Models the time between independent events or process time that is memoryless.
Dependent on parameter ( \lambda ) (failure rate in units per hour).
Range is always positive; cannot be less than zero.
Parameter of interest:
( \lambda = \frac{1}{\beta} ) or vice versa.
( \beta ) represents the mean.
Mean time between failures: ( \text{Mean} = \beta ).
Probability function: ( \frac{1}{\beta} e^{-x/\beta} ).
( \mu = \beta ), ( \sigma^2 = \beta^2 ).
Example Problem: Washing Machine Repair
Scenario: Time ( Y ) in years before a major repair is required for a washing machine, modeled as an exponential random variable with ( \mu = 4 ) years.
Mean time between failures is 4 years.
Probability Calculations
Bargain Determination: Unlikely to require repair before 6th year.
Calculate ( P(Y > 6) ) using integration.
Integral setup: ( \int_6^\infty \frac{1}{4} e^{-y/4} dy ).
Substitution: ( u = y/4 ), ( du = \frac{1}{4} dy ).
Solution: ( e^{-6/4} = 0.2231 ) or 22% probability of no major repair before 6th year.
Alternative method: ( 1 - \int_0^6 f(x) dx ).
Probability of repair before 6th year: ( 1 - 0.2231 = 0.7769 ) or 78%, meaning 78% are not bargains.
First Year Repair Probability:
Calculate ( P(0 < Y < 1) ).
Integral setup: ( \int_0^1 \frac{1}{4} e^{-y/4} dy ).
Solution: ( 1 - e^{-1/4} = 0.221 ) or 22% probability of requiring a major repair in the first year.
Key Takeaways on Exponential Distribution
Focus on time between independent events.
Consider failure rate as per context.
Application-oriented: Identify where exponential distribution is applicable.