<p><span style="color:#8e44ad;"><span style="font-size:20px;">Deductions from the Given Information</span></span></p>
<p>We can see that the data is not evenly distributed, since there are more values to the left of the middle.</p>
<p><img alt="" src="https://gregmatapi.s3.amazonaws.com/media/misc/files/_class_pp1-shorter-quant-section-2-medium-q3.png" style="width: 400px; height: 138px;" /></p>
<p><span style="color:#27ae60;"><span style="font-size:20px;">Solving the Problem</span></span></p>
<p><strong>Method 1 (Conceptual)</strong></p>
<p>We can see that the data is distributed in the following way:</p>
<ul>
<li>The peak (head) is to the left since there are more values to the left of the middle</li>
<li>The tail is to the right since there are fewer values to the right of the middle</li>
</ul>
<p>So, naturally, the median will be to the left, where more values are, and the mean will be to the right, where the extreme values are.</p>
<p>So, QA will be bigger.</p>
<p><strong>Method 2 (Algebra)</strong></p>
<p>We can declare the following variables:</p>
<ul>
<li>$s$ = sam's height</li>
<li>$s + 1$ = lin's height</li>
<li>$s + 3$ = ray's height</li>
</ul>
<p>Now to calculate QA:</p>
<ul>
<li>$mean = \frac{sum}{\#\,values}$</li>
<li>$sum = s + (s+1) + (s+3) = 3s+4$</li>
<li>$\#\,values = 3$</li>
<li>$mean = \frac{3s+4}{3} = s + \frac{4}{3}$</li>
</ul>
<p>Now to get QB:</p>
<p>- The median is the middle value<br />
- In this set of data, it would be $s+1$</p>
<p>Now, we can <strong><span style="color:#8e44ad;">Simplify</span></strong> by taking away $s$ from both quantities, so we are left with the following:</p>
<p>QA = $\frac{4}{3}$</p>
<p>QB = $1$</p>
<p>So, we know that QA is bigger.</p>
<p>Thus, the answer must be <span style="color:#27ae60;">A</span>.</p>