Find the mean.
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"}], "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "This question provides a list of data to the student. They are asked to find the mean, median, mode and range.
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", "definition": "repeat(random(0..8), 20)", "name": "a1", "group": "Ungrouped variables"}, "a_s": {"templateType": "anything", "description": "Sorted list.
", "definition": "sort(a)", "name": "a_s", "group": "final list"}, "modea2": {"templateType": "anything", "description": "", "definition": "mode(a2)", "name": "modea2", "group": "Ungrouped variables"}, "a3": {"templateType": "anything", "description": "Option 3 for the list. Ensures there is only one mode (2) while still randomising the data.
", "definition": "shuffle([ random(0..1),\n 2, \n random(4..6),\n random(0..3 except 2), \n random(0..3 except 2),\n random(4..6),\n 2,\n 2,\n random(4..6),\n random(7..8),\n random(0..3 except 2 except 1), \n random(4..6),\n 2,\n random(1..3 except 2), \n random(7..8),\n 2,\n random(7..8),\n random(4..6), \n random(0..3 except 2), \n 2\n])", "name": "a3", "group": "Ungrouped variables"}, "mean": {"templateType": "anything", "description": "", "definition": "mean(a)", "name": "mean", "group": "final list"}, "modetimes": {"templateType": "anything", "description": "The vector of number of times of each value in the data.
", "definition": "map(\nlen(filter(x=j,x,a)),\nj, 0..8)", "name": "modetimes", "group": "final list"}, "range": {"templateType": "anything", "description": "", "definition": "max(a) - min(a)", "name": "range", "group": "final list"}, "mode1": {"templateType": "anything", "description": "Mode as a value.
", "definition": "mode[0]", "name": "mode1", "group": "final list"}, "mode": {"templateType": "anything", "description": "Mode as a vector.
", "definition": "mode(a)", "name": "mode", "group": "final list"}, "a": {"templateType": "anything", "description": "The final list.
", "definition": "if(len(modea1) = 1, a1, if(len(modea2) = 1, a2, a3))", "name": "a", "group": "final list"}}, "rulesets": {}, "extensions": ["stats"], "functions": {}, "ungrouped_variables": ["modea1", "modea2", "a1", "a2", "a3"], "statement": "A random sample of 20 residents from Newcastle were asked about the number of times they went to see a play at the theatre last year.
\nHere is the list of their answers:
\n| $\\var{a[0]}$ | \n$\\var{a[1]}$ | \n$\\var{a[2]}$ | \n$\\var{a[3]}$ | \n$\\var{a[4]}$ | \n$\\var{a[5]}$ | \n$\\var{a[6]}$ | \n$\\var{a[7]}$ | \n$\\var{a[8]}$ | \n$\\var{a[9]}$ | \n
| $\\var{a[10]}$ | \n$\\var{a[11]}$ | \n$\\var{a[12]}$ | \n$\\var{a[13]}$ | \n$\\var{a[14]}$ | \n$\\var{a[15]}$ | \n$\\var{a[16]}$ | \n$\\var{a[17]}$ | \n$\\var{a[18]}$ | \n$\\var{a[19]}$ | \n
The mean is the sum of all the responses ($\\sum x$) divided by the number of responses ($n$).
\nHere, $n = 20$.
\n\\begin{align}
\\sum x &= \\var{a[0]} + \\var{a[1]} +\\var{a[2]} +\\var{a[3]} +\\var{a[4]} +\\var{a[5]} +\\var{a[6]} +\\var{a[7]} +\\var{a[8]} +\\var{a[9]} + \\var{a[10]} + \\var{a[11]} +\\var{a[12]} +\\var{a[13]} +\\var{a[14]} +\\var{a[15]} +\\var{a[16]} +\\var{a[17]} +\\var{a[18]} +\\var{a[19]} \\\\
&= \\var{sum(a)} \\text{.}
\\end{align}
Therefore we calculate the mean
\n\\begin{align}
\\overline{x} &= \\frac{\\sum x}{n} \\\\[0.5em]
&= \\frac{\\var{sum(a)}}{20} \\\\[0.5em]
&= \\var{mean} \\text{.}
\\end{align}
\n
The median is the middle value. We need to sort the list in order:
\n\\[ \\var{a_s[0]}, \\quad \\var{a_s[1]}, \\quad \\var{a_s[2]}, \\quad \\var{a_s[3]}, \\quad \\var{a_s[4]}, \\quad \\var{a_s[5]}, \\quad \\var{a_s[6]}, \\quad \\var{a_s[7]}, \\quad \\var{a_s[8]}, \\quad \\var{a_s[9]}, \\quad \\var{a_s[10]}, \\quad \\var{a_s[11]}, \\quad \\var{a_s[12]}, \\quad \\var{a_s[13]}, \\quad \\var{a_s[14]}, \\quad \\var{a_s[15]}, \\quad \\var{a_s[16]}, \\quad \\var{a_s[17]}, \\quad \\var{a_s[18]}, \\quad \\var{a_s[19]} \\]
\nThere is an even number of responses, so there are two numbers in the middle (10th and 11th place). To find the median, we need to find the mean of these two numbers $\\var{a_s[9]}$ and $\\var{a_s[10]}$:
\n\\begin{align}
\\frac{\\var{a_s[9]} + \\var{a_s[10]}}{2} &= \\frac{\\var{a_s[9] + a_s[10]}}{2} \\\\
&= \\var{median} \\text{.}
\\end{align}
\n
The mode is the value that occurs the most often in the data.
\nTo find a mode, we can look at our sorted list:
\n$\\var{a_s[0]}, \\var{a_s[1]}, \\var{a_s[2]}, \\var{a_s[3]}, \\var{a_s[4]}, \\var{a_s[5]}, \\var{a_s[6]}, \\var{a_s[7]}, \\var{a_s[8]}, \\var{a_s[9]}, \\var{a_s[10]}, \\var{a_s[11]}, \\var{a_s[12]}, \\var{a_s[13]}, \\var{a_s[14]}, \\var{a_s[15]}, \\var{a_s[16]}, \\var{a_s[17]}, \\var{a_s[18]}, \\var{a_s[19]}$.
\nWe notice that $\\var{mode1}$ occurs the most ($\\var{modetimes[mode1]}$ times) so $\\var{mode1}$ is the mode.
\n\n
Range is the difference between the highest and the lowest value in the data.
\nTo find this, we subtract the lowest value from the highest value:
\n\\[ \\var{max(a)} - \\var{min(a)} = \\var{range} \\text{.}\\]
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