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  • STA258 Lecture 04

STA258 Lecture 04 Raw

• ^[1]

  • How to visually see the CLT

• Central Limit Theorem

• $\overline{X}$ is our sample mean

  • $X_1,\dots ,X_n$ are iid
  • With $\mathrm{E}[X_i]=\mu$ and $\mathrm{Var}(X_i)={\sigma }^2$

• $\mathrm{E}[\overline{X}]=\mathrm{E}\left[\frac{1}{n}\sum _{i=1}^n{X}_i\right]=\frac{1}{n}[\mathrm{E}[X_1]+\cdots +\mathrm{E}[X_n]]=\frac{1}{n}[\mu +\mu +\cdots +\mu ]=\frac{1}{n}n\mu =\mu$

• Sample Mean and Sample Variance

• $$\mathrm{Var}(\overline{X})=\mathrm{Var}\left(\frac{1}{n}\sum _{i=1}^n{X}_i\right)=\frac{1}{n^2}\mathrm{Var}\left(\sum _{i=1}^n{X}_i\right)=\frac{1}{n^2}\mathrm{Var}(X_1+X_2+\cdots +X_n)=\frac{1}{n^2}[\mathrm{Var}(X_1)+\mathrm{Var}(X_2)+\cdots +\mathrm{Var}(X_n)]=\frac{1}{n^2}[{\sigma }^2+{\sigma }^2+\cdots +{\sigma }^2]=\frac{1}{n^2}(n{\sigma }^2)=\frac{{\sigma }^2}{n}$$

• Standardizing:

  • ${\displaystyle \frac{\overline{X}-\mathrm{E}[\overline{X}]}{\sqrt{\mathrm{Var}(\overline{X})}}}\sim N(0,1)⟹{\displaystyle \frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}}\sim N(0,1)$

• Suppose we have $Y\sim N(\mu ,{\sigma }^2)⟹kY\sim N(k\mu ,k^2{\sigma }^2)$

• So if we have $\left({\displaystyle \frac{1}{\frac{\sigma }{\sqrt{n}}}}\right)(\overline{X}-\mu )\sim N(0,1)⟹\overline{X}-\mu \sim N(0,\frac{{\sigma }^2}{n})$

• $\overline{X}\sim N(\mu ,\frac{{\sigma }^2}{n})$

• Another way to show this theorem:

  • $\overline{X}\sim N(\mu ,\frac{{\sigma }^2}{n})$
  • $\sum _{i=1}^n{X}_i\sim N(n\mu ,n{\sigma }^2)$

• Ex:

  • Number of accidents per week
  • $\mu =2.2$
  • $\mathrm{SD}[X]=1.4$
  • a:
    • $\overline{X}$ is the mean number of accidents per week
    • What is the approximate distribution of $\overline{X}$ according to CLT
    • It is normal
    • $n=52$
    • $\overline{X}\sim N(2.2,\frac{{1.4}^2}{\sqrt{52}})$ [COMPLETED] $⟹\overline{X}\sim N(2.2,\frac{1.96}{52})$
    • Sampling Distribution
    • Standard Error
  • b:
    • So we have $P(\overline{X}<2)$
    • $P({\displaystyle \frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}}<\frac{2-2.2}{\frac{1.4}{\sqrt{52}}})$
    • $P(Z<-1.0303)=0.1515$
  • c:
    • What is the prob there are $<100$ accidents
    • $X_i=\text{\# of accidents during week i}$
    • $P(\text{Total}<100)$
      • $=P(\sum _{i=1}^{52}{X}_i<100)$
      • $=P(\frac{1}{52}\sum _{i=1}^{52}{X}_i<\frac{100}{52})$
      • $=P(\overline{X}<\frac{100}{52})$
      • $=P(\frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{\frac{100}{52}-2.2}{\frac{1.4}{\sqrt{52}}})$
      • $=P(Z<-1.427)=0.0768$

• Ex:

  • Insurance
  • Population mean loss is $\mu =75$
  • $\sigma =300$
  • Distribution is right skewed
  • Most policies have $0$ loss
    • Mode
  • $\overline{X}\sim N(75,{\left(\frac{300}{\sqrt{n}}\right)}^2)$
  • $P(\overline{X}>85)$ with $n=10000$ [COMPLETED]
  • $=P(Z>\frac{85-75}{\frac{300}{\sqrt{10000}}})$
  • $=P(Z>\frac{85-75}{3})$
  • $=P(Z>3.3)=0.0004$
  • So it's a $1-0.0004=0.9996=99.96\mathrm{\%}$ chance that average losses do not exceed $85$.

• Ex:

  • Automotive batteries
  • $\mu =54$
  • $\mathrm{SD}[X]=6$
  • Suppose someone has $n=50$
  • a:
    • Assume claimed moments are true
    • Describe the Sampling Distribution of the mean lifetime of $50$ batteries
    • ${\sigma }_{\overline{X}}=\frac{\sigma }{\sqrt{n}}=\frac{\sigma }{\sqrt{50}}=0.8485$ months
    • ${\mu }_{\overline{X}}=54$
    • $\overline{X}\sim N(54,{0.8485}^2)$
    • #tk little confused on when to use each type of $\frac{{\sigma }^2}{n}$ or $\frac{\sigma }{\sqrt{n}}$…
    • $P(\overline{X}<52)=P(\frac{\overline{X}-{\mu }_{\overline{X}}}{{\sigma }_{\overline{X}}}<\frac{52-54}{0.8485})$
    • $P(Z<-2.357)=0.0081$

• Normal Approximation to Binomial Data

  • $X\sim \mathrm{Bin}(n,p)$
  • CLT tells us that $\frac{X-\mathrm{E}[X]}{\sqrt{\mathrm{Var}(X)}}\sim N(0,1)$
  • $\mathrm{E}[Y]=np$ [COMPLETED] (Variable switch to $Y$ in context of sums, or $X$)
  • $\mathrm{Var}(Y)=np(1-p)$
  • $\frac{X-np}{\sqrt{np(1-p)}}\sim N(0,1)$
    • Why we don't do $\frac{\sigma }{\sqrt{n}}$ because CLT doesn't necessarily require that, it just requires $\sqrt{\mathrm{Var}(X)}$
    • If we have a sample $\mathrm{Var}(X)=\mathrm{Var}(\overline{X})=\frac{{\sigma }^2}{n}$
    • So then $\sqrt{\mathrm{Var}(\overline{X})}=\frac{\sigma }{\sqrt{n}}$
  • We're approximating a discrete dist with a continuous distribution.
    • Because of this we need to correct
    • Instead of $P(X=b)$ we'll have $P(b-0.5\le X\le b+0.5)$
    • Instead of $P(Y\le b)$ we'll have $P(Y\le b+0.5)$
    • Instead of $P(b\le Z)$ we'll have $P(b+0.5\le Z)$ [COMPLETED] (Likely meant $P(X\ge b)\approx P(X\ge b-0.5)$, context implies direction matters)
  • Ex:
    • A Factory produces $300$ light bulbs per day
    • Each bulb has a $5\mathrm{\%}$ chance of being defective
    • What is the prob. that more than $20$ light bulbs are defective
    • $X=\text{\# of defective light bulbs}$
    • $P(X>20)$
    • Each is a binomial trial.
    • Since $X\sim \mathrm{Bin}(n=300,p=0.05)$
    • $\mathrm{E}[X]=np=(300)(0.05)=15.0$
    • $\mathrm{Var}(X)=np(1-p)=(15)(1-0.05)=14.25$
    • $\frac{X-15}{\sqrt{14.25}}\sim N(0,1)$
    • $Z\sim N(0,1)$
    • $P(X>20)$
    • Discrete so $P(X\ge 21)$
    • $P(Y\ge 21-0.5)$
    • $P(Y\ge 20.5)$
    • $P(Z\ge \frac{20.5-15}{\sqrt{14.25}})$
    • $\frac{20.5-15}{\sqrt{14.25}}=1.45698559277155$
    • $P(Z\ge 1.45698559277155)=0.0722$

• Footnotes