To find the value of the normalization constant \( c \), we need to ensure that the total probability over the domain of the probability density function (PDF) integrates to 1. The PDF is defined as follows:
\[ f_X(x) = \begin{cases} c & \text{if } 1 < x \leq 3 \ 2c & \text{if } 0 \leq x \leq 1 \ 0 & \text{otherwise} \end{cases} \]
a) Finding \( c \):
To find \( c \), we will find the integral of the PDF over its defined intervals and set it equal to 1.
- For the interval \( 1 < x \leq 3 \):
\[ \int_{1}^{3} c , dx = c \cdot (3 - 1) = 2c \]
- For the interval \( 0 \leq x \leq 1 \):
\[ \int_{0}^{1} 2c , dx = 2c \cdot (1 - 0) = 2c \]
- Adding up both integrals:
\[ \int_{-\infty}^{\infty} f_X(x) , dx = 2c + 2c = 4c \]
Setting this equal to 1 for normalization:
\[ 4c = 1 \implies c = \frac{1}{4} \]
So, \( c = \frac{1}{4} \).
b) Finding \( P \left( \frac{1}{2} \leq X \leq \frac{3}{2} \right) \):
To find this probability, we can integrate the PDF over the interval \( \left[ \frac{1}{2}, \frac{3}{2} \right] \).
This interval covers two parts:
- The interval from \( \frac{1}{2} \) to \( 1 \), where \( f_X(x) = 2c = \frac{2}{4} = \frac{1}{2} \).
- The interval from \( 1 \) to \( \frac{3}{2} \), where \( f_X(x) = c = \frac{1}{4} \).
Calculating each integral:
- From \( \frac{1}{2} \) to \( 1 \):
\[ P\left( \frac{1}{2} \leq X \leq 1 \right) = \int_{\frac{1}{2}}^{1} \frac{1}{2} , dx = \frac{1}{2} \cdot (1 - \frac{1}{2}) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \]
- From \( 1 \) to \( \frac{3}{2} \):
\[ P\left( 1 \leq X \leq \frac{3}{2} \right) = \int_{1}^{\frac{3}{2}} \frac{1}{4} , dx = \frac{1}{4} \cdot \left( \frac{3}{2} - 1 \right) = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8} \]
- Adding both probabilities:
\[ P\left( \frac{1}{2} \leq X \leq \frac{3}{2} \right) = P\left( \frac{1}{2} \leq X \leq 1 \right) + P\left( 1 \leq X \leq \frac{3}{2} \right) = \frac{1}{4} + \frac{1}{8} = \frac{2}{8} + \frac{1}{8} = \frac{3}{8} \]
Final Answer:
\[ P\left( \frac{1}{2} \leq X \leq \frac{3}{2} \right) = \frac{3}{8} \]