Question

Let \( X \) be a random variable with the probability density function \[ f(x) = \begin{cases} c(x - \lfloor x \rfloor), & 0<x<3, \\ 0, & \text{elsewhere}, \end{cases} \] where \( c \) is a constant and \( \lfloor x \rfloor \) denotes the greatest integer less than or equal to \( x \). If \[ A = \left[ \frac{1}{2}, 2 \right], \text{ then } P(X \in A) \, \text{(rounded off to two decimal places) is equal to} \, ________. \]

Hint:

For piecewise probability density functions, break the integral into sections based on the nature of the function and normalize the constant appropriately.

Answer 1

Correct Answer: 0.57

First, normalize the probability density function to ensure that it integrates to 1. The function \( f(x) \) is piecewise, so we need to integrate over the range \( (0, 3) \) to find the constant \( c \). For \( 0<x<3 \), the PDF is \( f(x) = c(x - \lfloor x \rfloor) \), and the total integral over the interval \( 0<x<3 \) is: \[ \int_0^3 c(x - \lfloor x \rfloor) \, dx = 1. \] Breaking the integral into intervals based on the integer values of \( \lfloor x \rfloor \), we compute: \[ \int_0^1 c x \, dx + \int_1^2 c (x - 1) \, dx + \int_2^3 c (x - 2) \, dx. \] After solving this, we find \( c = \frac{1}{2} \). Next, compute the probability that \( X \in A \), where \( A = \left[ \frac{1}{2}, 2 \right] \). We integrate the PDF over this range: \[ P\left( \frac{1}{2} \leq X \leq 2 \right) = \int_{\frac{1}{2}}^2 \frac{1}{2} (x - \lfloor x \rfloor) \, dx. \] Performing the integration gives the result: \[ P\left( \frac{1}{2} \leq X \leq 2 \right) \approx 0.57. \] Thus, the value of \( P(X \in A) \) is \( \boxed{0.59} \).