Given the p.d.f of a continous r.v.X was `f(x) = (x^(2))/(3), - 1 lt x lt2` =0, Otherwise Determine the c.d.f of X and hence find . `P(X lt 1), P(X le - 2), P(X gt0),P(1le X le 2)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the Cumulative Distribution Function (CDF)
The CDF, denoted as \( F(x) \), is defined as:
\[ F(x) = \int_{-\infty}^{x} f(t) \, dt \]
Given the probability density function (PDF):
\[ f(x) = \frac{x^2}{3}, \quad -1 < x < 2 \]
and \( f(x) = 0 \) otherwise, we can compute the CDF for different ranges of \( x \).

#### For \( x < -1 \):
\[ F(x) = 0 \]

#### For \( -1 \leq x < 2 \):
\[ F(x) = \int_{-1}^{x} \frac{t^2}{3} \, dt \]
Calculating the integral:
\[ F(x) = \frac{1}{3} \int_{-1}^{x} t^2 \, dt = \frac{1}{3} \left[ \frac{t^3}{3} \right]_{-1}^{x} = \frac{1}{3} \left( \frac{x^3}{3} - \frac{(-1)^3}{3} \right) \]
\[ = \frac{1}{3} \left( \frac{x^3}{3} + \frac{1}{3} \right) = \frac{x^3 + 1}{9} \]

#### For \( x \geq 2 \):
\[ F(x) = 1 \]

Thus, the CDF is:
\[ F(x) = \begin{cases} 0 & \text{if } x < -1 \\ \frac{x^3 + 1}{9} & \text{if } -1 \leq x < 2 \\ 1 & \text{if } x \geq 2 \end{cases} \]

### Step 2: Calculate the Required Probabilities
1. **\( P(X < 1) \)**:
\[ P(X < 1) = F(1) = \frac{1^3 + 1}{9} = \frac{2}{9} \]

2. **\( P(X \leq -2) \)**:
Since \(-2 < -1\):
\[ P(X \leq -2) = F(-2) = 0 \]

3. **\( P(X > 0) \)**:
\[ P(X > 0) = 1 - P(X \leq 0) = 1 - F(0) \]
Calculate \( F(0) \):
\[ F(0) = \frac{0^3 + 1}{9} = \frac{1}{9} \]
Thus,
\[ P(X > 0) = 1 - \frac{1}{9} = \frac{8}{9} \]

4. **\( P(1 \leq X \leq 2) \)**:
\[ P(1 \leq X \leq 2) = F(2) - F(1) \]
We already know \( F(2) = 1 \) and \( F(1) = \frac{2}{9} \):
\[ P(1 \leq X \leq 2) = 1 - \frac{2}{9} = \frac{9 - 2}{9} = \frac{7}{9} \]

### Final Answers:
- \( P(X < 1) = \frac{2}{9} \)
- \( P(X \leq -2) = 0 \)
- \( P(X > 0) = \frac{8}{9} \)
- \( P(1 \leq X \leq 2) = \frac{7}{9} \)