MEAN,MEDIAN AND PARTITION

VALUES

MEAN

DEFINETION OF CENTRAL TENDENCYIT IS DEFINED AS THE REPRESENTATIVE OF A GIVEN DATA.SOME Eg. OF CTs ARE

MEANMEDIANMODELOWER QUARTILEUPPER QUARTILEDECILEPERCENTILE

TO FIND THE MEAN OF A RAW OR UNGROUPED DATA.

FORMAT : x1, x2, x3…………xn

MEAN x = ∑xi/n

Eg. 1, 2, 3, 4, 5, 6X= = 21 = 3.5

6 6

TO FIND THE MEAN OF UNGROUPED FREQUENCY DISTRIBUTIONFORMAT :

x f fxx1 f1 f1x1

x2 f2 f2x2

x3 f3 f3x3

. . .

. . . xn fn fnxn

∑fi ∑fixi

MEAN x = ∑fixi/∑fi

TO FIND THE MEAN OF GROUPED FREQUENCY DISTRUBUTION(WHERE CI IS NON-CONTINUOUS)

FORMAT :C.I f mid value(x) fx0-4 2 2 45-9 3 7 2110-14 5 12 6015-19 2 17 34 ∑fi ∑fixi

MEAN: x = ∑fixi/∑fi

TO FIND THE MEAN OF GROUPED FREQUENCY DISTRUBUTION(WHERE CI IS CONTINUOUS)FORMAT :

C.I f mid value(x) fx -0.5-4.5 2 2 4 4.5-9.5 3 7 21 9.5-14.5 5 12 60 14.5-19.5 2 17 34

∑fi ∑fixi

MEAN: x = ∑fixi/∑fi

TO FIND THE MEAN WHEN CF IS GIVENFORMAT 1

MARKS NO. OF STUDENTS(c.f) fbelow 10 5 5-0 = 5below 20 9 9-5 = 4below 30 17 17-9 = 8below 40 29 29-17 = 12below 50 45 45-29 = 16

C.I f x fx0-10 5 5 2510-20 4 15 6020-30 8 25 20030-40 12 35 42040-50 16 45 720∑f ∑fxUSE X= ∑fixi/∑fi

TO FIND THE MEAN WHEN CF IS GIVEN

FORMAT 2marks no. of students(c.f) fabove 50 36 5above 60 31 10above 70 21 3above 80 18 11above 90 7 7above 100 0 0 C.I f x fx50-60 5 55 27560-70 10 65 65070-80 3 75 22580-90 11 85 93590-100 7 95 665∑f ∑fx

MEAN x =∑fx/∑f

CHANGE IN A MEANIF a IS ADDED, SUBTRACTED MULTIPLIED OR DIVIDED TO EACH OBSERVATION THEN THE MEAN CHANGES ACCORDINGLY ie, a IS ADDED, SUBTRACTED MULTIPLIED OR DIVIDED TO THE MEANeg. X1, X2 ……………………….Xn X1+a,X2+a…………..Xn+a

X= x1+x2+……………………….xn X= X+a

nEg.1,2,3,4,5,6 1+1,2+1,3+1,4+1,5+1,6+1 X= 3.5 X= 3.5+1 =4.5

MEAN BY SHORTCUT METHODFORMAT

c.i f mid value(x) di=xi-Afidi

0-10 7 5 -20-140

10-20 10 15 -10-100

20-30 15 A=25 0 0 30-40 8 35 10

80 40-50 10 45 20

200∑fi

∑fidi

USE MEAN x = A+ ∑fidi/ ∑fi

MEAN BY STEP DEVIATION METHOD

FORMATc.i f mid value(x) di=xi-A/hfidi

0-10 7 5 -2 -14 10-20 10 15 -1

-10 20-30 15 A=25 0 0 30-40 8 35 1

8 40-50 10 45 2

20∑fi

∑fidi

USE MEAN x = A+ h(∑fidi/ ∑fi)

COMBINED MEANLET, n1 AND n2 BE THE NO OF OBJECTS IN TWO GROUPS,LET, X1 AND X2 BE THE MEAN OF THE TWO GROUPS THEN THE COMBINED MEAN OF BOTH THE GROUPS IS GIVEN BY,

X = n1x1+n2x1/n1+n2

MEDIAN AND OTHER

PARTITION VALUES

MEDIAN FOR UNGROUPED DATAFORMAT 1

X1, X2, X3……………………………Xnn= odd

ARRANGE X1, X2, …………Xn IN ASCENDING OR DESCEDING ORDER

FIND THE VALUE OF OBSERVATION. THIS IS THE MEDIAN.Eg. 1 3 1 3 2 5 6 4 5

n=9(odd). )= 5th OBSERVATION AFTER ARRENGING IN ASCENDING/DESCENDING ORDER

1 1 2 3 3 4 5 5 6 5TH OBSERVATION

MEDIAN = 3

MEDIAN FOR UNGROUPED DATAFORMAT 2

IF n=EVENFIND THE VALUE OF th OBSERVATION AFTER

ARRANGING IN ASCENDING/DESCENDING ORDER. THE MEAN OF th AND THE NEXT OBSERVATION GIVES YOU THE MEDIANEg. 1 2 1 3 4 5 n=6

1 1 2 3 4 5 = 3rd OBSERVATION MEDIAN = = 2.5

LOWER AND UPPER QUARTILE OF UNGROUPED DATA

IF n = odd LOWER QUARTILE (Q1)= th OBSERVATIONUPPER QUARTILE (Q3)= OBSERVATIONIF n= evenLOWER QUARTILE (Q1)= th OBSERVATIONUPPER QUARTILE (Q3)= OBSERVATION

DECILES AND PERCENTILES OF UNGROUPED DATADECILE (Dx) = IF, X=oddDECILE (Dx) = IF, X=EVEN

DECILE CAN BE BETWEEN 1 AND 9D1,D2 ………….D9

PERCENTILE (Px) = IF, X=oddPERCENTILE (Px) = IF, X=EVEN

PERCENTILE CAN BE BETWEEN 1 AND 99P1,P2 ………….P99

PARTITION VALUES(Q2) OF UNGROUPED FREQUENCY DISTRIBUTIONFORMAT

x f <c.fx1 f1 .

x2 f2 m. . .. . .xn fn .

∑fi=N

FOR MEDIAN FIND AND LOOK FOR A NO. JUST GRATER THAN IN THE <c.f COLUM SAY(m)NOW, X VALUE CORRESPONDING TO m IS THE MEDIAN

PARTITION VALUES(Q1) OF UNGROUPED FREQUENCY DISTRIBUTION

FORMATx f <c.fx1 f1 .

x2 f2 m. . .. . .xn fn .

∑fi=N

FOR LOWER QUARTILE FIND AND LOOK FOR A NO. JUST GRATER THAN IN THE <c.f COLUM SAY(m)NOW, X VALUE CORRESPONDING TO m IS THE LOWER QUARTILE

PARTITION VALUES(DX) OF UNGROUPED FREQUENCY DISTRIBUTION

FORMATx f <c.fx1 f1 .

x2 f2 m. . .. . .xn fn .

∑fi=N

FOR DECILE X FIND AND LOOK FOR A NO. JUST GRATER THAN IN THE <c.f COLUM SAY(m)NOW, X VALUE CORRESPONDING TO m IS THE DESILE X

PARTITION VALUES(PX) OF UNGROUPED FREQUENCY DISTRIBUTIONFORMAT

x f <c.fx1 f1 .

x2 f2 m. . .. . .xn fn .

∑fi=N

FOR PERCENTILE X FIND AND LOOK FOR A NO. JUST GRATER THAN IN THE <c.f COLUM SAY(m)NOW, X VALUE CORRESPONDING TO m IS THE PERCENTILR X

TO FIND MEDIAN OF GROUPED FREQUENCY DISTRUBUTION

FORMAT c.i f <c.f 0-5 7 7 5-10 18 2510-15 25 5015-20 30 8020-25 20 100

∑f = N = 100N=100 =50

NO. JUST GREATER THAN 50 IN c.f COLUM IS 80MEDIAN CLASS IS 15-20

MEDIAN = L+ ×c.w

TO FIND D4 OF GROUPED FREQUENCY DISTRUBUTION

FORMAT c.i f <c.f 0-5 7 7 5-10 18 2510-15 25 5015-20 30 8020-25 20 100 ∑f = N = 100N=100 4 =40

NO. JUST GREATER THAN 40 IN c.f COLUM IS 50D4 CLASS IS 10-15D4 = L+ ×c.w

TO FIND P21 OF GROUPED FREQUENCY DISTRUBUTIONFORMAT

c.i f <c.f 0-5 7 7 5-10 18 2510-15 25 5015-20 30 8020-25 20 100

∑f = N = 100N=100 21=21

NO. JUST GREATER THAN 21 IN c.f COLUM IS 25P21 CLASS IS 5-10

P21 = L+ ×c.w

TO FIND THE MODETO FIND THE MODE OF UNGROUPED DATA JUST FIND

THE MAX FREQUENCY.OBSERVATION CORRESPONDING TO THE MAX

FREQUENCY IS THE MODE.Eg. 11, 9, 2, 2, 11, 15, 9, 2, 3, 12THE MODE FOR ABOVE DATA IS 2.

MODE FOR GROUPED FREQUENCY DATAFOR THIS A HISTOGRAM IS REQUIRED.ALSO, THE FOLLOWING FORMULA CAN BE USED

MODE = L +

Eg.

TO FIND PARTITION VALUES USING OGIVE CURVES

TO FIND MEDIAN USING BOTH OGIVE CURVES

THANK YOU