Question

Let \( X \) be a continuous random variable with the probability density function
\[ f(x) = \begin{cases} x^3, & 0 < x \leq 1 \\ 3x^5, & x > 1 \\ 0, & \text{otherwise} \end{cases} \] Then \( P\left( \frac{1}{2} \leq X \leq 2 \right) \) equals

• A. \( \frac{11}{16} \)
• B. \( \frac{15}{16} \)
• 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 B

Step 1: Write the probability formula.
We are given the probability density function (pdf) of the random variable \( X \): \[ f(x) = \begin{cases} x^3, & 0 < x \leq 1 \\ 3x^5, & x > 1 \\ 0, & \text{otherwise} \end{cases} \] We need to find the probability \( P\left( \frac{1}{2} \leq X \leq 2 \right) \), which is given by: \[ P\left( \frac{1}{2} \leq X \leq 2 \right) = \int_{\frac{1}{2}}^{2} f(x) \, dx \]

Step 2: Split the integral based on the definition of \( f(x) \).
Since the pdf changes at \( x = 1 \), we split the integral into two parts: \[ P\left( \frac{1}{2} \leq X \leq 2 \right) = \int_{\frac{1}{2}}^{1} x^3 \, dx + \int_{1}^{2} 3x^5 \, dx \]

Step 3: Compute the first integral.
\[ \int_{\frac{1}{2}}^{1} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{\frac{1}{2}}^{1} = \frac{1^4}{4} - \frac{\left( \frac{1}{2} \right)^4}{4} = \frac{1}{4} - \frac{1}{64} = \frac{16}{64} - \frac{1}{64} = \frac{15}{64} \]

Step 4: Compute the second integral.
\[ \int_{1}^{2} 3x^5 \, dx = \left[ \frac{x^6}{2} \right]_{1}^{2} = \frac{2^6}{2} - \frac{1^6}{2} = \frac{64}{2} - \frac{1}{2} = \frac{63}{2} \]

Step 5: Add the results.
Now add the two parts of the integral: \[ P\left( \frac{1}{2} \leq X \leq 2 \right) = \frac{15}{64} + \frac{63}{2} = \frac{15}{64} + \frac{2016}{64} = \frac{2031}{64} = \frac{15}{16} \]

Step 6: Final Answer.
The probability is \( P\left( \frac{1}{2} \leq X \leq 2 \right) = \frac{15}{16} \).

Final Answer: \( \frac{15}{16} \)