Fixed a small error

Author: bcaffo

06_StatisticalInference/homework/hw1.Rmd

@@ -45,7 +45,7 @@ Consider influenza epidemics for two parent heterosexual families. Suppose that
 1. 15%
 2. 6%
 3. 5%
-4. _1%_
+4. _2%_
 
 *** .hint
 $A = Father$, $P(A) = .06$, $B = Mother$, $P(B) = .05$ 
@@ -54,6 +54,9 @@ $P(A\cup B) = .15$,
 *** .explanation
 $P(A\cup B) = P(A) + P(B) - 2 P(AB)$ thus
 $$.15 = .06 + .05 - 2 P(AB)$$
+```{r}
+(0.15 - .06 - .05) / 2
+```
 
 ---  &radio
 

06_StatisticalInference/homework/hw1.html

@@ -69,7 +69,7 @@ <h2>About these slides</h2>
 <li>15%</li>
 <li>6%</li>
 <li>5%</li>
-<li><em>1%</em></li>
+<li><em>2%</em></li>
 </ol>
 
   <button class="quiz-submit btn btn-primary">Submit</button>
@@ -86,6 +86,12 @@ <h2>About these slides</h2>
   <p>\(P(A\cup B) = P(A) + P(B) - 2 P(AB)\) thus
 \[.15 = .06 + .05 - 2 P(AB)\]</p>
 
+<pre><code class="r">(0.15 - .06 - .05) / 2
+</code></pre>
+
+<pre><code>[1] 0.02
+</code></pre>
+
 </div>
 </div>
   </article>

06_StatisticalInference/homework/hw1.md

@@ -30,7 +30,7 @@ Consider influenza epidemics for two parent heterosexual families. Suppose that
 1. 15%
 2. 6%
 3. 5%
-4. _1%_
+4. _2%_
 
 *** .hint
 $A = Father$, $P(A) = .06$, $B = Mother$, $P(B) = .05$ 
@@ -40,6 +40,15 @@ $P(A\cup B) = .15$,
 $P(A\cup B) = P(A) + P(B) - 2 P(AB)$ thus
 $$.15 = .06 + .05 - 2 P(AB)$$
 
+```r
+(0.15 - .06 - .05) / 2
+```
+
+```
+[1] 0.02
+```
+
+
 ---  &radio
 
 A random variable, $X$, is uniform, a box from $0$ to $1$ of height $1$. (So that it's density is $f(x) = 1$ for $0\leq x \leq 1$.) What is it's median expressed to two decimal places? </p>