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MANE 3332.03

Lecture 14, March 11

Agenda

  • Complete Chapter 4 lecture
  • Today's Topics: Weibull distribution, Tech Report One Assignment, and start R
  • New: Exponetial Quiz (assigned 3/11/2025, due 3/13/2025)
  • New: Weibull Practice Problems (assigned 3/11/2025, due 3/13/2025)
  • New: Tech Report One Assignment (assigned 3/6/2025, due 4/3/2025)

Handouts

Schedule

  • March 11: Weibull distribution, Tech Report One and R (not on midterm)
  • March 13: Mid-term review
  • March 18: Spring Break
  • March 20: Spring Break
  • March 23: Chapter 6
  • March 25: Midterm

Lack of Memory Property

  • The mathematical definition is
\[ P(X<t_1+t_2|X>t_1)=P(X<t_2) \]

  • That is "the probability of a failure time that is less than \(t_1+t_2\) given the failure time is greater than \(t_1\) is the probability that the item's failure time is less than \(t_2\)

  • This property is unique to the exponential distribution

  • Often used to model the reliability of electronic components.

Problem 4--80

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Relationship to the Poisson Distribution

  • Let \(Y\) be a Poisson random variable with parameter \(\lambda\). Note: \(Y\) represents the number of occurrences per unit

  • Let \(X\) be a random variable that records the time between occurrences for the same process as \(Y\)

  • \(X\) has an exponential distribution with parameter \(\lambda\)

Lognormal Distribution

  • Let \(W\) have a normal distribution with mean \(\theta\) and variance \(\omega^2\); then \(X=\exp(W)\) is a lognormal random variable with pdf
\[ f(x)=\frac{1}{x\omega\sqrt{2\pi}}\exp\left[-\frac{\left(\ln(x)-\theta\right)^2}{2\omega^2}\right]\;\;0<x<\infty \]
  • The mean of \(X\) is
\[ E(X)=e^{\theta+\omega^2/2} \]
  • The variance of \(X\) is
\[ V(X)=e^{2\theta+\omega^2}\left(e^{\omega^2}-1\right) \]

Example Problem

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Gamma Distribution

  • The random variable \(X\) with pdf
\[ f(x)=\frac{\lambda^rx^{r-1}e^{-\lambda x}}{\Gamma(r)},\;\;\mbox{for }x>0 \]

is a gamma random variable with parameters \(\lambda>0\) and \(r>0\).

  • The gamma function is
\[ \Gamma(r)=\int_0^\infty x^{r-1}e^{-x}\,dx\;\;\mbox{for }r>0 \]

with special properties:

  • \(\Gamma(r)\) is finite

  • \(\Gamma(r)=(r-1)\Gamma(r-1)\)

  • For any positive integer \(r\), \(\Gamma(r)=(r-1)!\)

  • \(\Gamma(1/2)=\pi^{1/2}\)

Gamma Distribution

  • The mean and variance are
\[ \mu=E(X)=r/\lambda\mbox{ and }\sigma^2=V(X)=r/\lambda^2 \]
  • We will not work any probability problems using the gamma distribution

Gamma Tables

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Weibull Distribution

  • The random variable \(X\) with pdf

$$ f(x)=\frac{\beta}{\delta}\left(\frac{x}{\delta}\right)^{\beta-1}\exp\left[-\left(\frac{x}{\delta}\right)^\beta\right],\;\; \mbox{ for }x>0 $$ is a Weibull random variable with scale parameter \(\delta>0\) and shape parameter \(\beta>0\)

  • The CDF for the Weibull distribution is
\[ F(x)=1-\exp\left[-\left(\frac{x}{\delta}\right)^\beta\right] \]
  • The mean of the Weibull distribution is
\[ \mu=E(X)=\delta\Gamma\left(1+\frac{1}{\beta}\right) \]
  • The variance of the Weibull distribution is
\[ \sigma^2=V(X)=\delta^2\Gamma\left(1+\frac{2}{\beta}\right)-\delta^2\left[\Gamma\left(1+\frac{1}{\beta}\right)\right]^2 \]

Weibull Problem

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Weibull Practice Problems