The probability density function of the time to failure of an electronic component in a copier (in hours) is f(x) = (e^-x/1076)/(1076) 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.)

(a) 0.06154

(b) 0.2389

(c) 0.6052

(d) 2478

Step-by-step explanation:

probability density function of the time to failure of an electronic component in a copier (in hours) is

P(x) = 1/1076e^−x/1076

λ = 1/1076

A) A component lasts more than 3000 hours before failure:

P(x>3000) = 1 − e^−3000/1076

= 0.06154

B) A component fails in the interval from 1000 to 2000 hours:

P(1000>x>2000) =1 − e^−2000/1076 − 1 +e^−1000/1076 = e^−1000/1076 − e^−2000/1076 = 0.3948 − 0.1559

= 0.2389

C) A component fails before 1000 hours:

P(x<1000) = 0.6052

D) The number of hours at which 10% of all components have failed:

e^−x/1076 = 0.1

= −x/1076

= ln(0.1)

x =(2.3026)×(1076)

x = 2478

its d but looking at that hurts my eyes frfr

step-by-step explanation:

the answer is 175.4379

step-by-step explanation: