Setting up Exam 1 Review
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<xi:include href="./notes/1-22.ptx" />
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<xi:include href="./notes/1-27.ptx" />
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<xi:include href="./notes/1-29.ptx" />
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<xi:include href="./notes/2-3.ptx" />
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</chapter>
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<chapter xml:id="quizzes">
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<xi:include href="./recitations/calculus-review.ptx"/>
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</chapter>
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<chapter xml:id="review">
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<title>Exam Review</title>
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<xi:include href="./review/Exam-1-Review.ptx"/>
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</chapter>
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</book>
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</pretext>
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<?xml version="1.0" encoding="UTF-8"?>
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<section xml:id="notes-02-03">
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<title>Tuesday, Feb 3</title>
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<introduction>
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<p>
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This is an outline of the topics we covered in class.
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These notes are <em>not</em> a substitute for your own note-taking.
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I highly recommend that you take your own notes during class.
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If you ever miss a class for any reason, reach out to another student in class to get a copy of their notes.
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</p>
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</introduction>
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<subsection>
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<title>Variance</title>
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<p>
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Question: How spread out are <m>X</m> values? One answer we might try is to measure the average distance from the average value:
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<md>
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<mrow> \E\left[|X - \E(X)|\right] </mrow>
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</md>
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To simplify notation, we'll write <m>\mu = \E(X)</m>.
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Also, it's often usefull to square a term rather than take absolute value when we want to ensure a positive output, so we'll define...
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</p>
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<definition xml:id="def-variance">
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<statement>
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<p>
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Let <m>X</m> be a random variable with <m>\E(X) = \mu</m>.
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The <term>variance</term> of <m>X</m> is:
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<md>
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<mrow> \sigma^2 = \Var(x) = \E\left[ (X - \mu)^2\right] </mrow>
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</md>
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<m>\sigma = \sqrt{\Var(X)}</m> is called the <term>standard deviation</term>.
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</p>
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</statement>
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</definition>
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</subsection>
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</section>
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@@ -140,7 +140,16 @@
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<solution>
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<p>
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<m>\Pr(X = 2, Y = 0) = 0.2 </m>, and <m> \Pr(X = 2)\Pr(Y = 0) = (0.4)(0.25) = 0.1</m>. Since <m>0.2 \neq 0.1</m>, <m>X</m> and <m>Y</m> are not independent.
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Checking each cell in the table:
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<md>
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<mrow> \Pr(X = 0, Y = 0) \amp = 0.1 = (0.4)(0.25) = \Pr(X = 0)\Pr(Y = 0) </mrow>
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<mrow> \Pr(X = 1, Y = 0) \amp = 0.05 = (0.2)(0.25) = \Pr(X = 1)\Pr(Y = 0) </mrow>
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<mrow> \Pr(X = 2, Y = 0) \amp = 0.1 = (0.4)(0.25) = \Pr(X = 2)\Pr(Y = 0) </mrow>
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<mrow> \Pr(X = 0, Y = 1) \amp = 0.3 = (0.4)(0.75) = \Pr(X = 0)\Pr(Y = 1) </mrow>
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<mrow> \Pr(X = 1, Y = 1) \amp = 0.15 = (0.2)(0.75) = \Pr(X = 1)\Pr(Y = 1) </mrow>
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<mrow> \Pr(X = 2, Y = 1) \amp = 0.3 = (0.4)(0.75) = \Pr(X = 2)\Pr(Y = 1) </mrow>
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</md>
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So <m>X, Y</m> are independent.
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</p>
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</solution>
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</task>
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<?xml version="1.0" encoding="UTF-8"?>
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<section xml:id="Exam-1-Review">
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<title>Exam 1 Review</title>
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<introduction>
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<p>
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Use the following problems to prepare for the exam.
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There will be in-class review on Thursday, February 12.
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Your recitation this week will also be exam review.
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</p>
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</introduction>
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<subsection>
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<title>Allowed Materials</title>
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<p>
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You will be allowed to use a scientific calculator (<em>not</em> a graphing calculator, <em>not</em> a calculator app on your phone).
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You may not share a calculator with another student; you must use your own calculator.
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</p>
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<p>
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You may bring a standard 3 in x 5 in index card with prepared notes.
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You may use both sides of the notecard.
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You must put your full name in the top right corner of the card, and turn it in along with your exam.
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</p>
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</subsection>
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<exercises>
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<exercise>
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<p>
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</p>
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</exercise>
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</exercises>
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</section>
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