The p.d.f of a continous random variable X is
`f(x)=x/8, 0 lt x lt 4`
`=0`, otherwise
Then the value of `P(X gt 3)` is

Options

A. `3/16`
B. `5/16`
C. `7/16`
D. `9/16`

Answer

To solve the problem of finding \( P(X > 3) \) for the given probability density function (p.d.f) of the continuous random variable \( X \), we will follow these steps:

### Step 1: Understand the p.d.f
The given p.d.f is:
\[ f(x) = \begin{cases} \frac{x}{8} & \text{for } 0 < x < 4 \\ 0 & \text{otherwise} \end{cases} \]
This means that the function is defined and non-zero only in the interval \( (0, 4) \).

### Step 2: Set up the integral for \( P(X > 3) \)
To find \( P(X > 3) \), we need to integrate the p.d.f from 3 to 4 (since \( f(x) = 0 \) for \( x > 4 \)):
\[ P(X > 3) = \int_{3}^{4} f(x) \, dx \]

### Step 3: Substitute the p.d.f into the integral
Substituting the expression for \( f(x) \):
\[ P(X > 3) = \int_{3}^{4} \frac{x}{8} \, dx \]

### Step 4: Perform the integration
Now we will integrate \( \frac{x}{8} \):
\[ P(X > 3) = \frac{1}{8} \int_{3}^{4} x \, dx \]
The integral of \( x \) is \( \frac{x^2}{2} \), so:
\[ P(X > 3) = \frac{1}{8} \left[ \frac{x^2}{2} \right]_{3}^{4} \]
Calculating the limits:
\[ = \frac{1}{8} \left( \frac{4^2}{2} - \frac{3^2}{2} \right) = \frac{1}{8} \left( \frac{16}{2} - \frac{9}{2} \right) = \frac{1}{8} \left( \frac{16 - 9}{2} \right) = \frac{1}{8} \left( \frac{7}{2} \right) \]
\[ = \frac{7}{16} \]

### Step 5: State the final answer
Thus, the value of \( P(X > 3) \) is:
\[ \boxed{\frac{7}{16}} \]

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