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Measures of dispersion are descriptive statistics
that show how similar or varied the data are for
a particular variable (or data item).
Measures of spread include the range, quartiles
and the interquartile range, variance, standard
deviation and coefficient of variation.

Measure of Dispersion - Grade 8 Statistics.ppt

The mode, median, and mean summarise the
data into a single value that is typical or
representative of all the values in the dataset.
But this is only part of the 'picture' that
summarises a dataset.
Measures of spread summarise the data in a way
that shows how scattered the values are and how
much they differ from the mean value.
Batsman A has four innings and scores 25, 25, 25, 25
Batsman B has four innings and scores 0, 0, 0, 100
They both average 25 but they are very different scores.

Measures of dispersion are sometimes referred
to as variation or spread.
The main measures of dispersion are:
◦ Range
◦ Quartile deviation
◦ Mean deviation
◦ Standard deviation
◦ Variance
◦ Coefficient of variation

Measures the difference between the highest and
the lowest item of the data.
Range = highest observation – lowest observation
While easy to calculate and understand, the range
can easily be distorted by extreme values.

The quartiles divide the set of measurements into four equal parts.
•Twenty-five per cent of the measurements are less than the lower quartile
•Fifty per cent of the measurements are less than the median
•Seventy-five per cent of the measurements are less than the upper quartile.
So, fifty per cent of the measurements are between the lower quartile and the
upper quartile.
The lower quartile, median and upper quartile are often denoted by Q1, Q2 and
Q3 respectively.
The median is also denoted by m.
.

A quartile is found by dividing by dividing
the arrayed data into four quarters.
There will be three quartiles (not four!).

To determine the interquartile range
deduct Q1 from Q3

Let n = the number of observations
Where n/4 is not a whole number -
let m= the next whole number larger than n/4
the lower quartile is the mth observation of the
sorted data counting from the lower end.
the upper quartile is the mth observation of the
sorted data counting from the upper end.

Where n/4 is a whole number - let m= n/4
the lower quartile is halfway between the mth
observation and the (m + 1)th observation of the
sorted data counting from the lower end.
the upper quartile is similarly defined counting
from the upper end

The median of an even data set is calculated as
the average of n/2 and [(n/2) +1]

By measuring the middle 50% of values only, the
interquartile range overcomes the problem of
outlying observations.
It may be calculated from grouped frequency
distributions that contain open-ended class
intervals

Deviation is the difference between each item
of data and the mean.
The mean deviation measures the average
distance of each observation away from the
mean of the data.
Mean deviation gives an equal weight to each
observation and is generally more sensitive
than either the range or interquartile range,
since a change in any value will affect it.

1. Calculate the mean of the data
2. Subtract the mean from each observation
and record the difference
3. Write down the absolute value of each of
the differences (i.e. ignore positive and
negative signs)
4. Calculate the mean of the absolute values

The four steps for mean deviation are written as
1. Find x
̅
2. For each x, find x – x
̅
3. Now find Ix - x
̅ I for each x
4. Find ΣIx - x
̅ I and divide by n

The batting score of two cricketers, Joe and John were recorded
over their 10 completed innings to date. Their scores were
Joe 32 27 38 25 20 32 34 28
40 29
John 3 80 64 5 11 87 0 2
53 0
1. For each cricketer calculate the batting average (mean
score) and the mean deviation
2. There is only one batting position left on the team for
the next match.
Would you pick Joe or John? Why?

x
̅ = 32+27+38+25+20+32+34+28+40+29
10
= 30.5 runs

x
̅ = 3+80+64+5+11+87+0+2+53+0
10
x
̅ = 30.5 runs

Mean Deviation calculations for Joe
Score
( x )
Deviation from mean
( x - x
̅ )
Absolute value of
deviation
I x - x
̅ I
32 +1.5 1.5
27 -3.5 3.5
38 +7.5 7.5
25 -5.5 5.5
20 -10.5 10.5
32 +1.5 1.5
34 +3.5 3.5
28 -2.5 2.5
40 +9.5 9.5
29 -1.5 1.5
Σ( x - x
̅ ) = 0 ΣI x - x
̅ I = 47.0
Mean = 30.5

Joe = ΣIx - x
̅ I
n
= 47.0
10
= 4.7

Mean Deviation calculations for John
Score
( x )
Deviation from mean
( x - x
̅ )
Absolute value of
deviation
I x - x
̅ I
3 -27.5 27.5
80 +49.5 49.5
64 +33.5 33.5
5 -25.5 25.5
11 -19.5 19.5
87 +56.5 56.5
0 -30.5 30.5
2 -28.5 28.5
53 +22.5 22.5
0 -30.5 30.5
Σ( x - x
̅ ) = 0 ΣI x - x
̅ I = 324.0
Mean = 30.5

John = ΣI x - x
̅ I
n
= 324.0
10
= 32.4

It depends on your priorities!
If you are looking for a consistent batter, the
choice will be Joe, since he has a much smaller
mean deviation.
While he probably would not make a large score,
his past record indicates he can be relied on to
make a score fairly close to his average (the mean
deviation of his score is less than 5).

If you are looking for a batter who could
possibly obtain a large score (and in
doing so considerably help to win a
match) then John will be the choice.
However there also seems a high risk that
he would get a very low score.

The standard deviation measures the average
distance each item of data is from the mean.
It differs from the mean deviation in that it
squares each deviation and then finds the square
root of this rather than taking the absolute value.
Standard deviation is the most commonly used
measure of dispersion for statisticians.

.
. _____
√ Σ ;ǆ
- ǆ
ำ
Ϳ
Ϯ
E

In practice, it is rare to calculate the value of mu
since populations are usually very large. Instead,
it is far more likely that the sample standard
deviation (denoted by S) will be required.
The formula for calculating S is not the same as
simply substituting S for and n for N. There
are good theoretical reasons for not doing so.

If we did this, and used the value of S to estimate
the value of , the result would be too small.
To correct this error, instead of dividing by n we
divide by (n-1). This results in the following
formula for S:

A market researcher, Gavin, was interested in the
discrepancy in the prices charged by supermarkets
for a leading brand of pet food. To check this he selected
a random sample of 12 stores and recorded the
price displayed for the same 400 gram can.
The prices in cents were
89 72 77 78 82 94
80 88 85 73 78 76
Find
a) the mean
b) the range of prices
c) the mean deviation of prices
d) the standard deviation of prices

Now use the Financial Calculator to Find the Mean and Standard
Deviation… check the question to see if it a sample or a population.

 The standard deviation can not be negative
 The more scattered the data, the greater the
standard deviation
 The standard deviation of a set of data is zero if, and
only if, the observations are of equal value
 A rough guide to whether a calculated answer is
‘reasonable’ is for the standard deviation to be
approximately 30% of the range

Note for this data set is the standard deviation around 30% of the range?
Range is …… 94 – 72 = 22
Standard Deviation is 6.7 … 22 x .3 = 6.6 …. It won’t always be this close

 The standard deviation can never exceed the range of
data
 Due to the squaring operation involved in its calculation,
the standard deviation is more influenced by extreme
values than is the mean deviation and is usually slightly
larger than the mean deviation
 The square of the standard deviation is called variance

Variance measures the spread (in total) of
the data.
Variance is equal to the square of the
standard deviation so
Variance = (Standard Deviation) 2

Batsman A has four innings & scores 25, 25, 25, 25
Batsman B scores 0, 0, 0, 100
What are their averages ?
What are their Standard Deviations?
Example using standard deviation

Using the calculator Stat Mode 1,1 then
25, xy, 0, ENT,
25,xy, 0, ENT,
25, xy, 0, ENT,
25, xy, 100, ENT
RCL 4 and RCL 7 will give the calculation for
the mean score for each batsman.

 What is the difference between the
Population and a Sample?
 How can I remember that on my calculator?
Sample smaller than the population 5<6 and
8<9? OR “S” for sample

Back to our batsmen ….
Batsman A has four innings and scores 25, 25, 25, 25
Batsman B scores 0, 0, 0, 100
What are their Standard Deviations? If we took a sample of
their batting scores – perhaps there were 20 innings and we
sampled 4 innings – or the population that is they had only
batted 4 times – these were the complete scores
Batsman A has a standard deviation of 0 whether it is a
sample or not (RCL 5, RCL 6) and Batsman B has a Standard
Deviation of 50 if it was a sample (RCL 8) and 43.3 if it was
the population (total data) (RCL 9)
Long Hand calculation : -

Long Hand calculation :
Sample for A (0^2 + 0^2 + 0^2 + 0^2) / 3 = 0
Population for A (0^2 + 0^2 + 0^2 + 0^2) / 4 = 0
Dev Dev
Scores B From mean Squared
1 0 -25 625
2 0 -25 625
3 0 -25 625
4 100 75 5625
Total 7500
Sum of deviations divided by 3 2500
Now find the square root 50
Sum of deviations divided by 4 1875
Now find the square root 43.30127

This is a measure of relative variability.
It is used to measure the changes that have taken
place in a population over time, or to compare
the variability of two populations that are
expressed in different units of measurement.
It is expressed as a percentage rather than in
terms of the units of the particular data.

The formula for the coefficient of variation,
denoted by V is:
V = 100 multiplied by S and divided by x
̅
Where x
̅ = the mean of the sample
S = the standard deviation of the sample
V = 100 . S. %
x
̅

This is the Standard Deviation divided by the mean – that is the ratio of
the standard deviation to the mean – the higher the figure the greater the
deviation
Back to Batsman B we would have a Coefficient of variation of 50 / 25 =
2 – quite a significant variation

Using the calculator for the Standard Deviation – Mode 1,0 , then 10, ENT, 15,
ENT………. Then RCL 5 since the question said it was a sample ( not RCL 6)
Answer is 4.1231

Using Calc – Mode, 1,0 (2nd
f , Alpha,0,0 – to clear just in case
36, xy, 3, ENT, 37, xy, 3, ENT ………. Then RCL 4 for the mean and RCL 5 for
sample deviation = 1.70

Note we will get the calculator to calculate the standard deviation – just
demo long hand calculation here – also shouldn’t be asked for the Mean
Deviation in a class test.

Suggested Questions from Textbook……
Select a range of questions from the Problems in this chapter – enough so that you feel
comfortable with this topic

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Measure of Dispersion - Grade 8 Statistics.ppt

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