representation of data note

8/6/2019 Representation of Data Note

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epresentation ofdataVariables,ode median and mean,ean Variance and standard deviation

Coding- -tem and leaf diagrams

rouped dataHistogram

uartileOutlierox plotsSkewness


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,ariable X, ,or example shoe size weight and

.ationality are all variables

,ariable X easurement , xihoe size , . , , .6 5 7 7 5Weight . , .21 kg 7 9 kg

Nationality , ,ritish FrenchGerman


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Quantitatvevariable

iscretevariable

ContinuosvariableQualitatievariable

Variableas numerical values

as no numerical values

ntegers only

eal numbers


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From a sample of Jan 2010 group of A Level students, the followingdata are collected with height in cm ( x ) and weight in kg ( y ).

ID Home town Sex x , Height in cm y , Weight in kg1 Sarikei M 168.8 48.0

2 Sibu M 175.9 79.5

3 Kuching F 155.5 49.0

4 Kuching M 171.7 70.1

5 Kuching M 188.2 90.2

6 Kuching F 164.5 54.5

7 KK F 156.0 60.5

8 Kuching M 173.0 79.5

9 Sarikei M 161.8 45.5

10 Kuching M 175.9 56.8


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alitative data


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frequency

Hometown

6

2

1

%0

%0%0 %0

ualitative data


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alitative data


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Height

168.8

175.9

155.5

171.7

188.2

164.5

156

173

161.8

175.9

Sum = 1691.3

Mean = 169.13

,antitative datantinuous variable


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easure oflocationMode : s the value that occursost often .Median : s the iddle value henhe data is

.rranged in ascending orderMean : n

xx =

o find medianf n is odd , ake.aluef n is even , verage .alue and the following value

( )121 +n

n2

1


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.xample 1

,ind the mode median and mean) , , , , , , ,7 9 6 8 26 3 3

rranged in ascending order fir


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.xample 2

( )he monthly salary RM for the executives of:ompany ABZ are, ,000 2200 8000

) .ind the mean and median) .omment on the mean and median


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Coding ( )a

n

ax

x +

=

.xample 1: , ,ata 102 104106 104=x

( ) 12100 = x 1041003

12=+=x


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.xample 2= , ( ) = .iven n 80 x 160 240 find.he mean


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- -tem and leafdiagrams- nable the shape of the distribution.o be revealed- how all the data- .ll raw data values are not lost


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.xample 1raw a stem and leaf diagram:ata 60 51 53 42 51 652 50

.ind the median


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4 5 1 1 36 2 5

| 2 means 42

stem leaf

=edian 52


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.xample 2raw a stem and leaf diagram:ata. . . . . . .7 3 6 3 8 4 7 4 1 2 2 3 6. . .0 4 4 5 0

. . . . . . .7 4 6 4 8 3 7 3 2 2 5 3 6. . .5 4 7 5 2

. . . . . . .7 4 2 3 8 5 1 1 4 2 1 3 5. . .2 2 4 5 1.ind the mode and median


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.xample 3raw a stem and leaf diagram:ata 360 351 353 342356

.ind the median


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- -ack to back stem and leafdiagrams .xample 4 There are 24 children in class A .ach child is given the same problem to solvehe time each child takes to solve it is:ecorded in minutes

. . . . . . . .4 8 2 6 1 9 3 7 4 8 5 7 2 6 8

. . . . . . . .4 7 7 9 6 8 8 8 9 7 2 7 3 7 0

. . . . . . . .9 8 9 7 6 9 3 7 4 7 9 9 1 5 7he 27 children in class B are also given theame problem and their time for solving it:re

. . . . . . . .2 7 3 6 9 5 5 6 1 9 3 8 1 7 2.7

. . . . . . . .7 7 6 8 4 8 8 5 8 6 9 8 2 9 5.1

. . . . . . . .4 6 4 7 7 5 2 7 1 6 6 5 9 6 7


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rouped data, , -lasses Class boundaries Mid,oint Class width

Length of wire( cm )

Frequency, f

10 x 14 2

15 x 19 6

20 x 24 425 x 29 3

ype1 rouped frequencydistribution

014 59 04 59

0 50 59


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. 4 5 15.9 19 5

or class 15 19lass width

owerlassboundary pperlassboundary

ower classlimit pper classlimit

-idpoint17

=lass width 5


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Length of wire (cm )

Frequency, f

10 x < 16 3

16 x < 22 7

22 x < 34 5

34 x < 40 3

ype2

Length of wire( cm )

Frequency, f

10 < x 16 4

16 < x 22 6

22 < x 34 6

34 < x 40 2

ype3


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622

or class 16 22lass width

owerlassboundary pperlassboundary-idpoint19

=lass width 6


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Histogram- .o space between the bars- ives a good picture of how data.re distributed- rea is proportional to the.requency

:isadvantage Raw data values are.ost .xample 1 Give a reason to support the use.f a histogram to represent these dataecause the data is .ontinuous variable


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=requency density Frequencylass width=rea k frequency

=ormally we use k 1o =rea frequency. , ,ut k can be 1 5 2 3, . . .


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Time(minutes )

Frequency,f

Class width Frequencydensity

1 10 2

11 20 521 30 6

31 40 4

41 50 3

.xample 1random sample of 20 students was asked howong it took them to complete their. ) .omework a Draw a histogram

Class Boundaries Frequency


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0.1.2.3.4

10 20 30 40 50

Class Boundaries Frequencydensity1 10 0.5 10.5 0.2

11 20 10.5 20.5 0.5

21 30 20.5 30.5 0.6

31 40 30.5 40.5 0.4

41 50 40.5 50.5 0.3.5.6

ime in minutes

equency density

0


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) From the histogram estimate how many .tudents recorded times were between 20 5 and. .0 5 minutes) What is the modal class?) What is the probability of the students.aking more than 30 5 minutes to complete.heir homework


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umulative frequency graphshe umulative frequencies re plotted

gainst the pper class boundaries .


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Time (minutes)

Frequency, f Upper classboundaries

Cumulativefrequencies

0.5 0

1 10 2 10.5 2

11 20 5 20.5 7

21 30 6 30.5 1331 40 4 40.5 17

41 50 3 50.5 20

.xample 1random sample of 20 students was asked how .ong it took them to complete their homework.raw a cumulative frequency graph Estimate the.edian What is the probability of studentsaking more than 35 minutes to complete their.omework

C l ti


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10 20 30 40 50ime in minutes

0

510

1520


**

**

**

= .edian 25 5

= / = .robability 5 20 0 25


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Range = ighest valueowest value.xample 1 Find the.ange

, , , , ,15 26 34 4653

= ange 53 5= 48


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Quartiles,ower quartile Q1 :%5

,edian Q2 :%0,pper quartile Q3 :%5

Q1 Q2 Q3%00

%0

%25


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nterquartilerange=QR Q3 Q1


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inding the quartiles- .irst arrange the data in ascending order

(ase 1 even )umber of data- plit the data into upper half and lower.alf- .pply the median rule(ase 2 od )number of data- ind the median and delete t from the.ist- plit the data into upper half and lower.alf

- .pply the median rule


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.xample 1ind Q1 , 2 , 3 nd IQR. :ata 25 35 30 24 324 29


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Smallesvalue largesvalueQ1 Q2 Q3

( - -ox plots Box and)hisker plots


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.xample 1= ,iven smallest value 24 Q 1 = ,5 Q2 = ,0 Q 3 = , = .4 largest value 35.lot the box plot

20 25 30 35 x


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.xample 2ind Q1 , 2 , 3 .nd IQR Plot

.he box plot:ata 24 25 29 30 32 346 38


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Outliern outlier is an extreme value that

ies outside the overall pattern of the.ata, >pper fence Q3 + .5IQR,


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1 ( )2 2 4 5 ( )3 5 6 6 6 7 7 88 ( ) 0 1 2 2 4 5 6 7 7 7 8 ( )2 0 1 1 2( )

.| 1 means 2 1

.xample 3ind Q1 , 2 , 3 .nd IQR Identify the.utliers


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10 20 30 40 50ime in minutes

05

10

1520


**

**

**

2 = .5 5

.xample 4 Find Q1 , 2 ,3 .nd IQR

1 3


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,ean variance and standarddeviation

n

xx =

2

2)(

efiniti:n Variancen

xx

=ean

tandard =eviation 2

2

2

2 )(xn

x=


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.xample 1

,ind the mean variance and standardeviation of, , ,5 9 11


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Coding ( )a

n

ax

x +

=

( ) ( )22

2

=

n

ax

n

ax

= x a

y


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.xample 1= , ( ) =iven n 80 x 160 240 and

( )160 2 = .720 .ind the mean and standard deviation


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,ean variance and standarddeviationor rouped frequency

distributionn

fxx

=ean

tandard =eviation 2

2

2

2 )(xn

fx=

= fn


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.xample 1,he following table summarizes the distance,o the nearest mile traveled to work by a.andom sample of commuters Estimate the mean.nd standard deviation

Distances,

miles

Number of

commuters, f

0 - 9 15

10 - 19 38

20 - 29 2230 - 39 15

40 - 49 8

50 - 59 2

Total 100


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.xample 2company runs two manufacturing lines A and B.hich make rods 2 cm in diameter Random

.amples are taken and summarized belowDiameter( cm )

Standarddeviation

A 2 0.015

B 2 0.05

he company wishes to close one of the.ines down State which one and give a.eason


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.xample 3, ,he table summarizes the mark out of 75 for.wo different examinations

Mean mark Standarddeviation

Statistics 55 16

Mechanics 55 4

tate which paper is better for you to set.air grade boundaries


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hape of a datasetSkewness


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ositive skew

+ail in the ve xdirectionodeMedian mean

: <


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egative skew

-ail in the vedirection

modemedianmean

: > >egative skew mode median mean


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ymmetrical distribution

= =ode mean median


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.xample 1= , = ,iven mean 52 mode 56 median

= , .4 comment on the skewness


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:ymmetrical Q2 Q1 = Q3 Q2 :ositive skew Q2 Q

1 < Q3 Q2 :egative skew Q2 Q1 > Q3 Q2 Q1 Q2

Q3


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.xample 1iven Q1 = ,4 Q2 = ,1 Q3 = ,3.omment on the skewness


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kewness= tandarddeviation( )mean median


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.xample 1= , =iven mean 46 standard deviation 17= , .nd median 53 comment on the skewness

representation of data note

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