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Sample Mean (2 dp)Sample Standard Deviation (2 dp)Median (exact value)Interquartile Range (exact value)
[[0]][[1]][[2]][[3]]
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Sample of size $24$ is given in a table. Find sample mean, sample standard deviation, sample median and the interquartile range.

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The following data are taken from a survey of primary school class sizes

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
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$\\var{r0[0]}$

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\n

$\\var{r0[1]}$

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$\\var{r0[2]}$

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$\\var{r0[3]}$

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\n

$\\var{r0[4]}$

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$\\var{r0[5]}$

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$\\var{r0[6]}$

\n
\n

$\\var{r0[7]}$

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\n

$\\var{r0[8]}$

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\n

$\\var{r0[9]}$

\n
\n

$\\var{r0[10]}$

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\n

$\\var{r0[11]}$

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\n

$\\var{r0[12]}$

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$\\var{r0[13]}$

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\n

$\\var{r0[14]}$

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$\\var{r0[15]}$

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\n

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Sample mean

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The sample mean is $\\frac{\\var{sum(r0)}}{\\var{len(r0)}} = \\var{mean} $, to 2 decimal places.

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Sample standard deviation

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The formula required may be written as

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\\[s = \\sqrt{\\frac{ \\Sigma(x-\\bar{x})^2}{n-1}}\\]

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  \\text{where }\\bar{x}\\text{ is the sample mean and n is the  number of items in the data set (16 in this example). }

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This is very time consuming for all but the simplest of questions so you should use your calculator, in statistics mode, to find this.

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If you rearrange the data in ascending order you 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
$\\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]}$
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Denote the ordered data by $x_j$, thus $x_{10}=\\var{r1[9]}$ for example.

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Median: The median is the middle of the data once it has been ordered, it therefore lies between the 8th and 9th entries in the ordered table and is given by:

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\\[ \\frac{x_{8}+ x_{9}}{2}  =  \\frac{\\var{r1[7]}+ \\var{r1[8]}}{2}=\\var{median}\\]

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Interquartile range:   There is not one definitive method for finding the the interquartile range (IQR). We will use the following: as there is an even number of values, the Lower Quartile will be the 'median' of the first 8 values. Its position is calculated by finding

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\\[\\frac{n+1}{2}=\\frac{\\var{8+1}}{2}=4\\frac{1}{2}\\]

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Hence the Lower Quartile lies between the 4th and 5th entries in the ordered table.

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It is \\[x_4 + \\frac{1}{2}(x_5 - x_4) = \\var{r1[3]}+ \\frac{1}{2}(\\var{r1[4]} - \\var{r1[3]})=\\var{lower_quartile}\\]

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Once again as there is an even number of values, the Upper Quartile will lie between two values and its position is calculated by finding

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\\[x_{12} + \\frac{1}{2}(x_{13} - x_{12}) = \\var{r1[11]}+ \\frac{1}{2}(\\var{r1[12]} - \\var{r1[11]})=\\var{upper_quartile}\\]

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The interquartile range (IQR) is defined to be

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\\[ \\text{Upper Quartile} – \\text{Lower Quartile} \\]

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and so in this case we have:

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\\[ \\text{Interquartile range} = \\var{upper_quartile}-\\var{lower_quartile}=\\var{iqr} \\]

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