The probability density function of the time to failure of an
electronic component in a copier(in hours)is f(x)=exp(-x/3000)/3000
for x>0 and f(x)=0 for x<0 or x=0.Determine the probability that (A)A
component lasts more than 1000 hours before failure (B)A component
fails in the interval from 1000 to 2000 hours (C)A component fails
before 3000 hours (D)Determine the number of hours at which 10% of all
components have faild (E)Determine the mean and variance

wjs,

I believe that the answer to this question requires a small knowledge
of calculus, so I will use that to solve the question. If you have a
table that gives you the cumulative distribution of the exponential
distribution, you can use that to answer some of the question.

So to review :

       exp(-x/3000)
f(x) = ------------
           3000

As a general rule :

P(a < X < b) = Integral from a to b of f(x)

so:

Part A :

P(X > 1000) = Integral from 1000 to infinity of f(x)
            = 1 - Integral from 0 to 1000
               (because from 0 to infinity integrates to 1)
            = 1 - Integral(0,1000) of {exp(-x/3000)/3000}
            = 1 + exp(-x/3000) evaluated at 0 and 1000
            = 1 + exp(-1/3) - exp(0)
            = exp(-1/3)
            = 0.716

Part B :
P(1000 < X < 2000) = Integral from 1000 to 2000 of f(x)
                   = Integral from 1000 to infinity of f(x)
                     - Integral from 2000 to infinity of f(x)
                   = 0.716 - 1 - exp(-2000/3000) + exp(0)
                   = 0.716 - 0.513
                   = 0.203

Part C :
P(X < 3000)        = Integral from 0 to 3000 of f(x)
                   = exp(0) - exp(-3/3)
                   = 1 - 0.367
                   = 0.632

Part D :
P(X < c) = 0.1     = Integral from 0 to c of f(x)
                   = exp(0) - exp(-c/3000)
ie.        0.9     = exp(-c/3000)
ie.     ln(0.9)    = -c/3000
ie.  -ln(0.9)*3000 = c
ie.   0.10536*3000 = c
ie. c = 316 hours.

Part E :

From mathworld :
http://mathworld.wolfram.com/ExponentialDistribution.html

E(x) = 1/(1/3000) = 3000
Var(x) = [E(x)]² = 9,000,000



The best way of thinking about parts A through D is to draw the
picture of f(x) (see the mathworld page) and then shade of the
relevant area. If  you use the fact that the whole area adds up to 1,
you can solve any of these type of problems by reduction to a similar
problem you have already answered.

Hope this helps

Regards

CalebU2-ga

Resources :

Distribution Functions from Mathworld.com :
http://mathworld.wolfram.com/DistributionFunction.html