Question

The mean of the density function is \(f(x) = \lambda e^{-\lambda x}, x > 0\) is ____ .

• A. \( \frac{1}{\lambda} \)
• B. \( \lambda^2 \)
• C. \( \frac{1}{\lambda^2} \)
• D. 1

Hint:

For exponential distributions, the mean is the reciprocal of the rate parameter \( \lambda \). This property is important in many applications, including reliability analysis and queuing theory.

Answer 1

The Correct Option is A

The given function \( f(x) = \lambda e^{-\lambda x}, x > 0 \) is an exponential density function with rate parameter \( \lambda \). The mean \( \mu \) of an exponential distribution is the expected value of the random variable \( X \), which is calculated as: \[ \mu = \int_0^\infty x \lambda e^{-\lambda x} dx \] This is a standard integral in probability theory, and the result is: \[ \mu = \frac{1}{\lambda} \] Thus, the mean of the given exponential density function is \( \frac{1}{\lambda} \).