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MANE 3332.05

Lecture 12

Agenda

  • Continue Chapter 4 lectures - Normal Practice Problems
  • Old Midterm posted
  • Standard Normal Quiz (assigned 10/7/2025, due 10/9/2025)
  • Normal Practice Problems (assigned 10/9/2025, due 10/14/2025)
  • Schedule

Handouts

Tuesday Date and Topic(s) Thursday Date and Topic(s)
10/7: normal distribution 10/9: normal practice problems
10/14: Exponential and Weibull distributions 10/16: Chapter 5 (not on midterm)
**10/21: ** Midterm Review 10/23: Midterm Exam
Cumulative Standard Normal Distribution

page A-8

Cumulative Standard Normal Distribution

page A-9

Standardizing (the \(z\)-transform)

  • Suppose \(X\) is a normal random variable with mean \(\mu\) and variance \(\sigma^2\)
\[ P(X\leq x)=P\left(\frac{X-\mu}{\sigma}\leq\frac{x-\mu}{\sigma}\right)=P(Z\leq z) \]

  • The \(z\)-value is \(z=(x-\mu)/\sigma\)

  • Result allows the standard normal tables to be used to calculate probabilities for any normal distribution

Normal Practice Problems

Normal Approximation to the Binomial Distribution

  • If \(X\) is a binomial random variable,
\[ Z=\frac{X-np}{\sqrt{np(1-p)}} \]

is approximately a standard normal random variable. Consequently, probabilities computed from \(Z\) can be used to approximate probabilities for \(X\)

  • Usually holds when
\[ np>5\;\;\mbox{ and }n(1-p)>5 \]

Problem

image

  • How good are the approximations?

Continuity Correction Factor

  • Is a method to improve the accuracy of the normal approximation to the binomial

  • Examine Figure 6.22 from Walpole, Myers, Myers & Ye. Note that each rectangle is centered at \(x\) and extends from \(x-0.5\) to \(x+0.5\)

  • This table should help formulate problems

Binomial Probability with Correction Factor Normal Approximation
\(P(X\geq x)\) \(P(X\geq x-0.5)\) \(P\left(Z>\frac{x-0.5-np}{\sqrt{np(1-p)}}\right)\)
\(P(X\leq x)\) \(P(X\leq x+0.5)\) \(P\left(Z<\frac{x+0.5-np}{\sqrt{np(1-p)}}\right)\)
\(P(X=x)\) \(P(x-0.5\leq X\leq x+0.5)\) \(P\left(\frac{x-0.5-np}{\sqrt{np(1-p)}}<Z<\frac{x+0.5-np}{\sqrt{np(1-p)}}\right)\)

Normal Approximation - Figure

image

Rework Problem using Continuity Correction Factor

  • Are the approximations improved?

Normal Approximation to the Poisson Distribution

  • If \(X\) is a Poisson random variable with \(E(X)=\lambda\) and \(V(X)=\lambda\),
\[ Z=\frac{X-\lambda}{\sqrt{\lambda}} \]

is approximately a standard normal random variable.

Exponential Distribution

  • The exponential distribution is widely used in the area of reliability and life-test data.

  • Ostle, et. al. (1996) list the following applications of the exponential distribution

    • the number of feet between two consecutive erroneous records on a computer tape,

    • the lifetime of a component of a particular device,

    • the length of a life of a radioactive material and

    • the time to the next customer service call at a service desk

Exponential Distribution

  • The PDF for an exponential distribution with parameter \(\lambda >0\) is
\[ f(x)=\lambda e^{-\lambda x},\;\;\mbox{ for }0\leq x < \infty \]
  • The mean of \(X\) is
\[ \mu=E(X)=\frac{1}{\lambda} \]
  • The variance of \(X\) is
\[ \sigma^2=V(X)=\frac{1}{\lambda^2} \]

Note that other authors define \(f(x)=\frac{1}{\theta}e^{-x/\theta}\). Either definition is acceptable. However one must be aware of which definition is being used.

The Exponential CDF

The CDF for the exponential distribution is easy to derive

\[ \begin{aligned} F(x)&=&P(X\leq x)=\int_{-\infty}^x\lambda e^{-\lambda y}\,dy\\ &=&\int_{0}^x\lambda e^{-\lambda y}\,dy\\ &=&\left.\left(-e^{-\lambda y}\right)\right|_{y=0}^x\\ &=& -e^{-\lambda x}-(-e^0)\\ &=&-e^{-\lambda x}+1\\ &=&1-e^{-\lambda x}\end{aligned} \]

Problem 4-79

image

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

image

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

image

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

image

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

image

Weibull Practice Problems