// 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": [{"name": "Calculate the measures of central tendency for a sample", "tags": [], "metadata": {"description": "

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"}, "statement": "

A random sample of 10 residents from Newcastle were asked about the number of times they went to see a play at the theatre last year.

Here is the list of their answers:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\var{a[0]}$

$\\var{a[1]}$

$\\var{a[2]}$

$\\var{a[3]}$

$\\var{a[4]}$

$\\var{a[5]}$

$\\var{a[6]}$

$\\var{a[7]}$

$\\var{a[8]}$

$\\var{a[9]}$

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a)

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

Here, $n = 20$.

\\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

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

b)

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

\\[ \\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]} \\]

There is an even number of responses, so there are two numbers in the middle (10^th and 11^th place). To find the median, we need to find the mean of these two numbers $\\var{a_s[9]}$ and $\\var{a_s[10]}$:

\\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}

c)

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

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

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

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

d)

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

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

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

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Mode as a vector.

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

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

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Sorted list.

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

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Mode as a value.

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Find the difference between the mean and the median. Give your answer as a positive number correct to one decimal place.

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Find the median.

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Find the mode.

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