Probability Greater Than In Weibull Distribution

Question

X observes a Weibull distribution with shape parameter

β=1.2

and characteristic life

η=0.3

Find the probability of P(X>1.8)

I have used the following equation but I keep getting the answer as equal to 1, which is definitely incorrect.

equation

Answer 1

Assuming your interpretation of shape and characteristic life parameters is correct:

$$F(x)=P(X\leq x)=1-e^{-(\frac x \eta)^\beta}, \text{ for } x>0$$

$$\Rightarrow 1-F(x)=P(X > x)=e^{-(\frac x \eta)^\beta}$$

$$\Rightarrow 1-F(1.8)=P(X > 1.8)=e^{-(\frac {1.8} \eta)^\beta}$$

When $\beta = 1.2, \eta=0.3$ we get:

$$P(X > 1.8)=e^{-(\frac {1.8} {0.3})^{1.2}} \simeq 0.000186736$$