This document discusses measures of central tendency and introduces various types of means including arithmetic, geometric, and harmonic means. It provides definitions and formulas for calculating each type of mean. It also discusses how to handle both discrete and continuous data, including how to convert discrete class intervals into continuous intervals in order to calculate the mean. Examples are provided to demonstrate calculating the mean for both individual/ungrouped and grouped data.

1. @ Chaitali C. Dongaonkar 1
BIOSTATISTICS AND RESEARCH
METHODOLOGY
UNIT – I

2. @ Chaitali C. Dongaonkar 2
Contents
1. Learning Objectives
2. Introduction to Measures of Central Tendencies
3. Types of Statistical Averages
4. Mean – Arithmetic Mean
5. Mean – Combined Mean
6. Taking Logarithm of any number
7. Taking Antilogarithm of any number
8. Mean – Geometric Mean
9. Mean – Harmonic Mean

3. Onit 1L
MMeasures of Centrau Tendencies:
Leaning ObJectives:
To Know meaning of Mean, Meclidn & Mode
Tounderstand Calculations of vamous tupes of
mean like Amthmetic. Geometic an od Hoamonic
Todescibe meits & demerits of mean,median
&mode
Tnmoluction
AverageSingle wod/Single fiqure which gives
Complete plcture of pheno menon
eg If we collect dlata of heights ot
2000 Students in College
we will unable to emember al
2000 numbers
SO, we Calculate one single number
nown as Average"
2. Avemage Ke presentatrve closer to most
member3 ofdata
Hence
This mepresentative known osAvexage"
3.Central Tendency: This Average Lies more at the
Cente of the data
also known as
A
Centsa lendloney
-@CcD

4. 2
To Obtain Single number which wi
Tepres ent whole cdata.
4.0blective
5. CompaTislon: Compamslon of lbulkydata is cntical
So
taking average and compamsion
of data made simpler &guicker
6 Taking averrage: NOT last stage of dlatoa analysis
but
Stating point of Calculations
heeded fo further analysis.
Statistical Avemrage
A
Posttional Average
Mathematical Average
Median
CMe)
Mode
Hamoni c
Anthmatic
Mean
CX)
Geometic
Mean
CG)
CM.)
Mean
H)
MEAN
Three types of Means O Amthmatic Mean C)
Geome.tricMean C)
Hamonic Mean CH)
-ccp

5. lbO Amthmatic Me.an
A.M.of Set of observations is defined as the
Sum ofallo bseavacfions divided by the total
number 0f obsevations
Denotedl by o u
Athmahc meah tor an ungvouped data:
Tndividual semes
Tf ,2,A3,, . .
-,4n be the values of
n' obsevations then orithmartic ean
X =
=
2i
where ; =
Aj2,3,, - n
O gn=numberof Items/0bservaTOns
Thing to be Remember
Sum
of given data
No.0f given data
AM XK
ExamplesS
OFind Mean of given data 12,19,15,22,17
Ans
n
X =219+1S+22 +I17
5
Mean of given dato 1S
- cCD

6. (2) he a.bsorbance values ot aspisin Solution were
obtained in 9 folols using UV Specto photometer as
0-273,0:275, 0 271, 0.27S, 0.274,0. 27s,0-279,
O.278 0and 0:28I. find mean value of absorbance
Ans
n
X =U=0:273+0.27S +O.271 +0 27S+
O.274 t0 27S +0.279 +0.278 +
0. 281
2 481
0 27G
Mean value of absorbance was foundto be
0 2+6
Tryyourself
Example1: The Telativehumidity in the tablef
prduction department as given below. Calculate The
mean percent melative humidity. CAns: 66)
Days
Monday
Tuesday
Wednesda4
Thurs day
fiday
Saturoay
Percentreahve Humiclity
G 2
61
75
G5
-@cCD

7. DisCTete Series oY UhgaDupec requency distribatio
CWeighted AMD
0Direct Method:
A,2,3,M, ----Rn
fi,fa,fa,fhr fS-Voluesoffrequencies
(-Values of
OT weights
X f i CEf =n)
fi
Note if weights ov frequencies ave same
Ci.e. f, f2 f,= fn)
then
Ose simple mean. CAM)
Examples
OCalculate AM of maves obtained in Phama.ceutics
by Students of T: Y. B.Phom class
Marks 40 8 52 58 64 61 A7
No.of
Studentss
5 2 85 3 2 1
Ans Marks
ObtaIned
C)
No.0f
Stuclents
Cf)
f
5 200
=877
33
X = 56 37
52 364
58 46
64 320
69 207
148
+8 7-8
n f =33 PA =1977
Average marks secured by stuclents are 56.87
- CCD

8. Shovtcut Method:
X a +2fd (d--a
where
a- assumed mea /ATbitary Mecan
Middle Value of"x
d= deviation
n Number ofitems/obsemvations
f =tTeguency
EXample: C Previous Question)
No of d =-a
MaTks
Obtained Students)
C)
fd
Cf) Ca =58)
-18 90
48 -10 20
52 -42
Jo-58
64 5 30
61 33
7h 16 32
78 2 0 20
n=33 f d
f 33 = - 5I
X = a+ 2Fo
n
= 59 t
(3
= 58 + C-1:54)
X 56:46y
-CCD

9. it) Contihuous Senes oT ouped treguency distibution:
Step deviation/Assumecd/Tndlirect Method:
when class imtervals aTe in continuous Seies.
X = A + (2T)h
N
where A = asSumed mean /Arbitary mean
Class mark in which (Nl2)th
tem
falls orSimply micldle
Value of X.
h class width /5i2e Ch- UL-LL)
N Total frequency Both are
f Class treguency
Step deviation
egual
X = Mid Value
d =XA
h X =
UL FLL
(6Direct Method:
i m;
fi
where
f = treguency
m mid point of class
m = L + L L
2
UL Upper Limit ofclass)
LL Lower Limit of class )
- CCD

10. Example Calculate Avrithmatic mean of following
distibution of patient of Covicd-19 as per their
we ights
We ight CX) 30-40 40-So 50-60 60-7o70-20 80 -70
No.of patients s 12 1 4 7
Ans DStepdeviation /Tndimect Method
No.of X d
Patients
Cf)
Weight f d
Cwiiddpt (d -XA)
30-40 5 3S -IS
h0 -SO -16
SO 6 S
O
2 65A O
607O0
7-0 80 7S
80-90 I7 S5 2 34
N - f=63 Zfd 10
- At( h
h = UL LL = 0 3 0 = 10
:. Y A
+( h
N
A Middle tem of xCmicd point)
A = G5
6 5 + x 1o
= 65 + 100
6 3
X = 66.59
-cCD
X=65 + | 58

11. 2) Direct Methocl
No.of
Pattents
CP)
Mid poimt
weight
CA)
f m
Cm)
30-40 35 175
40-So 360
SO 60 5S 38S
60-70 | 2 6S 730
70-8 0 7s Oso
80-9o 17 85 144s
N-2f63 Zf.m-41s
fm .m -OLtLL
2
epuiteo 63
GG 59
Averagewe.ight of patientsof Covid -19
S 66.S8.
TryitYourSelf
EXample1: Calculate the mean Overal fartality Rate
InSmall pox trom age -wise fatality Rate qiven below.
Agegopup: O- 10 10-2.0 20-30
30-40 40-SO 50-60
FatalityRate 8ls 12
30 0 6
- @CCD

12. ) Discrete Semes in Groupeodl freguencydistmbution
when class intevals are in discrete semes
i.e. NoT Continuous.
Ose Same fovmula as continous class intervals
but Cohvet discTe te class intervals into Continuouc
class intervals.
Thing to be remember
9. How to Convet discrete class intervals
into Continuousclass intemvals 2
Ans: StepT Tdentify class intervals in Continuous
Series oT in discrete Semes.
eq Class Tntevals
30-3S
3-60
o)-4
6S-S0
0 - S S
Here, Upper limit of
fietclass is Lows
limit ot second class,
So,1t is considereol
LS ContinuouS Class
Class
1.30,31,32,39,3 4
2. 35,36,37,38,39
Co tinuous
Upper limit
NOT included
6O interval
i.e.35Uclass
-LLS
35
This is Cons id ere.d as
Continuous qroupecl Semes
ClassLntervals Here, Upper limt ot
31-3
e
Class first class is olifferent
Than Lower limt of
Second class. so, it js
Considered as "oliscvete
class infteval
i.e. 35 UL
36 LL
1.31,32,33,34,35 (36)-% o
2. 36,37,38,5140 6 -4S
6 5Oo
disconthnuous 5D- 55
L.LUL both
included
-cCD

13. Step Canvemting Discontinuous class intemials O
into continüous class intervals.
Take o Considew any two class interrvals
find out gab between 0L Of one class &
LL ofnext class
UL Of fiet class = 35
LL Of Next class = 36
difference 3G - 35
Divide difference obtainedl by 2.
acdol clifference/2 value in UL Subtract
difference! 2 Value fom LL
Class intemvals
-05 31 3 S +05
New class intervals
3 0 5 35 5
355-40 S
-0.5 36 -40 +0.5
h hSy +o.S 0 S 45 5
50 5
-0.546 0+0.5 hoAS 5
0 . 5
S
55
5+05
Toy tt youTself
50.5 55 5
Example 1: Convert diScontinuous class intenials
Into Continuous class ihtervals.
Class intemvals
class intervals
IS - 2O
25 3 0
35-40
c 4S 5 0
20
3 0
-ho
5o
235 5 -60
- CCD

14. Lmpotat
while Considemng class intervals treguency
UL value of any class should be considered
tn next Class interval
e Class inteTVal Values to be
included
O-9 CO 1,2,3,,S,6,7,
10-19 ULOf Fi33t
20-21 class to be
2t020
320 30
50 3 9
h04 1innext class
30-0 COnsidered
50
as LL
Example 1: Find the mean of frollowing clata
13S 139 140- 144 14S -14 1
Class: I30 - 134
freg
1S 2S S
Ans: 1. Observe class intervals in Continuo us or
in discontinuous semes. Here 1is disConhinuous
Semes. s0,Convertit thto Cohtimuous semes
oifference -L
next
Etass-
lass
L of fiat class = 34
LL Of Second class 135
, difference LL of second
Class
DLOf
fia clasS
=13 S 134
difference 1
2. Divide diffevence by 2.
o 5 -cD

15. 3. Now, Subtract o. 5 from LL and odd o. 5 in
UL.SO, It will give continous class interrval.
then find out meah by using formulah
Mid
Pointn)
.f
class New class
fm
130-1341215-134 5 13 2 5 6G0
13S-139 154S139,S37 S
140-154 139-S 14hS 14 2
20 55
2S 3550
14S-1491455 149 5 47 1S 220S
Zf=60 Efm-9 470
Mid point = LL of Neuw UL Of New
C tLss Class
f : m; ( f =N)
f
6O
141 16
Mean 0f giveh d ata is A116
EXample 2: Following an aray of 5 paTacetamol
Tablets pmduced by ponduction manager in dif Ferent 6 5
OuTches. Prepavme freguency distibution table anol
tindout mean.
5 27 50 4S 32 36 413 4 4 8 27 46
3 t 34 42- 4S 31 8 27 9 18 17 32 93
3S 37 47 29 46 2G G 3 35 33 42 31 0
5 42 44 41 36 27 39 5 56 53 88 $5 31
2 38 54 36
3 3 8 56 596I 65 64 2 64.
-@ CCD

16. Ans
.Step1 Deciding class width for given data.
Any suitable class width can be taken.
Here Cosider class width as 5
Step 2 Observe qiven clata and identify lowest
Value and highest value.
Here, Lowest value is 25L Highest
Value is 74.
Step3 Now, set classintervals Statingtoom
Lowest value to Highest value, fietly
Class intervals will be in dis crete dlishibuton
ahd then convevt it into Continuous one
Here, class width considered as 5.so,iet
Class intemval wil| become Cstasting tom
Lowest value 25) 25 29.
1.e. Class width = 05
1Sclass interval = 25 -29
i.e. 25,26,27,29,29 = 05
Step Consicler class intervalsHI Highestvalue
iven in distobution
Here.it is 74.So,we wil prepaTe
Class intevals starting tom Lowest
Value. i.e 25 to Highest value i.e. 74.
Important
Always PTe pare Cliscrete/ discontinuous class
Intervals when distibution data is given& then
Convet it into Continuous class intervals.s
it should include all values qiven
CCD

17. Now, we wil| wnte class intervals. &
we wil| fino freguency of same by
Considemng tally marks.
chep 6 We will convet cliscrete class intevale
into Coninuous classintervals then
we can easily tind out mean.
Step5
New class Mid
Lntervals Pointm) m
OG 2h 5 295 27 162
Freq
class Tally
Intervals Marks
25-29
30-34 HH1
34-31 H I 14
60-44
29 S 34:5 32 2 88
34 5-39.5 37 513
08 395 44: 5
hh:Sh1S 4
2 3 36
659
50-5hH 06 41.S - 54:5 52
55-59 I1
Go-6411
GS69 0665-9:567G7
t0 th O
03 5h:559 5
595 64 S
3 12
57 71
03 62 1 96
69-5 74:5 72 2
f-65 Efm 2-
Zfi mi e g
2f
2770
fog 4 2 61
Sd,Average tablets pmcduced by produdion
manager in different G5 batthes are 4261
l.e. 43 ehe
-CCD

18. Trick
1. while Converting discrete classes into
ContinuoUS classes Just add class width
into previous class's VL & LL
2. Calculating midpoint Just find out mid point
Of fiet class & aded class width into previous
Class's midpoint
Merts of AM:
I. Simple to understano
2. Easy fo calculate as based on obsenvations
3. Determined by rigid tomula 2Hence, itis
exacttigure
4 Considers all scOTes in disti buthions & means of
different distmbuctions are useful tor Compazative
PUTposes
5. Least affectecl by tluctuations as compared with
Other measures
Dememts OF AM:
1. Only guantitative measure, NOT forgqualitate
analysis
2. NOT used in case ot open end classes
3. Preguencyof any unit missing,nmean cannot be
Calculated
. Sometimes mean notbe an actual item in the semies
and may be meaningless tigure
eg Aveoage number 0f Chlldren per family
1S 2 22
- CCD

19. Exdmples on Combined Mean
Example 1 Calculate Combined AM of medical &
paramedical StatF wosleing in govemment hospitals
of Pune distict in Last flve year3.
Phamacist NuTSeSMidwives ard boys Drs
HI3 3122 26S
28
Yea 817
21S
2.015
24S 354
217 917
2S
912
2 016 296 27
2.017 208
298 1190
229 1o12 313
2018
2h0 1342 324 3 02 1420
2019
Ans: Step1 Calculate To tal & Mean of each
Category
To tal Il09 314 IS 2S 1388 5139
Cn Cn) ns)
3 0S 2 776 1038
Cn.) Cn2)
Mean 221 8 962.8
C,) C2)
n, +n, M tngs tns, +nsM
.CombIned Mean
C Combined) 0,tna t ha+htns
CIo9 K221:) + C48)4 x 962-8)+
CIS2S X305) t (1388 X
2t7 6)+
C5139 X 1038 6)
1o9+ 4814 + 1S2S+1388 + S139
2,45,946 2+40,34,119 2 +4,6S,125 +
3,85,3088+ 5393,449 8
2545
13975r
11,124,+79
13,97S
76716 0 4
X
Combined AM of medical & pavameolical staff
woreing in gov. h0spitals of pune odismct in
ast tive years1S 79G 04.
- CcD

20. GIy itYousself
Example: Th one of Phavmaceutical Company, 120
phaumacists are working,out of uwhich 50 gis &
70 boys with mean heights 0f14SL 148cm resp
ind out mean fov boys & qirls height togethe
CAns: 14h6 75cm)
aking Log of any number
Log value
Con sists
Charaetemstic MantisSa
Step 4 Finding of Chavra ctei stic
Charactemstic =K-T
where
K Total No.ofF ddigit in
q Iven Value/ number
Note: Tf No.contain /value contain
decimal point fhen Cons]der
oligits which are on Left Side
of decimal point Ci.e. befoe
decimal point only
Characteristic = Z +I
where,
Z Total No.0f 2eros Come
oufter clecimal Pointli.e on
right sde of decimal poin t)
iofe I NOte Only when value / numbr conta) ns
rte en0. Values.i.e.zen before
decimoal point.
- CCD

21. Examplesind out chaacteri stic
O 5h67 K - I 4 3
2 I=
k-
2.8 73 k-
= 1 - 1 o
034 51 =
Z+ =
O+T =
33 57
)0-05 791 = 2 +T = E
G) 0003257-z+ =
2 +1T 3
step2 Finding of Matissa
1.Make pair of Staring 2 digits
eq 54 6 7 54 Find in T column having
values 0 -
99
GFind valuein I column
8601/7 aving values o -G.
th0emrbc 7Find mean difference
Ci3i.e. in IL column having
Values c4to 1)
.. 5h67
Search 54Thvalue
in G Cin columnTT)
G Mean dff. in 7
Cin column LE)
Mantissa =73 72
F378
LOg C5467). =
Chara &Mantissa
373 78
o 05467 ZtI -T+T-2
Log (o.o5467) = 27378
3 054G7 =k-I1- o
Log CO64¢7)= 37 O0233
O233
-ccD

22. laking Antilog otAny value
Step1 Tdentify Characteidic &Mantissa
tom given number or Value.
eg. Antilog C264
5 3
Mantiss a
Characteristic
Step 2 Consider Mantissoa &Take firrt tuo
digits & Search for 3 digit along
wIth mean difference for 4th oligit.in
Antilogarithm Tabe
eg G4 TST column value
Value fom I Column
3 Mean difference fovm 3.
Cmentioned in mean olH4
Column).
5
64 41G
3
Mean
diff.in 3 hh19
Step 3: Now add 1 into Charactenstie
eg 2 G 4S3 2+1
=
3
So,here wegot 3 digit.
Step 4 Put decimal point in Anti log value as
per obtalned digit
eg. Here value is 44 19.
'Antilog (2-6453) =
441 9
- CCo

23. 2
Example 0 2 521
Anti loq (2 921) AntfilogC0. 402
F 2
1SIX Io-3
+100O
OS29
471 215I give decimalpointafter
fir3tdigit & wite 10
ol which number yo have
3
added. C-2 2=
Antilog (-2. 2797) = 5 252x 1o
+IO O0o
O 21
O.7203
3 Antilog Co 7525) =5.G56
5649
Antilog Co-07s2 = 1 190
1189
T190
- ccD

24. 2
Geomehic Mean CG
Geometmc mean of a set'n"observations is
defined as the n th root of their product
i.e. Geometncivean nn,1,
Actual Calculation
Antilog
log logi
z) TF ,, n
ff
9iven
. . n
G Antilog : logi
N Ef
f freguency
=0bservati on
where
3 G29
-
Two geometmc means ot two
groups ot Sizesni&n
then
geometmc mean ot Combination
is
Antilog loaatn,:log$2
+h2
A) ,292 wo geomeme mean of fwo gooups
Of sizes neach then Combination
GM is
-@CCD

25. M
g
M
M

26. Where
Tf,20,be non-2ero postive values of
two obsevations
CM AM) XCHM)
Example : The weights of 10 tablets Cin gm) are
given belouw 20,1 18,1 23,117,1-19, 1. 24, L25,1 2),
1:22,2C.
Calculate Havomonicmean.
1/
AnS
2 0
O.9333
0 947
0 8130
O 8547
0 9403 o
O 9064
I18
2 3
117
19
124
1 2S
O 8 26 4
2
22
126 O793
Total = 2347
to y, H.M=
8 23 4 7
H.M. =I:2143 gm
Hammonic Mean ofweght for 10 tablets
- CCD