Q74 / 110 · VITEEE 2023 · Mathematics
For the probability density function
\(f(x)= \begin{cases}2 e^{-2 x} & , x > 0 \\ 0 & , x \leq 0\end{cases}\)
\(F(2)\) is equal to
- Correct answerA\(\frac{e^4-1}{e^4}\)
- B\(\frac{e^3-1}{e^3}\)
- C\(\frac{\mathrm{e}^{5 / 2}-1}{\mathrm{e}}\)
- D\(\frac{e^5-1}{e^5}\)
Solution
\(\begin{aligned}
& F(2)=F(X \leq 2)=\int_{-\infty}^2 f(x) d x \\
& =\int_0^2 2 e^{-2 x} d x=2 \cdot\left[\frac{e^{-2 x}}{-2}\right]_0^2 \\
& =-\left[e^{-4}-1\right]=1-e^{-4}=\frac{e^4-1}{e^4}
\end{aligned}\)
Hence, the correct answer is option: 'A'.