Session
Session
– –5
5
Measures of Central Tendency
Measures of Central Tendency
Combined Mean
Combined Mean
Combined arithmetic mean can be computed if we know the mean a
Combined arithmetic mean can be computed if we know the mean a nd numbernd number of items in each groups of the data.
of items in each groups of the data. The following equation is used
The following equation is used to compute combined mean.to compute combined mean. Let
Let xx11 && xx22 are the mean of first and second group of data containing Nare the mean of first and second group of data containing N11 &&
N
N22items respectively.items respectively.
Th
Thenen,, cocombmbinined ed memean an ==
2 2 1 1 2 2 2 2 1 1 1 1 12 12 N N N N x x N N x x N N x x
If there are 3 groups then If there are 3 groups then
3 3 2 2 1 1 3 3 3 3 2 2 2 2 1 1 1 1 123 123 N N N N N N x x N N x x N N x x N N x x
Ex Ex - - 11:: a)a) Find thFind the means fe means for the enor the entire grtire group of woup of workerorkers for the s for the followinfollowing data.g data.
Group
Group – – 11 GGrroouupp – – 22 M
Meeaann wwaaggeess 7755 6600
N
Noo.. ooffwwoorrkkeerrss 11000000 11550000 G
Giivveen n ddaattaa:: NN11==11000000 NN22= 1500= 1500
60 60 x x & & 75 75 x x11
22
G Grroouup p MMeeaann == 2 2 1 1 2 2 2 2 1 1 1 1 12 12 N N N N x x N N x x N N x x
= = 1500 1500 1000 1000 60 60 x x 1500 1500 75 75 x x 1000 1000
= = xx1212
RsRs..6666 ExEx - - 22: Compute mean for entire : Compute mean for entire group.group.
M
Meeddiiccaal l eexxaammiinnaattiioonn NNoo. . eexxaammiinneedd MMeeaan n wweeiigghht t ((ppoouunnddss)) A A 5500 111133 B B 6600 112200 C C 9900 111155
Combined mean (grouped mean weight) Combined mean (grouped mean weight)
3 3 2 2 1 1 3 3 3 3 2 2 2 2 1 1 1 1 N N N N N N x x N N x x N N x x N N
)) 90 90 60 60 50 50 (( )) 115 115 x x 90 90 120 120 x x 60 60 113 113 x x 50 50 (( x x123123
pounds pounds 116 116 weight weight Mean Mean x x123123
Merits of Arithmetic Mean Merits of Arithmetic Mean
1.
1. It iIt is sims simple ple and and easy easy to cto compompute.ute. 2.
2. It It is is ririgigidldly dy defiefinened.d. 3.
3. It caIt can be un be used fsed for fuor furthrther caer calculculatilation.on. 4.
4. It is baIt is based osed on all obn all observservatiations ions in the sen the serieries.s. 5.
5. It hIt helpelps fos for dr direcirect cot compamparisrison.on. 6.
6. It is morIt is more stable me stable measureasure of cente of central tendral tendency (ency (ideal avideal average)erage)..
Limitat
Limitationsions / Demeri/ Demerits of Meants of Mean 1.
1. It is uIt is undunduly afly affecfected bted by exty extremreme itee items.ms. 2.
2. It It is is somsometietimes mes un-un-rearealislistictic.. 3.
3. It mIt may lay leadeads to cs to cononfufusisionon.. 4.
4. SuitabSuitable onlle only foy for quar quantitantitative tive data (data (for vfor variablariables).es). 5.
5. It can noIt can not be loct be located by ated by graphgraphical metical method or hod or by obsby observatervations.ions.
Geometric Mean (GM)
Geometric Mean (GM)
The GM is n
The GM is nthth root of product of quantities of the series. It is observed byroot of product of quantities of the series. It is observed by multiplying the values of items together and extracting the root of the product multiplying the values of items together and extracting the root of the product corresponding to the number o
corresponding to the number of items. f items. Thus, square root of the prodThus, square root of the products of two itemsucts of two items and cub
and cubee root oroot of the prof the products oducts of thef the threethree items aritems are the Geome the Geometric Meetric Mean.an. Usually, geometric mean
Usually, geometric mean is never largis never larger than arithmetic mer than arithmetic mean. ean. If there arIf there aree zero and negative numb
zero and negative number in the series. If there arer in the series. If there are zeros and negative nue zeros and negative numbersmbers in thein the series, the geometric means cannot be used logarithms can be used to find geometric series, the geometric means cannot be used logarithms can be used to find geometric mean to reduce large number and to save time.
mean to reduce large number and to save time. In the
In the field of businfield of business management vess management various problems oarious problems often arise relating ften arise relating toto average percentage rate of
average percentage rate of change over a period change over a period of time. of time. In such cases, the arithmeticIn such cases, the arithmetic mean is not an
mean is not an approappropriatpriatee averaaverage to emploge to employ, so, that we can usy, so, that we can use geometre geometric mean inic mean in such case.
such case. GM are highly GM are highly useful in the construuseful in the construction of index nuction of index numbers.mbers. Geometric Mean (GM) =
Geometric Mean (GM) = nn xx11 xx xx22 xx ...xx xxnn When th
When the numbee number of items in thr of items in thee series is laseries is larger thrger than 3, the pran 3, the process of ocess of computing GM
The log of all the value added
The log of all the value added up and divided by up and divided by number of items. number of items. The antilog of The antilog of quotient obtained is the required GM.
quotient obtained is the required GM. (GM
(GM) =) = AntAntilogilog
NN x x log log log log Anti Anti n n log log ... ... ... ... log log log log ii 1 1 ii n n 2 2 1 1 Merits of GM Merits of GM a.a. It is baIt is based osed on all thn all the obse observervatioations in thns in the serie series.es. b.
b. It iIt is ris rigigidldly dy defiefinened.d. c.
c. It is bIt is best suest suiteited fod for aver averagrages and es and ratratiosios.. d.
d. It is It is less less affaffectected by ed by extrextreme veme valualues.es. e.
e. It is It is useful useful for for studystudying ing social social and and econoeconomics mics data.data.
Demerits of GM Demerits of GM a.
a. It iIt is nos not sit simplmple to e to undunderserstantand.d. b.
b. It rIt requequireires cos compumputattationional sal skilkill.l. c.
c. GM cannGM cannot be ot be compucomputed ited if any f any of item of item is zero is zero or negor negative.ative. d.
d. It hIt has ras restrestricticted aed applpplicaicatiotion.n.
Ex Ex - - 11::
a.
a. FiFind nd ththe Ge GM oM of df data ata 2, 2, 4, 4, 88 x x11= 2,= 2, x x22= 4,= 4, x x33= 8= 8 n = 3 n = 3 GM = GM = nn xx11 xx xx22 xx xx33 G GMM== 33 22 xx 44 xx 88 GM = GM = 33 6464
44 GM = 4 GM = 4 b.b. FinFind GM of dd GM of data 2, ata 2, 4, 8 u4, 8 usinsing logg logariarithmthms.s. D Daattaa:: xx11= 2= 2 x x22= 4= 4 x x33= 8= 8 N = 3 N = 3
x x llooggxx 2 2 00..330011 4 4 00..660022 8 8 00..990033 Σ Σlogx = 1.806logx = 1.806 GM
GM == AntAntiloilogg
N N x x log log GM = Antilog GM = Antilog
3 3 806 806 .. 1 1 GM = Antilog (0.6020) GM = Antilog (0.6020) = 3.9997 = 3.9997 GM GM
44 Ex Ex - - 22::Compare the previous year the Over Head (OH) expenses which went up to Compare the previous year the Over Head (OH) expenses which went up to 32% in year 2003, then increased by 40% in next year and 50% increase in the 32% in year 2003, then increased by 40% in next year and 50% increase in the following year.
following year. Calculate average increase Calculate average increase in over head in over head expenses.expenses. Let 100% OH Expenses at base year
Let 100% OH Expenses at base year
Y
Yeeaarr OOH H EExxppeennssees s ((xx)) lloog g xx 2 2000022 BBaasse e yyeeaarr – – 2 2000033 113322 22..112266 2 2000044 114400 22..114466 2 2000055 115500 22..117766 Σ Σlolog x g x == 6.6.444488 GM = Antilog GM = Antilog
N N x x log log GM = Antilog GM = Antilog
3 3 448 448 .. 6 6 GM = 141.03 GM = 141.03GM for discrete series GM for discrete series GM for discrete
GM = Antilog GM = Antilog
NN x x log log ii 1 1 ii Ex Ex - - 33:: ConsidConsider follower following timeing time series for moseries for monthly santhly sales of ABC compales of ABC company for 4ny for 4 months.
months. Find average rate Find average rate of change per of change per monthly sales.monthly sales.
M Moonntthh SSaalleess II 1100000000 IIII 88000000 IIIIII 1122000000 IIVV 1155000000
Let Base year = 100% sales. Let Base year = 100% sales.
Solution Solution::
M
Moonntthh BBaasse e yyeeaarr SalesSales (Rs) (Rs) Increase / Increase / decrease decrease %ge %ge Conversion Conversion (x) (x) log (x)log (x) II 110000%% 1100000000 – – –– – – II II – – 2200%% 88000000 8800 8800 11..990033 IIIIII + + 5500%% 1122000000 113300 113300 22..111133 IIVV ++2255%% 1155000000 115555 115555 22..119900 Σ Σloglogx =x = 6.6.206206 GM = Antilog GM = Antilog
3 3 206 206 .. 6 6 = 117.13 = 117.13 Average sales = 117.13 Average sales = 117.13 – – 100 = 14.46%100 = 14.46% ExEx - - 44: Find: Find GM for follGM for following dowing data.ata.
Marks Marks (x) (x) No. of students No. of students (f) (f) lloog g xx f f lloog g xx 1 13300 33 22..111133 66..333399 1 13355 44 22..113300 88..5522 1 14400 66 22..114466 1122..887766 1 14455 66 22..116611 1122..999966 1 15500 33 22..117766 66..552288 Σ Σf = f = N =N = 2222 ΣΣf log x =47.23f log x =47.23
GM = Antilog GM = Antilog
N N x x log log f f GMGM = A= Antintiloglog
22 22 23 23 .. 47 47 GM = 140.212 GM = 140.212Geometric Mean for continuous series Geometric Mean for continuous series Steps
Steps:: 1.
1. Find mFind mid valid value m and ue m and take ltake log of og of m for m for each mieach mid vald value.ue. 2.
2. Multiply log m with frequency ‘f’ of each class to get f log m and sum up toMultiply log m with frequency ‘f’ of each class to get f log m and sum up to obtain
obtain
f log m.f log m. 33.. DDiivviiddee
ff log m by N and taklog m by N and take antiloe antilog to get GM.g to get GM.Ex:
Ex:Find out GM for given data belowFind out GM for given data below
Yield of wheat Yield of wheat in in MT MT No. of farms No. of farms frequency frequency (f) (f) Mid value Mid value ‘m’ ‘m’ lloog g mm f f lloog g mm 1 1 – – 1100 33 55..55 00..774400 22..222200 11 11 – – 2200 1166 1155..55 11..119900 1199..004400 21 21 – – 3300 2266 2255..55 11..440066 3366..555566 31 31 – – 4400 3311 3355..55 11..555500 4488..005500 41 41 – – 5500 1166 4455..55 11..665588 2266..552288 51 51 – – 6600 88 5555..55 11..774444 1133..995544 Σ Σf = N = 100f = N = 100 ΣΣf log m = 146.348f log m = 146.348 GM = Antilog GM = Antilog
N N m m log log f f GM = Antilog GM = Antilog
100 100 348 348 .. 146 146 GM GM == 29.29.0707Harmonic Mean
Harmonic Mean
It is the total number of items of a value divided by the sum of reciprocal of It is the total number of items of a value divided by the sum of reciprocal of values of v
values of variable. ariable. It is It is a specified a specified average which average which solves problemsolves problems involvings involving variables expressed in within ‘Time rates’ that vary according to time.
Ex
Ex: Speed in km/hr, min/day, price/unit.: Speed in km/hr, min/day, price/unit.
Harmonic Mean (HM) is suitable only when time factor is variable and the act being Harmonic Mean (HM) is suitable only when time factor is variable and the act being performed remains constant.
performed remains constant. HM = HM = x x 1 1 N N
Merits of Harmonic Mean Merits of Harmonic Mean 1.
1. It iIt is bas based sed on aon all oll obsebservarvatiotions.ns. 2.
2. It It is is ririgigidldly dy defiefinened.d. 3.
3. It is suIt is suitabitable in cale in case of sese of serieries havs havinging widwide dispe dispersersionion.. 4.
4. It is It is suitabsuitable fole for fur further rther mathemmathematical atical treatmtreatment.ent.
Demerits of Harmonic Mean Demerits of Harmonic Mean 1.
1. It iIt is nos not eat easy tsy to coo compmputute.e. 2.
2. CanCannot unot used wsed when ohen one one of the if the item item is zers zero.o. 3.
3. It caIt cannonnot rept represresent dent distristribuibutiotion.n.
Ex Ex::
1.
1. The daThe daily incily income oome of 05 faf 05 families imilies in a vern a very rury rural villagal village are gie are given beven below. low. CompComputeute HM.
HM. F
Faammiillyy IInnccoomme e ((xx)) RReecciipprrooccaal l ((11//xx)) 1 1 8855 00..00111177 2 2 9900 00..0011111111 3 3 7700 00..00114422 4 4 5500 00..0022 5 5 6600 00..001166 x x 1 1
= 0.0738= 0.0738 HM = HM = x x 1 1 N N
= = 0738 0738 .. 0 0 5 5 = 67.72 = 67.72 HM = HM = 6767.7.7222.
2. A man traA man travel by vel by a car foa car for 3 dayr 3 days he covs he covered 48ered 480 km ea0 km each daych day. . On the fOn the first dairst day hey he drives for 10 hrs at the rate of 48 KMPH, on the second day for 12 hrs at the rate drives for 10 hrs at the rate of 48 KMPH, on the second day for 12 hrs at the rate of 40 KMPH, and on the 3
of 40 KMPH, and on the 3rdrd day for day for 15 hrs 15 hrs @ 32 K@ 32 KMPH. MPH. Compute HM Compute HM andand weighted mean and compare them.
weighted mean and compare them. Har
Harmonmonicic MeaMeann x x 11xx 4 488 00..00220088 4 400 00..002255 3 322 00..00331122 x x 1 1
= 0.0770= 0.0770 Data: Data: 10 hrs @ 48 KMPH 10 hrs @ 48 KMPH 12 hrs @ 40 KMPH 12 hrs @ 40 KMPH 15 hrs @ 32 KMPH 15 hrs @ 32 KMPH HM = HM = x x 1 1 N N
= = 0770 0770 .. 0 0 3 3 HM = 38.91 HM = 38.91 Weighted Mean Weighted Mean w w xx wwxx 1 100 4488 448800 1 122 4400 448800 1 155 3322 448800
w = 37w = 37 ΣΣwx = 1440wx = 1440 Weighted Mean = Weighted Mean = w w wx wx x x
= = 37 37 1440 1440 91 91 .. 38 38 x x
Both the same HM and WM are same. Both the same HM and WM are same.
3.
3. FinFind HM d HM for for the the folfollowlowing ing datadata.. C
Cllaasss s ((CCII)) FFrreeqquueennccy y ((ff)) MMiid d ppooiinnt t ((mm)) ReciprocalReciprocal
m m 1 1 f f
m m 1 1 0 0 – – 1100 55 55 00..22 11 10 10 – – 2200 1155 1155 00..00666666 00..999999 20 20 – – 3300 2255 2255 00..0044 11 30 30 – – 4400 88 3355 0..000228855 00..222288 40 40 – – 5500 77 4455 0..000222222 00..11555544 Σ Σf f == 6600
f f
m m 1 1 = 3.3824 = 3.3824 HM = HM =
m m 1 1 f f N N = = 3824 3824 .. 3 3 60 60 HM = 17.73 HM = 17.73Relationship between Mean, Geometric Mean
Relationship between Mean, Geometric Mean and Harmonic Mean.
and Harmonic Mean.
1.
1. If all thIf all the items ie items in a varn a variable ariable are the sae the same, thme, the arithe arithmetic mmetic mean, hean, harmonarmonic meaic meann andand Geometric mean
Geometric mean are equal. are equal. i.e.,i.e., xx
GMGM
HMHM.. 2.2. If the sizIf the size varye vary, mean wil, mean will be grel be greater thaater than GM and n GM and GM will be gGM will be greater threater than HM.an HM. This is because of the property that geometric mean to give larger weight to This is because of the property that geometric mean to give larger weight to smaller item and of the HM to give largest weight to smallest item.
smaller item and of the HM to give largest weight to smallest item. Hence,
Hence, xx
GMGM
HMHM..Median
Median
Median is the value of that item in a series which divides the array into two Median is the value of that item in a series which divides the array into two equal parts, one consisting of all the values less than it and other consisting of all the equal parts, one consisting of all the values less than it and other consisting of all the values more than
values more than it. it. Median is a Median is a positional averagpositional average. e. The number oThe number of items below f items below it isit is equal to the number.
equal to the number. The number of items below The number of items below it is equal to the number of it is equal to the number of itemsitems above it.
above it. It occupiIt occupies central es central position.position. Thus,
Thus, Median Median is defis defined ined as the as the mid vmid value oalue of the f the variavariants.nts. If thIf the value values arees are arranged in ascending or descending order of their magnitude, median is the middle arranged in ascending or descending order of their magnitude, median is the middle value of the number of variant is odd and average of two middle values if the number value of the number of variant is odd and average of two middle values if the number of variants is even.
of variants is even.
Ex
Ex: If: If 9 students are stand 9 students are stand in the order oin the order of their heights; the 5f their heights; the 5thth student from either sidestudent from either side
shall
shall be the be the one whose one whose height will height will be Median be Median height oheight of the f the students grstudents group. oup. Thus,Thus, median of group is given by an
Median = Median =
2 2 1 1 N N Ex Ex 1.1. FinFind the md the mediedian foan for folr followlowing ing datadata.. 2
222 2200 2255 3311 2266 2244 2233 Arrange the given data in a
Arrange the given data in array form (either in ascending or descending order).rray form (either in ascending or descending order). 2 200 2222 2233 2424 2255 2266 3311 Median is given by Median is given by
2 2 1 1 N N thth item = item =
2 2 1 1 7 7 = = 4 4 8 8 Median = 4Median = 4ththitem.item. 2.
2. FinFind med median dian for for folfollowlowing ing datdata.a.
2 200 2211 2222 2424 2288 3322 Median is given by Median is given by
2 2 1 1 N N thth item = item =
2 2 1 1 6 6 MedMedian ian == 3.53.5ththitem.item.
The item lies between 3
The item lies between 3rdrdand 4.and 4. So, there are two values 22 and 24. So, there are two values 22 and 24. The median value
The median value will be the mean will be the mean valuesvalues of these two vof these two values.alues. Median = Median =
2 2 24 24 22 22 = 23 = 23 Discrete SeriesDiscrete Series – – MedianMedian
In discrete series, the values are (already) in the form of array and the In discrete series, the values are (already) in the form of array and the frequencies are
frequencies are recorded against recorded against each value. each value. However, to However, to determine the size determine the size of of median median
2 2 1 1 N N ththitem, a separate column is to be prepared for cumulative item, a separate column is to be prepared for cumulative frequencies.
frequencies. The mediThe median size is an size is first located first located with reference with reference to the to the cumulativecumulative frequency which
frequency which covers the size covers the size first. first. Then, against thThen, against that cumulative frat cumulative frequency, theequency, the value will be located as median.
Ex
Ex: Find the median for the students’ marks.: Find the median for the students’ marks.
Obtained in statistics Obtained in statistics Marks (x)
Marks (x) No. of No. of students (f) students (f) Cumulative Cumulative frequency frequency 1 100 55 55 2 200 55 1100 3 300 33 1313 4 400 1155 2288 50 50 3300 5588 6 600 1100 6688 N = 68 N = 68 Ex
Ex:: In a In a class 1class 15 stu5 studentsdents, 5 , 5 studenstudentsts werewere failed failed in a in a test. test. The The marks marks of 10 of 10 studenstudentsts
who have
who have passed were passed were 9, 6, 79, 6, 7, 8, , 8, 9, 6, 9, 6, 5,5, 4, 7, 4, 7, 8. 8. Find the Find the Median marks Median marks of 15of 15 students.
students.
M
Maarrkkss NNoo. . oof f ssttuuddeenntts s ((ff)) ccf f 0 0 5 5 1 1 2 2 3 3 4 4 11 66 5 5 11 77 6 6 22 99 7 7 22 1111 8 8 22 1133 9 9 22 1155 Σ Σf = 15f = 15 Me Medidianan == 2 2 1 1 N N
thth item item Me = Me = 2 2 1 1 15 15
= 8 = 8thth Me 8Me 8ththitem covers in cf of 9. item covers in cf of 9. the marks against cf 9 is 6 and henthe marks against cf 9 is 6 and hencece Median = 6
Median = 6
Just above 34 Just above 34 is 58
is 58.. AgaAgainsinstt 58 c.f. the 58 c.f. the value is 50 value is 50 which is which is median value median value
Continuous Series Continuous Series
The procedure
The procedure is different to is different to get median get median in continuouin continuous series. s series. The classThe class intervals are already in the form of array and the frequency are recorded against each intervals are already in the form of array and the frequency are recorded against each class interval.
class interval. For determining For determining the size, the size, we should we should taketake
th th 2 2 n n
item and median class item and median class located accordingly with reference to the cumulative frequency, which covers the size located accordingly with reference to the cumulative frequency, which covers the size first.
first. When the median class is located, the mWhen the median class is located, the median value is to be interpoedian value is to be interpolated usinglated using formula given below.
formula given below. Median = Median =
CC
2 2 N N f f h h Where Where 2 2 1 1 0 0
where,where, 00 is left end point of N/2 class and lis left end point of N/2 class and l11is right endis right endpoint of previous class. point of previous class.
h = Class width, f = frequency of median clas h = Class width, f = frequency of median clas C
C = Cumulative frequ= Cumulative frequency of class precedinency of class preceding the median g the median class.class.
Ex
Ex: Find the median for following d: Find the median for following data. ata. The class marks obtained by 50 students arThe class marks obtained by 50 students aree
as follows. as follows.
C
CII FFrreeqquueennccy y ((ff)) Cum.Cum. frequency (cf) frequency (cf) 10 10 – – 1155 66 66 15 15 – – 2200 1188 2244 20 20 – – 2255 99 3333N/2 classN/2 class 25 25 – – 3300 1100 4433 30 30 – – 3355 44 4477 35 35 – – 4400 33 5500 Σ Σf = f = N =N = 5050 25 25 2 2 50 50 2 2 N N
CumCum. fr. frequequencency juy just ast abovbovee 2525 is 3is 333 andand henhence, ce, 2020 – – 25 is median class.25 is median class.
2 2 1 1 0 0
20 20 2 2 20 20 20 20
20 20
h = 20 h = 20 – – 15 = 515 = 5f = 9 f = 9 c = 24 c = 24 Median = Median =
CC
2 2 N N f f h h Median = Median =
2525 2424
9 9 5 5 20 20
= = 9 9 5 5 20 20
Median = 20.555 Median = 20.555 ExEx:: Find Find the mthe median edian for ffor followiollowingng data.data.
M
Miid d vvaalluuees s ((mm)) 111155 112255 113355 114455 115555 116655 117755 118855 119955 F
Frreeqquueenncciiees s ((ff)) 66 2255 4488 7722 111166 6600 3388 2222 33
The interval of mid-values of CI and magnitudes of class intervals are same The interval of mid-values of CI and magnitudes of class intervals are same i.e. 10.
i.e. 10. So, half of 10 is deducted So, half of 10 is deducted from and added to mid-values will give us the lowerfrom and added to mid-values will give us the lower and upper limits. Thus, classes are.
and upper limits. Thus, classes are. 115
115 – – 5 = 110 (lower limit)5 = 110 (lower limit) 115
115 – – 5 = 120 (upper limit) similarly for all mid values we can get CI.5 = 120 (upper limit) similarly for all mid values we can get CI. C
CII FFrreeqquueennccy y ((ff)) Cum.Cum. frequency (cf) frequency (cf) 110 110 – – 112200 66 66 120 120 – – 113300 2255 3311 130 130 – – 114400 4488 7799 140 140 – – 115500 7722 115511 150 150 – – 116600 111166 226677 N/2 classN/2 class 160 160 – – 117700 6600 332277 170 170 – – 118800 3388 336655 180 180 – – 119900 2222 338877 190 190 – – 220000 33 339900 Σ Σf = f = N =N = 393900 2 2 390 390 2 2 N N
195 195
Cum. frequency just above 195 is 267. Cum. frequency just above 195 is 267.
Median class = 150 Median class = 150 – – 160160 == 2 2 150 150 150 150
= 150 = 150 h = h = 111166 N N//22 == 119955 C C == 115511 h = 10 h = 10 Median = Median =
CC
2 2 N N f f h h Median = Median =
195195 151151
116 116 10 10 150 150
Med Median =ian = 153153.8.8 Merits of Median Merits of Median a.a. It is simIt is simpleple, easy t, easy to como computpute and une and underderstanstand.d. b.
b. It’s vaIt’s value is not affected by extreme variables.lue is not affected by extreme variables. c.
c. It is capIt is capablable for fure for furthether algr algebrebraic treaic treatmatmentent.. d.
d. It can It can be dbe determietermined bned by insy inspectiopection for n for arrayarrayed dated data.a. e.
e. It cIt can ban be foe found und grgraphaphicaically lly alsoalso.. f.
f. It iIt indindicatecates ths the ve valualue of e of midmiddle dle iteitem.m.
Demerits of Median Demerits of Median a.
a. It may It may not bnot be represe representatientative vave value as it lue as it ignoignores extres extreme vreme values.alues. b.
b. It can’t be determined precisely when its size falls between the two values.It can’t be determined precisely when its size falls between the two values. c.
