// Numbas version: exam_results_page_options {"name": "Calculate the measures of central tendency for a sample", "extensions": ["stats"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"statement": "

A random sample of 20 residents from Dublin were asked about the number of times they went to see a film at the cinema last year.

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Here is the list of their answers:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\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]}$
\n

", "extensions": ["stats"], "ungrouped_variables": ["modea1", "modea2", "a1", "a2", "a3"], "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "

a)

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The mean is the sum of all the responses ($\\sum x$) divided by the number of responses ($n$).

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Here, $n = 20$.

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\\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}

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Therefore we calculate the mean

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\\begin{align}
\\overline{x} &= \\frac{\\sum x}{n} \\\\[0.5em]
&= \\frac{\\var{sum(a)}}{20} \\\\[0.5em]
&= \\var{mean} \\text{.}
\\end{align}

\n

 

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b)

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The median is the middle value. We need to sort the list in order:

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\\[ \\var{as[0]}, \\quad \\var{as[1]}, \\quad \\var{as[2]}, \\quad \\var{as[3]}, \\quad \\var{as[4]}, \\quad \\var{as[5]}, \\quad \\var{as[6]}, \\quad \\var{as[7]}, \\quad \\var{as[8]}, \\quad \\var{as[9]}, \\quad \\var{as[10]}, \\quad \\var{as[11]}, \\quad \\var{as[12]}, \\quad \\var{as[13]}, \\quad \\var{as[14]}, \\quad \\var{as[15]}, \\quad \\var{as[16]}, \\quad \\var{as[17]}, \\quad \\var{as[18]}, \\quad \\var{as[19]} \\]

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There 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 $\\var{as[9]}$ and $\\var{as[10]}$:

\n

\\begin{align}
\\frac{\\var{as[9]} + \\var{as[10]}}{2} &=  \\frac{\\var{as[9] + as[10]}}{2} \\\\
&= \\var{median} \\text{.} 
\\end{align}

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c)

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The mode is the value that occurs the most often in the data.

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To find a mode, we can look at our sorted list:

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$\\var{as[0]}, \\var{as[1]}, \\var{as[2]}, \\var{as[3]}, \\var{as[4]}, \\var{as[5]}, \\var{as[6]}, \\var{as[7]}, \\var{as[8]}, \\var{as[9]}, \\var{as[10]}, \\var{as[11]}, \\var{as[12]}, \\var{as[13]}, \\var{as[14]}, \\var{as[15]}, \\var{as[16]}, \\var{as[17]}, \\var{as[18]}, \\var{as[19]}$.

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We notice that $\\var{mode1}$ occurs the most ($\\var{modetimes[mode1]}$ times) so $\\var{mode1}$ is the mode.

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d)

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Range is the difference between the highest and the lowest value in the data.

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To find this, we subtract the lowest value from the highest value:

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\\[ \\var{max(a)} - \\var{min(a)} = \\var{range} \\text{.}\\]

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Mode as a vector.

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Option 1 for the list. Only used if there is only one mode.

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The vector of number of times of each value in the data.

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The final list.

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Option 2 for the list. Only used if there is only one mode and option 1 was not used.

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Mode as a value.

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Option 3 for the list. Ensures there is only one mode (2) while still randomising the data.

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Sorted list.

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Find the mean.

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Find the median.

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Find the mode.

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Find the range.

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This question provides a list of data to the student. They are asked to find the mean, median, mode and range.

", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "contributors": [{"name": "Hugh O'Donnell", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2158/"}]}]}], "contributors": [{"name": "Hugh O'Donnell", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2158/"}]}