Skip to main content

Exponential: semiconductor service life

Short answer

Exponential distribution has pdf f(x)=λe^(−λx), x≥0, with mean 1/λ. For mean life 60 h, λ=1/60. a) P(X>90)=e^(−1.5)≈0.2231. b) P(X≤120)=1−e^(−2)≈0.8647.

Define exponential distribution. Suppose that the service life of a semiconductor is exponentially distributed with an average of 60 hours. Find the probability that a semiconductor will a) still working after 90 hours b) fail within 120 hours

Exponential distribution: a continuous random variable X (waiting time or life until an event) has an exponential distribution with parameter λ > 0 if its pdf is

f(x) = λ e^(−λx), x ≥ 0 (and 0 otherwise)

  • Distribution function: F(x) = P(X ≤ x) = 1 − e^(−λx)
  • Reliability: P(X > x) = e^(−λx)
  • Mean = 1/λ, variance = 1/λ², standard deviation = 1/λ
  • It has the memoryless property: P(X > s + t | X > s) = P(X > t).

Given data

Mean life = 1/λ = 60 hours, so λ = 1/60 = 0.01667 per hour.

a) Still working after 90 hours

P(X > 90) = e^(−λx) = e^(−90/60) = e^(−1.5) = 0.2231

b) Fail within 120 hours

P(X ≤ 120) = 1 − e^(−120/60) = 1 − e^(−2) = 1 − 0.1353 = 0.8647

Answer: a) 0.2231 (about 22.31%) b) 0.8647 (about 86.47%)

✓ Answer checked by Claude Sonnet 5.5

Related questions

All Probability and Statistics chapter 3 questions →