The lifespan, in years, of a certain computer is exponentially distributed. The probability that its lifespan exceeds four years is 0.30.

Let [math]f(x)[/math] represent the density function of the computer’s lifespan, in years, for [math]x \gt 0[/math].

Determine which of the following is an expression for [math]f(x)[/math].

• [math]1 − (0.3)^{x/4}[/math]
• [math]1 − (0.7)^{x/4}[/math]
• [math]1 − (0.3)^{x/4}[/math]
• [math]-\frac{\ln 0.7}{4}(0.7)^{x/4}[/math]
• [math]-\frac{\ln 0.3}{4}(0.3)^{x/4}[/math]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: E

The cumulative distribution function for the exponential distribution of the lifespan is

[[math]] F(x) = 1-e^{-\lambda x}, \quad \textrm{for positive }x. [[/math]]

The probability that the lifespan exceeds 4 years is [math]0.3 = 1 - F(4) = e^{-4\lambda} [/math]. Thus [math]\lambda = −(\ln 0.3) / 4 [/math]. For positive [math]x[/math], the probability density function is

[[math]] f(x) = \lambda e^{-\lambda x} = - \frac{\ln(0.3)}{4}e^{(\ln(0.3))x/4} = - \frac{\ln(0.3)}{4}(0.3)^{x/4}. [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.