STA260 Lecture 03

• If we have $x_1,x_2,\dots ,x_n\sim N(\mu ,{\sigma }^2)$
• $\overline{X}=\frac{1}{n}\sum _{i=1}^n{X}_i$ where ($\overline{X}\sim N(\mu ,\frac{{\sigma }^2}{n})$)-> Central Limit Theorem. This is just one case
• If $X_i\sim N({\mu }_i,{\sigma }_i^2)$
• Then $\sum _{i=1}^n{a}_i{X}_i\sim N(\sum _{i=1}^n{a}_i{\mu }_i,\sum _{i=1}^n{a}_i^2{\sigma }_i^2)$
  • #tk expected to remember
• A consequence of $\overline{X}\sim N(\mu ,\frac{{\sigma }^2}{n})$
• $\overline{X}\sim N(\mu ,\frac{{\sigma }^2}{n})$
• Difference in $\mu$ affects the mean median mode.
• ${\sigma }^2$ obv affects dispersion
• Normal Distribution
• $\overline{X}-\mu$ we shift it towards the middle
• So $\overline{X}-\mu \sim N(0,\frac{{\sigma }^2}{n})$
• This is how we standardize the Normal Distribution to become a Standard Normal Distribution.
• $\frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}\sim N(0,1)$
• So now we have the final Standard Normal Distribution
• So now $Z\sim N(0,1)$
• $Z={\displaystyle \frac{\overline{X}-\mu }{\frac{{\sigma }^2}{n}}}$
• $X_1,X_2,\dots ,X_n\stackrel{iid}{\sim }N(\mu ,{\sigma }^2)$
• For $\overline{X}=\frac{1}{n}\sum _{i=1}^n{X}_i$
• Since our standardization is $Z={\displaystyle \frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}}\sim N(0,1)$
• Used to we can use z-score tables to evaluate probabilities.
• Ex:
  • Original problem has: $\overline{X}\sim N(\mu ,\frac{{\sigma }^2}{n})$
  • We want $P(a\le \overline{X}\le b)$
  • Then we convert it into $P(c\le Z\le d)$ where $Z\sim N(0,1)$
  • Then use the Z-Score table to find the output.
• Example:
  • $\overline{X}\sim N(\mu ,1)$
  • $n=9$
  • $P(\mu -0.3\le \overline{X}\le \mu +0.3)$
    • You can rewrite it into:
    • $P(|\overline{X}-\mu |\le 0.3)$
    • $P(-0.3\le \overline{X}-\mu \le 0.3)$
    • $P(-\frac{0.3}{\frac{\sigma }{\sqrt{n}}}\le \frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}\le \frac{0.3}{\frac{\sigma }{\sqrt{n}}})$
    • Because $\frac{\sigma }{\sqrt{n}}>0$, it's positive, we don't need to worry about flipping the inequality signs.
    • $P(-\frac{0.3}{\frac{\sigma }{\sqrt{n}}}\le Z\le \frac{0.3}{\frac{\sigma }{\sqrt{n}}})$
    • $P(-\frac{0.3}{\frac{1}{3}}\le Z\le \frac{0.3}{\frac{1}{3}})$
    • $P(-0.9\le Z\le 0.9)$
    • $1-(2)(0.1841)=0.6318$

  \begin{document}
    \begin{tikzpicture}[>=stealth, scale=3]
   % Define the area to be filled (the centre)
   \fill[color=blue!15] plot[domain=-0.9:0.9, samples=100] (\x, {exp(-0.5*\x*\x)}) -- (0.9,0) -- (-0.9,0) -- cycle;
  
   % Draw the normal distribution curve
   \draw[thick, color=blue!80!black] plot[domain=-3:3, samples=100] (\x, {exp(-0.5*\x*\x)});
  
   % Draw the horizontal axis
   \draw[->] (-3.5,0) -- (3.5,0) node[right] {$z$};
  
   % Vertical dashed lines at z = -0.9 and z = 0.9
   \draw[dashed, thin] (-0.9,0) -- (-0.9, {exp(-0.5*0.81)});
   \draw[dashed, thin] (0.9,0) -- (0.9, {exp(-0.5*0.81)});
  
   % Labels for the z-axis coordinates
   \node[below] at (-0.9,0) {\small $-0.9$};
   \node[below] at (0.9,0) {\small $0.9$};
   \node[below] at (0,0) {\small $0$};
  
   % Label for the central area
   \node at (0, 0.4) {\scalebox{0.9}{$0.6318$}};
  
   % Labelling the symmetrical tails
   % Left tail
   \draw[<-] (-1.3, 0.1) -- (-2, 0.5) node[above] {\small $0.1841$};
   % Right tail
   \draw[<-] (1.3, 0.1) -- (2, 0.5) node[above] {\small $0.1841$};
  
   % Symmetry indicator arrow
   \draw[<->, bend left=30, color=gray] (-1.1, 1.1) to node[above, color=black] {\footnotesize Symmetry} (1.1, 1.1);
  
   % Title or formula context
   \node[above] at (0, 1.3) {\footnotesize $P(-0.9 \leq Z \leq 0.9) = 1 - 2(0.1841)$};
    \end{tikzpicture}
  \end{document}
  

• How big of a sample size do we want if the sample mean should be within $0.3$ of $\mu$ with probability $0.95$
  • $P(-0.3\le \overline{X}-\mu \le 0.3)=0.95$
  • $P(-\frac{0.3}{\frac{1}{\sqrt{n}}}\le Z\le \frac{0.3}{\frac{1}{\sqrt{n}}})=0.95$
  • $P(-0.3\sqrt{n}\le Z\le 0.3\sqrt{n})=0.95$
  • $P(\left|Z\right|\le 0.3\sqrt{n})=0.95$
  • Let $0.3\sqrt{n}=a$
  • $P(\left|Z\right|\le a)=0.95$
  • $1-2a=0.95$
  • $a=0.025$
  • $0.3\sqrt{n}=1.96$ because the values at the $\pm a$ on the z-score table results in this.
  • $0.3\sqrt{n}=1.96$
  • $n\approx 42.684$
  • Find ceiling so $n=43$, always round up with sampling.
• Central Limit Theorem#STA260
• Review from STA256:
• CDF is $F_X(x)=P(X\le x)$
• Convergence in Distribution
• Let $X_1,X_2,\dots$ be a sequence of random variables.
• With corresponding CDFs
  • $\begin{array}{l}{X}_1\sim F_{X_1}(x)\\ X_2\sim F_{X_2}(x)\\ \dots \end{array}$
  • Let $X$ have a CDF: $F_X(x)$
  • $X_1,X_2,\cdots \stackrel{D}{\to }X$ if the limit as $\underset{n\to \mathrm{\infty }}{lim}{F}_{X_n}(x)=F_X(x)\mathrm{\forall }x\in \mathbb{R}$
  • Where $F_X(x)$ is continuous.
  • Denoted as $X_n\stackrel{D}{\to }X$
• Showing how Central Limit Theorem shows us that the distribution of means is normal.
• Let $X_1,X_2,\dots ,X_n$ be an iid sequence of $n$ random variables with $\mathrm{E}[X_i]=\mu$ and $\mathrm{Var}(X_i)={\sigma }^2$
• Let $\overline{X}=\frac{1}{n}\sum _{i=1}^n{X}_i$
• The Central Limit Theorem tells us $\overline{X}\stackrel{D}{\to }N(\mu ,\frac{{\sigma }^2}{n})$
• Also $\frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}\stackrel{D}{\to }N(0,1)$
• Example:
  • $\mu =1.5$
  • $\sigma =1$
  • Approximate the probability that $n=100$ can be served in $2$ hours
  • $2\mathrm{h}⟹120\mathrm{mins}$
  • $P(\sum _{i=1}^n{X}_i<120)$
  • Since we don't know the distribution of each $X_i$, we have to use CLT.
  • $X_i\stackrel{iid}{\sim }?(\mu =1.5,\sigma =1)$
  • $P(\sum _{i=1}^{100}{X}_i<120)$
  • By Central Limit Theorem: $\overline{X}\sim N({\mu }_{\overline{X}}=\mu ,{\sigma }_{\overline{X}}^2=\frac{{\sigma }^2}{n})$
  • $P(n\overline{X}<120)$
  • $P(\overline{X}<\frac{120}{n})$
  • $P(\overline{X}<1.2)$
  • $P(\overline{X}-\mu <1.2-\mu )$
  • $P(\frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{1.2-\mu }{\frac{\sigma }{\sqrt{n}}})$
  • $P(10(\overline{X}-1.5)<-3)$
  • $P(Z<-3)$
  • $\stackrel{\text{By Symmetry}}{=}0.00135$
• Chi-squared Distribution
• Gamma Function
• Gamma Distribution
  STA260 Post-Lecture 03
  STA260 Lecture 03