Adding some answers

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2026-01-29 06:35:50 -05:00
parent a4f71931a9
commit 2d3e7d69cc
3 changed files with 155 additions and 5 deletions
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<exercises xml:id="exercises-Continuous-RVs">
<exercise>
<statement>
<introduction>
<p>
A continuous random variable <m>X</m> taking values in <m>[1, 4]</m> has p.d.f.
<m>f(x) = k(x - \sqrt{x})</m> for some constant <m>k</m>.
What is the value of <m>k</m>?
</p>
</statement>
</introduction>
<task>
<statement>
<p>
What is the value of <m>k</m>?
</p>
</statement>
<answer>
<p>
<m>k = \frac{6}{17}.</m>
</p>
</answer>
</task>
<task>
<statement>
<p>
Find <m>\Pr(2 \leq X \leq 3).</m>
</p>
</statement>
<answer>
<p>
<m>\Pr(2 \leq X \leq 3) \approx 0.325</m>
</p>
</answer>
</task>
</exercise>
<exercise>
@@ -318,6 +347,12 @@
to find <m>\Pr\left(1 \leq X \leq \frac{3}{2}\right)</m>.
</p>
</statement>
<answer>
<p>
<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>
</p>
</answer>
</exercise>
<exercise>
@@ -329,8 +364,29 @@
<m>f(x)</m>.
</p>
</statement>
<answer>
<p>
<m>f(x) = x^2 - 2x.</m>
</p>
</answer>
</exercise>
<exercise>
<statement>
<p>
Let <m>T \sim \Exp(3)</m>.
Find <m>\Pr(1 \leq T \leq 3)</m> and <m>\Pr(T \geq 0.5)</m>.
</p>
</statement>
<answer>
<p>
<m>\Pr(1 \leq T \leq 3) \approx 0.050</m>. <m>\Pr(T \geq 0.5) \approx 0.223</m>.
</p>
</answer>
</exercise>
<!--
<exercise>
<statement>
<p>
@@ -350,5 +406,5 @@
</p>
</statement>
</exercise>
</exercises>
--> </exercises>
</section>
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Which flip has the highest chance of being the first flip to come up heads?
</p>
</statement>
<solution>
<p>
<m>\Pr(T = k) = 0.6^{k-1} \cdot 0.4</m>, so every additional flip multiplies the probability by 0.6.
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>
</p>
</solution>
</task>
</exercise>
<!--
@@ -418,6 +425,18 @@
During a 4-hour afternoon shift, what is the probability of serving 80 customers.
</p>
</statement>
<solution>
<p>
Let <m>N</m> be the number of customers seen during the afternoon shift.
Since the store averages 20 customers per hour, it will average 80 per 4-hours.
So <m>N \sim \Poiss(20, 4)</m> will have distribution <m>\Pr(N = k) = \frac{80^{k}}{k!} e^{-80}</m>.
Therefore:
<md>
<mrow> \Pr(N = 80) = \frac{80^{80}}{80!} e^{-80} \approx 0.045. </mrow>
</md>
</p>
</solution>
</exercise>
<!--
<exercise>
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<exercise>
<statement>
<p>
TODO
Consider <m>X, Y</m> with the joint distribution table below.
Are <m>X, Y</m> independent?
</p>
<table>
<title>Joint distribution for <m>X, Y</m></title>
<tabular halign="center">
<row bottom="minor">
<cell right="minor"></cell>
<cell right="minor"><m>X = 0</m></cell>
<cell><m>X = 1</m></cell>
</row>
<row bottom="minor">
<cell right="minor"><m>Y = 0</m></cell>
<cell right="minor">0.2</cell>
<cell>0.3</cell>
</row>
<row>
<cell right="minor"><m>Y = 1</m></cell>
<cell right="minor">0.4</cell>
<cell>0.1</cell>
</row>
</tabular>
</table>
</statement>
<answer>
<p>
No.
For example, <m>\Pr(X = 0, Y = 0) \neq \Pr(X = 0)\Pr(Y = 0).</m>
</p>
</answer>
</exercise>
<exercise>
<statement>
<p>
Suppose <m>X, Y</m> have the distributions:
<md>
<mrow> \Pr(X = 0) \amp = 0.1 \amp \Pr(Y = 0) \amp = 0.4 </mrow>
<mrow> \Pr(X = 1) \amp = 0.4 \amp \Pr(Y = 1) \amp = 0.6 </mrow>
<mrow> \Pr(X = 2) \amp = 0.5 </mrow>
</md>
Assuming <m>X, Y</m> are independent, write a joint distribution table.
</p>
</statement>
<answer>
<table>
<title>Joint Distribution</title>
<tabular halign="center">
<row bottom="minor">
<cell right="minor"></cell>
<cell right="minor"><m>X = 0</m></cell>
<cell right="minor"><m>X = 1</m></cell>
<cell><m>X = 2</m></cell>
</row>
<row bottom="minor">
<cell right="minor"><m>Y = 0</m></cell>
<cell right="minor">0.04</cell>
<cell right="minor">0.16</cell>
<cell>0.2</cell>
</row>
<row>
<cell right="minor"><m>Y = 1</m></cell>
<cell right="minor">0.06</cell>
<cell right="minor">0.24</cell>
<cell>0.3</cell>
</row>
</tabular>
</table>
</answer>
</exercise>
</exercises>
</section>