Uniform distribution

• $X$: uniformly distributed aver the interval $(a,b)$ if pdf is given by: $$f_X(x)=\{\begin{array}{ll}\frac{1}{b-a}& a<x<b\\ 0& \text{otherwise}\end{array}$$
  • denoted by: $X\sim U(a,b)$
  • every neighbor: equally likely
• CDF: $$F_X(x)=\underset{-\infty }{\overset{x}{\inf }}{f}_X(y)dy=\{\begin{array}{ll}0& x<a\\ \frac{x-a}{b-a}& a\le x<b\\ 1& \text{otherwise}\end{array}$$
• similarly shows that: $E(X)=\frac{a+b}{2}$, $Var(X)=\frac{(b-a{)}^2}{12}$

Problems

• buses: arrive at stop on 15-min intervals from 7:15 am
  • passenger: arrives at stop uniformly at random time between $7:00,7:30$.
  • find probability: the man arriving in less that 5 mins after arrival
    • i.e. $7:10-15,7:25-30$
    • $P(10<X<15\text{ or }25<X<30)$
    • $10/30=1/3$
  • find probability: the man arriving more than 10 minutes late
    • 👨‍🏫 exercise!

Normal distribution

• $X$: normally distributed w/ parameters $\mu ,{\sigma }^2$, pmf given by: $$f_X(x)=\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{1}{2}(\frac{x-\mu }{\sigma }{)}^2}$$
• notation: $X\sim N(\mu ,{\sigma }^2)$
  • $E(X)=\mu ,Var(X)={\sigma }^2$
• density function: bell-shaped, always positive. symmetric on and peaks at $\mu$

Standard normal r.v.

• $Z\sim N(0,1)$, pdf: $\varphi (x)$, cdf: $\Phi$
  • $\varphi (x)=\frac{1}{\sqrt{2\pi }}{e}^{-x^2/2}$
  • $\Phi (x)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^x{e}^{-y^2/2}dy$
• theorem: for $X\sim N(\mu ,{\sigma }^2)$, $Z\sim N(0,1)$ $$\begin{array}{rl}Z& =\frac{X-\mu }{\sigma }\sim N(0,1)\\ P(a<X\le b)& =P(\frac{a-\mu }{\sigma }<Z\le \frac{b-\mu }{\sigma })\\ & =\Phi (\frac{b-\mu }{\sigma })-\Phi (\frac{a-\mu }{\sigma })\end{array}$$
• note: don't try to evaluate cdf of general normal distribution by hand
  • simple impossible :p
• some intuitions
  • bell curve's left half: same size as right
  • total area / region: 1
  • for $X\sim N(\mu ,{\sigma }^2)$, $P(X\le \mu )=P(X\ge \mu )=\frac{1}{2}$
• properties of the standard normal
  1. $P(Z\ge 0)=P(Z\le 0)=0.5$ (aligned by center)
  2. $-Z\sim N(0,1)$ (symmetric)
  3. $P(Z\le x)=1-P(Z>x)$ for $x\in (-\infty ,\infty )$
  4. $P(Z\le -x)=P(Z\ge x)$ for $x\in (-\infty ,\infty )$
  5. ⭐ if $Y\sim N(\mu ,{\sigma }^2)$. then $Z=\frac{Y-\mu }{\sigma }\sim N(0,1)$
  6. ⭐ if $Z\sim N(0,1)$. then $Y=aZ+b\sim N(b,1^2)$ for $a,b\in \mathbb{R}$
     • 👨‍🏫 more generally: linear transformation of a normal r.v. is still a normal r.v.
  • $\forall x,\Phi (-x)=1-\Phi (x)$
• $q$-th quantile of a r.v. $X$: $z_q$ s.t. $P(X\le z_q)=q$
  • for $Z\sim N(0,1)$, $\Phi (z_q)=q$

• when $X\sim N(65,5^2)$, compute $P(47.5<X\le 80)$
  • $=P(X\le 80)-P(X\le 47.5)=P(Z\le 15/5)-P(Z\le -17.5/5)=\Phi (3)-(1-\Phi (3.5))\approx 1$
• width of a slot of duralumin in forging: normally distributed with $(\mu ,\sigma )=(0.9000,0.0030)$
  • specification limits: $0.9000\pm 0.0050$
  • what percentage of forgings will be defective?
    • $P(0.9000-0.0050>X)+P(0.9000+0.0050<X)=P(-5/3>Z)+P(5/3<Z)=2\Phi (-5/3)$
    • $=2(1-\Phi (5/3))=9.5\%$
  • what is the maximum allowable value of $\sigma$ to permit no more than 1 in 100 defectives, when $\mu =0.9000$?
    • $2\Phi (-0.0050/\sigma )\le 0.01$,$z_{0.005}\approx -2.576$, $\sigma =-0.0050/(-2.576)=0.00194$