Question

Let \( X \) be a continuous random variable with the probability density function 

Then \( P\left( \frac{1}{2} \leq X \leq 2 \right) \) equals 
 

• A. \( \frac{15}{16} \)
• B. \( \frac{11}{16} \)
• C. \( \frac{7}{12} \)
• D. \( \frac{3}{8} \)

Hint:

To compute probabilities for continuous random variables, always integrate the probability density function over the desired interval.

Answer 1

The Correct Option is A

Step 1: Understanding the problem.
The probability is given by the integral of the probability density function \( f(x) \) over the interval \( \left[ \frac{1}{2}, 2 \right] \). Hence, the required probability is: \[ P\left( \frac{1}{2} \leq X \leq 2 \right) = \int_{\frac{1}{2}}^{1} x^3 \, dx + \int_{1}^{2} 3x^5 \, dx \]
Step 2: Solving the integrals.
First, solve \( \int_{\frac{1}{2}}^{1} x^3 \, dx \): \[ \int_{\frac{1}{2}}^{1} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{\frac{1}{2}}^{1} = \frac{1}{4} - \frac{1}{64} = \frac{15}{64} \] Next, solve \( \int_{1}^{2} 3x^5 \, dx \): \[ \int_{1}^{2} 3x^5 \, dx = \left[ \frac{x^6}{2} \right]_{1}^{2} = \frac{64}{2} - \frac{1}{2} = \frac{63}{2} = \frac{2016}{64} \]
Step 3: Final calculation.
Thus, the total probability is: \[ P\left( \frac{1}{2} \leq X \leq 2 \right) = \frac{15}{64} + \frac{2016}{64} = \frac{2031}{64} = \frac{15}{16} \]
Step 4: Conclusion.
The correct answer is (A) \( \frac{15}{16} \), as the total probability is \( \frac{15}{16} \).