Question

Compute the probability that an exponential random variable, i.e. $f_x(x) = \lambda \cdot e^{-\lambda \cdot x}$ , takes a value more than two standard deviations more than it's mean.

Answer 1

Mean $\mu$ = 1 / $\lambda$

Standard deviation, $\sigma$ = 1 / $\lambda$

Probability that exponential random variable takes a value more than two standard deviations more than it's mean is,

P[X > ($\mu + 2 \sigma$)] = P[X > ((1 / $\lambda$ )+ (2 / $\lambda$)) ]

= P[X > 3 / $\lambda$]

$= e^{-\lambda * (3/\lambda)}$

$= e^{-3 } = 0.04978707$