Six Transistor
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Calculus question involving integrals?
a.) The probably of a transistor failing between t = a months and t = b months is given by "c ∫[a, b] e^(-ct)dt", for some constant c. If the probability of failure within the first six months is 10%, what is c?
b.) Given the value of c in part (a), what is the probability that transistor fails within the second six months?
a) c S e^(-ct)dt = c / -c e^(-ct) = - e^(- ct)
from 0 to 6 months: it becomes - (e^-6t - 1) = 0.1
so e^-6c = 0.9
- 6c = ln (0.9)
c = - ln(0.9) / 6
b) use the value of c found in (a) and solve the integral:
c ∫[6, 12] e^(-ct)dt
