---
title: "Exponential: semiconductor service life"
subject: "Probability and Statistics"
chapter: 3
key: "semester_5/probability_and_statistics:exponential-semiconductor-service-life"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/exponential-semiconductor-service-life
site: BCT Engineering
---

# Exponential: semiconductor service life

Probability and Statistics · Chapter 3: Continuous Probability Distributions

- Asked in: 2074 Chaitra, 5 marks
- Video: [The Exponential Distribution](https://www.youtube.com/watch?v=WKFk0NXyO50) — LearnChemE
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/exponential-semiconductor-service-life
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-3

## Question

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

## 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.

## Answer

**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%)**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. A semiconductor has exponentially distributed life with mean 60 hours. What is the rate parameter λ?
   A. 60 per hour
   B. 1/60 per hour
   C. 1/3600 per hour
   D. sqrt(60) per hour
   Answer: B. For the exponential distribution the mean is 1/λ, so λ = 1/60 = 0.01667 per hour. (Key point: Parameter from mean)

2. Which expression gives the probability that the semiconductor is still working after 90 hours?
   A. 1 − e^(−90/60)
   B. e^(−90/60)
   C. e^(−60/90)
   D. 90/60
   Answer: B. Reliability is P(X > x) = e^(−λx), so P(X > 90) = e^(−1.5) = 0.2231. (Key point: Reliability formula)

3. P(fail within 120 hours) is found as:
   A. e^(−2) = 0.1353
   B. 1 − e^(−2) = 0.8647
   C. e^(−0.5) = 0.6065
   D. 1 − e^(−0.5) = 0.3935
   Answer: B. Failing within 120 h is P(X ≤ 120) = 1 − e^(−120/60) = 1 − e^(−2) = 0.8647. (Key point: Distribution function)

4. What does the memoryless property of the exponential distribution say?
   A. P(X > s + t | X > s) = P(X > t)
   B. P(X > s + t) = P(X > s) + P(X > t)
   C. Mean equals variance
   D. The pdf is symmetric about the mean
   Answer: A. Having survived s hours does not change the chance of surviving a further t hours. (Key point: Memoryless property)

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