Example # 01: Consider following frequency distribution: Class 18 19 20 21 20
Freque ncy 10 20 30 20 10
1. Find out out Mean Mean Media Median n and Mode! Mode! ". Find out out arianc ariance e and standard standard de$iati de$iation! on!
%olution: For Median: To To fnd out median median we always fnd fnd median group group frst.
&ow to 'nd median group! To To fnd out median median group calculate ()f*"+ and then fnd this value in (C.f+ column where it lies that class would be our required median group. ,ote: )f*" - / lies in rd class (tose $alues wic lie in between 1 to 20 lies in rd class+
Class 18 19
Freque ncy (f+ 10 20
Cumulati$e frequency (C.f+ 10 30
30 20 10
0 80 90
Median group
"0 21 20 3otal ()+
)f - 40
Calculation of Median: !bove "requency distribution is simple "requency distribution where we don#t have class intervals so$ to fnd out median simply pic%ed class value o" median group i.e.$ 20.
Median - "0
For Mode: To fnd out median we always fnd median group frst.
&ow to 'nd mode group! To fnd out mode group fnd ma&imum "requency$ where it lies that class would be our required mode group. ,ote: if " or more classes a$e a same maximum frequency ten tere would be " or more mode groups wic mean " or more mode lies in data subsequently.
Class 18 19
Frequen cy (f+ 10 20 Mode group
"0 21 20
30 20 10
Calculation of Mode: !bove "requency distribution is a simple "requency distribution where we don#t have class intervals so$ to fnd out mode simply pic%ed class value o" mode group i.e.$ 20.
Mode - "0
For Mean $ariance and standard de$iation:
Mean=
∑ fx ∑ f
∑f x variance =σ = ∑ f
2
2
−
( ) ∑ fx ∑ f
2
standard deviation =σ =√ variance =σ
2
Freque ncy (f+ 10 20 30 20 10
Class (x+ 18 19 20 21 20 3otal ()+
Mean=
)f - 40
180 380 00 '20 200 )fx 1560
32'0 (220 12000 8820 '000 )fx" /"60
( )
∑ f x − ∑ fx σ = ∑ f ∑ f 2
σ =
fx"
∑ fx = 1780 =19.78 ∑ f 90
2
2
fx
35280 90
2
( )
−
1780 90
2
=392−( 19.78 ) =0.8395 2
standard deviation =σ =√ variance =√ 0.8395 = 0.916 2
2
Mean=19.78, σ =0.839 5∧σ =0.9162
Example # 0": Consider following frequency distribution: Class 7nter$al 1)))* )))10 11)))1* 1)))20 21)))2*
frequenc y * 10 1* 12 8
1. Find out Mean Median and Mode! ". Find out ariance and standard de$iation! For Median: To fnd out median we always fnd median group frst.
&ow to 'nd median group! To fnd out median group calculate +,"-2 and then fnd this value in +/." column where it lies that class would be our required median group. ,ote: )f*" - "/ lies in rd class (tose $alues wic lie in between 12 to 0 lies in rd class+
Class 7nter$al (C87+ 1)))* )))10
Frequen cy (f+ * 10
Cumulati$e frequency (C.f+ * 1/ C
Class 9oundary (C89+ 0.*)))*.* *.*)))10.* Median group
11)))1* 1)))20 21)))2* 3otal ()+
1/ f 12 8 )f - /0
30 '2 *0
10./8881/./ 1*.*)))20.* 20.*)))2*.*
Calculation of Median: !bove "requency distribution is grouped "requency distribution where we have class intervals so$ to fnd out median use "ormula to calculate median.
Formula:
[(
Median=l .b +
) ( )]
∑ f −C × 2
h f
here
l. b → lower boundryof median group .
h → height of median group ( u . b−l . b ) . f → frequency of median group.
C → above C . f of median group.
l. b =10.5, h=u . b −l . b =15.5− 10.5 =5
f =15 ,C =
∑ f =25
15 ∧
2
[(
Median =l .b +
[
) ( )]
∑ f −C × 2
Median=10.5 + ( 25 −15 ) ×
h f
( )] 5
15
Median=10.5 + 3.33 =13.83
Median - 1.6
For Mode: To fnd out median we always fnd median group frst.
&ow to 'nd mode group! To fnd out mode group fnd ma&imum "requency$ where it lies that class would be our required mode group. ,ote: if " or more classes a$e a same maximum frequency ten tere would be " or more mode groups wic mean " or more mode lies in data subsequently.
Class 7nter$al (C87+ 1)))*
Frequen cy (f+ * 10 f 1 1/ f m 1" f " 8
Class 9oundary (C89+ 0.*)))*.*
)))10
11)))1*
*.*)))10.* Mode 10.*)))1*.* group
1)))20 21)))2*
1*.*)))20.* 20.*)))2*.*
Calculation of Mode: !bove "requency distribution is a grouped "requency distribution where we have class intervals so$ to fnd out mode use "ormula to calculate mode.
Formula:
M ode =l . b +
[(
) ]
f m −f 1 ×h 2 f m− f 1− f 2
here
l. b → lower boundryof mode group .
h→heightof mode group ( u .b −l . b ) . f m → max. frequency ∨frequency of mode group. f 1 → above frequency of mode group.
f 2 → below frequency of mode group .
l. b =10.5, h=u . b −l . b =15.5−10.5 =5 f m=15 , f 1= 10 ∧f 2=12
