Mean, SD, median and IQR

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Mario Orsi 5 years, 10 months ago

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Name	Status	Author	Last Modified
Mean, SD, median and IQR	draft	Mario Orsi	13/06/2019 10:15
Leonardo's copy of Mean, SD, median and IQR	draft	Leonardo Juliano	13/09/2019 10:02

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Question statement

Find the mean, SD, median and IQR for the following sample of $\var{n}$ data values:

{r0[0]}	{r0[1]}	{r0[2]}	{r0[3]}	{r0[4]}	{r0[5]}	{r0[6]}	{r0[7]}	{r0[8]}	{r0[9]}	{r0[10]}	{r0[11]}
{r0[12]}	{r0[13]}	{r0[14]}	{r0[15]}	{r0[16]}	{r0[17]}	{r0[18]}	{r0[19]}	{r0[20]}	{r0[21]}	{r0[22]}	{r0[23]}

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			Name	Type	Generated Value

uquartile	number	40
r0	list	List of 24 items
description	string	from a new kind of tomato pla
things	string	tomatoes
median	number	35
interq	number	8
whatever	string	weights
l	integer	16
var	number	58.6938405797
lquartile	number	32
u	integer	49
mean	number	35.2083333333
stdev	number	7.66
units	string	grams
n	integer	24
r1	list	List of 24 items

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Name

Data type

Value

xxxxxxxxxx
 
0.25*r1[17]+0.75*r1[18]

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Mean (1 dp)	SD (3 sf)	Median (exact value)	IQR (exact value)
Gap 0.	Gap 1.	Gap 2.	Gap 3.

Advice

Sample mean: The sample mean is $\frac{\var{sum(r0)}}{\var{len(r0)}} = \var{precround(mean(r0),1)}$ to 1 decimal place.

Sample standard deviation: The sample standard deviation is $\var{stdev(r0,true)}=\var{siground(stdev(r0,true),3)}$ to 3 significant figures.

If you order the data in increasing order you get the following table:

$\var{r1[0]}$	$\var{r1[1]}$	$\var{r1[2]}$	$\var{r1[3]}$	$\var{r1[4]}$	$\var{r1[5]}$	$\var{r1[6]}$	$\var{r1[7]}$
$\var{r1[8]}$	$\var{r1[9]}$	$\var{r1[10]}$	$\var{r1[11]}$	$\var{r1[12]}$	$\var{r1[13]}$	$\var{r1[14]}$	$\var{r1[15]}$
$\var{r1[16]}$	$\var{r1[17]}$	$\var{r1[18]}$	$\var{r1[19]}$	$\var{r1[20]}$	$\var{r1[21]}$	$\var{r1[22]}$	$\var{r1[23]}$

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

Median: The median lies between the 12th and 13th entries in the ordered table and is given by:

\[0.5\times x_{12}+0.5\times x_{13} = 0.5\times\var{r1[11]}+0.5\times \var{r1[12]}=\var{median}\]

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

\[\frac{n+1}{4}=\frac{\var{n+1}}{4}=6\frac{1}{4}\]

Hence the Lower Quartile lies between the 6th and 7th entries in the ordered table.

It is \[0.75\times x_6+0.25\times x_7 = 0.75\times\var{r1[5]}+0.25\times \var{r1[6]}=\var{lquartile}\]

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

\[\frac{3(n+1)}{4}=\frac{\var{3*(n+1)}}{4}=18\frac{3}{4}\]

Hence the Upper Quartile lies between the 18th and 19th entries in the ordered table.

We find it is \[0.25\times x_{18}+0.75\times x_{19} = 0.25\times\var{r1[17]}+0.75\times \var{r1[18]}=\var{uquartile}\]

The interquartile range is defined to be

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

and so in this case we have:

\[ \text{Interquartile range} = \var{uquartile}-\var{lquartile}=\var{interq} \]

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This question is used in the following exams:

Descriptive statistics test by Mario Orsi in Mario's workspace.
MA5636 Week 1 by Ben Parker in Ben's workspace.

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Mario Orsi 5 years, 10 months ago

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Generated value: `number`

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Mean, SD, median and IQR

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Mario Orsi 5 years, 10 months ago

Error in variable testing condition

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Generated value: number

Parts

Gap-fill

Generated value: `number`