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 _________ .
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 _________ .
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To calculate the probability between two values for a continuous probability density function, integrate the PDF over the desired range.
Updated On: Nov 27, 2025
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Correct Answer: 0.11
Solution and Explanation
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.
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