Answer:
E[X]= [tex]2 + 3 e^{(-2/3)[/tex]
Step-by-step explanation:
The objective of this question is to determine E[X].
T is defined (0,infinity)
X=max(c,T)
where; c=constant
E[X]=c+function (c,infinity) Sf(t)dt
E[X] =[tex]e^{-t/3[/tex]
E[X]=2+function (2,infinity)[tex]e^{-t/3[/tex]dt
E[X] =[tex](2+e^{-t/3})/(1/3)[/tex] function (2,infinity)
E[X]= [tex]2 + 3 e^{(-2/3)[/tex]
If X = T if T ≥ 2 and X = 2 if 0 ≤ T < 2,
So Since T is exponentially distributed with mean 3, the density function of T is [tex]f(t) = (1/3)e^{(-t/3)[/tex]
