quantitative techniques

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1 QDM notes for MBA Second Edition 2013 - 2014

E-mail :- [email protected]

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Arithmetic Mean:-Arithmetic Mean is the most widely used measurement which represents the

entire data. Generally it is temed as an Average to a layman. It is the quantity obtained bydividing the sum of the values of the items in a variable by their number. It is denoted by asymbol X. Arithmetic mean by be either:

(i) Simple Arithmetic Mean or (ii) Weighted Arithmetic Mean

(i) Simple Arithmetic Mean :- The simple arithmetic mean is the quotient obtained bydividing the sum of the values by the number of items. Algebraically we can have theformula as under:a) For individual observations

x1+x2+x3+.......+ xn xN N

b) For Discrete and Continuous Seriesfx1+fx2+fx3+.......+ xn fx

N N(Note:- In continuous Series x is the mid-value of the class interval)

Ex:- Find out the Simple Arithmetic Mean by direct methodEx (1) :- 21, 8, 31, 9, 18, 25 (Tip: write the values in increasing (ascending) order)Ans :- 8 , 9 , 18 , 21 , 25 , 31 x

1 2 3 4 5 6 N

x1+x2+x3+ x4+ x5+ x6 xN N

8+9+18+21+25+31 1126 6

Simple Arithmetic Mean (X) = 18.67

Ex (2) :- 3, 8, 2, 9, 6, 1, 12, 14, 16 (Tip: write the values in ascending order)Ans:- 1 2 3 6 8 9 12 14 16 x

1 2 3 4 5 6 7 8 9 N

x1+x2+x3+ x4+ x5+ x6 + x7 + x8 + x9 x N N

1+2+3+6+8+9+12+14+16 71 9 9

Simple Arithmetic Mean (X) = 7.89

Index01) Mean -- 0102) Median -- 1003) Mode -- 1504) Mean Deviation -- 1605) Quartile Deviation -- 1906) Standard Devaition -- 2107) Coefficient of Correlation -- 2708) Rank Correlation Coefficient -- 3309) Regression -- 3510) Trend Value -- 4011) Linear Programming : Graphic method -- 4812) LPP : Transportation Method (NWM, LCM, VAM) -- 6113) LPP : Assignment Problem -- 76

Important Tip1. Notes under development, so some calculation mistakes present. Student can

solve all problem again for verfied the answers.2. Missing frequency problem see the page number 5 & 133. See page number 25 & 26 for how the 5 types of problem can solved from one

example like mean, median, quartile deviation, mean deviation &standard deviation.

4. See page number 74 & 75 for direct method to solve the NWM, LCM & VAM.5. LPP : Assignment Problem - see page number 80 to 82

X = =

X = =

X = =

X = = = 18.6666

X = =

X = = = 7.8888

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Ex : Find out the Simple Arithmetic Mean For Discrete Series by direct method

fx N

Here, f = Frequency, x = variable, N = f

Ex:- Income 10 20 30 40 50 60 70No. of Families 5 7 2 3 1 6 9

Ans :-Income No. of Families x f fx10 5 10x5 = 5020 7 20x7 = 14030 2 30x2 = 6040 3 40x3 = 12050 1 50x1 = 5060 6 60x6 = 36070 9 70x9 = 630

N = f = 33 fx = 1410

fx 50+140+60+120+50+360+630 1410N 5+7+2+3+1+6+9 33

Arithmetic Mean for discrete series by direct method (X) = 42.73

Ex:- Find out the Simple Arithmetic Mean For Continous Series by direct methodf . Xm N

Here, f = Frequency, Xm = mid value, N = f

C.I. f Xm f . Xm16 - 17 5 16.5 16.5 x 5 = 82.514 - 15 4 14.5 14.5 x 4 = 5812 - 13 4 12.5 12.5 x 4 = 5010 - 11 8 10.5 10.5 x 8 = 84 8 - 9 5 8.5 8.5 x 5 = 42.5 6 - 7 4 6.5 6.5 x 4 = 26 4 - 5 2 4.5 4.5 x 2 = 9

N=32 f.Xm = 82.5+58+50+84+42.5+26+9 = 352

f.Xm 82.5+58+50+84+42.5+26+9 352 N 5 + 4 + 4 + 8 + 5 + 4 + 2 32

Arithmetic Mean for Continous series by direct method (X) =11

X = = = = 42.7272

Simple Arithmetic Mean by short cut method :-If the number of items is large and values of variable big in size, a short cut

method of computing x is adopted. This method is based on the following property of thearithmetic average, The algebric sum of the devations of values of variable from theirmean is always to Zero.

Let us assume the difference figure A as assumed mean. Take deviations ofvalues of variable from the assumed mean denoted as d. Obtain the sum of thesedevations denote as d. Substitute the values of these symbols in the short-cut formulaas under.

dN

fd N

(Note:- fd is the product of frequency and deviation)

Ex (1):- 52, 47, 37, 32, 42, 27 (Short cut method)Ans:- x direct method x d=x - A Assumed Mean (A) = 37

27 27 27-37= -10 d32 x 32 32-37 = -5 N37 N 37 37-37= 0 1542 237 42 42-37= 5 647 6 47 47-37 = 10 = 37 + 2.552 X = 39.5 52 52 - 37 = 15 X = 39.5

x = 237 d = 15

x d=x - A Since A = 32 d=x - A Since A = 4227 27-32= -5 d 27-42= -15 d32 32-32 = -0 N 32-42= -10 N37 37-32= 5 45 37-42 = -5 (-15)42 42-32= 10 6 42-42= 0 647 47-32 = 15 = 32 + 7.5 47-42= 5 X = 42 - 2.552 52 - 32 = 20 X = 39.5 52-42= 10 X = 39.5

d = 45 d = -15

(Tip:- From above example, you observed that any value of assumed mean taken thearithmetic mean answer is comes same. So did not confuse and assume any value asassumed mean (A), i.e. from given example and solve the example.)

X =

X = = = = 11

X =

X = A + ............. for individual observations

X = A + ............. for discrete & continous series

X =

X =

X = A +

X = 37 +

X = A +

X = 32 +

X = A +

X = A +

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Ex (2):- 10 20 30 40 50 60 707 11 31 17 16 5 3

x f d= x-A f.d10 7 10-40= -30 7 * -30= -21020 11 20-40= -20 11 * -20= -22030 31 30-40= -10 31 * -10= -31040 17 40-40= 0 17 * 0= 050 16 50-40= 10 16 * 10= 16060 5 60-40= 20 5 * 20= 10070 3 70-40= 30 3 * 30= 90 N=90 fd = -210-220-310+0+160+100+90 = -390

fd (-390)N 90

OR

x f d= x-A f.d10 7 10-30= -20 7 * -20= -14020 11 20-30= -10 11 * -10= -11030 31 30-30= 0 31 * 00= 040 17 40-30= 10 17 * 10= 17050 16 50-30= 20 16 * 20= 32060 5 60-30= 30 5 * 30= 15070 3 70-30= 40 3 * 40= 120 N=90 fd = -140-110+0+170+320+150+120 = 510

fd 510N 90

(Tip :- From above example, it observed that you will solve any taking assumed meanfrom given example and solve it and answer comes as same.)

Ex (3) :- C.I. f Xm d=Xm- A f.d16-17 5 (16+17)/2= 33/2 = 16.5 16.5-10.5= 6 5 * 6 = 3014-15 4 (14+15)/2= 29/2 = 14.5 14.5-10.5= 4 4 * 4 = 1612-13 4 (12+13)/2= 25/2 = 12.5 12.5-10.5= 2 4 * 2 = 810-11 8 (10+11)/2= 21/2 = 10.5 10.5-10.5= 0 8 * 0 = 08-9 5 (8+9)/2= 17/2 = 8.5 8.5-10.5= -2 5 * -2 = -106-7 4 (6+7)/2= 13/2 = 6.5 6.5-10.5= -4 4 * -4 = -164-5 2 (4+5)/2= 9/2 = 4.5 4.5-10.5= -6 2 * -6 = -12

N=32 fd = 16

fd 16N 32

OR C.I. f Xm d=Xm- A f.d16-17 5 (16+17)/2= 33/2 = 16.5 16.5-12.5= 4 5 * 4 = 2014-15 4 (14+15)/2= 29/2 = 14.5 14.5-12.5= 2 4 * 2 = 812-13 4 (12+13)/2= 25/2 = 12.5 12.5-12.5= 0 4 * 0 = 010-11 8 (10+11)/2= 21/2 = 10.5 10.5-12.5= -2 8 * -2 = -168-9 5 (8+9)/2= 17/2 = 8.5 8.5-12.5= -4 5 * -4 = -206-7 4 (6+7)/2= 13/2 = 6.5 6.5-12.5= -6 4 * -6 = -244-5 2 (4+5)/2= 9/2 = 4.5 4.5-12.5= -8 2 * -8 = -16

N=32 fd = -48

fd (-48)N 32


The Step Deviation Method:- The step deviation method is the only additional adjust-ment to the short cut method. Under this method, the devations are devided by a singlecommon factor to reduce the figures to the minimum size. It is, in a way, a deliberate errorand it is compensated by multiplying the quotient. The adjusted or reduced value of d innow termed as d so the formula stands as under.

fd N

(Note:- c the common factor or width of the class interval is used for dividing andmultiplying)

Ex (1):- 90-99 80-89 70-79 60-69 50-59 40-49 30-39 20-29 2 9 12 17 20 9 6 5

C.I. f Xm d= (Xm-A)/c fd90-99 2 94.5 (94.5-54.5)/10= 4 2*4= 880-89 9 84.5 (84.5-54.5)/10= 3 9*3= 2770-79 12 74.5 (74.5-54.5)/10= 2 12*2= 2460-69 17 64.5 (64.5-54.5)/10= 1 17*1= 1750-59 20 54.5 (54.5-54.5)/10= 0 20*0= 040-49 9 44.5 (44.5-54.5)/10= -1 9*-1= -930-39 6 34.5 (34.5-54.5)/10= -2 6*-2= -1220-29 5 24.5 (24.5-54.5)/10= -3 5*-3= -15

N=80 fd = 40

94.5 - 54.5 40 84.5 - 54.5 3010 10 10 10

Same as remaining values of d are finds.

X = A + = 40 + = 40 + (-4.3333) = 40 - 4.3333 = 35.6667

X = A + = 30 + = 30 + 5.6666 = 35.6666 = 35.67

X = A + = 10.5 + = 10.5 + 0.5 = 11

X = A + = 10.5 + = 12.5 + (-1.5) = 12.5 - 1.5 = 11

X = A + x c

Q: How to find the value CA: Range 90-99, the numberstart from 0 to 9, it means0,1,2,3,4,5,6,7,8,9 = 10 intervals

So the common factor (c)value is 10

d = = = 4 , d = = = 3

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Since, N=80, A = 54.5, fd = 40, c = 10, then put the values on formula

4080

OR

C.I. f Xm d= (Xm-A)/c fd90-99 2 94.5 (94.5-44.5)/10= 5 2 * 5 = 1080-89 9 84.5 (84.5-44.5)/10= 4 9 * 4 = 3670-79 12 74.5 (74.5-44.5)/10= 3 12 * 3 = 3660-69 17 64.5 (64.5-44.5)/10= 2 17 * 2 = 3450-59 20 54.5 (54.5-44.5)/10= 1 20 * 1 = 2040-49 9 44.5 (44.5-44.5)/10= 0 9 * 0 = 030-39 6 34.5 (34.5-44.5)/10= -1 6 * -1 = -620-29 5 24.5 (24.5-44.5)/10= -2 5 * -2 = -10

N=80 fd = 120

120 80

Ex (2): - 7-11 12-16 17-21 22-26 27-31 32-36 37-41 42-46 2 3 5 4 8 7 2 4

C.I. f Xm d= (Xm-A)/c fd42-46 4 44 (44-29)/5= 3 4 * 3 = 1237-41 2 39 (39-29)/5= 2 2 * 2 = 432-36 7 34 (34-29)/5= 1 7 * 1 = 727-31 8 29 (29-29)/5= 0 8 * 0 = 022-26 4 24 (24-29)/5= -1 4 * -1 = -417-21 5 19 (19-29)/5= -2 5 * -2 = -1012-16 3 14 (14-29)/5= -3 3 * -3 = -97-11 2 9 (9-29)/5= -4 2 * -4 = -8

N=35 fd = -8

fd = 12 + 4 + 7 + 0 -4 - 10 - 9 - 8 = -8

Since, N=35, A = 29, fd = -8, c = 5, then put the values on formula

(-8)35

X = 27.86

C.I. f Xm d= (Xm-A)/c fd42-46 4 44 (44-24)/5= 4 4 * 4 = 1637-41 2 39 (39-24)/5= 3 2 * 3 = 632-36 7 34 (34-24)/5= 2 7 * 2 = 1427-31 8 29 (29-24)/5= 1 8 * 1 = 822-26 4 24 (24-24)/5= 0 4 * 0 = 017-21 5 19 (19-24)/5= -1 5 * -1 = -512-16 3 14 (14-24)/5= -2 3 * -2 = -67-11 2 9 (9-24)/5= -3 2 * -3 = -6

N=35 fd = 27

2735


Ex (3): - 4-5 6-7 8-9 10-11 12-13 14-15 4 10 20 15 8 3

C.I. f Xm d= (Xm-A)/c fd14-15 3 14.5 (14.5-8.5)/2= 3 3 * 3 = 912-13 8 12.5 (12.5-8.5)/2= 2 8 * 2 = 1610-11 15 10.5 (10.5-8.5)/2= 1 15 * 1 = 158-9 20 8.5 (8.5-8.5)/2= 0 20 * 0 = 06-7 10 6.5 (6.5-8.5)/2= -1 10 * -1 = -104-5 4 4.5 (4.5-8.5)/2= -2 4 * -2 = -8 N=60 fd = 22

2260

OR

C.I. f Xm d= (Xm-A)/c fd14-15 3 14.5 (14.5-10.5)/2= 2 3 * 2 = 612-13 8 12.5 (12.5-10.5)/2= 1 8 * 1 = 810-11 15 10.5 (10.5-10.5)/2= 0 15 * 0 = 08-9 20 8.5 (8.5-10.5)/2= -1 20 * -1 = -206-7 10 6.5 (6.5-10.5)/2= -2 10 * -2 = -204-5 4 4.5 (4.5-10.5)/2= -3 4 * -3 = -12 N=60 fd = -38

(-38) 60


X = A + x cfd N

X = 54.5 + x 10 = 54.5 + 0.5 * 10 = 54.5 + 5 = 59.5

X = 44.5 + x 10 = 44.5 + 1.5 * 10 = 44.5 + 15 = 59.5

Q: How to find the value CA: Range 42-46, the numberstart from 2 to 6, it means2,3,4,5,6 = 5 intervals


X = A + x cfd N

X = 29 + x 5 = 29 + (-0.2286 * 5) = 29 - 1.1429 = 27.8571

X = 24 + x 5 = 24 + (0.7714 * 5) = 24 + 3.8571 = 27.8571= 27.86

X = 8.5 + x 2 = 8.5 + (0.3666 * 2) = 8.5 + 0.7333 = 9.2333 = 9.23

X = 10.5 + x 2 = 10.5 + (-0.6333 * 2) = 10.5 - 1.2666 = 9.2334

Q: How to find the value CA: Range 14-15, the numberstart from 4 to 5, it means4, 5 = 2 intervals


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Ex (4): - Find the missing frequency from the following distribution if its mean is 15.25.x: 10 12 14 16 18 20f: 3 7 (?) 20 8 5

Soln:- Let us assume the missing frequency as A

x f fx10 3 10*3 = 3012 7 12*7 = 8414 (A) 14*A = 14A16 20 16*20 = 32018 8 16*8 = 14420 5 20*5 = 100

N=43+A fx = 678+14A

Since X = 15.25, N = 43+A, fx = 678+14A, put the values on formula

678 + 14A 43 + A

(43+A)* 15.25 = 678 + 14A655.75 + 15.25A = 678 + 14A 15.25A - 14A = 678 - 655.75

1.25A = 22.2522.251.25

Therefore, the missing frequency equal to 17.8 or 18 approximately.

Ex (5):- If the average wages paid to 25 workers is Rs. 79.60, find the missing frequencies.Wages(Rs.) (x): 50 60 70 80 90 100 110No.of workers (f): 1 3 (?) (?) 6 2 1

Soln:- Assume that the missing frequencies are A and B . Total frequencies are 25.So, 1 + 3 + A + B + 6 + 2 + 1 = 25

13 + A + B = 25A + B = 25 - 13 A + B = 12, B = 12 - A

x f fx50 1 50*1 = 5060 3 60*3 = 18070 (A) 70*A = 70 A80 12-A 80*(12-A) = 960-80 A90 6 90*6 = 540100 2 100*2 = 200110 1 110*1 = 110

N=25 fx = 2040 - 10 A

fx = 50+180+70A + 960 - 80A + 540+200+100 = 2040 - 80 A + 70 A = 2040 - 10A

Since X = 79.60, N = 25, fx = 2040 - 10A, put the values on formula

2040 - 10A 25 Now put the value A = 5

79.60 * 25 = 2040 - 10A 1990 = 2040 - 10A A + B = 12 10A = 2040 - 1990 5 + B = 12 10A = 50 B = 12 - 5A = 50/10 = 5 B = 7A = 5So, missing frequencies are A = 5 and B = 7

Ex (6):- From the following data calcuate the missing value when the mean is 115.86.Wages(Rs.) (x): 110 112 113 117 (?) 125 128 130No.of workers (f): 25 17 13 15 14 8 6 2

Soln:- Let us assume the missing value a A.

x f fx110 25 110*25 = 2750112 17 112*17 = 1904113 13 113*13= 1469117 15 117*15 = 1755A 14 A*14 = 14A125 8 125*8 = 1000128 6 128*6 = 768130 2 130*2 = 260

N=100 fx = 9906 + 14ASo, missing value is 120.

Ex (7):- You are given the following incomplete information and its mean 25. find out themissing frequencies.Class interval (x): 0-10 10-20 20-30 30-40 40-50 TotalFrequencies (f): 5 - 15 - 5 = 45

Soln:- Assume that the frequencies are A and B. Total frequencies are 45.So,

5+A+15+B+5= 45A + B + 25 = 45A + B = 45 - 25A + B = 20B = 20 - A

X =fx N

X =fx N

15.25 =

A = = 17.8

2040 - 10A 25

79.60 * 25 = 2040 - 10A 1990 = 2040 - 10A 10A = 2040 - 1990 10A = 50A = 50/10 = 5A = 5

79.60 =

79.60 =

X =fx N9906 - 14A 100

115.86*100 = 9906 - 14A 11586 = 9906 - 14A 14A = 11586 - 9906 14A = 1680

168014

A = 120

115.86 =

A = = 120

x f Xm fx00-10 5 (10+0)/2=10/2=5 5 * 5 = 2510-20 A (10+20)/2=30/2=15 A * 15 = 15A20-30 15 (20+30)/2=50/2=25 15 * 25 = 37530-40 20-A (30+40)/2=70/2=35 (20-A)*35 = 700-35 A40-50 5 (40+50)/2=90/2=45 5 * 45 = 225

N=45 fx = 1325 - 20 A

fx =25+15A+375+700-20A+225=1325-35A+15A = 1325-20A

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Since, put the values on formula i.e. N=45, X = 25, fx = 1325 - 20 A

, 1325 - 20A 45

25 * 45 = 1325 - 20 A Put A = 10 on eqn A + B = 201125 = 1325 - 20 A A + B = 20 20 A = 1325 - 1125 10 + B = 20 20 A = 200 B = 20 - 10

200 B = 10 20

A = 10

So, missing frequncies are A = 10 and B = 10

OR

Lets assume the frequencies as A and B

C.I. f Xm fx00-10 5 (10+0)/2=10/2=5 5 * 5 = 2510-20 A (10+20)/2=30/2=15 A * 15 = 15A N=5+A+15+B+520-30 15 (20+30)/2=50/2=25 15 * 25 = 375 45 = 25 + A + B30-40 B (30+40)/2=70/2=35 B*35 = 35 B 45 - 25 = A + B40-50 5 (40+50)/2=90/2=45 5 * 45 = 225 A + B = 20 ----- (1)

N=45 fx = 625+15A+35B

A + B = 20 ------ (1) fx =25+15A+375+35B+225=625+15A+35B

625+15A+35B (1) A + B = 20 multiply by 35 to eqn (1) 45 (2) 15A +35B = 500

25 * 45 = 625+15A+35B 1125 = 625+15A+35B 35A + 35 B = 700 15A+35B = 1125-625 _ 15A 35 B = 500 15A+35B = 500 ------ (2) 20 A = 200

A = 200 / 20 = 10

Now, substituting the value of A = 10 on eqn (1)

A + B = 20,10 + B = 20,B = 20 - 10,B = 10


(Tip :- From above example, it observed that you will solve the example both methodand answer comes as same, so student can deside which method is easy to solve.)

Ex (8):- Find out the missing frequencies from the following data, if the Mean of it 67.45.Height (x): 60-62 63-65 66-68 69-71 72-74 TotalNo of Student (f): 5 18 - - 8 = 100

Soln:- Assume that the frequencies are A and B. Total frequencies are 100.

C.I. f Xm fx60-62 5 (60+62)/2=122/2=61 5 * 61 = 30563-65 18 (63+65)/2=128/2=64 18 * 64 = 1152 N=5+18+A+B+866-68 A (66+68)/2=134/2=67 A * 67 = 67A 100 = 31 + A + B69-71 B (69+71)/2=140/2=70 B * 70 = 70B 100 - 31 = A + B72-74 8 (72+74)/2=146/2=73 8 * 73 = 584 A + B = 69 ----- (1)

N=100 fx = 2041+67A+70B

A + B = 69 ------ (1) fx =305+1152+67A+70B+584=2041+67A+70B

2041+67A+70B (1) A + B = 69 multiply by 70 to eqn (1)100 (2) 67A +70B = 4704

67.45 * 100 = 2041+67A+70B 6745 = 2041+67A+70B 70A + 70 B = 4830 6745 -2041 = 67A + 70 B _ 67A 70 B = 4704 67A+70B = 4704 ------ (2) 3 A = 126

A = 126 / 3 = 42

Now, substituting the value of A = 42 on eqn (1)

A + B = 69, 42 + B = 69, B = 69 - 42, B = 27


Ex (9):- Find mean for the following data.Class interval (x): 60-69 50-59 40-49 30-39 20-29 10-19 0-9 TotalFrequency (f): 2 4 5 10 6 9 7 = 43Soln:-

C.I. f Xm d= (Xm-A)/c fd60-69 2 64.5 (64.5-34.5)/10= 3 2 * 3 = 650-59 4 54.5 (54.5-34.5)/10= 2 4 * 2 = 840-49 5 44.5 (44.5-34.5)/10= 1 5 * 1 = 530-39 10 34.5 (34.5-34.5)/10= 0 10 * 0 = 020-29 6 24.5 (24.5-34.5)/10= -1 6 * -1 = -610-19 9 14.5 (14.5-34.5)/10= -2 9 * -2 = -180-9 7 4.5 (4.5-34.5)/10= -3 7 * -3 = -21

N=43 fd = -26

X =fx N 25 =

A = = 10

25 =

67.45 =

Q: How to find the value CA: Range 60-69, the numberstart from 0 to 9, it means0,1,2,3,4,5,6,7,8,9 = 10 intervals


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Since, N=43, A = 34.5, fd = -26, c = 10, then put the values on formula

(-26) 43

Ex (10): -10-19 20-29 30-39 40-49 50-59 60-69 70-79 80-89 90-99 5 6 7 8 11 6 4 2 1

C.I. f Xm d= (Xm-A)/c fd10-19 5 14.5 (14.5-54.5)/10= -4 5 * -4 = -2020-29 6 24.5 (24.5-54.5)/10= -3 6 * -3 = -1830-39 7 34.5 (34.5-54.5)/10= -2 7 * -2 = -1440-49 8 44.5 (44.5-54.5)/10= -1 8 * -1 = -850-59 11 54.5 (54.5-54.5)/10= 0 11 * 0 = 060-69 6 64.5 (64.5-54.5)/10= 1 6 * 1 = 670-79 4 74.5 (74.5-54.5)/10= 2 4 * 2 = 880-89 2 84.5 (84.5-54.5)/10= 3 2 * 3 = 690-99 1 94.5 (94.5-54.5)/10= 4 1 * 4 = 4

N=50 fd = -36


(-36) 50

Ex (11): - 121-140 101-120 81-100 61-80 41-60 21-40 1-201 3 9 20 8 5 4

C.I. f Xm d= (Xm-A)/c fd1-20 4 10.5 (10.5-70.5)/20= -3 4 * -3 = -1221-40 5 30.5 (30.5-70.5)/20= -2 5 * -2 = -1041-60 8 50.5 (50.5-70.5)/20= -1 8 * -1 = -861-80 20 70.5 (70.5-70.5)/20= 0 20 * 0 = 081-100 9 90.5 (90.5-70.5)/20= 1 9 * 1 = 9101-120 3 110.5 (110.5-70.5)/20= 2 3 * 2 = 6121-140 1 130.5 (130.5-70.5)/20= 3 1 * 3 = 3

N=50 fd = -12


(-12) 50

Ex (12):- Find the Simple Arithmetic Mean by using different methodsa) 86, 70, 96, 93, 97, 94Ans:- Short cut Method x direct method x d=x - A Assumed Mean (A) = 9370 70 70-93 = -23 d86 x 86 86-93 = -7 N93 N 93 93-93 = 0 -2294 536 94 94-93 = 1 696 6 96 96-93 = 3 = 93 - 3.666797 X = 89.33 97 97-93 = 4 X = 89.33x = 536 d = 22

(Tip:- Observe the above example, by using both the method i.e. Direct as will asShort cut method, the answer will be same i.e. 89.33 , So student can be solve theexample any method.)

b) x: 5 10 15 20 25 30 f: 4 5 9 15 10 7Ans:-

Direct Method Short cut Methodx f fx x f d= x-A f.d5 4 5*4=20 5 4 5-20=-15 4* -15= -6010 5 10*5=50 10 5 10-20=-10 5* -10= -5015 9 15*9=135 15 9 15-20=-5 9* -5= -4520 15 20*15=300 20 15 20-20=0 15* 0=025 10 25*10=250 25 10 25-20=5 10* 5=5030 7 30*7=210 30 7 30-20=10 7* 10=70 N=f =50 fx = 965 N=50 fd = -35

fx fd N N965 -35 50 50

X = 19.3 X = 20 - 0.7 = 19.3

(Tip:- Observe the above example, by using both the method i.e. Direct as will asShort cut method, the answer will be same i.e. 19.3 , So student can be solve theexample any method.)

c) x: 80-84 75-79 70-74 65-59 60-64 55-59 50-54 45-49 40-44 f: 2 0 10 17 24 19 7 4 1Ans:-

Above example solve by both the methods i.e. direct as will as short cut method.

X = A + x cfd NX = 34.5 + x 10 = 34.5 + (-0.6046 * 10) = 34.5 - 6.0465 =28.4534 = 28.45

X = A + x cfd N

X = 54.5 + x 10 = 54.5 + (-0.72 * 10) = 54.5 - 7.2 = 47.30

X = A + x cfd N

X = 70.5 + x 20 = 70.5 + (-0.24 * 20) = 70.5 - 4.8 = 65.70

X = A +

X = 93 +X =

X =

X =

X =

X = A +

X = 20 +

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Direct Method Short cut Method

C.I. f Xm f . Xm x f Xm d=Xm- A f.d80-84 2 82 164 80-84 2 82 20 2*20=4075-79 0 77 0 75-79 0 77 15 0*15=070-74 10 72 720 70-74 10 72 10 10*10=10065-69 17 67 1139 65-69 17 67 5 17*5=8560-64 24 62 1488 60-64 24 62 0 24*0=055-59 19 57 1083 55-59 19 57 -5 19 * -5= -9550-54 7 52 364 50-54 7 52 -10 7 * -10= -7045-49 4 47 188 45-49 4 47 -15 4 * -15= -6040-44 1 42 42 40-44 1 42 -20 1 * -20= -20

N=84 fXm=5188 N=84 fd = -20

(Note: Above table the value of Xm, f.Xm and d = Xm- A are directly calculate due to comparethe both method. If any doubt to how the values are calculate, then see the before exampleto check.)

fXm fd N N5188 (-20) 84 84

X = 61.76 X = 62 - 0.2380 = 61.7619X = 61.76 X = 61.76

By Using Step Deviation Method

C.I. f Xm d= (Xm-A)/c fd80-84 2 82 (82-62)/5 = 4 2 * 4 = 875-79 0 77 (77-62)/5 = 3 0 * 3 = 070-74 10 72 (72-62)/5 = 2 10 * 2 = 2065-69 17 67 (67-62)/5 = 1 17 * 1 = 1760-64 24 62 (62-62)/5 = 0 24 * 0 = 055-59 19 57 (57-62)/5 = -1 19 * -1 = -1950-54 7 52 (52-62)/5 = -2 7 * -2 = -1445-49 4 47 (47-62)/5 = -3 4 * -3 = -1240-44 1 42 (42-62)/5 = -4 1 * -4 = -4

N=84 fd = -4

(-4) 84

(Tip:- Observe the above example, by using Direct , Short cut as will as The StepDeviation method, the answer will be same i.e. 61.76 , So student can be solve theexample any method.)

X = 62 + x 5 = 62 + (-0.0476 * 5) = 62 - 0.2380 =61.7619 = 61.76

X =

X =

X = A +

X = 62 +

d) x: 18-19 16-17 14-15 12-13 10-11 8-9 f: 3 8 15 20 10 4Ans:-

Direct Method Short cut Method

C.I. f Xm f . Xm x f Xm d=Xm- A f.d18-19 3 18.5 55.5 18-19 3 18.5 6 3*6 = 1816-17 8 16.5 132 16-17 8 16.5 4 8*4 = 3214-15 15 14.5 217.5 14-15 15 14.5 2 15*2 = 3012-13 20 12.5 250 12-13 20 12.5 0 20*0 = 010-11 10 10.5 105 10-11 10 10.5 -2 10* -2= -20 8-9 4 8.5 34 8-9 4 8.5 -4 4 * -4 = -16

N=60 fXm=794 N=60 fd = 44

(Note: Above table the value of Xm, f.Xm and d = Xm- A are directly calculate due to comparethe both method. If any doubt to how the values are calculate, then see the before exampleto check.)

fXm fd N N 794 44 60 60

X = 13.2333 X = 12.5 + 0.7333 = 13.2333X = 13.23 X = 13.23

By Using Step Deviation Method

C.I f Xm d= (Xm-A)/c fd18-19 3 18.5 (18.5-14.5)/2 = 2 3 * 2 = 616-17 8 16.5 (16.5-14.5)/2 = 1 8 * 1 = 814-15 15 14.5 (14.5-14.5)/2 = 0 15 * 0 = 012-13 20 12.5 (12.5-14.5)/2 = -1 20 * -1 = -2010-11 10 10.5 (10.5-14.5)/2 = -2 10 * -2 = -20 8-9 4 8.5 (8.5-14.5)/2 = -3 4 * -3 = -12

N=60 fd = -38

(-38) 60

(Tip:- Observe the above example, by using Direct , Short cut as will as The StepDeviation method, the answer will be same i.e. 13.23 , So student can be solve theexample any method.)

X = 14.5 + x 2 = 14.5 + (-0.6333 * 2) = 14.5 - 1.2666 = 13.2333 = 13.23

X =

X =

X = A +

X = 12.5 +

X = A + x cfd N

X = A + x cfd N

Formula Formula

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Ex (13): - From the follwoing information relating to a certain industry consisting of 32firms, calculate the average salary paid in the whole industry.

Income Group: 50-75 75-100 100-150 150-200 200-300No. of Firms: 30 26 24 24 28Average No. ofWorkers : 5 8 5.5 6.5 1

IncomeGroup

C.I. f Xm f . Xm50-75 30*5 160 (50+75)/2=125/2=62.5 160*62.5 = 1000075-100 26*8 208 (75+100)/2=175/2=87.5 208*87.5 = 18200100-150 24*5.5 132 (100+150)/2=250/2=125 132*125 = 16500150-200 24*6.5 130 (150+200)/2=350/2=175 130*175 = 22750200-300 28*1 28 (200+300)/2=500/2=250 28*250 = 7000

N=658 f.Xm= 74450

f.Xm 74450 N 658

So, the average salary paid in the whole industry is Rs. 113.15

(ii) Weighted Arithmetic Mean:

In computing simple arithmetic mean, it is assumed that all the items are ofequal importance. This may not always be the case. When items vary in importance theymust be assigned some weights in proportion to their importance. The value of each itemis then multiplied by its weight. The products are summated and divided by the number ofweights, and not by the number of items. the quotient is the weighted arithmetic mean oraverage.Symbolically,

WX W

Weight are numbers or percentages which stand for the relative importance ofitems. Such a relative importance may be real or estimated, or actual or approximated.These weights are assigned to each item under the following circumstance:(i) When the results of the series are studied comparatively,(ii) When the computation involves ratios or percentages i.e. Death Rates and Birth Rates.(iii) When we compute the Index Numbers involving prices of essential commodities

having different importance to each one.Thus weight of a variate is numerical multiplier assigned to it in order to indicate its

relative importance.

Ex (1) :- Calculate Weighted average price of coal purchased by an industry.Month: Jan Feb March April May JunePrice per Ton (Rs) 42.50 51.25 50.00 52.00 44.25 54.00Tons Purchased 25 30 40 50 10 45Ans :-

Prices Per TonsTon (Rs) Purchased

C.I. W Wx42.50 25 42.50 * 25 = 1062.5051.25 30 51.25 * 30 = 1537.5050.00 40 50.00 * 40 = 2000.0052.00 50 52.00 * 50 = 2600.0044.25 10 44.25 * 10 = 0442.5054.00 45 54.00 * 45 = 2430.00

W = 200 Wx = 10072.50

The Weighted Arithmetic average price of coal purchased is 50.36

Ex (2) :- The following table gives the marks of two candidated:Subject Weights Marks of Candidates

X YA 1 70 80B 2 65 64C 3 58 56D 4 63 60

Find the Weighted average marks of each candidate.Ans:-

Marks bySubjects Weights(W) X Y WX WY

A 1 70 80 1 * 70 = 70 1 * 80 = 80B 2 65 64 2 * 65 = 130 2 * 64 = 128C 3 58 56 3 * 58 = 174 3 * 56 = 168D 4 63 60 4 * 63 = 252 4 * 60 = 240

W = 10 WX =626 WY = 616

X average marks is Y average marks is

The Weighted arithmetic mean for X Candidate is 62.60 and The Weighted arithmeticmean for Y Candidate is 61.60

X = = = 113.1458

XW = , WX = X * W , Where W = Weight

WxW10072.50 200

Xw = 50.3625Xw = 50.36

Xw =

Xw =

WxW626 10

Xw = 62.60Xw = 62.60

Xw =

Xw =

WxW616 10

Xw = 61.60Xw = 61.60

Xw =

Xw =

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Median :- Median is the value of that item in a series which divides the array into twoequal parts, one condisting of all the values less than it and the other consisting of all thevalues more than it. It means, when the values of variable are arranged in an array form (inan ascending or descending order of their magnitude), the value of middle item of thearray is the median and it is denote as Me.

Me is the size of th item where Me represents the median and N the numberof items

(A) Individual Observation

Ex (1) :- Find the median value of the following dataX: 7, 12, 37, 32, 17, 22, 27 (Note: Arrange the data in increasingAns:- (ascending) order)

7 12 17 22 27 32 37 X1 2 3 4 5 6 7 N

N + 1 27 + 1 8 2 2

Median (Me) = 22

Ex (1) :- Find the median value of the following dataX: 52, 47, 37, 32, 42, 27 (Note: Arrange the data in ascending order)Ans:-

27 32 37 42 47 52 X 1 2 3 4 5 6 N

N + 1 26 + 1 7 2 2

The Item lies between 4th & 5th item. So there are two values i.i. 37 & 42.Then the median value will be mean of these two values i.e.

(3)rd item + (4)th item 37 + 42 792 2 2

Median (Me) = 39.5

(B) Discrete Series

In discrete series, the value are already in the form of array and the frequenciesare recorded against each value.

However, for determining the size of item, a separate column is to be preparedfor cumulative frequencies (cf). The median size is first located with reference to thecumulative frequency (cf) which cover the size first. Then, against that cumulative fre-quency (cf) , the value will be located as the median value.

Ex (1) x : 62, 60, 66, 78, 75, 72, 67, 80f : 2 10 17 24 19 7 4 1

Ans:- Arrange the data in an array form with serial numbers (ascending order).

x f cf60 10 1062 2 2+10 =1266 17 17+12 = 2967 4 4+29 = 3372 7 7+33 = 4075 19 19+40 = 5978 24 24+59 = 8380 1 1+83 = 84

N=84

In above example Me = 42.5th item, it means that find such c.f. value which is greater thanfind th item i.e. c.f. value. Here, we find 42.5th item and from our table value 59 is greaterthan value 42.5. So, it mean that the Median value is from 59 c.f. and thats value x = 75.Therefor, the Median value is 75

Ex (2) x : 71, 66, 78, 74, 60, 63, 69, 61, 75f : 4 5 6 7 4 3 2 3 1

Ans: - Arrange the data in an array form with serial numbers (ascending order).x f cf60 4 461 3 3+4 = 763 3 3+7 = 1066 5 5+10 = 1569 2 2+15 = 1771 4 4+17 = 2174 7 7+21 = 2875 1 1+28 = 2978 6 6+29 = 35

N=35

In above example Me = 18th item, it means that find such c.f. value which is greater thanfind th item i.e. c.f. value. Here, we find 18th item and from our table value 21 is greaterthan value 18. So, it mean that the Median value is from 21 c.f. and thats value x = 71.Therefor, the Median value is 71

N + 1 2

Median (Me) = th item

Median (Me) = th item = th item = 4 th item = 22

Median (Me) = = = = 39.5

Median (Me) = th item

Median (Me) = th item = th item = 3.5 th item

N + 1 2

Median is the size of th item84 + 1 2

It lies in 59 c.f.Against 59 c.f. the value is 75So, Median Marks is 75(Note: Median value is identified by locatingthe respective c.f. in which item falls.)

N + 1 2Me = )th = 42.5 th item

)thN + 1 2

Median is the size of th item35 + 1 2

It lies in 21 c.f.Against 21 c.f. the value is 71So, Median Marks is 71(Note: Median value is identified by locatingthe respective c.f. in which item falls.)

N + 1 2Me = )th = 18 th item

)thN + 1 2

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(c) Continous SeriesA different procedure is adopted to find out the Median value under continous

series. The class intervals are already in the form of array and the frequencies are re-corded against each class interval. For determining the size, we should take N/2th itemand median class is locaed accordingly with reference to the cumulative frequency, whichcovers the size first. When the Median class is located, the median value is to be interpo-lated with the help of the following formula:

L - Lower limit of the median class, N - Total number of the frequencies,c - Magnitude of the median class, f - frequency of the median class,c.f. - Cumulative frequency of the class precding the median class.

Ex (1) 6-9 10-13 14-17 18-21 22-25 26-29 30-33 34-37 38-41 42-45 2 9 7 8 4 5 0 1 1 1

Ans:- Since there is an inclusive class interval distribution, we shall convert it into exclu-sive class interval series.

C.I. f c.f.41.5-45.5 1 37 + 1 = 3837.5-41.5 1 36 + 1 = 3733.5-37.5 1 35 + 1 = 3629.5-33.5 0 35 + 0 = 3525.5-29.5 5 30 + 5 = 3521.5-25.5 4 26 + 4 = 3017.5-21.5 8 18 + 8 = 2613.5-17.5 7 11 + 7 = 189.5-13.5 9 2 + 9 = 115.5-9.5 2 2

N=38

Here, N = 38 then N/2 = 38/2 = 19th item.It is lies in 26 c.f. , and class interval is 17.5 - 21.5 L = 17.5, f = 8, c.f. = 18 and c = 4 (find it in note)Put the values in formulas and find the Median.

19 - 18 1 8 8

Me = 17.5 + 0.125 x 4 = 17.5 + 0.5 = 18 Me = 18Median of Class interval is 18

Ex (2) C.I. : 10-15 15-20 20-25 25-30 30-35 35-40f: 6 18 9 10 4 3

Ans:- Above series is already exclusive class interval series.

C.I. f c.f.10-15 6 615-20 18 6 + 18 = 2420-25 9 9 + 24 = 3325-30 10 10 + 33 = 4330-35 4 4 + 44 = 4735-40 3 3 + 48 = 50

N=50

Here, N = 50 then N/2 = 50/2 = 25th item.It is lies in 33 c.f. , and class interval is 20 - 25 L = 20, f = 9, c.f. = 24 and c = 5 (find it in note)Put the values in formulas and find the Median.

25 - 24 1 9 9

Me = 20 + 0.1111 x 5 = 20 + 0.5555 = 20.5555 Me = 20.56Median of Class interval is 20.56

Ex Marks C.I. : 20-40 40-60 60-80 80-100 100-120 120-140 140-160No. of Student f: 4 6 10 16 12 7 3

Marks No. of C.I. Student (f) Marks L.T. c.f. Marks M.T. c.f20-40 4 Less than 40 4 More than 20 4+54=5840-60 6 Less than 60 4+6=10 More than 40 6+48=5460-80 10 Less than 80 10+10=20 More than 60 10+38=4880-100 16 Less than 100 16+20=36 More than 80 16+22=38100-120 12 Less than 120 12+36=48 More than 100 12+10=22120-140 7 Less than 140 7+48=55 More than 120 7+3=10140-160 3 Less than 160 3+55=58 More than 140 3

From above table, observe that, how the cumulative frequency (c.f) find when class intervalis in the form of Less than or More than.

Median (Me) = L + ( ) x cN/2 - c.f. f

Note: How to identified the exclusive classinterval series.=> Series always start from 0. i.e. 0-5, 0-10.means when lower class sustract from up-per class then find the class interval.Our example series start from 6-9.

means 6,7,8,9 = 4. The class interval valueis 4, but when lower class substract fromupper class then it find the value 3 (9 - 6 = 3).then this class is an inclusive.So, convert an inclusive class into exclusiveclass interval series. That is 5.5 - 9.5.Then 9.5 - 5.5 = 4. It is correct class interval.


Me = 17.5 + ) x 4 = 17.5 + ) x 4


Note:Lower interval (L1) = 10Upper interval (L2) = 15c = L2 - L1c = 15 - 10c = 5

Me = 20 + ) x 5 = 50 + ) x 5

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Ex (3):- Calculate the Median of the following data.Less than : 10 20 30 40 50 60 70frequency: 2 6 11 21 27 36 43Ans:- Let us convert the data from less than frequency distribution into normal distribution.

C.I. c.f. C.I. f c.f.10 2 Less than 10 2 220 6 10 - 20 6-2=4 630 11 20 - 30 11-6=5 1140 21 30 - 40 21-11=10 2150 27 40 - 50 27-21=6 2760 36 50 - 60 36-27=9 3670 43 60 - 70 43-36=7 43

N=43

Here, N = 43 then N/2 = 43/2 = 21.5th item.It is lies in 27 c.f. , and class interval is 40 - 50 L = 40, f = 6, c.f. = 21 and c = 10Put the values in formulas and find the Median.

21.5 - 21 0.5 6 6

Me = 40 + 0.08333 x 10 = 40 + 0.8333 = 40.8333 Me = 40.83Median of Class interval is 40.83

Ex (4):- Calculate the Median of the following data.Less than : 10 20 30 40 50 60 70frequency: 50 47 41 29 21 7 2Ans:- Let us convert the data from more than frequency distribution into normal distribu-tion.

C.I. c.f. C.I. f c.f.10 50 More than 10 50-47=3 5020 47 20 - 30 47-41=6 4730 41 30 - 40 41-29=12 4140 29 40 - 50 29-21=8 2950 21 50 - 60 21-7=14 2160 7 60 - 70 7-2=5 770 2 70 - 80 2 2

N=50

Here, N = 50 then N/2 = 50/2 = 25th item.It is lies in 29 c.f. , and class interval is 40 - 50 L = 40, f = 8, c.f. = 21 and c = 10Put the values in formulas and find the Median.

25 - 21 4 8 8

Me = 40 + 0.5 x 10 = 40 + 5 = 45 Me = 45Median of Class interval is 45

Ex (5):- In a group of 1000 wage earners, the monthly wages of 4% are below Rs. 60 andthose of 15% are under Rs. 62.50. 15% earned Rs. 95 over, and 5% got Rs. 100 and over.Find the median wage.Ans: Let us prepare the table showing different types of % frequencies and then convert itinto the original frequencies.

In problem Monthly C.I. s.f. F(%) fgiven wages (Rs.)below 60 Less than 60 0 - 60 4% 4% 40under 62.50 Less than 62.50 60 - 62.5 15% 15-4=11% 11062.50 - 95 62.50 - 95 62.5 - 95 Diff 100-30=70% 700over 95 More than 95 95 - 100 15% 15-5=10% 100100 & over 100 & over 100 & over 5% 5% 50

Note :- 100 = 4 + 11 + Diff + 10 + 5 = Diff + 30100 - 30 = DiffDiff = 70

For Less than and More than Class interval, see the before examples

Less than 60 0-60 4% More than 95 15-5=10%Less than 62.50 60-62.50 15-4=11% 100 & over 5%

After converting, The actual table is given belowC.I. f c.f.

0-60 40 4060-62.50 110 40+110=15062.5-95 700 110+150=85095-100 100 850+100=950100 & over 50 950+50=1000

N=1000

After Converting


Me = 40 + ) x 10 = 40 + ) x 10


Me = 40 + ) x 10 = 40 + ) x 10

After Converting

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Here, N = 1000 then N/2 = 1000/2 = 500th item.It is lies in 850 c.f. , and class interval is 62.5 - 95So, L = 62.5, f = 700, c.f. = 150 and c = 95 - 62.5 = 32.5Put the values in formulas and find the Median.

500-150 350 700 700

Me = 62.5 + 0.5 x 32.5 = 62.5 + 16.25 = 78.75 Me = 78.75Median of Class interval is 78.75

Ex (6):- 10 percent of the workers in a firm, employing a total of 1000 workers, earnbetween Rs. 5 and 9.99, 30 percent between Rs. 10 and Rs. 14.99, 250 workers betweenRs. 15 and Rs. 19.99 and the rest Rs. 20 and above. What is the median wage?Ans:- Since there is an inclusive class interval distribution, we shall convert it into exclu-sive class interval series.

Inclusive Class Interval

Wages Rs. (C.I.) No. of workers f5 - 9.99 10% 10010 - 14.99 30% 30015 - 19.99 250 25020 & above (Diff) (Diff)

N=1000

After converting, the Exclusive Class Interval is given belowC.I. f c.f.

4.995 - 9.995 100 1009.995 - 14.995 300 100+300 = 40014.995 - 19.995 250 250+400 = 65019.995 & above 350 650+350 = 1000

N=1000Here, N = 1000 then N/2 = 1000/2 = 500th item.It is lies in 650 c.f. , and class interval is 14.995 - 19.995 L = 14.995, f = 250, c.f. = 400 and c = 19.995 - 14.995 = 5Put the values in formulas and find the Median.

500-400 150 250 250

Me = 14.995 + 0.4 x 5 = 14.995 + 2 = 16.995 = 17Median of Class interval is 17

Ex (7):- From the following data find out the missing frequency if the median is 50.Class interval: 10-20 20-30 30-40 40-50 50-60 60-70Frequencies: 2 8 6 --- 15 10

Ans: Let us assume the missing frequency as A

C.I. f c.f. Since, Median = 50 (Given)10 - 20 2 2 Class interval 50-6020 - 30 8 2+8 = 10 L = 50, f = 15, c = 10 &30 - 40 6 6+10 = 16 c.f. = 16 + A40 - 50 A 16+A = 16 + A Now put the values in formula50 - 60 15 15+16+A = 31+A60 - 70 10 10+31+A = 41+A

N=41+A

41+A 41+AMe = 2 , 50 = 2

15 15

41+A 41*10 + 10A50 - 50 = 2 , 0 = 2

15 15

410 + 10A , 205 + 5A - 160 - 10A 0 = 2 15

15 , 0 * 15 = 205 + 5A - 160 - 10A

0 = 45 - 5A , 5A = 45 , A = 45 / 5 = 9

Missing frequency is 9

Ex (8):- An incomplete distribution is as follows:Class interval: 10-20 20-30 30-40 40-50 50-60 60-70 70-80Frequencies: 12 30 --- 65 --- 25 18Complete the distribution, if its median is 46 and Total is 229.

Ans: Let us assume the missing two frequency as A & B.C.I. f c.f.

10 - 20 12 1220 - 30 30 30+12 = 4230 - 40 A A+42 = 42 + A40 - 50 65 65+42+A = 107 + A50 - 60 B B+107+A = 107+A+B60 - 70 25 25+107+A+B = 132+A+B70-80 18 18+132+A+B = 150+A+B

N=229


Me = 62.5 + ) x 32.5 = 62.5 + ) x 32.5

1000=100+300+250+Diff1000=650+DiffDiff=1000-650Diff=350

Me =14.995+ ) x 5 = 14.995 + ) x 5


( ) - ( 16 + A )50 + x 10

( ) - ( 16 + A ) x 10

( ) - (16*10 + 10A )

( ) - ( 16 + A )50 + x 10

( ) - 160 - 10A 0 =

229 = 12+30+A+65+B+25+18229 = 150 + A + B229 - 150 = A + B79 = A + BA + B = 79 ---- (1)

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(Tip :- Median is given and from defination of Median the c.f. value is less than N / 2 )

Since, Median = 46 (Given). So, from table, class interval is (40 - 50)

L = 40, N = 229, f = 65 c.f. = 42 + A Put these value in formula

N 229Me = 2 , 46 = 2

f 65

114.5 - (42+A) , 114.5 - 42 - A65 65

72.5 - A , 72.5 * 10 - 10 A 65 65

6 * 65 = 725 - 10 A Put, A = 33.5 on equation A + B = 79390 = 725 - 10 A 33.5 + B = 7910A = 725 - 390 B = 79 - 33.510 A = 335 B = 45.5A = 335 / 10A = 33.5 B = 45.5

Missing frequencies are A = 33.5 and B = 45.5 As the frequencies are interger, then A = 33 and B = 46 or A = 34 and B = 45.

Ex (9):- Calculate the Median of the following data.Height : 5.1-6.0 6.1-7.0 7.1-8.0 8.1-9.0 9.1-10 10.1-11 11.1-12No.of Plants: 3 8 27 25 17 11 9Ans:- Since there is an inclusive class interval distribution, we shall convert it into exclu-sive class interval series.

C.I. C.I. f c.f.5.1-6.0 5.05-6.05 3 36.1-7.0 6.05-7.05 8 8+3 = 117.1-8.0 7.05-8.05 27 27+11 = 388.1-9.0 8.05-9.05 25 25+38 = 639.1-10.0 9.05-10.05 17 17+63 = 8010.1-11.0 10.05-11.05 11 11+80 = 9111.1-12.0 11.05-12.05 9 9+91 = 100

N=100N / 2 = 100 / 2 = 50, L = 8.05, f = 25, c.f. = 38 & c = 1, put th value on formula

50 - 38 12 25 25

Me = 8.05 + 0.48 * 1 , Me = 8.05 + 0.48 , Me = 8.53Median of Class interval is 8.53

Ex (10):- Calculate the Median of the following data.C.I. : 90-100 80-90 70-80 60-70 50-60 40-50 30-40 20-30f : 2 9 12 17 20 9 6 5

Ans:C.I. f c.f.

90-100 2 2+78 = 8080-90 9 9+69 = 7870-80 12 12+57 = 6960-70 17 17+40 = 5750-60 20 20+20 = 4040-50 9 9+11 = 2030-40 6 6+5 = 1120-30 5 5

N=80

Median of Class interval is 60

Ex (11):- Calculate the Median of the following data.C.I. : 35-40 30-35 25-30 20-25 15-20 10-15 5-10f : 1 5 12 15 10 4 3

Ans:C.I. f c.f.

35-40 1 1+49 = 5030-35 5 5+44 = 4925-30 12 12+32 = 4420-25 15 15+17 = 3215-20 10 10+7 = 1710-15 4 4+3 = 75-10 3 3

N=50

Median of Class interval is 22.67

(Note:- Above examples, class interval takes descreasing order and defination of me-dian, the cumulative frequency (c.f.) is find from summation of frequency with increas-ing order of class interval. Thats why summation of frequency start from bottom oftable means increasing order of class interval.)

( ) - c.f.L + x c

( ) - ( 42 + A )40 + x 10

46 - 40 = x 10

6 = x 10 6 =

6 = x 10

After Converting

Me = 8.05 + ( ) x 1 , Me = 8.05 + ( ) x 1

N / 2 = 80 / 2 = 40

L = 50 , f = 20, c.f. = 20, c = 10

40-20 202020

Me = 50 + 1 x 10 Me = 50 + 10 Me = 60

Me = 50 + ( ) x 10Me = 50 + ( ) x 10

N / 2 = 50 / 2 = 25

L = 20 , f = 15, c.f. = 17, c = 5

25-17 15 815

Me = 20 + 0.5333 x 5 Me = 20 + 2.6666 Me = 22.6666

Me = 20 + ( ) x 5Me = 20 + ( ) x 5

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Mode :-It is the value which occurs with the maximum frequency. It is most typical or

common value that receives the highest frequency. The modal class of a frequency distri-bution is the class with the highest frequency. It is denoted b a symbol Z.

Mode is the value of variable which is repeated the greatest number of times inthe series.

Ex (1):- Find the mode from following values of variable.13, 9, 8, 7, 13, 12, 15, 13

Ans:- Arrange the values in increasing (ascending) order7, 8, 9, 12, 13, 13, 13, 15

The mode is 13 as it occurs the highest number of times (i.e. 3 times) in theseries. So the mode is clearly defined.

When the number of items increase, we may convert the data into a descreteform. Then, the value having the maximum frequency will be treated as the Mode.

Ex (2):- Find the mode from following values of variable.65, 63, 68, 66, 61, 62, 65, 63, 67

Ans:- Arrange the values in increasing (ascending) order61, 62, 63, 63, 65, 65, 66, 67, 68

There are two modes i.e. 63 and 65 each occurring 2 times.

When the number of items increase, we may convert the data into a descreteform. Then, the value having the maximum frequency will be treated as the Mode.

Continous Series:- Under the continous series, the modal class is located with the help ofhighest frequencies or grouping or inspection (maxium concentration of frequencies).The same procdure is adopted as it is followed in case discret series. After locating themodal class, we have to interpolate the value of the mode within the modal class by usingthe following formulas:

f1 - f0 f1 - f0 2f1 - f0 - f2 (f1 - f0) - (f2 - f1)

Z = Mode, L = Lower limit of the modal class interval,f0 = Frequency of the class interval preceding the modal class interval,f1 = Fequency of the modal class interval,f2 = Fequency of the class interval succeeding the modal class interval,c = magnitude of the modal class interval

Other also formula of Mode i.e. Z = 3 Me - 2 X

Z = Mode,Me = Medain X = Mean

Ex (3):- C.I. : 90-100 80-90 70-80 60-70 50-60 40-50 30-40 20-30f : 2 9 12 17 20 9 6 5

Ans:C.I. f c.f.

90-100 2 2+78 = 8080-90 9 9+69 = 7870-80 12 12+57 = 6960-70 17 17+40 = 5750-60 20 20+20 = 4040-50 9 9+11 = 2030-40 6 6+5 = 1120-30 5 5

20 - 9 11 (20 - 9) - (17-20) (11) - (-3)

11 11 * 10 110 14 14 14

Z = 50 + 7.8571, Z = 57.8571, Z = 57.86

Mode ( Z ) = 57.86

Ex (4):- C.I. : 35-40 30-35 25-30 20-25 15-20 10-15 5-10f : 1 5 12 15 10 4 3

Ans:C.I. f c.f.

5-10 3 310-15 4 4+3 = 715-20 10 10+7 = 1720-25 15 15+17 = 3225-30 12 12+32 = 4430-35 5 5+44 = 4935-40 1 1+49 = 50

15 - 10 5 (15 - 10) - (12-15) (5) - (-3)

5 5 * 5 25 8 8 8

Z = 20 + 3.125 , Z = 23.125 , Z = 23.13

Mode ( Z ) = 23.13

Z = L + x c or Z = L + x c

The modal class is located with the help ofhighest frequencies. i.e. 20 and modal classis (50-60)

f0 = 9, f1 = 20, f2 = 17, L = 50, c = 10

f1 - f0 (f1 - f0) - (f2 - f1)

Z = L + x c

Z = 50 + x 10 , Z = 50 + x 10

Z = 50 + x 10 , Z = 50 + , Z = 50 +

The modal class is located with the help ofhighest frequencies. i.e. 15 and modal classis (20-25)

f0 = 10, f1 = 15, f2 = 12, L = 20, c = 5

f1 - f0 (f1 - f0) - (f2 - f1)

Z = L + x c

f0f1f2

f2f1f0

Z = 20 + x 5 , Z = 20 + x 5

Z = 20 + x 5 , Z = 20 + , Z = 20 +

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Ex (5):- Find the mode , Mean = 49.2 and Median = 48.37?Mode (Z) = 3 x Me - 2 * X

= 3 x 48.37 - 2 x 49.2 = 145.11 - 98.4 = 46.71

Mode (Z) = 46.71

Ex (6):- Find the mode , Mean = 48.40 and Median = 48.67?Mode (Z) = 3 x Me - 2 * X

= 3 x 48.67 - 2 x 48.40 = 146.01 - 96.8 = 49.21

Mode (Z) = 49.21

Range:-Range repersents the difference between the values of the etremes-- the larges

value and the smallest value. The value in between the two extremes are not at all takeninto consideration. It is denoted symbolically by R.

(i) Range = Largest value - Smallest valueR = L - S

(ii) Coefficient of Range (R) = =

Ex (1):- Compute the range and the Coefficient of Range of the series, and state which oneis more dispersed and which one is more uniform.

Series Values of variables i. 13, 14, 15, 16, 17 ii 9, 12, 15, 18, 21 iii 1, 8, 15, 22, 29Ans: (I) (II) (III)

Range R = L - S R = L - S R = L - S = 17 - 13 = 21 - 9 = 29 - 1 = 4 = 12 = 28

L - S L - S L - SL+ S L+ S L+ S17 - 13 21 - 9 29 - 117+ 13 21+9 29+ 1 4 12 28 30 30 30

= 0.1333 = 0.4 = 0.933

Series (I) is less dispersed and more uniform.Series (III) is less uniform and more dispersed.

Quartilies:-The Quartiles are also positional averages like the median. As the median value

divides the entire distribution into two equal parts, the quartiles (Q1 , Q2 and Q3 ) divide theentire distribution into four equal parts.

First Quartile (Lower Quartile) Q1 is the value below which there are one fourth of the itemsand above which there are three fourth of the items.

Second Quartile (Median) Q2 divides the total distribution into two halves.

Third Quartile (Upper Quartile) Q3 is the value below which there are three fourth of theitems and above which there are one fourth of the items.

n + 1 2 (n + 1) 3 (n + 1) 4 4 4

Quartile Deviation:-In distribution, we consider Q3 as the largest value and Q1 as the smallest. It

means, the items below the lower quartile and the items above the upper quartile are notat all included in the computation. Thus we are considering only the middle half portion ofthe distribution. The range so obtained is divided by two as we are considering only half ofthe data. Thus the Quartil deviation measures the difference between the value of Q1 andQ3 . It is denoted symbolically by Q.D.

Ex :- Find the Quartile and Coefficient of Q.D. of the following values

(1):- 19, 23, 9, 27, 3, 1, 31 Arrange increasing (ascending) order

Ans:- 1 3 9 19 23 27 31 X1 2 3 4 5 6 7 N

Q3 - Q1 n + 1 3 (n + 1) 2 4 4

7 + 1 3 ( 7 + 1) 4 4

8 3 * 8 4 4

Largest Value - Smallest Value L - RLargest Value + Smallest Value L + R

Coeff. of R. = C.R. = C.R. =



Q1 = ( )th item Q2 = ( )th item Q3 = ( )th item

Coefficient of C of Q.D. =

Q.D. = , Q1 = ( )th item , Q3 = ( )th item

Q1 = ( )th item , Q3 = ( )th item


Quartile Deviation(QD) =

Q1 = L + ( ) x c & Q3 = L + ( ) x c

Q3 - Q1 2

N/4 - F f

3N/4 - F f

Q3 - Q1Q3 + Q1

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8 24 4 4

Q1 = ( 2 )th item Q3 = ( 6 )

th itemQ1 = 3 Q3 = 27

Q3 - Q1 27 - 3 24 2 2 2

Quartile Deviation (Q.D.) = 12

Q3 - Q1 27 - 3 24Q3 + Q1 27+ 3 30

Coffiecient of Q.D. = 0.8

(2):- 4, 7, 17, 24, 13, 30, 32, 39 Arrange increasing (ascending) order

Ans:- 4 7 13 17 24 30 32 39 X1 2 3 4 5 6 7 8 N

Q3 - Q1 n + 1 3 (n + 1) 2 4 4

8 + 1 3 ( 8 + 1) 4 4

9 3 * 9 4 4

9 27 4 4

Q1 = ( 2.25 )th item , Q3 = ( 6.75 )

th itemQ1 = ( 2 )

nd item + 0.25 x (3rd item - 2nd item) , Q3 = ( 6 )th item + 0.75 x (7th item - 6th item)

Q1 = 7 + 0.25 x ( 13 - 7 ) , Q3 = 30 + 0.75 x ( 32 - 30 )Q1 = 7 + 0.25 x ( 6 ) , Q3 = 30 + 0.75 x ( 2 )Q1 = 7 + 1.25 , Q3 = 30 + 1.5Q1 = 8.25 , Q3 = 31.5

Q3 - Q1 31.5 - 8.25 23.25 2 2 2

Quartile Deviation (Q.D.) = 11.63

Q3 - Q1 31.5 - 8.25 23.25Q3 + Q1 31.5+ 8.25 39.75


(3):- 3, 10, 12, 5 Arrange increasing (ascending) order

Ans:- 3 5 10 12 X1 2 3 4 N

Q3 - Q1 n + 1 3 (n + 1) 2 4 4

4 + 1 3 ( 4 + 1) 4 4

5 3 * 5 4 4

5 15 4 4

Q1 = ( 1.25 )th item , Q3 = ( 3.75 )

th itemQ1 = ( 1 )

st item + 0.25 x (2nd item - 1st item) , Q3 = ( 3 )rd item + 0.75 x (4th item - 3rd item)

Q1 = 3 + 0.25 x ( 5 - 3 ) , Q3 = 10 + 0.75 x ( 12 - 10 )Q1 = 3 + 0.25 x ( 2 ) , Q3 = 10 + 0.75 x ( 2 )Q1 = 3 + 0.25 , Q3 = 10 + 1.5Q1 = 3.5 , Q3 = 11.5

Q3 - Q1 11.5 - 3.5 8 2 2 2

Quartile Deviation (Q.D.) = 4

Q3 - Q1 11.5 - 3.5 8Q3 + Q1 11.5+ 3.5 15


(4):- 1, 3, 10, 15, 27, 35, 42, 46, 49, 50, 55 Arrange ascending order

Ans:- 1 3 10 15 27 3 42 46 49 50 55 X1 2 3 4 5 6 7 8 9 10 11 N

Q3 - Q1 n + 1 3 (n + 1) 2 4 4

11 + 1 3 ( 11 + 1) 4 4

12 3 * 12 4 4

12 36 4 4

Q.D. = = = = 12


Coffiecient of Q.D. = = = = 0.8

Q.D. = = = = 11.625






Q.D. = = = = 11.625










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Q1 = ( 3 )th item , Q3 = ( 9 )

th itemQ1 = 10 , Q3 = 49

Q3 - Q1 49 - 10 39 2 2 2


Q3 - Q1 49 - 10 39Q3 + Q1 49 + 10 59


Ex (5):- 0-9 10-19 20-29 30-39 40-49 50-59 60-69 7 9 6 10 5 4 2


C.I. C.I. f c.f.60-69 59.5-69.5 2 2+41= 4350-59 49.5-59.5 4 4+37 = 4140-49 39.5-49.5 5 5+32 = 3730-39 29.5-39.5 10 10+22 = 3220-29 19.5-29.5 6 6+16 = 2210-19 9.5-19.5 9 9+7 = 160-9 0.5-9.5 7 7

N=43

N/4 - cf 3N/4- cf f f

N = 43, N 43 3N 3 x 43 129 4 4 4 4 4

L = 9.5, cf = 7, f = 9, c = 10 L = 39.5,cf = 32, f = 5, c = 10

10.75-7 32.25 - 32 9 5 3.75 0.25 9 5

= 9.5 + 0.41666 x 10 = 39.5 + 0.05 x 10= 9.5 + 4.1666 = 39.5 + 0.5= 13.6666 = 40

Q1 = 13.67 Q3 = 40

Q3 - Q1 40 - 13.67 26.33 2 2 2


Q3 - Q1 40 - 13.67 26.33Q3 + Q1 40 + 13.67 53.67


Ex (6):- 7-11 12-16 17-21 22-26 27-31 32-36 37-41 42-46 2 3 5 4 8 7 2 4


C.I. C.I. f c.f.7-11 6.5-11.5 2 212-16 11.5-16.5 3 3+2 = 517-21 16.5-21.5 5 5+5 = 1022-26 21.5-26.5 4 4+10 = 1427-31 26.5-31.5 8 8+14 = 2232-36 31.5-36.5 7 7+22 = 2937-41 36.5-41.5 2 2+29 = 3142-46 41.5-46.5 4 4+31 = 35

N=35

N/4 - cf 3N/4- cf f f

N = 35, N 35 3N 3 x 35 129 4 4 4 4 4

L = 16.5, cf = 5, f = 5, c = 5 L = 31.5,cf = 22, f = 7, c = 5

8.75-5 26.25 - 22 5 7 3.75 4.25 5 7

= 16.5 + 0.75 x 5 = 31.5 + 0.6071 x 5= 16.5 + 3.75 = 31.5 + 3.0357= 20.25 = 34.5357

Q1 = 20.25 Q3 = 34.54

Q3 - Q1 34.54-20.25 14.29 2 2 2


Q3 - Q1 34.54 -20.25 14.29Q3 + Q1 34.54+20.25 54.79


Q.D. = = = = 19.5


After Converting

Quartile Deviation(QD) = Q1 = L + ( ) x c & Q3 = L + ( ) x c3N/4-cf fN/4-cf f

Q3 - Q1 2

Q1 = L + x c Q3 = L + x c

= = 10.75 = = = 32.25

Q1 = 9.5 + x 10 Q3 = 39.5 + x 10

Q1 = 9.5 + x 10 Q3 = 39.5 + x 10

Q.D. = = = = 13.16


QD = Q1 = L + ( ) x c & Q3 = L + ( ) x c3N/4-cf fN/4-cf fQ3 - Q1 2


Q1 = L + x c Q3 = L + x c

= = 8.75 = = = 26.25

Q1 = 16.5 + x 5 Q3 = 31.5 + x 5

Q1 = 16.5 + x 5 Q3 = 31.5 + x 5

Q.D. = = = = 7.145

After Converting

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Mean Deviation:-Mean Devaiation is the average difference among the items in a series from the

mean itself or median or mode of that series. It is concerned with the extent to which thevalues are despersed about the mean or the median or the mode. It is found by theaveraging all the deviations from the central tendency. These deviations are taken intocomputations with regard to negative sign (i.e. all the deviations assumed as possitive).

In aggregating the devations, the algebraic negative signs are not taken intoaccount. It means all the devations are treated as positive ignoring the negative sign.

Mean deviation or average deviation is denoted symbolically by the Greak smallalphabet (delta)

d n

Me x z

fd n

(Note: d is the sum of the deviations and fd is the sum of the products f and d).

When the averages are in fractions, the calculation of mean deviation becomes a tediousjob. So, to make the things more simplified, a shor-cut method or formula is used asunder:

xA - xB - (A - B) Me* n

fxA - fxB - (fA - fB) Me* n

(Note: In place of Me* we can take x or z as the deviations are concerned.)

Steps to be followed in the short cut method:

(i) Make two sections in the entire distribution, as A and B, so that all the items greater thanaverage (including average) should fall in A section and all the items smaller than aver-age should fall in the B section.(ii) Terms used in the formula:xA : the sum of the values greater than average.xB : the sum of the values smaller than average.A : the total number of items greater than average.B : the total number ot items smaller than average.n : the total number of items.fxA : the sum of the fx greater than the average.fxB : the sum of the fx smaller than the average.fA : the sum of the frequencies greater than averagefB : the sum of the frequencies smaller than average

Ex (1):- From the following variables find the Mean Deviation and Coefficient of MeanDeviation from the mean. ====> 10, 18, 3, 9, 6, 4Ans: Arrange the data in an ascending order to have the short cut method applicable.

x d=x-X3 3-8.34= 5.34 x 504 4-8.34= 4.34 n 66 6-8.34= 2.34 d 249 9-8.34= 0.66 n 610 10-8.34= 1.66 418 18-8.34= 9.66 x 8.34

x=50,n=6 d = 24

x Put all these values in formula34 B=3 xB = 3+4+6=13 xA -xB - (A - B) X6 B n

n=6, X = 8.34 37 - 13 - (3 - 3) * 8.349 610 A=3 xA = 9+10+18=37 24 - ( 0 ) * 8.3418 A 6

24 6

From above observation, the value of Mean Deviation = 4, So student can solve theproblem both the formula, but use easy one.

Ex (2):- From the following variables find the Mean Deviation and Coefficient of MeanDeviation from the mean. ====> 15, 4, 3, 10, 12, 5, 1, 2, 6, 12Ans: Arrange the data in an ascending order to have the short cut method applicable.

x d=x-X1 1-7= 6 x 702 2-7= 5 n 103 3-7= 4 d 424 4-7= 3 n 105 5-7= 2 4.26 6-7= 1 x 710 10-7= 312 12-7= 512 12-7= 515 15-7= 8 (Tip: the devations (d) are treated as positive

x=70,n=10 d = 42 ignoring the negative sign.)

(A) Individual Observations: = ............. Absolute Measure

Coefficient of = or or .......... Relative Measure

(B) Discrete and Continous Series: = ............. Absolute Measure

(A) Individual Observations: =

(B) Discrete and Continous Series: =

X = = = 8.3333 = 8.34

= = = 4

Coefficient of = = = 0.4796

Coefficient of = 0.48

=

=

=

= = 4

X = = = 7

= = = 4.2

Coefficient of = = = 0.6


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Ex (3):- Following are the runs scored by the btsmen in different innings of cricket tests.Runs: 20 40 60 80 100 120 140 160 180No of Batsmen: 6 19 40 23 65 83 55 20 9Compute the Mean Deviation from mode and its Coefficient.Ans: Arrange the data in an ascending order to have the short cut method applicable.

x f d=x-z fd20 6 20-120=100 6*100=600 From the defination of mode40 19 40-120=80 19*80=1520 The highest frequency is 8360 40 60-120=60 40*60=2400 z = 12080 23 80-120=40 23*40=920 fd 9180100 65 100-120=20 65*20=1300 n 320120 83 120-120= 0 83*0= 0140 55 140-120=20 55*20=1100160 20 160-120=40 20*40=800 28.69180 9 180-120=60 9*60=540 z 120

n=6 fd = 9180

x f fx20 6 20*6=120 B=6+19+40+23+65=15340 19 40*19=760 xB = 120+760+2400+1840+6500=1162060 40 60*40=240080 23 80*23=1840 A=83+55+20+9=167100 65 100*65=6500 xA = 9960+7700+3200+1620=22480

Put all these values in formula120 83 120*83= 9960 xA -xB - (A - B) z140 55 140*55=7700 n160 20 160*20=3200 22480-11620-(167-153) * 120180 9 180*9=1620 320

10860 - (14*120) 10860 - 1680 320 3209180 320

From above observation, the value of Mean Deviation = 28.69, So student can solve theproblem both the formula, but use easy one.

Ex:- Find the Mean Deviation from following data.

4):- 50-59 60-69 70-79 80-89 90-99 100-109 110-119 120-129 130-1394 8 14 16 20 16 14 8 4

5):- 4 - 5 6 - 7 8 - 9 10 - 11 12 - 13 14 - 15 16 - 17 2 4 5 8 4 4 5

4) Ans: Arrange the data in an descending order.

C.I. f Xm f.Xm X = f.Xm/ n d=Xm - X f.d130-139 4 134.5 4*134.5=538 134.5-94.5=40 4*40=160120-129 8 124.5 8*124.5=996 124.5-94.5=30 8*30=240110-119 14 114.5 14*114.5=1603 114.5-94.5=20 14*20=280100-109 16 104.5 16*104.5=1672 9828 104.5-94.5=10 16v10=16090-99 20 94.5 20*94.5=1890 104 94.5-94.5=0 20*0=080-89 16 84.5 16*84.5=1352 =94.5 84.5-94.5=10 16*10=16070-79 14 74.5 14*74.5=1043 74.5-94.5=20 14*20=28060-69 8 64.5 8*64.5=516 64.5-94.5=30 8*30=24050-59 4 54.5 4*54.5=218 54.5-94.5=40 4*40=160

N=104 f.Xm=9828 fd = 1680

fd 1680 n 104

Mean Deviation ( ) = 16.15

5) Ans: Arrange the data in an descending order.

C.I. f Xm f.Xm X = f.Xm/ n d=Xm - X f.d16-17 5 16.5 5*16.5=82.5 16.5-11=5.5 5*5.5= 27.514-15 4 14.5 4*14.5=58 14.5-11=3.5 4*3.5= 1412-13 4 12.5 4*12.5=50 352 12.5-11=1.5 4*1.5= 610-11 8 10.5 8*10.5=4 32 10.5-11=0.5 8*0.5= 48-9 5 8.5 5*8.5=42.5 =11 8.5-11=2.5 5*2= 12.56-7 4 6.5 4*6.5=26 6.5-11=4.5 4*4.5= 184-5 2 4.5 2*4.5=9 4.5-11=6.5 2*6.5= 13

N=32 f.Xm=352 fd = 95

fd 95 n 32

Mean Deviation ( ) = 2.97

(Tip: the devations (d) are treated as positive ignoring the negative sign.)

= = = 28.6875

= 28.69

Coefficient of = =


A

B

=

=

= =

= = 28.6875 = 28.69

= = = 16.1538 = 16.15

*

= = = 2.9687 = 2.97

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Standard Deviation ():-Standard Deviation is the root of the sum of the squares of the deviations devided

by their number. it is also called Mean Error Deviation, Mean Square Deviation or RootMean Square Deviation. It is a second moment of a dispersion. Since the sum of thesquare of the deviation from the mean is minimum, the deviations are taken only frommean (but not from median or mode).

Standard Deviation is the root-mean-square average of all the deviations fromthe mean and it is denoted by (sigma).

A: Individual Observations:

d2 (x - X)2 n n

X

B: Discrete and Continuous series:

fd2 f(x - X)2 n n

X

Short Cut Method:-Sometimes the mean will be a fractional figure. Then we should take the devia-

tions from the assumed mean and the direct method formula will be having some adjust-ment.. As the deviations are not taken from the actual mean we get the d as some valueinstead zero. The short cut formula works as under.

d2 d 2 n n

fd2 fd 2 n n

Step-deviation Method:-The deviations are further divided by the common factor in case of assumed

mean. This deliberate error is compensated by multiplying the entire formula by the samefactor. The formula works as under.

d2 d 2 n n

fd2 fd 2 n n

c - Common factor

Ex (1):- From the following variables find the Standard Deviation and Coefficient ofvariation(c.v.) ====> 43, 48, 60, 30, 58, 23Ans: Arrange the data in an ascending order

x d=x-X d2 d=(x- A) d223 23-43.67= -20.67 (-20.67)2=427.25 23-48= -25 (-25)2= 62530 30-43.67= -13.67 (-13.67)2=186.87 30-48= -18 (-18)2= 32443 43-43.67= -0.67 (-0.67)2=0.45 43-48= -5 (-5)2= 2548 48-43.67= 4.33 (4.33)2=18.75 48-48= 0 (0)2= 058 58-43.67= 14.33 (14.33)2=205.35 58-48= 10 (10)2= 10060 60-43.67= 16.33 (16.33)2=266.67 60-48= 12 (12)2= 144

x=262 d2 = 1105.34 d = -26 d2 = 1218

x 262 d2 d 2 1218 -26 2 n 6 n n 6 6

d2 1105.34 n 6 = 203 - ( -4.33)2 = 203 - 18.77

c.v. = * 100 = * 100 = 0.3107*100 = 184.23 = 13.57

c.v. = 31.07%

From above observation, it seen than example solve by any method the answer is comesame. So student solve easy method, which they are suitable to solve the problems.

Ex (2):- From the following variables find the Standard Deviation and Coefficient ofvariation(c.v.) ====> 50, 26, 37, 35, 34Ans: Arrange the data in an ascending order

x d=x-X d2 d=(x- A) d226 26-36.40 = -10.4 (-10.4)2=108.16 26-35= -9 (-9)2 = 8134 34-36.40 = -2.4 (-2.4)2=5.76 34-35= -1 (-1)2 = 135 35-36.40 = -1.4 (-1.4)2=1.96 35-35= 0 (0)2 = 037 37-36.40 = 0.6 (0.6)2=0.36 37-35= 2 (2)2 = 450 50-36.40 = 13.6 (13.6)2=184.96 50-35= 15 (15)2 = 225

x=182 d2 = 301.20 d = 7 d2 = 311

x 182 d2 d 2 311 7 2 n 5 n n 5 5

d2 301.20 n 5 = 62.2 - ( 1.4)2 = 62.2 - 1.96

c.v. = * 100 = * 100 = 0.2131*100 = 60.24 = 7.76

c.v. = 21.31%

= or ----- Absolute MeasureCoefficient of Variation = * 100 ----- Relative Measure

= or ----- Absolute MeasureCoefficient of Variation = * 100 ----- Relative Measure

A: Individual Observation = - ( )

B: Discrete & Continous series = - ( )

A: Individual Observation = - ( ) x c

B: Discrete & Continous series = - ( ) x c

X = = = 43.6666 = 43.67 = - ( ) = - ( ) = = = 184.22 = 13.57

13.57 X 43.67

X = = = 36.40 = - ( ) = - ( ) = = = 60.24 = 7.76

7.76 X 36.4

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Ex (3):- Following are the runs scored by the two batsmen named NEKO and DECO in teninnings. Find who is better scorer and who is more consistent.NEKO: 101 22 0 36 82 45 7 13 65 14DECO: 97 12 40 96 13 8 85 8 56 16Ans: Arrange the data in an ascending order.

NECO DECOx d=x - X d2 = d*d x d=x-X d2 = d*d0 0-38.5= -38.5 (-38.5)2=1482.25 8 8-43.1= -35.1 (-35.1)2=1232.017 7-38.5= -31.5 (-31.5)2=992.25 8 8-43.1= -35.1 (-35.1)2=1232.0113 13-38.5= -25.5 (-25.5)2=650.25 12 12-43.1= -31.1 (-31.1)2=967.2114 14-38.5= -24.5 (-24.5)2=600.25 13 13-43.1= -30.1 (-30.1)2=906.0122 22-38.5= -16.5 (-16.5)2=272.25 16 16-43.1= -27.1 (-27.1)2=734.4136 36-38.5= -2.5 (-2.5)2=6.25 40 40-43.1= -3.1 (-3.1)2=9.6145 45-38.5= 6.5 (6.5)2=42.25 56 56-43.1= 12.9 (12.1)2=166.4165 65-38.5= 26.5 (26.5)2=702.25 85 85-43.1= 41.9 (41.9)2=1755.6182 82-38.5= 43.5 (43.5)2=1892.25 96 96-43.1= 52.9 (52.9)2=2798.41101 101-38.5= 62.5 (62.5)2=3906.25 97 97-43.1= 53.9 (53.9)2=2905.21

x=385 d2 = 10546.50 x=431 d2 = 12706.90

x 385 x 431 n 10 n 10

d2 10546.50 d2 12706.90 n 10 n 10

32.48 35.65 X 38.50 X 43.10

CV = 0.8436 * 100 = 84.36% CV = 0.8271 * 100 = 82.71%

DECO is a better run scorer and more consistent player than NEKO. (because his aver-age is more and variation is less)

(Tip:- The word-> more consistent, more stable, more uniform then check the coefficient ofvariation value and if coefficient value (cv) is less, then that group is more consistent,more stable and more uniform.)

Ex (4):- Following are the marks obtained by two students A and B in 10 tests of 100marks each.Tests: 1 2 3 4 5 6 7 8 9 10 A : 44 80 76 48 52 72 72 51 60 54 B : 48 75 54 60 63 69 72 51 57 66Find who is better in studies and if consistency is the criterion for awarding a prize, whoshould get the prize.

Ans: Arrange the data in an ascending order.

(A) x d=x - X d2 = d*d (B) x d=x-X d2 = d*d44 44-60.9= -16.9 (-16.9)2=285.61 48 48-61.5= -13.5 (-13.5)2=182.2548 48-60.9= -12.9 (-12.9)2=166.41 51 51-61.5= -10.5 (-10.5)2=110.2551 51-60.9= -9.9 (-9.9)2=98.01 54 54-61.5= -7.5 (-7.5)2=56.2552 52-60.9= -8.9 (-8.9)2=79.21 57 57-61.5= -4.5 (-4.5)2=20.2554 54-60.9= -6.9 (-6.9)2=47.61 60 60-61.5= -1.5 (-1.5)2=2.2560 60-60.9= -0.9 (-0.9)2=0.81 63 63-61.5= 1.5 (1.5)2=2.2572 72-60.9= 11.1 (11.1)2=123.21 66 66-61.5= 4.5 (4.5)2=20.2572 72-60.9= 11.1 (11.1)2=123.21 69 69-61.5= 7.5 (7.5)2=56.2576 76-60.9= 15.1 (15.1)2=228.01 72 72-61.5= 10.5 (10.5)2=110.2580 80-60.9= 19.1 (19.1)2=364.81 75 75-61.5= 13.5 (13.5)2=182.25

x=609 d2 = 1516.90 x=615 d2 = 742.50

x 609 x 615 n 10 n 10

d2 1516.90 d2 742.50 n 10 n 10

12.32 8.62 X 60.90 X 61.5

CV = 0.2022 * 100 = 20.22% CV = 0.1401 * 100 = 14.01%

B is better in studies and he should get a prize also as his average is more and varianceis less.

Ex (5):- Prices of particular commodity in five yearsin two cities are given below.Tests: 1 2 3 4 5City A : 20 22 19 23 16City B : 10 20 18 12 15Find from the table which city had more stable prices.Ans: City A City B

x d=x - X d2 = d*d x d=x-X d2 = d*d16 16-20= -4 (-4)2 =16 10 10-15 = -5 ( -5 )2 = 2519 19-20= -1 (-1)2 = 1 12 12-15 = -3 ( -3 )2 = 920 20-20= 0 (0)2 = 0 15 15-15 = 0 ( 0 )2 = 022 22-20= 2 (2)2 = 4 18 18-15 = 3 ( 3 )2 = 923 23-20= 3 (3)2 = 9 20 20-15 = 5 ( 5 )2 = 25

x=100 d2 = 30 x=75 d2 = 68x 100 x 75 n 5 n 5

d2 30 d2 68 n 5 n 5

X = = = 38.5 X = = = 43.1

= = = 1054.65 = = = 1270.69

= 32.475 = 32.48 = 35.647 = 35.65

CV = * 100 = * 100 CV = * 100 = * 100

X = = = 60.90 X = = = 61.5

= = = 151.69 = = = 74.25

= 12.3162 = 12.32 = 8.6168 = 8.62

CV = * 100 = * 100 CV = * 100 = * 100

X = = = 20 X = = = 15

= = = 6 = 2.45 = = = 13.6 = 3.6878 = 3.69

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2.45 3.69 X 20 X 15

CV = 0.1225 * 100 = 12.25% CV = 0.246 * 100 = 24.60%

Coefficient of variation is less in prices of City A: the prices of City A are more stable thanthe City B.

Ex (6):- The goals scored by two teams A and B in the football matches were as follows.Goals: 0 1 2 3 4Matches A: 27 9 8 4 5 B: 17 9 6 5 3Find the team which is more consistent.Ans: Match A

x f fx d=x-A fd fd20 27 0*27=0 -2 27* -2 = -54 -2 * -54 = 1081 9 1*9=9 -1 9* -1 = -9 -1* -9 = 92 8 2*8=16 0 8* 0 = 0 0 * 0 = 03 4 3*4=12 1 4* 1 = 4 1 * 4 = 44 5 4*5=20 2 5* 2 = 10 2 * 10 = 20

n=53 x=57 fd= -49 fd2 =141

fd2 d 2 141 -49 2 n n 53 53

= 1.8044 = 1.3432 = 1.34

1.34 X 1.08

Match Bx f fx d=x-A fd fd20 17 0*17=0 -2 17* -2 = -34 -2 * -34 = 681 9 1*9=9 -1 9* -1 = -9 -1* -9 = 92 6 2*6=12 0 6* 0 = 0 0 * 0 = 03 5 3*5=15 1 5* 1 = 5 1 * 5 = 54 3 4*3=12 2 3* 2 = 6 2 * 6 = 12

n=40 x=48 fd= -32 fd2 =94

fd2 d 2 94 -32 2 n n 40 40

= 1.71 = 1.3076 = 1.31

1.31 X 1.2

Team B is more consistent as it has less variation.

Ex (7):- The following table gives the age distribution of boys and girls in a high school.Find which of the tow groups is more variable in age.Age in Years: 13 14 15 16 17No.of Stu (boys): 12 15 15 5 3 (girls): 13 10 12 2 1Ans: Boys

x f fx d=x-A fd fd213 12 13*12=156 -2 12* -2 = -24 -2 * -24 = 4814 15 14*15=210 -1 15* -1 = -15 -1* -15 = 1515 15 15*15=225 0 15* 0 = 0 0 * 0 = 016 5 16*5=80 1 5* 1 = 5 1 * 5 = 517 3 17*3=51 2 3* 2 = 6 2 * 6 = 12

n=50 x=722 fd= -28 fd2 =80

fd2 d 2 80 -28 2 n n 50 50

= 1.2864 = 1.1341 = 1.134

1.134 X 14.44

Girlsx f fx d=x-A fd fd213 13 13*13=169 -2 13* -2 = -26 -2 * -26 = 5214 10 14*10=140 -1 10* -1 = -10 -1* -10 = 1015 12 15*12=180 0 12* 0 = 0 0 * 0 = 016 2 16*2=32 1 2* 1 = 2 1 * 2 = 217 1 17*1=17 2 1* 2 = 2 2 * 2 = 4

n=38 x=538 fd= -32 fd2 =68

fd2 d 2 68 -32 2 n n 38 38

= 1.0804 = 1.0394 = 1.04

1.04 X 14.16

Age of boys is more variable as it variation is more.

Ex (8):- An agent obtained samles of bulbs from the 2 companies. He had them tested fordurability and got the followig results.Durability in 00 hr: 17-19 19-21 21-23 23-25Company A: 100 160 260 80Company B: 30 420 120 30Which company bulbs are more uniform?

CV = * 100 = * 100 CV = *100 = * 100

fx 57 n 53

X = =

X = 1.0755 = 1.08

= - ( ) = - ( ) = 2.66 - ( -0.925) 2 = 2.66 - 0.8556

CV = * 100 = * 100 = 1.24074 * 100 = 124.07%

fx 48 n 40

X = =

X = 1.2

= - ( ) = - ( ) = 2.35 - ( -0.8) 2 = 2.35 - 0.64

CV = * 100 = * 100 = 1.08972 * 100 = 108.97%

fx 722 n 50

X = =

X = 14.44

= - ( ) = - ( ) = 1.6 - ( -0.56) 2 = 1.6 - 0.3136

CV = * 100 = * 100 = 0.0785 * 100 = 7.85%

fx 538 n 38

X = =

X = 14.1578 = 14.16

= - ( ) = - ( ) = 1.7895 - ( -0.8421) 2 = 1.7895- 0.7091

CV = * 100 = * 100 = 0.07340 * 100 = 7.34%

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Ans: A Company B Company

CI Xm d f fd fd2 f fd fd2

17-19 18 -1 100 -1*100= -100 -1* -100=100 30 -1*30=-30 -1*-30=3019-21 20 0 160 0*160= 0 0*0 = 0 420 0*420=0 0*0 = 021-23 22 1 260 1*260=260 1*260=260 120 1*120=120 1*120=12023-25 24 2 80 2*80=160 2*160=320 30 2*30=60 2*60=120

n=600 fd=320 fd2=680 n=600 fd=150 fd2=270

17+19 36 Xm- 20 18-20 -2 2 2 2 2 2

Similarly, all Xm and d values find from above table.

fd fd2 fd 2 n n n X

Put the values from above table and A = 20, c = 2

320 150600 600

= 20 + 0.5333 * 2 = 20 + 0.25 * 2 = 20 + 1.0667 = 21.0667 = 20 + 0.5 = 20.5 XA = 21.07 XB = 20.5

680 320 2 270 150 2600 600 600 600

A = 1.1333 - (0.5333)2 * 2 B = 0.45 - (0.25)

2 * 2

A = 1.1333 - 0.2844 * 2 B = 0.45 - 0.0625 * 2

A = 0.8489 * 2 = 0.9213 * 2 B = 0.3875 * 2 = 0.6225 * 2

A = 1.8426 = 1.85 B = 1.2449 = 1.25

A 1.85 B 1.25

XA 21.07 XB 20.5

CVA = 0.08780 * 100 = 8.78% CVB = 0.06097 * 100 = 6.097%

The Bulbs of B company are more Uniform and durable than the bulbs of A companyas the variation in B bulbs is less.

Ex (9):- A purchasing agent obtained samples of lamps from two suppliers A and B withthe following information.Length of the Life hr: 500-700 700-900 900-1100 1100-1300Supplier A: 10 16 30 8Supplier B: 3 42 12 4Which Suppliers lamps are more uniform?

Ans: A Supplier B Supplier

CI Xm d f fd fd2 f fd fd2

500-700 600 -1 10 -1*10= -10 -1* -10=10 3 -1*3=-3 -1*-3=3700-900 800 0 16 0*16= 0 0*0 = 0 42 0*42=0 0*0 = 0900-1100 1000 1 30 1*30=30 1*30=30 12 1*12=12 1*12=121100-1300 1200 2 8 2*8=16 2*16=32 4 2*4=8 2*8=16

n=64 fd=36 fd2=72 n=61 fd=17 fd2=31

500+700 1200 Xm- 800 600-800 -2002 2 200 200 200

Similarly, all Xm and d values find from above table.

fd fd2 fd 2 n n n X

Put the values from above table and A = 800, c = 200

36 1764 61

= 800 + 0.5625 * 200 = 800 + 0.2787 * 200 = 800 + 112.5 = 912.5 = 800 + 55.7377 = 855.7377 XA = 912.5 XB = 855.74

72 36 2 31 17 2 64 64 61 61

A = 1.125 - (0.5625)2 * 200 B = 0.5081 - (0.2787)

2 * 200

A = 1.125 - 0.31641 * 200 B = 0.5081 - 0.07767 * 200

A = 0.8086 * 200 = 0.8992 * 200 B = 0.43043 *200 = 0.6560 * 200

A = 179.8432 = 179.84 B = 131.2143 = 131.22

A 179.84 B 131.22

XA 912.5 XB 855.74

CVA = 0.197084 * 100 = 19.71% CVB = 0.15334 * 100 = 15.34%

B Suppliers lamps are more Uniform as their variation is less than the A Supplierslamps.

Ex (10):- An analysis of the monthly wages paid to workers in two firms A and B belong-ing to the same industries givs the following results: Firm A Firm B

No. of wage earners 586.00 648.00Average monthly wage Rs. 52.50 Rs. 47.50Variance of the distribution Rs. 100.00 Rs. 121.00

quantitative techniques

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