The p.d.f. of a continuous random variable X is `f(x) = K/(sqrt(x)), 0 lt x lt 4`
= 0 , otherwise
Then `P(X ge1)` is equal to

Options

A. 0.2
B. 0.3
C. 0.4
D. 0.5

Answer

To solve the problem, we need to find the probability \( P(X \geq 1) \) for the given probability density function (p.d.f.) of a continuous random variable \( X \).

### Step 1: Identify the p.d.f. and find the constant \( K \)

The given p.d.f. is:

\[ f(x) = \frac{K}{\sqrt{x}}, \quad 0 < x < 4 \]
\[ f(x) = 0, \quad \text{otherwise} \]

To find the constant \( K \), we use the property that the total probability must equal 1:

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

Since \( f(x) = 0 \) outside the interval \( (0, 4) \), we only need to integrate from 0 to 4:

\[ \int_0^4 \frac{K}{\sqrt{x}} \, dx = 1 \]

### Step 2: Calculate the integral

We can rewrite the integral:

\[ \int_0^4 \frac{K}{\sqrt{x}} \, dx = K \int_0^4 x^{-1/2} \, dx \]

The integral of \( x^{-1/2} \) is:

\[ \int x^{-1/2} \, dx = 2x^{1/2} \]

Thus, we have:

\[ K \left[ 2x^{1/2} \right]_0^4 = K \left( 2\sqrt{4} - 2\sqrt{0} \right) = K(4) = 4K \]

Setting this equal to 1 gives:

\[ 4K = 1 \implies K = \frac{1}{4} \]

### Step 3: Set up the probability \( P(X \geq 1) \)

Now, we want to find \( P(X \geq 1) \):

\[ P(X \geq 1) = \int_1^4 f(x) \, dx \]

Substituting \( f(x) \):

\[ P(X \geq 1) = \int_1^4 \frac{1/4}{\sqrt{x}} \, dx = \frac{1}{4} \int_1^4 x^{-1/2} \, dx \]

### Step 4: Calculate the integral for \( P(X \geq 1) \)

Calculating the integral:

\[ \int_1^4 x^{-1/2} \, dx = \left[ 2x^{1/2} \right]_1^4 = 2\sqrt{4} - 2\sqrt{1} = 4 - 2 = 2 \]

Now substituting back into the probability:

\[ P(X \geq 1) = \frac{1}{4} \cdot 2 = \frac{1}{2} \]

### Final Answer

Thus, the probability \( P(X \geq 1) \) is:

\[ \boxed{\frac{1}{2}} \]