Quartile Deviationsand Standard Deviation

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Quartiles

Quartiles are the values that divide a list of numbers into quarters.

•

First put the list of numbers in order • Then cut the list into four equal parts

• The Quartiles are at the "cuts"

Like this:

Example: 5, 8, 4, 4, 6, 3, 8

Put them in order: 3, 4, 4, , !, ,

#ut the list into quarters:

$nd the result is:

• Quartile % &Q%' ( 4

• Quartile ) &Q)', *hich is also the +edian, ( 5

• Quartile 3 &Q3' ( 8

ometimes a "cut" is bet*een t*o numbers ... the Quartile is the avera-e of the t*o numbers.

Example: 1, 3, 3, 4, 5, 6, 6, 7, 8, 8

The numbers are alread in order

#ut the list into quarters:

http://www.mathsisfun.com/median.html

http://www.mathsisfun.com/median.html



/n this case Quartile ) is half *a bet*een and !:

Q) ( &0!'1) ( 5.5

$nd the result is:

• Quartile % &Q%' ( 3

• Quartile ) &Q)' ( 5.5

• Quartile 3 &Q3' ( 7

Interquartile Rane

The "/nterquartile 2an-e" is from Q% to Q3:

To calculate it ust su!tra"t Quartile 1 #r$m Quartile 3, like this:

Example:



The Interquartile Rane is:

Q3 Q% ( 4 ( 4

%$x an& 'his(er )l$t

5ou can sho* all the important values in a "6o7 and 8hisker Plot", like this:

$ final e7ample coverin- everthin-:

Example: %$x an& 'his(er )l$t an& Interquartile Rane #$r

4, %9, 9, %4, %, %), 3, %!, %, 4, 4, %%

Put them in order:

3, 4, 4, 4, 9, %, %%, %), %4, %!, %9, %

#ut it into quarters:

3, 4, 4 ; 4, 9, % ; %%, %), %4 ; %!, %9, %

/n this case all the quartiles are bet*een numbers:

• Quartile % &Q%' ( &404'1) ( 4

• Quartile ) &Q)' ( &%0%%'1) ( 1*.5

• Quartile 3 &Q3' ( &%40%!'1) ( 15

$lso:

• The Lo*est <alue is 3,

• The =i-hest <alue is 18

o no* *e have enou-h data for the %$x an& 'his(er )l$t:



$nd the Interquartile Rane is:

Q3 Q% ( % 4 ( 11

QUESTION : What are the quartiles for the following set of numbers?

8, 11, 20, 10, 2, 17, 15, 5, 16, 15, 25, 6

First arrange the numbers in order: 2, 5, 6, 8, 10, 11, 15, 15, 16, 17, 20, 25

his list !an be s"lit u" into four equal grou"s of three:

herefore:

#1 is the mean of 6 and 8 $ %6 & 8' ( 2 $ 7

#2 is the mean of 11 and 15 $ %11 & 15' ( 2 $ 1)

#) is the mean of 16 and 17 $ %16 & 17' ( 2 $ 16*5

QUESTION : What is the interquartile range for the following set of numbers?

+, 5, 6, 8, , 11, 1), 16, 16, 18, 20, 21, 25, )0, )1, )), )6, )7, +0, +1



his list !an be s"lit u" into four equal grou"s of -.e:

herefore:

#1 is the mean of and 11 $ % & 11' ( 2 $ 10

#2 is the mean of 18 and 20 $ %18 & 20' ( 2 $ 1

#) is the mean of 16 and 17 $ %)1 & ))' ( 2 $ )2

herefore the interquartile range $ #) / #1 $ )2 / 10 $ 22

QUESTION : What is the interquartile range for the information shown in the abo.e

bo and whiser "lot,?

he u""er quartile, #) is

he lower quartile, #1 is +

herefore, the interquartile range $ / + $ 5



QUESTION : For the information shown in the abo.e bo and whiser "lot, what is

the range and what is the interquartile range?

he highest .alue $ 18

he lowest .alue $ +

herefore the range $ 18 / + $ 1+

he u""er quartile, #) is 15*5

he lower quartile, #1 is 5*5

herefore, the interquartile range $ 15*5 / 5*5 $ 10

QUESTION : For the information shown in the abo.e bo and whiser "lot, what are

the median, range and interquartile range?

edian $ #2 $ 6*5



3ange $ 4ighest .alue / owest .alue $ 1+ / 0 $ 1+

nterquartile range $ #) / #1 $ 12 / 2 $ 10

QUESTION : he "o"ulation of a town was re!orded e.er twent ears from 100

to 2000* he results are shown in the line gra"h*

stimate the interquartile range o.er that "eriod*



he lower quartile %#1' was the "o"ulation in 125 9 1*+ 1,000 $ 1,+00

he u""er quartile %#)' was the "o"ulation in 175 9 )*8 1,000 $ ),800

; he interquartile range 9 ),800 / 1,+00 $ 2,+00



QUESTION : he gra"h shows how the "o"ulation of rabbits on an island !hanged

from 150 to 2010*

stimate the interquartile range o.er that "eriod*



he lower quartile %#1' was the "o"ulation in 165 9 1 1,000 $ 1,000

he u""er quartile %#)' was the "o"ulation in 15 9 6*5 1,000 $ 6,500

; he interquartile range 9 6,500 / 1,000 $ 5,500



QUESTION : he bar gra"h shows the s!ores obtained from 20 throws of a die*

What is the lower quartile?

First write out the numbers in order: 1, 1, 2, 2, 2, 2, 2, 2, ), +, +, +, 5, 5, 5, 5, 5, 6, 6,

6

his list !an be s"lit u" into four equal grou"s of -.e:

herefore:

#1 is the mean of 2 and 2 $ 2



stan&ar& &e+iati$n

68 of values are *ithin

1 stan&ar& &e+iati$n of the mean

-5 are *ithin stan&ar& &e+iati$ns

--.7 are *ithin 3 stan&ar& &e+iati$ns

Question : !> of the marks in a test are bet*een % and !4

$ssumin- this data is normall distributed, *hat are the mean and standard deviation?

The mean is half*a bet*een % and !4:+ean ( &% 0 !4'1) ( 9.

!> is % standard deviation either side of the mean &a total of ) standard deviations' so:

% standard deviation ( &!4 %'1) ( %31) ( !.



$nd this is the result :

Question : he Fresha ea <om"an "a! tea in bags mared as 250 g

= large number of "a!s of tea were weighed and the mean and standard

de.iation were !al!ulated as 255 g and 2*5 g res"e!ti.el*

=ssuming this data is normall distributed, what "er!entage of "a!s are

underweight?

he following diagram shows 1, 2 and ) standard de.iations on either side ofthe mean:

250 g is two standard de.iations below the mean

>in!e 5 of "a!s are within 2 standard de.iations of the mean, it follows

that 5 ( 2 $ 2*5 of "a!s were underweight*

Questi$n : tudents pass a test if the score > or more.

The marks of a lar-e number of students *ere sampled and the mean and standard deviation

*ere calculated as 4)> and > respectivel.

$ssumin- this data is normall distributed, *hat percenta-e of students pass the test?

he following diagram shows 1, 2 and ) standard de.iations on either side of

the mean:



50 is one standard de.iation abo.e the mean

>in!e 68 of mars are within 1 standard de.iation of the mean, it follows

that )2 will be more than 1 standard de.iation from the mean*

=nd the half of those that are on the high side will "ass*

>o )2 ( 2 $ 16 of students "ass*

Question : = !om"an maes "arts for a ma!hine* he lengths of the "arts must

be within !ertain limits or the will be re@e!ted*

= large number of "arts were measured and the mean and standard de.iation were

!al!ulated as )*1 m and 0*005 m res"e!ti.el*

=ssuming this data is normall distributed and *7 of the "arts were a!!e"ted,

what are the limits?

*7 is ) standard de.iations on either side of the mean*

he following diagram shows 1, 2 and ) standard de.iations on either side of

the mean:



>o the lengths of the "arts must be between )*085 m and )*115 m

Question : he mean Aune midda tem"erature in Besert.ille is )6C< and

the standard de.iation is )C<

=ssuming this data is normall distributed, how man das in Aune would ou

e"e!t the midda tem"erature to be between )C< and +2C<?)C< is one standard de.iation abo.e the mean and +2C< is two standard

de.iations abo.e the mean:

herefore we would e"e!t the tem"erature to be between )C< and +2C< on

1)*6 of the das

%*2 & +*+ $ 1)*6'

here are )0 das in Aune

1)*6 of )0 $ +*08 $ + to the nearest da



Question he heights of male adults are Dormall distributed with mean

1*7 m and standard de.iation 0*2 m

n a "o"ulation of +00 male adults, how man would ou e"e!t to ha.e a

height between 1*+ m and 1*6 m?

Eou !an use this >tandard Dormal Bistribution !ur.e:

1*6 m is half a standard de.iation below the mean and 1*+ m is one and a half

standard de.iations below the mean:



herefore the "er!entage of male adults with heights between 1*+ m and 1*6

m

$ 15*0 & *2 $ 2+*2

2+*2 of +00 $ 6*8

herefore we would e"e!t 7 male adults with heights between 1*+ m and

1*6 m

he mean Aul dail rainfall in Water.ille is 10 mm and the standard de.iation

is 1*5 mm

=ssuming this data is normall distributed, how man das in Aul would ou

e"e!t the dail rainfall to be less than 8*5 mm?

8*5 mm is one standard de.iation below the mean:

herefore we would e"e!t the rainfall to be less than 8*5 mm on 15* of

the das

%*2 & +*+ & 1*7 & 0*5 & 0*1 $ 15*'

here are )1 das in Aul

15* of )1 $ +*2 $ 5 to the nearest da

Question : he ages of the "o"ulation of a town are Dormall distributed

with mean +) and standard de.iation 1+



f the town has a "o"ulation of 5,000, how man would ou e"e!t to be aged

between 22 and 57?

Eou !an use this >tandard Dormal Bistribution !ur.e:

22 is one and a half a standard de.iation below the mean and 57 is one

standard de.iations abo.e the mean:

herefore the "er!entage of the "o"ulation aged between 22 and 57

$ *2 & 15*0 & 1*1 & 1*1 & 15*0 $ 77*+

77*+ of 5,000 $ ),870

herefore we would e"e!t about ),870 "eo"le to be aged between 22 and

57*

Quartile Deviationsand Standard Deviation

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