Added Exam 1 Review answers

source/review/Exam-1-Review.ptx

@@ -1924,28 +1924,14 @@
           coin coming up heads on a flip?
         </p>
       </statement>
+
+      <answer>
+        <p>
+          <m>5/27 \approx 0.227</m>.
+        </p>
+      </answer>
     </exercise>
 
-    <!-- <exercise>
-      <statement>
-        <p>
-          Suppose a coin has an unknown probability of coming up heads.
-          We perform the experiment in <m>n</m> independent trials, during which
-          it takes <m>k_1, k_2, \dotsc, k_n</m> flips to see our first heads in
-          each trial.
-          Find a "common sense" MLE formula for the geometric distribution.
-        </p>
-      </statement>
-
-      <hint>
-        <p>
-          You <em>could</em> set up a calculation analogous to
-          <xref ref="example-exponential-MLE"/>.
-          Or, you could consider <xref ref="fact-MLE-binomial"/>.
-        </p>
-      </hint>
-    </exercise> -->
-
     <exercise>
       <statement>
         <p>
@@ -1956,6 +1942,12 @@
           rate?
         </p>
       </statement>
+
+      <answer>
+        <p>
+          <m>20</m>.
+        </p>
+      </answer>
     </exercise>
 
     <exercise>
@@ -1970,6 +1962,13 @@
           What is the maximum likelihood estimation for <m>\lambda</m>?
         </p>
       </statement>
+
+      <answer>
+        <p>
+          <m>\mathcal{L}(\lambda) = \left( \lambda e^{-1.1\lambda} \right) \left( \lambda e^{-1.7\lambda} \right) \left( \lambda e^{-1.3\lambda} \right) \left( \lambda e^{-2.2\lambda} \right) \left( \lambda e^{-1.9\lambda} \right) \left( \lambda e^{-1.8\lambda} \right)</m>.
+          The MLE is <m>0.6</m>.
+        </p>
+      </answer>
     </exercise>
 
     <exercise>
@@ -1981,6 +1980,12 @@
           Find the maximum likelihood estimation of <m>\theta</m>.
         </p>
       </statement>
+
+      <answer>
+        <p>
+          <m>\left(\frac{1}{4}\right)^{2/3} \approx 0. 37</m>.
+        </p>
+      </answer>
     </exercise>
   </exercises>
 </section>