Setting up Exam 1 Review

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2026-02-10 08:19:40 -05:00
parent 3102cc41b5
commit 52180c4466
4 changed files with 97 additions and 1 deletions
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<xi:include href="./notes/1-22.ptx" />
<xi:include href="./notes/1-27.ptx" />
<xi:include href="./notes/1-29.ptx" />
<xi:include href="./notes/2-3.ptx" />
</chapter>
<chapter xml:id="quizzes">
@@ -44,5 +45,14 @@
<xi:include href="./recitations/calculus-review.ptx"/>
</chapter>
<chapter xml:id="review">
<title>Exam Review</title>
<xi:include href="./review/Exam-1-Review.ptx"/>
</chapter>
</book>
</pretext>
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<?xml version="1.0" encoding="UTF-8"?>
<section xml:id="notes-02-03">
<title>Tuesday, Feb 3</title>
<introduction>
<p>
This is an outline of the topics we covered in class.
These notes are <em>not</em> a substitute for your own note-taking.
I highly recommend that you take your own notes during class.
If you ever miss a class for any reason, reach out to another student in class to get a copy of their notes.
</p>
</introduction>
<subsection>
<title>Variance</title>
<p>
Question: How spread out are <m>X</m> values? One answer we might try is to measure the average distance from the average value:
<md>
<mrow> \E\left[|X - \E(X)|\right] </mrow>
</md>
To simplify notation, we'll write <m>\mu = \E(X)</m>.
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...
</p>
<definition xml:id="def-variance">
<statement>
<p>
Let <m>X</m> be a random variable with <m>\E(X) = \mu</m>.
The <term>variance</term> of <m>X</m> is:
<md>
<mrow> \sigma^2 = \Var(x) = \E\left[ (X - \mu)^2\right] </mrow>
</md>
<m>\sigma = \sqrt{\Var(X)}</m> is called the <term>standard deviation</term>.
</p>
</statement>
</definition>
</subsection>
</section>
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<solution>
<p>
<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.
Checking each cell in the table:
<md>
<mrow> \Pr(X = 0, Y = 0) \amp = 0.1 = (0.4)(0.25) = \Pr(X = 0)\Pr(Y = 0) </mrow>
<mrow> \Pr(X = 1, Y = 0) \amp = 0.05 = (0.2)(0.25) = \Pr(X = 1)\Pr(Y = 0) </mrow>
<mrow> \Pr(X = 2, Y = 0) \amp = 0.1 = (0.4)(0.25) = \Pr(X = 2)\Pr(Y = 0) </mrow>
<mrow> \Pr(X = 0, Y = 1) \amp = 0.3 = (0.4)(0.75) = \Pr(X = 0)\Pr(Y = 1) </mrow>
<mrow> \Pr(X = 1, Y = 1) \amp = 0.15 = (0.2)(0.75) = \Pr(X = 1)\Pr(Y = 1) </mrow>
<mrow> \Pr(X = 2, Y = 1) \amp = 0.3 = (0.4)(0.75) = \Pr(X = 2)\Pr(Y = 1) </mrow>
</md>
So <m>X, Y</m> are independent.
</p>
</solution>
</task>
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<?xml version="1.0" encoding="UTF-8"?>
<section xml:id="Exam-1-Review">
<title>Exam 1 Review</title>
<introduction>
<p>
Use the following problems to prepare for the exam.
There will be in-class review on Thursday, February 12.
Your recitation this week will also be exam review.
</p>
</introduction>
<subsection>
<title>Allowed Materials</title>
<p>
You will be allowed to use a scientific calculator (<em>not</em> a graphing calculator, <em>not</em> a calculator app on your phone).
You may not share a calculator with another student; you must use your own calculator.
</p>
<p>
You may bring a standard 3 in x 5 in index card with prepared notes.
You may use both sides of the notecard.
You must put your full name in the top right corner of the card, and turn it in along with your exam.
</p>
</subsection>
<exercises>
<exercise>
<p>
</p>
</exercise>
</exercises>
</section>