Adding some answers
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@@ -298,13 +298,42 @@
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<exercises xml:id="exercises-Continuous-RVs">
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<exercise>
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<statement>
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<introduction>
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<p>
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A continuous random variable <m>X</m> taking values in <m>[1, 4]</m> has p.d.f.
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<m>f(x) = k(x - \sqrt{x})</m> for some constant <m>k</m>.
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What is the value of <m>k</m>?
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</p>
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</statement>
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</introduction>
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<task>
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<statement>
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<p>
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What is the value of <m>k</m>?
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</p>
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</statement>
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<answer>
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<p>
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<m>k = \frac{6}{17}.</m>
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</p>
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</answer>
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</task>
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<task>
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<statement>
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<p>
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Find <m>\Pr(2 \leq X \leq 3).</m>
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</p>
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</statement>
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<answer>
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<p>
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<m>\Pr(2 \leq X \leq 3) \approx 0.325</m>
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</p>
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</answer>
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</task>
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</exercise>
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<exercise>
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@@ -318,6 +347,12 @@
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to find <m>\Pr\left(1 \leq X \leq \frac{3}{2}\right)</m>.
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</p>
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</statement>
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<answer>
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<p>
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<m>\frac{1}{2}\left(\frac{x^2}{2} - \frac{1}{x}\right) + \frac{1}{4}</m>. <m>\Pr\left(1 \leq X \leq \frac{3}{2}\right) \approx 0.479.</m>
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</p>
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</answer>
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</exercise>
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<exercise>
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@@ -329,8 +364,29 @@
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<m>f(x)</m>.
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</p>
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</statement>
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<answer>
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<p>
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<m>f(x) = x^2 - 2x.</m>
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</p>
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</answer>
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</exercise>
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<exercise>
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<statement>
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<p>
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Let <m>T \sim \Exp(3)</m>.
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Find <m>\Pr(1 \leq T \leq 3)</m> and <m>\Pr(T \geq 0.5)</m>.
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</p>
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</statement>
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<answer>
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<p>
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<m>\Pr(1 \leq T \leq 3) \approx 0.050</m>. <m>\Pr(T \geq 0.5) \approx 0.223</m>.
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</p>
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</answer>
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</exercise>
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<!--
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<exercise>
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<statement>
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<p>
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@@ -350,5 +406,5 @@
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</p>
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</statement>
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</exercise>
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</exercises>
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--> </exercises>
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</section>
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@@ -398,6 +398,13 @@
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Which flip has the highest chance of being the first flip to come up heads?
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</p>
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</statement>
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<solution>
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<p>
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<m>\Pr(T = k) = 0.6^{k-1} \cdot 0.4</m>, so every additional flip multiplies the probability by 0.6.
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Therefore, the highest value for <m>\Pr(T = k)</m> occurs when <m>k = 1</m>, in which case <m>\Pr(T = 1) = 0.4.</m>
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</p>
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</solution>
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</task>
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</exercise>
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<!--
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@@ -418,6 +425,18 @@
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During a 4-hour afternoon shift, what is the probability of serving 80 customers.
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</p>
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</statement>
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<solution>
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<p>
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Let <m>N</m> be the number of customers seen during the afternoon shift.
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Since the store averages 20 customers per hour, it will average 80 per 4-hours.
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So <m>N \sim \Poiss(20, 4)</m> will have distribution <m>\Pr(N = k) = \frac{80^{k}}{k!} e^{-80}</m>.
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Therefore:
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<md>
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<mrow> \Pr(N = 80) = \frac{80^{80}}{80!} e^{-80} \approx 0.045. </mrow>
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</md>
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</p>
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</solution>
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</exercise>
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<!--
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<exercise>
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@@ -120,9 +120,84 @@
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<exercise>
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<statement>
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<p>
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TODO
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Consider <m>X, Y</m> with the joint distribution table below.
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Are <m>X, Y</m> independent?
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</p>
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<table>
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<title>Joint distribution for <m>X, Y</m></title>
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<tabular halign="center">
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<row bottom="minor">
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<cell right="minor"></cell>
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<cell right="minor"><m>X = 0</m></cell>
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<cell><m>X = 1</m></cell>
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</row>
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<row bottom="minor">
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<cell right="minor"><m>Y = 0</m></cell>
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<cell right="minor">0.2</cell>
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<cell>0.3</cell>
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</row>
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<row>
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<cell right="minor"><m>Y = 1</m></cell>
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<cell right="minor">0.4</cell>
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<cell>0.1</cell>
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</row>
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</tabular>
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</table>
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</statement>
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<answer>
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<p>
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No.
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For example, <m>\Pr(X = 0, Y = 0) \neq \Pr(X = 0)\Pr(Y = 0).</m>
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</p>
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</answer>
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</exercise>
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<exercise>
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<statement>
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<p>
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Suppose <m>X, Y</m> have the distributions:
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<md>
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<mrow> \Pr(X = 0) \amp = 0.1 \amp \Pr(Y = 0) \amp = 0.4 </mrow>
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<mrow> \Pr(X = 1) \amp = 0.4 \amp \Pr(Y = 1) \amp = 0.6 </mrow>
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<mrow> \Pr(X = 2) \amp = 0.5 </mrow>
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</md>
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Assuming <m>X, Y</m> are independent, write a joint distribution table.
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</p>
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</statement>
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<answer>
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<table>
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<title>Joint Distribution</title>
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<tabular halign="center">
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<row bottom="minor">
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<cell right="minor"></cell>
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<cell right="minor"><m>X = 0</m></cell>
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<cell right="minor"><m>X = 1</m></cell>
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<cell><m>X = 2</m></cell>
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</row>
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<row bottom="minor">
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<cell right="minor"><m>Y = 0</m></cell>
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<cell right="minor">0.04</cell>
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<cell right="minor">0.16</cell>
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<cell>0.2</cell>
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</row>
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<row>
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<cell right="minor"><m>Y = 1</m></cell>
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<cell right="minor">0.06</cell>
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<cell right="minor">0.24</cell>
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<cell>0.3</cell>
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</row>
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</tabular>
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</table>
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</answer>
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</exercise>
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</exercises>
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</section>
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