Continuous Random Variables & Probability Distributions – Detailed Study Notes

Chapter Outline

• 4.1 Probability Distributions & Probability Density Functions (PDFs)

• 4.2 Cumulative Distribution Functions (CDFs)

• 4.3 Mean & Variance of a Continuous Random Variable

• 4.4 Continuous Uniform Distribution

• 4.5 Normal Distribution

• 4.6 Normal Approximation to Binomial & Poisson

• 4.7 Exponential Distribution

• 4.8 Erlang & Gamma Distributions

• 4.9 Weibull Distribution

• 4.10 Lognormal Distribution

• 4.11 Beta Distribution

• Key terms list provided on “Important Terms & Concepts” slide (reinforce vocabulary for exam)

Learning Objectives (LO)

• LO-1 Determine probabilities directly from PDFs

• LO-2 Move interchangeably between PDFs & CDFs (integrate / differentiate)

• LO-3 Calculate means & variances for any continuous r.v.

• LO-4 Recognize core assumptions behind each family of continuous distributions

• LO-5 Select an appropriate model for real-world data / engineering applications

• LO-6 Compute $P(\cdot)$, $E(X)$, $V(X)$ for the common distributions in Ch. 4

• LO-7 Use a normal distribution to approximate binomial or Poisson when direct evaluation is unwieldy

4.1 Probability Distributions & PDFs

• A continuous random variable $X$ is fully described by a PDF $f(x)$ satisfying:

  • $f(x) \ge 0 \;\;\forall x$

  • $\int_{-\infty}^{\infty} f(x)\,dx = 1$

  • $P(a < X < b)=\int_{a}^{b} f(x)\,dx$

• Example 4.1 (Electric Current)

  • Current in thin Cu wire, domain $[4.9,5.1]$ mA, PDF $f(x)=5$ inside range, $0$ outside.

  • Probability current < $5$ mA:
    $P(X<5)=\int_{4.9}^{5}5\,dx = 5(5-4.9)=0.5$

  • Probability $4.95<X<5.1$ mA:
    $P(4.95<X<5.1)=\int_{4.95}^{5.1}5\,dx=5(5.1-4.95)=0.75$

  • Significance: uniform PDFs often arise from measurement tolerances / manufacturing specs.

4.2 Cumulative Distribution Functions (CDFs)

• Defined for every real number: $F(x)=P(X\le x)=\int_{-\infty}^{x} f(t)\,dt$

• Properties: non-decreasing, right-continuous, $\lim{x\to\infty}F(x)=1$, $\lim{x\to-\infty}F(x)=0$ .

• Example 4.3 (Current again): Piecewise CDF for uniform [4.9,5.1]
  F(x)=\begin{cases}0,&x<4.9\5(x-4.9),&4.9\le x\le5.1\1,&x>5.1\end{cases} ntiate → $f(x)=\frac{d}{dx}F(x)$

• Example 4.4 (Reaction Time) CDF:

• $ithin 200 ms:$ P(X<0.2)=F(0.2)=1-e^{-0.01(200)}=1-e^{-2}=0.8647

  • Practical tie-in: waiting‐time models for chemical kinetics.

4.3 Mean & Variance of a Continuous r.v.

• Definitions:
  \mu=E(X)=\int{-\infty}^{\infty}x f(x)\,dx$$\sigma^{2}=V(X)=\int{-\infty}^{\infty}(x-\mu)^{2}f(x)\,dx$</p></li><li><p>Expected value of a function$h(X)$:$E[h(X)]=\int h(x) f(x)\,dx$(crucial for engineering cost/energy calculations).</p></li><li><p>Example 4.6 (Power dissipated in resistor)</p><ul><li><p>$P=10^{-6} R I^{2}$, with$R=100$Ω, so$h(X)=10^{-6}(100)X^{2}=10^{-4}X^{2}$.</p></li><li><p>$E[P]=10^{-4}E[X^{2}]$, compute via uniform current PDF from Ex 4.1 → plug into integral to find expected power. (Exact numeric not given in slide; note method.)</p></li></ul></li></ul><h3 collapsed="false" seolevelmigrated="true">4.4 Continuous Uniform Distribution</h3><ul><li><p>PDF:$f(x)=\frac{1}{b-a}$for$a\le x\le b$; zero elsewhere.</p></li><li><p>Mean:$\mu=\frac{a+b}{2}$</p></li><li><p>Variance:$\sigma^{2}=\frac{(b-a)^{2}}{12}$</p></li><li><p>Example 4.7 (Current)</p><ul><li><p>$a=4.9$,$b=5.1$→$\mu=5\,\text{mA}$,$\sigma^{2}=\frac{0.2^{2}}{12}=0.0033$mA².</p></li><li><p>Desired probability$P(4.95<X<5.0)=\frac{5.0-4.95}{0.2}=0.25$.</p></li><li><p>Application: uniform models appear when every value in a spec range is equally likely (quality control).</p></li></ul></li></ul><h3 collapsed="false" seolevelmigrated="true">4.5 Normal Distribution</h3><ul><li><p>PDF:$f(x)=\frac{1}{\sqrt{2\pi\,\sigma^{2}}}\,e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}$,$-\infty<x<\infty$.</p></li><li><p>Notation:$X\sim N(\mu,\sigma^{2})$.</p></li><li><p>Properties: <br>$E(X)=\mu$,$V(X)=\sigma^{2}$.</p></li><li><p>Empirical Rule (68-95-99.7$P(\mu-\sigma<X<\mu+\sigma)\approx0.68$</p></li><li><p>$P(\mu-2\sigma<X<\mu+2\sigma)\approx0.95$</p></li><li><p>$P(\mu-3\sigma<X<\mu+3\sigma)\approx0.997$</p></li></ul></li><li><p>Standard Normal$Z$:$Z=\frac{X-\mu}{\sigma}\sim N(0,1); Appendix Table III & software supply \Phi(z)=P(Z\le z)$.</p></li><li><p>Example 4.9 (Table reading): Students practiced finding$P(Z<1.23)$and$P(Z>-0.56)$, reinforcing use of tables.</p></li><li><p>Example 4.11 (Normal current)</p><ul><li><p>$\mu=10$mA,$\sigma^{2}=4$→$\sigma=2$.</p></li><li><p>$P(X>13)=P\left(Z>\frac{13-10}{2}=1.5\right)=1-\Phi(1.5)=0.0668$.</p></li></ul></li><li><p>Example 4.12a$P(9<X<11)$: convert to$Z$bounds$-0.5,0.5$→ probability$\Phi(0.5)-\Phi(-0.5)=0.3829$.</p></li><li><p>Example 4.12b (Quantile)</p><ul><li><p>Find$x$s.t.$P(X<x)=0.98$→$z_{0.98}=2.05$→$x=\mu+z\sigma=10+2.05(2)=14.1$mA.</p></li></ul></li></ul><h3 collapsed="false" seolevelmigrated="true">4.6 Normal Approximation (Continuity Correction)</h3><ul><li><p>Binomial$B(n,p)$or Poisson$Pois(\lambda)$with large$n$/$\lambda$resemble normal: <br>$X\approx N(np, np(1-p))$or$X\approx N(\lambda,\lambda)$.</p></li><li><p>Continuity correction: include$\pm0.5$adjustment on discrete boundary.</p></li><li><p>Example 4.14 (Digital comms): demonstrated computational ease versus exact binomial.</p></li><li><p>Example 4.16 (Asbestos particles)</p><ul><li><p>$\lambda=1000$, want$P(X\le950)$.</p></li><li><p>Use correction$P(X\le950.5)$→$Z=\frac{950.5-1000}{\sqrt{1000}}=-1.57$→$P=0.058$.</p></li><li><p>Practical: industrial hygiene, assessing contamination counts rapidly.</p></li></ul></li></ul><h3 collapsed="false" seolevelmigrated="true">4.7 Exponential Distribution</h3><ul><li><p>Model for distance or waiting-time between Poisson events.</p></li><li><p>PDF:$f(x)=\lambda e^{-\lambda x},\;x\ge0$.</p></li><li><p>CDF:$F(x)=1-e^{-\lambda x}.

• Mean & variance: \mu=\frac{1}{\lambda}$,$\sigma^{2}=\frac{1}{\lambda^{2}}$.</p></li><li><p>Lack-of-Memory property:$P(X>s+t\mid X>s)=P(X>t)$, unique among continuous distributions.</p></li><li><p>Example 4.17a (Computer log-ons)</p><ul><li><p>Log-on rate$\lambda=25\,\text{h}^{-1}$.</p></li><li><p>$P(\text{no log-on in }0.1\,h)=e^{-25(0.1)}=0.082$.</p></li></ul></li><li><p>Example 4.17b</p><ul><li><p>$P(2\text{–}3\,\text{min})=0.152$.</p></li><li><p>Interval for$P(X>t)=0.90$gives$t=0.25$min (0.00421 h).</p></li><li><p>Mean = SD =$2.4$min.</p></li></ul></li><li><p>Example 4.18 (Geiger counter)</p><ul><li><p>$E(X)=1.4$min →$\lambda=\frac{1}{1.4}$.</p></li><li><p>$P(X<0.5)=1-e^{-0.5/1.4}=0.30.

• Lack-of-memory illustrated: waiting extra 3 min doesn’t change next-30-s probability.

4.8 Erlang & Gamma Distributions

• Gamma PDF: f(x)=\frac{\lambda^{r}x^{r-1}e^{-\lambda x}}{\Gamma(r)},\;x>0,\;r>0$.</p></li><li><p>Erlang = Gamma with integer$r$(often counts of Poisson events).</p></li><li><p>Mean$\mu=\frac{r}{\lambda}$, variance$\frac{r}{\lambda^{2}}$.</p></li><li><p>Example 4.19 (Processor failure)</p><ul><li><p>Poisson failures rate$0.0001\,h^{-1}$, want time to 4th failure.</p></li><li><p>$P(X>40{,}000)=P(N\le3)$with$N\sim Pois(4)$→$0.433$.</p></li></ul></li><li><p>Example 4.20 (Micro-array slides)</p><ul><li><p>Rate$\lambda=\frac{1}{2}$, r = 10.</p></li><li><p>$P(X>25)=P(Pois(12.5)\le9)=0.2014$(uses Poisson equivalence again).</p></li></ul></li><li><p>Gamma function reminder:$\Gamma(r)=(r-1)!$for integer$r$; extends factorial concept.</p></li></ul><h3 collapsed="false" seolevelmigrated="true">4.9 Weibull Distribution</h3><ul><li><p>PDF:$f(x)=\frac{\beta}{\delta}\left(\frac{x}{\delta}\right)^{\beta-1}e^{-(x/\delta)^{\beta}},\;x>0$.</p></li><li><p>CDF:$F(x)=1-e^{-(x/\delta)^{\beta}}$.</p></li><li><p>Mean$\mu=\delta\,\Gamma!\left(1+\frac{1}{\beta}\right)$.</p></li><li><p>Variance$\sigma^{2}=\delta^{2}\Gamma!\left(1+\frac{2}{\beta}\right)-\mu^{2}$.</p></li><li><p>Shape parameter$\beta<1$→ decreasing failure rate;$\beta=1$exponential;$\beta>1$increasing failure rate.</p></li><li><p>Example 4.21 (Bearing wear)</p><ul><li><p>$\beta=0.5$,$\delta=5000$h.</p></li><li><p>Mean$E(X)=5000\,\Gamma(1+2)=5000\,\Gamma(1.5)=4431.1$h.</p></li><li><p>$P(X>6000)=e^{-(6000/5000)^{0.5}}=0.237$→ only 23.7$W\sim N(\theta,\omega^{2})$, then$X=e^{W}$is lognormal:$\ln X\sim N(\theta,\omega^{2})$.</p></li><li><p>PDF:$f(x)=\frac{1}{x\omega\sqrt{2\pi}}e^{-\frac{(\ln x-\theta)^{2}}{2\omega^{2}}},\;x>0$.</p></li><li><p>Mean:$E(X)=e^{\theta+\omega^{2}/2}$.</p></li><li><p>Variance:$V(X)=\big(e^{\omega^{2}}-1\big)e^{2\theta+\omega^{2}}$.</p></li><li><p>Example 4.22a (Laser lifetime)</p><ul><li><p>$\theta=10$,$\omega=1.5$.</p></li><li><p>$P(X>10{,}000)=1-\Phi!\left(\frac{\ln10{,}000-10}{1.5}\right)=1-\Phi(-0.30)=0.701$.</p></li></ul></li><li><p>Example 4.22b (99$x$with$P(X>x)=0.01$→$\Phi\left(\frac{\ln x-10}{1.5}\right)=0.99$→$z=2.33$.</p></li><li><p>$\ln x=10+2.33(1.5)=13.495$→$x=\exp(13.495)=668.5$h.</p></li><li><p>Mean$E(X)=e^{10+1.125}=67{,}846.3$h, SD huge (≈197,662 h) illustrating skewness.</p></li></ul></li></ul><h3 collapsed="false" seolevelmigrated="true">4.11 Beta Distribution</h3><ul><li><p>Support on [0,1]. PDF:<br>$f(x)=\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\,x^{\alpha-1}(1-x)^{\beta-1}$.</p></li><li><p>Mean:$\mu=\frac{\alpha}{\alpha+\beta}$.</p></li><li><p>Variance:$\sigma^{2}=\frac{\alpha\beta}{(\alpha+\beta)^{2}(\alpha+\beta+1)}$.</p></li><li><p>Mode when$\alpha,\beta>1$:$\frac{\alpha-1}{\alpha+\beta-2}$.</p></li><li><p>Highly flexible for proportions, reliability, Bayesian priors.</p></li><li><p>Example 4.23 (CV joint service time)</p><ul><li><p>$\alpha=2.5$,$\beta=1$.</p></li><li><p>$P(X>0.7)=1-\int_{0}^{0.7} f(x)dx = 0.59$.</p></li><li><p>Interpretation: 59$Z.

• Normal approximations: to binomial & Poisson.

Connections & Practical Implications

• Selection of distribution grounded in mechanism (e.g., Poisson → exponential waiting, fatigue → Weibull, proportions → Beta).

• Approximations (normal) trade accuracy for computational ease; justify via large-sample theory.

• Engineering relevance: current tolerances, reaction completion, failure times, network log-ons all map naturally to presented distributions.

• Ethical/practical angle: choosing correct model affects safety margins (bearings, asbestos counts) & resource allocation (server capacity, staffing).

• Philosophical note: Lack-of-memory challenges intuition—past waiting doesn’t affect future risk in exponential setting.

Formula Summary (Cheat-Sheet)

• Uniform \mu=\frac{a+b}{2},\;\sigma^{2}=\frac{(b-a)^{2}}{12}$</p></li><li><p>Normal PDF$f(x)=\frac{1}{\sqrt{2\pi\sigma^{2}}}e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}$, standardization$Z=\frac{X-\mu}{\sigma}$</p></li><li><p>Exponential$f(x)=\lambda e^{-\lambda x},\;\mu=\sigma=1/\lambda$</p></li><li><p>Gamma$f(x)=\frac{\lambda^{r}x^{r-1}e^{-\lambda x}}{\Gamma(r)}$,$\mu=r/\lambda$</p></li><li><p>Weibull$f(x)=\frac{\beta}{\delta}(x/\delta)^{\beta-1}e^{-(x/\delta)^{\beta}}$</p></li><li><p>Lognormal mean$e^{\theta+\omega^{2}/2}$, var$\big(e^{\omega^{2}}-1\big)e^{2\theta+\omega^{2}}$</p></li><li><p>Beta mean$\alpha/(\alpha+\beta)$, var$\alpha\beta/[(\alpha+\beta)^{2}(\alpha+\beta+1)]$$

Keep these notes handy; they condense Chapter 4 into a quick-reference while preserving detailed worked examples and formulae for exam preparation.