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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 | 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 display = "block"> | <math display = "block"> | ||
Latest revision as of 19:38, 2 May 2023
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.