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## statistics cheat sheet

Sumemr 2008   Compiled on January 28, 2024 at 9:24pm

### 2 second cheat sheet

problem: phone calls received at rate $$\lambda =2$$ per hr. If person wants to take 10 min shower, what is probability a phone will ring during that time?

answer: ﬁrst change to $$\omega =\lambda \frac {10}{60}=2\frac {10}{60}=.3333$$, now we want $$P(X\geq 1)=1-P(X\leq 1)=1-P(0)$$

but $$P\left ( k\right ) =\frac {\lambda ^{k}}{k!}e^{-\lambda }$$, but remember, we are using $$\omega$$, so $$P\left ( k\right ) =\frac {\omega ^{k}}{k!}e^{-\omega }$$ so $$P\left ( 0\right ) =\frac {.3333^{0}}{0!}e^{-.3333}=0.777$$

so $$P(X\geq 1)=1-.777=0.283$$, so $$28\%$$ change the phone will ring.

How long can shower be if they wish probability of receiving no phone calls to be at most 0.5?

$$P\left ( 0\right ) =0.5=\frac {\omega 0}{0!}e^{-\omega }\rightarrow 0.5=e^{-\omega }$$ hence $$\ln 0.5=-\omega \rightarrow \omega =0.693$$, so $$\lambda \frac {x}{60}=0.693\rightarrow x=20.7$$ minutes

To ﬁnd quantile, say $$\frac {1}{4}$$, ﬁrst ﬁnd an expression for $$F\left ( x\right )$$ as function of $$x$$, then solve for $$x$$ in $$F\left ( x\right ) =.25$$

For median, solve for $$x$$ in $$F\left ( x\right ) =.5$$

properties of CDF: 1. Show $$F\left ( x\right ) \geq 0$$ for all $$x.$$ Do this by showing $$F^{\prime }\left ( x\right ) \geq 0$$, and show limit $$F\left ( x\right ) \rightarrow 1$$ as $$x\rightarrow \infty$$ and limit $$F\left ( x\right ) \rightarrow 0$$ as $$x\rightarrow -\infty$$. And $$P\left ( k_{1}\leq T<k_{2}\right ) =F\left ( k_{2}\right ) -F\left ( k_{1}\right )$$

properties of pdf:

1. piecewise continuous
2. pdf$$\left ( x\right ) \geq 0$$
3. $$\int _{-\infty }^{\infty }pdf\left ( x\right ) =1$$

remember $$\frac {d}{dx}\tan ^{-1}x=\frac {1}{1+x^{2}}$$

The geometric distribution is the only discrete memoryless random distribution. It is a discrete analog of the exponential distribution. continuous

Some relations  $\sum _{k=1}^{n}k=\frac {1}{2}n\left ( n+1\right )$ Geometric sum ${\sum \limits _{k=0}^{n}}r^{k}=\frac {1-r^{n+1}}{1-r}$ if $$-1<r<1$$, then ${\sum \limits _{k=0}^{\infty }}r^{k}=\frac {1}{1-r}$ if the sum is from 1 then ${\sum \limits _{k=1}^{n}}r^{k}=\frac {r\left ( 1-r^{n+1}\right ) }{1-r}$ if $$-1<r<1$$, then \begin {align*} {\sum \limits _{k=1}^{\infty }}r^{k} & =\frac {r}{1-r}\\ \Gamma \left ( x\right ) & =\int _{0}^{\infty }u^{x-1}e^{-u}du\\ \Gamma \left ( n\right ) & =\left ( n-1\right ) !\\ \frac {d}{dx}\ln \left ( x\right ) & =\frac {1}{x}\\ \int \ln \left ( y\right ) dy & =-y+y\ln \left ( y\right ) \\ \int \frac {1}{y}dy & =\ln \left ( y\right ) \end {align*}

And

$\begin {pmatrix} & n & \\ n_{1} & n_{2} & n_{3}\end {pmatrix} =\frac {n!}{n_{1}!\ n_{2}!\ n_{3}!}$ If given joint density $$f_{XY}\left ( x,y\right )$$ and asked to ﬁnd conditional $$P\left ( X|Y\right ) =\frac {f_{XY}\left ( x,y\right ) }{f_{Y}\left ( y\right ) }$$ so need to ﬁnd marginals. Marginal is found from $$f_{Y}\left ( y\right ) =\int _{x}f_{XY}\left ( x,y\right ) dx$$, and $$f_{X}\left ( x\right ) =\int _{y}f_{XY}\left ( x,y\right ) dy$$

To convert from $$x,y\,\$$to polar, example: given $$f\left ( x,y\right ) =c\sqrt {1-\left ( x^{2}+y^{2}\right ) }$$ ﬁnd $$c$$, where $$x^{2}+y^{2}\leq 1$$, then write

$c\int _{\theta =-\pi }^{\theta =\pi }\int _{r=0}^{r=1}\sqrt {1-r^{2}}rdrd\theta$ Use identity above.

law of total probablity: if we know $$Y|X$$ and $$X$$ and want to know distribution of $$Y$$, then $$f\left ( Y\right ) =\int _{-\infty }^{\infty }f_{Y|X}\left ( y|x\right ) f_{X}\left ( x\right ) dx$$

$Z=\frac {\bar {X}_{n}-\mu }{\sigma /\sqrt {n}}\rightarrow N(0,1)$

$T=\frac {\bar {X}_{n}-\mu }{S_{n}/\sqrt {n}}\rightarrow T(n)$ where $$S_{n}\$$is $$std$$ of the sample.

Note Var(sample) has chi square (n) distribution.

CI for T: \begin {align*} \Pr \left ( -A<\frac {\bar {X}_{n}-\mu }{S_{n}/\sqrt {n}}<A\right ) & =1-\alpha \\ \Pr \left ( \bar {X}_{n}-A\frac {S_{n}}{\sqrt {n}}<\mu <\bar {X}_{n}+A\frac {S_{n}}{\sqrt {n}}\right ) & =1-\alpha \end {align*}