// Numbas version: exam_results_page_options
{"name": "Measures of central tendency from a list 2", "extensions": ["stats"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Measures of central tendency from a list 2", "tags": [], "metadata": {"description": "<p>Given two distributions, calculate the measures of average and spread and make some decisions based on the results.</p>", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "<p><span size=\"3\" face=\"Calibri\" color=\"#1f497d\">There is subtle variation from one step to the next when you walk. When the neuromuscular control of walking deteriorates due to ageing or neurological disease, steps become less consistent. Two people were each asked to walk $10$ steps and the lengths of each of their steps have been recorded in the following table.&nbsp;</span></p>\n<table border=\"1\">\n<tbody>\n<tr>\n<th>Sally (cm)</th>\n<td>$\\var{a[0]}$</td>\n<td>$\\var{a[1]}$</td>\n<td>$\\var{a[2]}$</td>\n<td>$\\var{a[3]}$</td>\n<td>$\\var{a[4]}$</td>\n<td>$\\var{a[5]}$</td>\n<td>$\\var{a[6]}$</td>\n<td>$\\var{a[7]}$</td>\n<td>$\\var{a[8]}$</td>\n<td>$\\var{a[9]}$</td>\n</tr>\n<tr>\n<th>Kate (cm)</th>\n<td>$\\var{b[0]}$</td>\n<td>$\\var{b[1]}$</td>\n<td>$\\var{b[2]}$</td>\n<td>$\\var{b[3]}$</td>\n<td>$\\var{b[4]}$</td>\n<td>$\\var{b[5]}$</td>\n<td>$\\var{b[6]}$</td>\n<td>$\\var{b[7]}$</td>\n<td>$\\var{b[8]}$</td>\n<td>$\\var{b[9]}$</td>\n</tr>\n</tbody>\n</table>\n<p></p>", "advice": "<p>We&nbsp;denote Sally as $s$&nbsp;and&nbsp;Kate as $k$.</p>\n<h4>a)</h4>\n<p>We are going to&nbsp;start with completing&nbsp;the column for Sally.</p>\n<p>First, we need to find the sum of the step lengths:</p>\n<p>\\[\\begin{align} &nbsp;\\sum&nbsp;s &amp;=&nbsp;\\var{a[0]} +&nbsp;\\var{a[1]}&nbsp;+&nbsp;\\var{a[2]} +&nbsp;\\var{a[3]} +&nbsp;\\var{a[4]} +&nbsp;\\var{a[5]} +&nbsp;\\var{a[6]} +&nbsp;\\var{a[7]} +&nbsp;\\var{a[8]} +&nbsp;\\var{a[9]}&nbsp;&nbsp; \\\\&amp;= \\var{suma} \\text{.}<br/>\\end{align}\\]</p>\n<p>The total number of measurements $n$ is $10$.</p>\n<p>Therefore the <strong>mean</strong> is</p>\n<p><span face=\"Courier New\">\\[ \\begin{align} \\overline{s} &amp;= \\frac{\\sum s}{n}&nbsp;\\\\[3pt]&amp;= \\frac{\\var{suma}}{10}&nbsp;\\\\&amp;= \\var{meana} \\text{.} \\end{align}\\]</span></p>\n<p><span face=\"Courier New\">&nbsp;</span></p>\n<p><span face=\"Courier New\">The <strong>median</strong> is&nbsp;the middle value. We need to sort the list in order:</span></p>\n<table border=\"1\">\n<tbody>\n<tr>\n<th>Sally (cm)</th>\n<td>$\\var{asort[0]}$</td>\n<td>$\\var{asort[1]}$</td>\n<td>$\\var{asort[2]}$</td>\n<td>$\\var{asort[3]}$</td>\n<td>$\\var{asort[4]}$</td>\n<td>$\\var{asort[5]}$</td>\n<td>$\\var{asort[6]}$</td>\n<td>$\\var{asort[7]}$</td>\n<td>$\\var{asort[8]}$</td>\n<td>$\\var{asort[9]}$</td>\n</tr>\n</tbody>\n</table>\n<p>To find the middle position within a data set, we take the sample size, add $1$, then divide by $2$. For our data set, the middle position is</p>\n<p style=\"text-align: center;\">$\\displaystyle\\frac{n+1}{2}=\\frac{10+1}{2}=5.5.$</p>\n<p>As there is not actually a $5.5$<sup>th</sup>&nbsp;position, we need to find the mean of the&nbsp;$5$<sup>th</sup>&nbsp;and $6$<sup>th</sup>&nbsp;values:</p>\n<p>\\[ \\displaystyle \\begin{align} \\frac{\\var{asort[4]} + \\var{asort[5]}}{2} &amp;= &nbsp;\\frac{\\var{asort[4] + asort[5]}}{2} \\\\&amp;= \\var{mediana} \\text{.}&nbsp;\\end{align}\\]</p>\n<p>&nbsp;</p>\n<p>The <strong>mode</strong> is the value that occurs the most often in the data.</p>\n<p>To find a mode, we can look at our sorted list:</p>\n<table border=\"1\">\n<tbody>\n<tr>\n<th>Sally (cm)</th>\n<td>$\\var{asort[0]}$</td>\n<td>$\\var{asort[1]}$</td>\n<td>$\\var{asort[2]}$</td>\n<td>$\\var{asort[3]}$</td>\n<td>$\\var{asort[4]}$</td>\n<td>$\\var{asort[5]}$</td>\n<td>$\\var{asort[6]}$</td>\n<td>$\\var{asort[7]}$</td>\n<td>$\\var{asort[8]}$</td>\n<td>$\\var{asort[9]}$</td>\n</tr>\n</tbody>\n</table>\n<p>We&nbsp;notice that $\\var{modea}$ occurs the most times and so $\\var{modea}$ is the mode.</p>\n<p>&nbsp;</p>\n<p>The <strong>range</strong> is the difference between the highest and the lowest value in the data.</p>\n<p>To find this, we subtract the lowest value from the highest value:</p>\n<p>\\[ \\var{max(a)} - \\var{min(a)} = \\var{rangea} \\text{.}\\]</p>\n<p>&nbsp;</p>\n<p>So the first column is</p>\n<table border=\"1\">\n<tbody>\n<tr>\n<td></td>\n<th>Sally (cm)</th>\n</tr>\n<tr>\n<td>Mean length</td>\n<td>$\\var{meana}$</td>\n</tr>\n<tr>\n<td>Median length</td>\n<td>$\\var{mediana}$</td>\n</tr>\n<tr>\n<td>Modal length</td>\n<td>$\\var{modea}$</td>\n</tr>\n<tr>\n<td>Range</td>\n<td>$\\var{rangea}$</td>\n</tr>\n</tbody>\n</table>\n<p>&nbsp;</p>\n<p>Similarly for Kate,</p>\n<p>\\[\\begin{align} &nbsp;\\sum&nbsp;k &amp;=&nbsp;\\var{b[0]} +&nbsp;\\var{b[1]}&nbsp;+&nbsp;\\var{b[2]} +&nbsp;\\var{b[3]} +&nbsp;\\var{b[4]} +&nbsp;\\var{b[5]} +&nbsp;\\var{b[6]} +&nbsp;\\var{b[7]} +&nbsp;\\var{b[8]} +&nbsp;\\var{b[9]}&nbsp;&nbsp; \\\\&amp;= \\var{sumb} \\text{.}<br/>\\end{align}\\]</p>\n<p>The total number of measurements $n$ is&nbsp;$10$ again.</p>\n<p>Therefore the <strong>mean</strong> is</p>\n<p><span face=\"Courier New\">\\[ \\begin{align} \\overline{k} &amp;= \\frac{\\sum k}{n}&nbsp;\\\\[3pt]&amp;= \\frac{\\var{sumb}}{10}&nbsp;\\\\&amp;= \\var{meanb} \\text{.} \\end{align}\\]</span></p>\n<p><span face=\"Courier New\">&nbsp;</span></p>\n<p><span face=\"Courier New\">For <strong>median</strong>, we sort the list in order:</span></p>\n<table border=\"1\">\n<tbody>\n<tr>\n<th>Kate (cm)</th>\n<td>$\\var{bsort[0]}$</td>\n<td>$\\var{bsort[1]}$</td>\n<td>$\\var{bsort[2]}$</td>\n<td>$\\var{bsort[3]}$</td>\n<td>$\\var{bsort[4]}$</td>\n<td>$\\var{bsort[5]}$</td>\n<td>$\\var{bsort[6]}$</td>\n<td>$\\var{bsort[7]}$</td>\n<td>$\\var{bsort[8]}$</td>\n<td>$\\var{bsort[9]}$</td>\n</tr>\n</tbody>\n</table>\n<p>Agin, we need to find the mean of the&nbsp;$5$<sup>th</sup>&nbsp;and $6$<sup>th</sup>&nbsp;values:</p>\n<p>\\[ \\displaystyle \\begin{align} \\frac{\\var{bsort[4]} + \\var{bsort[5]}}{2} &amp;= &nbsp;\\frac{\\var{bsort[4] + bsort[5]}}{2} \\\\&amp;= \\var{medianb} \\text{.}&nbsp;\\end{align}\\]<span face=\"Courier New\"></span></p>\n<p><strong><span face=\"Courier New\">&nbsp; </span></strong></p>\n<p>For <strong>mode</strong>, we look at our sorted list:</p>\n<table border=\"1\">\n<tbody>\n<tr>\n<th>Kate (cm)</th>\n<td>$\\var{bsort[0]}$</td>\n<td>$\\var{bsort[1]}$</td>\n<td>$\\var{bsort[2]}$</td>\n<td>$\\var{bsort[3]}$</td>\n<td>$\\var{bsort[4]}$</td>\n<td>$\\var{bsort[5]}$</td>\n<td>$\\var{bsort[6]}$</td>\n<td>$\\var{bsort[7]}$</td>\n<td>$\\var{bsort[8]}$</td>\n<td>$\\var{bsort[9]}$</td>\n</tr>\n</tbody>\n</table>\n<p>We&nbsp;notice that $\\var{modeb}$ occurs the most times and so $\\var{modeb}$ is the mode.</p>\n<p>&nbsp;</p>\n<p>To find the <strong>range</strong>, we subtract the lowest value from the highest value:</p>\n<p>\\[ \\var{max(b)} - \\var{min(b)} = \\var{rangeb} \\text{.}\\]</p>\n<p>&nbsp;</p>\n<p>So the complete table is\u200b</p>\n<table border=\"1\">\n<tbody>\n<tr>\n<td></td>\n<th>Sally (cm)</th>\n<th>Kate (cm)</th>\n</tr>\n<tr>\n<td>Mean length</td>\n<td>$\\var{meana}$</td>\n<td>$\\var{meanb}$</td>\n</tr>\n<tr>\n<td>Median length</td>\n<td>$\\var{mediana}$</td>\n<td>$\\var{medianb}$</td>\n</tr>\n<tr>\n<td>Modal length</td>\n<td>$\\var{modea}$</td>\n<td>$\\var{modeb}$</td>\n</tr>\n<tr>\n<td>Range</td>\n<td>$\\var{rangea}$</td>\n<td>$\\var{rangeb}$</td>\n</tr>\n</tbody>\n</table>\n<p></p>\n<h4>b)</h4>\n<p><span size=\"3\" face=\"Calibri\" color=\"#1f497d\">Kate is the correct answer, because she has a larger range of values, therefore her step length is less consistent.</span></p>", "rulesets": {}, "extensions": ["stats"], "variables": {"b7": {"name": "b7", "group": "B", "definition": "meanab - 3", "description": "", "templateType": "anything"}, "a2": {"name": "a2", "group": "A", "definition": "meanab - 1 - random2", "description": "", "templateType": "anything"}, "random": {"name": "random", "group": "B", "definition": "random(2..10 #2)", "description": "", "templateType": "anything"}, "a8": {"name": "a8", "group": "A", "definition": "meanab + 1", "description": "", "templateType": "anything"}, "a10": {"name": "a10", "group": "A", "definition": "meanab + 2 + random2", "description": "", 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