The p. d. f. of a continuous r. v. Xis
`f(x) = {:((3x^2)/8 , 0 lt x lt 2),(0, "otherwise"):}`
Determine the c.d.f of X and hence find
(i) `P(X lt 1), (ii) P(1 lt X lt 2)`.

Answer

To solve the problem, we need to find the cumulative distribution function (c.d.f.) of the continuous random variable \(X\) given the probability density function (p.d.f.):

\[ f(x) = \begin{cases} \frac{3x^2}{8} & \text{for } 0 < x < 2 \\ 0 & \text{otherwise} \end{cases} \]

### Step 1: Determine the c.d.f. \(F(x)\)

The cumulative distribution function \(F(x)\) is defined as:

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

Since \(f(x) = 0\) for \(x \leq 0\), we have:

\[ F(x) = 0 \quad \text{for } x \leq 0 \]

For \(0 < x < 2\), we need to integrate \(f(t)\):

\[ F(x) = \int_{0}^{x} \frac{3t^2}{8} \, dt \]

Calculating the integral:

\[ F(x) = \frac{3}{8} \int_{0}^{x} t^2 \, dt \]

The integral of \(t^2\) is:

\[ \int t^2 \, dt = \frac{t^3}{3} \]

Thus,

\[ F(x) = \frac{3}{8} \left[ \frac{t^3}{3} \right]_{0}^{x} = \frac{3}{8} \cdot \frac{x^3}{3} = \frac{x^3}{8} \]

For \(x \geq 2\), since the total probability must equal 1, we have:

\[ F(x) = 1 \quad \text{for } x \geq 2 \]

Putting it all together, the c.d.f. \(F(x)\) is:

\[ F(x) = \begin{cases} 0 & \text{for } x \leq 0 \\ \frac{x^3}{8} & \text{for } 0 < x < 2 \\ 1 & \text{for } x \geq 2 \end{cases} \]

### Step 2: Calculate \(P(X < 1)\)

To find \(P(X < 1)\), we use the c.d.f.:

\[ P(X < 1) = F(1) \]

Since \(1\) is in the interval \(0 < x < 2\):

\[ F(1) = \frac{1^3}{8} = \frac{1}{8} \]

### Step 3: Calculate \(P(1 < X < 2)\)

To find \(P(1 < X < 2)\), we can use the c.d.f.:

\[ P(1 < X < 2) = F(2) - F(1) \]

We already calculated \(F(1)\):

\[ F(1) = \frac{1}{8} \]

Now, calculate \(F(2)\):

\[ F(2) = 1 \quad \text{(since \(x \geq 2\))} \]

Thus,

\[ P(1 < X < 2) = F(2) - F(1) = 1 - \frac{1}{8} = \frac{7}{8} \]

### Final Answers

1. \(P(X < 1) = \frac{1}{8}\)
2. \(P(1 < X < 2) = \frac{7}{8}\)