Transcript MEAN,MEDIAN AND PARTITION VALUES

DEFINETION OF CENTRAL TENDENCY
 IT IS DEFINED AS THE REPRESENTATIVE OF A GIVEN DATA.
 SOME Eg. OF CTs ARE
MEAN
MEDIAN
MODE
LOWER QUARTILE
UPPER QUARTILE
DECILE
PERCENTILE
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, 6
X= 1 + 2 + 3 + 4 + 5 + 6 = 21 = 3.5
6
6
TO FIND THE MEAN OF UNGROUPED
FREQUENCY DISTRIBUTION
FORMAT :
x
x1
x2
x3
.
.
xn
MEAN
f
f1
f2
f3
.
.
fn
fx
f1x1
f2x2
f3x3
.
.
fnxn
∑fi
∑fixi
x = ∑fixi/∑fi
TO FIND THE MEAN OF GROUPED
FREQUENCY DISTRUBUTION(WHERE CI
IS NON-CONTINUOUS)
FORMAT :
C.I
0-4
5-9
10-14
15-19
f
2
3
5
2
∑fi
MEAN: x = ∑fixi/∑fi
mid value(x)
2
7
12
17
fx
4
21
60
34
∑fixi
TO FIND THE MEAN OF GROUPED
FREQUENCY DISTRUBUTION(WHERE
CI
IS
CONTINUOUS)
FORMAT :
C.I
-0.5-4.5
4.5-9.5
9.5-14.5
14.5-19.5
MEAN: x = ∑fixi/∑fi
f
2
3
5
2
∑fi
mid value(x)
2
7
12
17
fx
4
21
60
34
∑fixi
TO FIND THE MEAN WHEN CF IS
GIVEN
FORMAT 1
MARKS
below 10
below 20
below 30
below 40
below 50
C.I
0-10
10-20
20-30
30-40
40-50
USE X= ∑fixi/∑fi
NO. OF STUDENTS(c.f)
5
9
17
29
45
f
5
4
8
12
16
∑f
f
5-0
9-5
17-9
29-17
45-29
x
5
15
25
35
45
=5
=4
=8
= 12
= 16
fx
25
60
200
420
720
∑fx
TO FIND THE MEAN WHEN CF IS
GIVEN
FORMAT 2
marks
above 50
above 60
above 70
above 80
above 90
above 100
no. of students(c.f)
36
31
21
18
7
0
C.I
50-60
60-70
70-80
80-90
90-100
f
5
10
3
11
7
∑f
MEAN
x =∑fx/∑f
f
5
10
3
11
7
0
x
55
65
75
85
95
fx
275
650
225
935
665
∑fx
CHANGE IN A MEAN
IF 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 MEAN
eg. X1, X2 ……………………….Xn
X1+a,X2+a…………..Xn+a
X= x1+x2+……………………….xn
X= X+a
n
Eg.1,2,3,4,5,6
X= 3.5
1+1,2+1,3+1,4+1,5+1,6+1
X= 3.5+1 =4.5
MEAN BY SHORTCUT METHOD
FORMAT
c.i
0-10
10-20
20-30
30-40
40-50
f
7
10
15
8
10
∑fi
mid value(x)
5
15
A=25
35
45
USE MEAN x = A+ ∑fidi/ ∑fi
di=xi-A
-20
-10
0
10
20
fidi
-140
-100
0
80
200
∑fidi
MEAN BY STEP DEVIATION
METHOD
FORMAT
c.i
0-10
10-20
20-30
30-40
40-50
f
7
10
15
8
10
∑fi
mid value(x)
5
15
A=25
35
45
di=xi-A/h
-2
-1
0
1
2
USE MEAN x = A+ h(∑fidi/ ∑fi)
fidi
-14
-10
0
8
20
∑fidi
COMBINED MEAN
LET, 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 FOR UNGROUPED DATA
FORMAT 1
X1, X2, X3……………………………Xn
n= odd
ARRANGE X1, X2, …………Xn IN ASCENDING OR DESCEDING ORDER
FIND THE VALUE OF
𝑛+1
( )th
2
OBSERVATION. THIS IS THE MEDIAN.
Eg. 1 3 1 3 2 5 6 4 5
n=9(odd).
9+1
)=
2
(
5th OBSERVATION AFTER ARRENGING IN
ASCENDING/DESCENDING ORDER
112334556
5TH OBSERVATION
MEDIAN = 3
MEDIAN FOR UNGROUPED DATA
FORMAT 2
IF n=EVEN
𝑛
( )th
2
FIND THE VALUE OF
OBSERVATION AFTER
ARRANGING IN ASCENDING/DESCENDING ORDER.
𝑛
THE MEAN OF ( )th AND THE NEXT OBSERVATION
2
GIVES YOU THE MEDIAN
Eg. 1 2 1 3 4 5
n=6
𝑛
2
112345
MEDIAN =
2+3
2
= 2.5
=
6
( )
2
= 3rd OBSERVATION
LOWER AND UPPER QUARTILE OF
UNGROUPED DATA
IF n = odd
𝑛+1
LOWER QUARTILE (Q1)=
th OBSERVATION
4
𝑛+1
UPPER QUARTILE (Q3)= 3
th OBSERVATION
4
IF n= even
LOWER QUARTILE (Q1)=
𝑛
4
UPPER QUARTILE (Q3)= 3
th OBSERVATION
𝑛
4
th OBSERVATION
DECILES AND PERCENTILES OF
UNGROUPED DATA
DECILE (Dx) =
DECILE (Dx) =
𝑛+1
×
10
𝑛
×X
10
X
IF, X=odd
IF, X=EVEN
DECILE CAN BE BETWEEN 1 AND 9
D1,D2 ………….D9
PERCENTILE (Px) =
PERCENTILE (Px) =
𝑛+1
×
10
𝑛
×X
10
X
IF, X=odd
IF, X=EVEN
PERCENTILE CAN BE BETWEEN 1 AND 99
P1,P2 ………….P99
PARTITION VALUES(Q2) OF UNGROUPED
FREQUENCY DISTRIBUTION
FORMAT
x
x1
x2
.
.
xn
f
f1
f2
THAN
.
m
.
.
.
.
fn
∑fi=N
FOR MEDIAN FIND
𝑁
( )
2
<c.f
𝑁
( )
2
.
AND LOOK FOR A NO. JUST GRATER
IN THE <c.f COLUM SAY(m)
NOW, X VALUE CORRESPONDING TO m IS THE MEDIAN
PARTITION VALUES(Q1) OF UNGROUPED
FREQUENCY DISTRIBUTION
FORMAT
x
x1
x2
.
.
xn
f
f1
f2
.
.
fn
∑fi=N
<c.f
.
m
.
.
.
𝑁
( ) AND
4
FOR LOWER QUARTILE FIND
LOOK FOR A NO. JUST
𝑁
GRATER THAN ( ) IN THE <c.f COLUM SAY(m)
4
NOW, X VALUE CORRESPONDING TO m IS THE LOWER
QUARTILE
PARTITION VALUES(DX) OF UNGROUPED
FREQUENCY DISTRIBUTION
FORMAT
x
x1
x2
.
.
xn
f
<c.f
f1
f2
.
m
.
.
.
.
fn
∑fi=N
𝑁
.
FOR DECILE X FIND
𝑋 AND LOOK FOR A NO. JUST
10
𝑁
GRATER THAN
𝑋 IN THE <c.f COLUM SAY(m)
10
NOW, X VALUE CORRESPONDING TO m IS THE DESILE X
PARTITION VALUES(PX) OF UNGROUPED
FREQUENCY DISTRIBUTION
FORMAT
x
x1
x2
.
.
xn
f
f1
f2
.
m
.
.
.
.
fn
∑fi=N
FOR PERCENTILE X FIND
GRATER THAN
<c.f
𝑁
100
.
𝑁
100
𝑋 AND LOOK FOR A NO. JUST
𝑋 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
0-5
5-10
10-15
15-20
20-25
N=100
f
7
18
25
30
20
∑f = N = 100
<c.f
7
25
50
80
100
𝑁
2
( ) =50
NO. JUST GREATER THAN 50 IN c.f COLUM IS 80
MEDIAN CLASS IS 15-20
MEDIAN = L+ [
𝑵
𝟐
−𝒄,𝒇
𝒇
] ×c.w
TO FIND D4 OF GROUPED
FREQUENCY DISTRUBUTION
FORMAT
c.i
0-5
5-10
10-15
15-20
20-25
f
7
18
25
30
20
∑f = N = 100
<c.f
7
25
50
80
100
𝑁
N=100
4 ( ) =40
10
NO. JUST GREATER THAN 40 IN c.f COLUM IS 50
D4 CLASS IS 10-15
𝑵
D4 = L+
𝟒 𝟏𝟎 −𝒄,𝒇
[
] ×c.w
𝒇
TO FIND P21 OF GROUPED
FREQUENCY DISTRUBUTION
FORMAT
c.i
0-5
5-10
10-15
15-20
20-25
f
7
18
25
30
20
∑f = N = 100
<c.f
7
25
50
80
100
𝑁
N=100
21( ) =21
100
NO. JUST GREATER THAN 21 IN c.f COLUM IS 25
P21 CLASS IS 5-10
𝑵
P21 = L+
𝟐𝟏 𝟏𝟎𝟎 −𝒄,𝒇
[
] ×c.w
𝒇
TO FIND THE MODE
 TO 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, 12
 THE MODE FOR ABOVE DATA IS 2.
MODE FOR GROUPED
FREQUENCY DATA
 FOR THIS A HISTOGRAM IS REQUIRED.
 ALSO, THE FOLLOWING FORMULA CAN BE USED
MODE = L +
𝑓1−𝑓0
2𝑓1−𝑓0−𝑓2
×𝑖
Eg.
TO FIND PARTITION VALUES USING
OGIVE CURVES
TO FIND MEDIAN USING BOTH
OGIVE CURVES
THANK YOU