Simple averages (frequency table)

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Ruth Hand

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3 years, 7 months ago

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Christian Lawson-Perfect 2 years, 6 months ago

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Ruth Hand 3 years, 7 months ago

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Ruth Hand 4 years, 1 month ago

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Question statement

This table gives demographic data for households in the North East of England (UK) in {year}.

People per household	1	2	3	4	5	6
Number of households (1000s)	{f1}	{f2}	{f3}	{f4}	{f5}	{f6}

Data (adapted) from the ONS website.

Give any introductory information the student needs.

			Name	Type	Generated Value

f1	integer	412
f2	integer	398
f3	integer	187
f4	integer	145
f5	integer	34
f6	integer	14
flist	list	[ 412, 398, 187, 145, 34, 14 ]
mf	number	412
mode_find	list	[ 0 ]
data	list	List of 12 items
data_yr	list	List of 7 items
mode	number	1

			Name	Type	Generated Value

cf2	integer	810
cf3	integer	997
tot	integer	1190
cf	list	List of 6 items

			Name	Type	Generated Value

mp	rational	595
m_val	list	[ 0 ]
median	number	2

			Name	Type	Generated Value

			sfx	integer	2603
			mn	rational	2603/1190

			Name	Type	Generated Value

year

integer

2016

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
data_yr[0]

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `integer`

412

← Depends on:

data_yr

→ Used by:

cf
cf2
flist
sfx

Statement
Advice

Advice

(a)

To find the mean use the formula $\frac{\Sigma fx}{\Sigma f}$

In other words

multiply each value by its corresponding frequency:
Work out $1\times\var{f1}, 2\times \var{f2}, 3\times\var{f3}$ etc
add up all of these:
$(1\times\var{f1})+(2\times \var{f2})+(3\times\var{f3}+ (4\times\var{f4}) + (5\times\var{f5}) + (6\times\var{f6})$
divide by the total number of households:
Total households = $\var{f1}+\var{f2}+\var{f3}+\var{f4}+\var{f5}+\var{f6} = \var{f1+f2+f3+f4+f5+f6}$

In other words
$\frac{(1\times\var{f1})+(2\times \var{f2})+(3\times\var{f3})+ (4\times\var{f4}) + (5\times\var{f5}) + (6\times\var{f6})}{\var{f1+f2+f3+f4+f5+f6}}$

$=\frac{\var{sfx}}{\var{tot}}=\var{precround(sfx/(tot),2)}$

(b)

The median is the "middle" value.

In a frequency table, the observations are already arranged in an ascending order. We can obtain the median by looking for the value in the middle position.

First add up the frequencies to find $n$ , for a large total frequency, the median is the value at the $\frac{n}{2}^{th}$ position (be careful working with small datasets).

We need to add up the frequencies until we reach this value and then the class we land in is the median.

In our example, the total frequency is $\var{tot}$ , the median is the value at the $\frac{n}{2}=\frac{\var{tot}}{2}^{th}$ position, i.e. value number $\var{(tot)/2}$

From our table, we can calculate the following cumulative frequencies:

People per household	1	2	3	4	5	6
Cumulative frequency	{cf[0]}	{cf[1]}	{cf[2]}	{cf[3]}	{cf[4]}	{cf[5]}

And by looking at our cumulative frequencies we can see that value number $\var{(tot)/2}$ occurs at some point in column $\var{median}$ , therefore the median number of people per household is $\var{median}$ .