Question

The probability density function of a random variable \( X \) is \( p(x) = 2e^{-2x} \). The probability \( P(1 \leq X \leq 2) \) (rounded off to two decimal places) is _________ .

Hint:

To calculate the probability between two values for a continuous probability density function, integrate the PDF over the desired range.

Answer 1

Correct Answer: 0.11

The probability \( P(1 \leq X \leq 2) \) is found by integrating the probability density function \( p(x) \) over the range from 1 to 2: \[ P(1 \leq X \leq 2) = \int_1^2 2e^{-2x} \, dx. \] To solve the integral: \[ \int 2e^{-2x} \, dx = -e^{-2x}. \] Now, evaluate the integral from 1 to 2: \[ P(1 \leq X \leq 2) = \left[ -e^{-2x} \right]_1^2 = -e^{-4} + e^{-2}. \] Substitute the values of \( e^{-4} \) and \( e^{-2} \): \[ P(1 \leq X \leq 2) = -(0.0183) + (0.1353) = 0.1170. \] Thus, the probability is approximately 0.11.