X is continuous random variable with probability density function `f(x)=(x^(2))/(8),0 le x le 1`. Then, the value of `P(0.2leXle0.5)` is

Options

A. `(0.117)/(24)`
B. `(0.112)/(24)`
C. `(0.113)/(36)`
D. `(0.112)/(36)`

Answer

To solve the problem of finding the probability \( P(0.2 \leq X \leq 0.5) \) for the continuous random variable \( X \) with the given probability density function \( f(x) = \frac{x^2}{8} \) for \( 0 \leq x \leq 1 \), we will follow these steps:

### Step 1: Set up the integral for the probability
The probability \( P(a \leq X \leq b) \) for a continuous random variable is given by the integral of the probability density function over the interval \([a, b]\). In this case, we want to find:
\[ P(0.2 \leq X \leq 0.5) = \int_{0.2}^{0.5} f(x) \, dx \]
Substituting the given \( f(x) \):
\[ P(0.2 \leq X \leq 0.5) = \int_{0.2}^{0.5} \frac{x^2}{8} \, dx \]

### Step 2: Calculate the integral
Now we will compute the integral:
\[ \int_{0.2}^{0.5} \frac{x^2}{8} \, dx = \frac{1}{8} \int_{0.2}^{0.5} x^2 \, dx \]
The integral of \( x^2 \) is \( \frac{x^3}{3} \), so we have:
\[ \frac{1}{8} \left[ \frac{x^3}{3} \right]_{0.2}^{0.5} \]

### Step 3: Evaluate the definite integral
Now we will evaluate the definite integral:
\[ = \frac{1}{8} \left( \frac{(0.5)^3}{3} - \frac{(0.2)^3}{3} \right) \]
Calculating \( (0.5)^3 \) and \( (0.2)^3 \):
\[ (0.5)^3 = 0.125 \quad \text{and} \quad (0.2)^3 = 0.008 \]
Substituting these values back:
\[ = \frac{1}{8} \left( \frac{0.125}{3} - \frac{0.008}{3} \right) = \frac{1}{8} \left( \frac{0.125 - 0.008}{3} \right) \]
Calculating \( 0.125 - 0.008 = 0.117 \):
\[ = \frac{1}{8} \left( \frac{0.117}{3} \right) = \frac{0.117}{24} \]

### Step 4: Final result
Thus, the probability \( P(0.2 \leq X \leq 0.5) \) is:
\[ P(0.2 \leq X \leq 0.5) = \frac{0.117}{24} \]