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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.

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Eine Zufallsstichprobe von 20 Einwohnern aus Newcastle wurde nach der Anzahl der Besuche im Theater im vergangenen Jahr gefragt.

\n

Hier ist die Liste ihrer Antworten:

\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

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a)

\n

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

\n

Here, $n = 20$.

\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}

\n

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

 

\n

b)

\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]} \\]

\n

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 $\\var{a_s[9]}$ 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

 

\n

c)

\n

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

\n

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

\n

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

\n

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

\n

 

\n

d)

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

\n

To 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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Bestimmen Sie das arithmetische Mittel.

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Bestimmen Sie den Median.

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Finden Sie den Modus (Modalwert).

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Bestimmen Sie die Spannweite.

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

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The final list.

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Sorted list.

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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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Mode as a value.

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Mode as a vector.

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