013.tex

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\title{Engineering Statistics Lecture XIII}
\author{Jonathan Bender}
\date{October 17, 2019}

\begin{document}
	
	\maketitle
	
	\begin{abstract}	
		HW \#3 is due October 31, 2019:
		\begin{itemize}
			\item Chapter 4 \#1-31 odd
		\end{itemize}
	
		HW \#4 is due October 20, 2019 at 8:00 PM:
		
		Submit 2 opportunity questions with, preferably, correct answers.
		\begin{itemize}
			\item Multiple guess questions
			\item Extended response questions
		\end{itemize}
	
		Opportunity \#0 Given October 22, 2019;
		
		Opportunity \#0 Due October 24, 2019.
	\end{abstract}
	
	\section{Brendan's concrete something or another}
		f(x) is a line from [0,2] over [0,1], then from [2,0] over [1,2]
		\subsection{Find A: A = 1}
		\subsection{Find $\mu_w:\ \mu_w = 1$}
	
	
		\pagebreak
		\subsection{Find $\sigma_w^2$}
			Variance: $$ \sigma_w^2 = \dfrac{\sum_{i=1}^{n}(f(x_i) - \mu)^2}{n-1} = \int_{all\ x}(f(x) - \mu)dx$$
		
			$$\to \sigma_w^2 = E[x^2] - E[x]^2$$
		
			Big brain time: $\mu = E[x] = \int xf(x)dx$.
	
			\begin{align*}
				E[w^2] = \int w^2f(w)dw &= \int_0^1 w^2*wdw + \int_1^2 w^2(2-w)dw \\
				&= \frac{w^4}{4}|_0^1 + \frac{2}{3}w^3|_1^2 - \frac{w^4}{4}|_1^2\\
				&= \frac{1-0}{4} + \frac{2}{3}(8-1) - \frac{1}{4}(16 - 1)\\
				&= \frac{7}{6}
			\end{align*}
	
		\subsection{Margin: 10w - 20}
		\subsection{Find E[Margin]}
			\begin{align*}
				E[Margin] &= \int(10W-20)f(w)dW \\
				&= 10\int(W-2)f(W)dW \\
			\end{align*}
	
			For a linear function f(x) of random variable X,
			E[f(x)] = $A\mu_x + b$ $$\to E[f(x)] = AE[x] + B$$
			$$\to E[10W - 20] = 10(1) - 20 = -10$$ (See 1.1)
	
			$$\to \sigma^2(f(x)) = A^2\sigma_x^2$$
	
	\pagebreak
		\subsection{Find cumulative distribution function (CDF)}
			CDF - Shows how probability mass accumulates as x goes from $-\infty$ to X.
	
			\begin{align*}
				f(X) CDF &= \int_{-\infty}^Xf(\tau)d\tau\\
				         &= \int_0^X\tau d\tau \iff x\leq 1 \\
				     OR\ &= \int_0^1\tau d\tau + \int_1^X (2-\tau )d\tau \iff X\in [1,2] \\
				         &= \frac{1}{2}\tau^2|_0^1 + [2\tau - \frac{1}{2}\tau^2]^X_1 \\
				         &= 2x - \frac{1}{2}x^2 - 1 \iff X\in [1,2]
			\end{align*}
	
	\subsection{Find p($\frac{2}{5}\leq W \leq \frac{4}{3})$}
		\begin{align*}
			&= CDF(\frac{4}{3}) - CDF(\frac{2}{5}) \\
			&= -\frac{1}{2}(\frac{4}{25}) + \frac{1}{2} + 2\frac{4}{3} - \frac{1}{2}\frac{16}{9} - 2 + \frac{1}{2} 
		\end{align*}
	
\end{document}