Topic: Engineering Mathematics
A random variable z, has a probability density function $f(z)$ where $f(z)=e^{-z}$ for $0 \le z < \infty$, the probability of $0 \le z \le 2$ will be approximately.
- A. 0.368
- B. 0.135
- C. 0.393
- D. 0.865
Answer: D. 0.865
Explanation: The probability is the integral of the PDF over the interval. $P(0 \le z \le 2) = \int_{0}^{2} e^{-z} dz$. Evaluating the integral gives $[-e^{-z}]_0^2 = -[e^{-2} - e^0] = 1 - e^{-2}$. Since $e^{-2} \approx 0.135$, the probability is $1 - 0.135 = 0.865$.