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Given 32 datapoints in a table find their minimum, lower quartile, median, upper quartile, and maximum.

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Below are the lengths (rounded to the nearest metre) thrown by $32$ javelin throwers in a competition.

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$\\var{r0[0]}$

$\\var{r0[1]}$

$\\var{r0[2]}$

$\\var{r0[3]}$

$\\var{r0[4]}$

$\\var{r0[5]}$

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$\\var{r0[8]}$

$\\var{r0[9]}$

$\\var{r0[10]}$

$\\var{r0[11]}$

$\\var{r0[12]}$

$\\var{r0[13]}$

$\\var{r0[14]}$

$\\var{r0[15]}$

$\\var{r0[16]}$

$\\var{r0[17]}$

$\\var{r0[18]}$

$\\var{r0[19]}$

$\\var{r0[20]}$

$\\var{r0[21]}$

$\\var{r0[22]}$

$\\var{r0[23]}$

$\\var{r0[24]}$

$\\var{r0[25]}$

$\\var{r0[26]}$

$\\var{r0[27]}$

$\\var{r0[28]}$

$\\var{r0[29]}$

$\\var{r0[30]}$

$\\var{r0[31]}$

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a)

If we sort the data in increasing order we get the following table:

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

$\\var{r1[0]}$

$\\var{r1[1]}$

$\\var{r1[2]}$

$\\var{r1[3]}$

$\\var{r1[4]}$

$\\var{r1[5]}$

$\\var{r1[6]}$

$\\var{r1[7]}$

$\\var{r1[8]}$

$\\var{r1[9]}$

$\\var{r1[10]}$

$\\var{r1[11]}$

$\\var{r1[12]}$

$\\var{r1[13]}$

$\\var{r1[14]}$

$\\var{r1[15]}$

$\\var{r1[16]}$

$\\var{r1[17]}$

$\\var{r1[18]}$

$\\var{r1[19]}$

$\\var{r1[20]}$

$\\var{r1[21]}$

$\\var{r1[22]}$

$\\var{r1[23]}$

$\\var{r1[24]}$

$\\var{r1[25]}$

$\\var{r1[26]}$

$\\var{r1[27]}$

$\\var{r1[28]}$

$\\var{r1[29]}$

$\\var{r1[30]}$

$\\var{r1[31]}$

Denote the ordered data by $x_j$, thus $x_{10}=\\var{r1[9]}$ for example.

Minimum value: The minimum value is $x_1=\\var{r1[0]}$.

Lower Quartile: As there is an even number of values, the Lower Quartile will lie between two values. Its position is calculated by finding

\\[\\frac{n+1}{4}=\\frac{\\var{n+1}}{4}=8\\frac{1}{4}.\\]

Hence the lower quartile lies a quarter of the way between the $8$th and $9$th entries in the ordered table. We need to find the difference between the entries in $8$th and $9$th place, quarter it, and add this back onto the $8$th entry.

\\[x_9 - x_8 = \\var{r1[8]}-\\var{r1[7]}=\\var{r1[8]-r1[7]}.\\]

\\[0.25\\times\\var{r1[8]-r1[7]}=\\var{lquartile-r1[7]}.\\]

\\[\\var{lquartile-r1[7]}+x_8 = \\var{lquartile-r1[7]}+\\var{r1[7]} = \\var{lquartile}.\\]

Median: The position of the median in the table is given by

\\[ \\frac{2(n+1)}{4} = \\frac{\\var{2*(n+1)}}{4} = 16 \\frac{1}{2}.\\]

The median lies halfway between the $16$th and $17$th entries in the ordered table and is given by:

\\[0.5\\times x_{16}+0.5\\times x_{17} = 0.5\\times\\var{r1[15]}+0.5\\times \\var{r1[16]}=\\var{median}.\\]

Upper Quartile: As there is an even number of values, the Upper Quartile will lie between two values. Its position is calculated by finding

\\[\\frac{3(n+1)}{4}=\\frac{\\var{3*(n+1)}}{4}=24\\frac{3}{4}.\\]

Hence the upper quartile lies three quarters of the way between the $24$th and $25$th entries in the ordered table. We need to find the difference between the entries in $24$th and $25$th place, find three quarters of this value, and add this back onto the $24$th entry.

\\[x_{25} - x_{24} = \\var{r1[24]}-\\var{r1[23]}=\\var{r1[24]-r1[23]}.\\]

\\[0.75\\times\\var{r1[24]-r1[23]}=\\var{uquartile-r1[23]}.\\]

\\[\\var{uquartile-r1[23]}+x_{24} = \\var{uquartile-r1[23]}+\\var{r1[23]} = \\var{uquartile}.\\]

Maximum value: The maximum value is $x_{32}=\\var{r1[31]}$.

b)

The interquartile range is defined to be

\\[ \\text{upper quartile} – \\text{lower quartile} \\]

and so in this case we have:

\\[ \\text{interquartile range} = \\var{uquartile}-\\var{lquartile}=\\var{uquartile-lquartile}. \\]

c)

The range is defined to be

\\[ \\text{maximum} – \\text{minimum} \\]

and so in this case we have:

\\[ \\text{range} = \\var{r1[31]}-\\var{r1[0]}=\\var{range}. \\]

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Complete the following table using the data given.

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

Minimum	Lower Quartile	Median	Upper Quartile	Maximum
[[0]]	[[1]]	[[2]]	[[3]]	[[4]]

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Enter the interquartile range: [[0]]

Enter the range: [[0]]

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