
2. Second integral: \( \int_{-1}^{1} x^2 \, dx = \frac{2}{3} \) (since the integral of \( x^2 \) from \( -1 \) to \( 1 \) is symmetric)

Now substitute these values:

\[
c \left( 2 - \frac{2}{3} \right) = 1
\]

Simplify:

\[
c \left( \frac{6}{3} - \frac{2}{3} \right) = 1 \quad \Rightarrow \quad c \cdot \frac{4}{3} = 1
\]

Solving for \( c \):

\[
c = \frac{3}{4}
\]

So, the value of \( c \) is \( \frac{3}{4} \).

---

### Part (b) - Finding the cumulative distribution function (CDF) \( F(x) \):

The CDF is defined as:

\[
F(x) = \int_{-\infty}^{x} f(t) \, dt
\]

We need to compute \( F(x) \) in different regions based on the support of the PDF:

- For \( x < -1 \), \( F(x) = 0 \) because the probability density is zero for values less than \( -1 \).
- For \( -1 \leq x \leq 1 \), we integrate \( f(t) = \frac{3}{4}(1 - t^2) \) from \( -1 \) to \( x \):

\[
F(x) = \int_{-1}^{x} \frac{3}{4}(1 - t^2) \, dt
\]

Let's compute this integral step by step:

\[
F(x) = \frac{3}{4} \left( \int_{-1}^{x} 1 \, dt - \int_{-1}^{x} t^2 \, dt \right)
\]

1. The first integral is:

\[
\int_{-1}^{x} 1 \, dt = x + 1
\]

2. The second integral is:

\[
\int_{-1}^{x} t^2 \, dt = \frac{x^3}{3} - \left( \frac{(-1)^3}{3} \right) = \frac{x^3}{3} + \frac{1}{3}
\]

Now substitute these into the expression for \( F(x) \):

\[
F(x) = \frac{3}{4} \left( (x + 1) - \left( \frac{x^3}{3} + \frac{1}{3} \right) \right)
\]

Simplify:

\[
F(x) = \frac{3}{4} \left( x + 1 - \frac{x^3}{3} - \frac{1}{3} \right)
\]
\[
F(x) = \frac{3}{4} \left( x + 1 - \frac{x^3}{3} - \frac{1}{3} \right)
\]
\[
F(x) = \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right)
\]

Thus, for \( -1 \leq x \leq 1 \):

\[
F(x) = \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right)
\]

- For \( x > 1 \), \( F(x) = 1 \) because the entire probability is accumulated.

---

### Final Answer:
- (a) \( c = \frac{3}{4} \)
- (b) The cumulative distribution function is:
  \[
  F(x) =
  \begin{cases}
  0, & \text{if } x < -1 \\
  \frac{3}{4} \left( x - \frac{x^3}{3} + \frac{2}{3} \right), & \text{if } -1 \leq x \leq 1 \\
  1, & \text{if } x > 1
  \end{cases}
  \]

---

Do you want more details or have any further questions?

Here are some additional related questions:

1. What is the mean of the random variable \( X \) based on this PDF?
2. How do you find the variance of a random variable with a given PDF?
3. Can the PDF take negative values? Why or why not?
4. How would you verify that the CDF is valid and continuous?
5. What are the properties of a probability density function?

**Tip**: When integrating to find the CDF, always ensure the result is continuous and correctly normalized for probabilities.

Let X be a random variable with probability density f(x) = c(1 - x^2), for -1 < x < 1 and 0 otherwise. (a) What is the value of c? (b) What is the cumulative distribution function of X?

This problem explores a random variable X with a given probability density function and asks to find the constant c and the cumulative distribution function. The solution involves integration and normalization of the PDF.

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