Suppose that the average lifetime of a plane is 100 working
hours and that the lifetime distribution is exponential.
(a) Estimate the probability that the plane will work at least 50
hours.
(b) Given that the plane has functioned for 50 hours, what is the
chance that it fails in the next 25 hours.
(c) Suppose that two planes, one in active, the other in reserve (if the primary one malfunctions, the second will be used at once). What is the chance that these two planes remain working more than 50 hours?
average lifetime = 100
the distribution of exponential.
so here probability distribution function is
f(x) = (1/100) e-x/100 ; x > 0
cumulative probability distribution
F(x) = 1- e-x/100 ; x > 0
(a) f(x > 50) = 1 - F(50) = 1 - (1 - e-50/100) = 0.6065
(b) As we have to find here.
P(x > 50 + 25 l x > 50)
as we know that exponential distribution has memoryless property
P(x > 50 + 25 l x > 50) = P(x > 25)
= 1 - (1 - e-25/100) = exp(-0.25) = 0.7788
(c) Here for one car P(x > 50 hours) = 0.6065
P(Both will work) = 0.6065 * 0.6065 = 0.3679
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