he probability density function of the time to failure of an electronic componen

Question

he probability density function of the time to failure of an electronic component in a copier (in hours) is e^-x/1056/1056 for x>0. Determine the probability that
a) A component lasts more than 3000 hours before failure. (Round the answer to 3 decimal places.)

b) A component fails in the interval from 1000 to 2000 hours. (Round the answer to 3 decimal places.)

c) A component fails before 1000 hours. (Round the answer to 3 decimal places.)

d) Determine the number of hours at which 10% of all components have failed. (Round the answer to the nearest integer.)

Explanation / Answer

here this is pdf of exponential distribution for which cumulative distribution function =P(X<x) =1-e-x/1056

a) P(X>3000)=1-P(X<3000)=1-(1-e-3000/1056) =e-3000/1056 =0.0584

b)P(1000<X<2000)=(1-e-2000/1056)-(1-e-1000/1056) =0.2374

c)

P(X<1000)=(1-e-1000/1056) =0.6120

d) P(X<x)=0.10 =1-e-x/1056

e-x/1056 =1-0.1=0.9

taking log and solving

x=111.26 Hours

please revert for any clarification required

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