a) A ct ua l M ea n M et ho d or D ir ec t M et ho d

Mean MedianMode Quartile deviation

Whilelookingattheearliermeasuresofdispersionallofthemsuffer fromoneortheotherdemeriti.e. Rangeitsufferfromaseriousdrawbackconsidersonly2valuesand neglectsalltheothervaluesoftheseries. Quartiledeviationconsidersonly50%oftheitemandignoresthe other50%ofitemsintheseries. Meandeviationnodoubtanimprovedmeasurebutignoresnegative signswithoutanybasis. 3

TheconceptofstandarddeviationwasfirstintroducedbyKarlPearsonin 1893. KarlPearsonafterobservingallthesethingshasgivenusamorescientific formulaforcalculatingormeasuringdispersion.WhilecalculatingSDwe takedeviationsofindividualobservationsfromtheirAMandthen eachsquares.ThesumofthesquaresisdividedbytheTotalnumberof observations.Thesquarerootofthissumisknowsasstandard deviation. Thestandarddeviationisthemostusefulandthemostpopularmeasureof dispersion. Itisalwayscalculatedfromthearithmeticmean,medianandmodeisnot considered. 4

D ef in iti on :

StandardDeviationisthepositivesquarerootoftheaverageofsquared deviationtakenfromarithmeticmean. ThestandarddeviationisrepresentedbytheGreekletter(sigma). Formula. Standarddeviation==

Standard deviation == Alternatively =

CA LC U LA TI O N O F ST AN D AR D D EV IA TI O N - IN D IV ID U A L O BS ER VA TI O N Tw o M et ho ds :- By ta ki ng d ev ia tio n of th e ite m s fr om th e ac tu al m ea n. By ta ki ng d ev ia tio n of th e ite m s fr om a n as su m ed m ea n.

CASE-I. When the deviation are taken from the actual mean. DIRECTMETHOD Standarddeviation== or =valueofthevariableofobservation, =arithmeticmean =totalnumberofobservations.

Example:Findthemeanrespirationrateperminuteanditsstandarddeviationwhenin4 casestheratewasfoundtobe:16,13,17and22. Solution: HereMean = 16 13 17 22 Standard deviation ====

-1 -4 0 5

1 16 0 25

Short-Cut Method Standard deviation ==

CASE-II. When the deviation are taken from the Assumed mean.

= = = = = 16.398

Example:Blood serum cholesterol levels of 10 persons are as under: 240, 260, 290, 245, 255, 288, 272, 263, 277, 251. calculation standard deviation with the help of assumed mean. Value A=264 240 260 290 245 255 288 272 263 277 251

576

16 676 361

81 576

64 1 169 169

-24 -4 26 -19 -9 24 8 -1 13 13

Here, Mean== = 9 = 263.9 is a fraction.

CA LC U LA TI O N O F ST AN DA RD D EV IA TI O N -D IS CE RE TE S ER IE S O R G RO U PE D D AT A T hr ee M et ho ds a) A ct ua l M ea n M et ho d or D ir ec t M et h od b) A ss um ed M ea n M et ho d or S ho rt -c u t M et ho d c) St ep D ev ia ti on M et ho d

a) A ct ua l M ea n M et ho d or D ir ec t M et ho d

The S.D. for the discrete series is given by the formula. = Whereis the arithmetic mean, is the size of items, is the corresponding frequency and

b) A ss um ed M ea n M et ho d or S ho rt -c ut M et ho d

Standard deviation== Whereis the assumed mean, is the corresponding frequency and

Ex am pl e:

Periods:10111213141516 No. of patients:2711151041

So lu tio n:

Period s:(x)No. of patients() 10 11 12 13 14 15 16

2 7 11 15 10 4 1 TotalN==50

-3 -2 -1 0 1 2 3

-6 -14 -11 0 10 8 3 =-10

9 4 1 0 1 4 9

18 28 11 0 10 16 9 =92

Mean== =13 =12.8isafraction. = = = = = 1.342

c) S te p D ev ia tio n M et ho d

We divide the deviation by a common class interval and use the following formula Standard deviation==× Wherecommon class interval, is assumed mean f is the respective frequency.

=× =× =× =× = 1.235× =4.94 mm Hg.

Ex am pl e:

B.P.(mmHg):102106110114118122126 No. of days:39253517101

So lu tio n :

B.P.(mmHg)No. of days () 102 106 110 114 118 122 126

3 9 25 35 17 10 1 TotalN=100

-3 -2 -1 0 1 2 3

-9 -18 -25 0 17 20 3 =-12

27 36 25 0 17 40 9 =154

A.M= × mm Hg

S.D. of Continues Series can be calculated by any one of the methods discussed for discrete frequency distribution But Step Deviation Method is mostly used. Standard deviation==× Wherecommon class interval, is assumed mean f is the respective frequency.

Ex am pl e:

I.Q.10-2020-3030-4040-5050-6060-7070-80 No. of students:51215201042

So lu tio n:

I.Q.No. of students:()Mid-value 10-20 20-30 30-40 40-50 50-60 60-70 70-80

5 12 15 20 10 4 2 Total=N=68

-3 -2 -1 0 1 2 3

-15 -24 -15 0 10 8 6 =-30

45 48 15 0 10 16 18 =152

Standard deviation= = = =

15 25 35 45 55 65 75

It is possible to compute combined mean of two or more than two groups. Combined Standard Deviation is denoted by = Wherecombined standard deviation ,

a) Combined S.D. = combined Mean = = == 55

Thefollowingaresomeoftheparticularsofthe distributionofweightofboysandgirlsinaclass: a)Findthestandarddeviationofthecombineddata b)whichofthetwodistributionsismorevariable

BoysGirls Numbers10050 Meanweight60kg45kg Variance()94 = = b) C.V (Boys)= C.V (Girls)=

M ER IT S O F ST AN DA RD D EV IA TI O N V er y po pu la r sc ie nt if ic m ea su re of di sp er si on Fr om SD w e ca n ca lc ul at e Sk ew ne ss ,C or re la ti on et c It co ns id er s al lt he it em s of th e se ri es T he sq ua ri ng of de vi at io ns m ak e th em po si tiv e an d th e di ff ic ul ty ab ou ta lg eb ra ic si gn s w hi ch w as ex pr es se d in ca se of m ea n de vi at io n is no tf ou nd he re .

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D EM ER IT S O F ST A N DA RD D EV IA TI O N C al cu la ti on is d if fi cu lt no t a s ea si er a s R an ge a nd Q D It a lw ay s de pe nd s on A M E xt re m e it em s ga in g re at im po rt an ce T he f or m ul a of S D is = Pr ob le m : C al cu la te S ta nd ar d D ev ia tio n of th e fo ll ow in g se ri es X 40 , 4 4, 5 4, 6 0, 6 2, 6 4, 7 0, 8 0, 9 0, 9 6

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U SE S O F ST AN D AR D D EV IA TI O N It is w id el y us ed in b io lo gi ca l s tu di es . It is u se d in f itt in g a no rm al c ur ve to a f re qu en cy d is tr ib ut io n. It is m os t w id el y us ed m ea su re o f di sp er si on .

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