Question

A device that continuously measures and records seismic activity is placed in a remote region. The time, T, to failure of this device is exponentially distributed with mean 3 years. Since the device will not be monitored during its first two years of service, the time to discovery of its failure is X = max(T, 2).

Answer 1

Answer:

E[X]= $2 + 3 e^{(-2/3)$

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] =$e^{-t/3$

E[X]=2+function (2,infinity)$e^{-t/3$dt

E[X] =$(2+e^{-t/3})/(1/3)$ function (2,infinity)

E[X]= $2 + 3 e^{(-2/3)$

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  $f(t) = (1/3)e^{(-t/3)$