QDM Notes for Calculating Statistical Measures

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1 QDM notes for MBA Second Edition 2013 - 2014
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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 by
dividing the sum of the values of the items in a variable by their number. It is denoted by a
symbol 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 by
dividing the sum of the values by the number of items. Algebraically we can have the
formula as under:
a) For individual observations
x1
+x2
+x3
+.......+ xn
x
N N
b) For Discrete and Continuous Series
fx1
+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 method
Ex (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
x
N N
8+9+18+21+25+31 112
6 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
Index
01) Mean -- 01
02) Median -- 10
03) Mode -- 15
04) MeanDeviation -- 16
05) QuartileDeviation -- 19
06) StandardDevaition -- 21
07) Coefficient of Correlation -- 27
08) Rank Correlation Coefficient -- 33
09) Regression -- 35
10) Trend Value -- 40
11) Linear Programming : Graphic method -- 48
12) LPP : Transportation Method (NWM, LCM, VAM) -- 61
13) LPP : Assignment Problem -- 76
Important Tip
1. 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 & 13
3. 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 70
No. of Families 5 7 2 3 1 6 9
Ans :-
Income No. of Families
x f fx
10 5 10x5 = 50
20 7 20x7 = 140
30 2 30x2 = 60
40 3 40x3 = 120
50 1 50x1 = 50
60 6 60x6 = 360
70 9 70x9 = 630
N = f = 33 fx = 1410
fx 50+140+60+120+50+360+630 1410
N 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 method
f . Xm
N
Here, f = Frequency, Xm
= mid value, N = f
C.I. f Xm
f . Xm
16 - 17 5 16.5 16.5 x 5 = 82.5
14 - 15 4 14.5 14.5 x 4 = 58
12 - 13 4 12.5 12.5 x 4 = 50
10 - 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 the
arithmetic average, “The algebric sum of the devations of values of variable from their
mean is always to Zero.”
Let us assume the difference figure ‘A’ as assumed mean. Take deviations of
values of variable from the assumed mean denoted as ‘d’. Obtain the sum of these
devations denote as ’d’. Substitute the values of these symbols in the short-cut formula
as under.
d
N
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 d
32 x 32 32-37 = -5 N
37 N 37 37-37= 0 15
42 237 42 42-37= 5 6
47 6 47 47-37 = 10 = 37 + 2.5
52 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 = 42
27 27-32= -5 d 27-42= -15 d
32 32-32 = -0 N 32-42= -10 N
37 37-32= 5 45 37-42 = -5 (-15)
42 42-32= 10 6 42-42= 0 6
47 47-32 = 15 = 32 + 7.5 47-42= 5 X = 42 - 2.5
52 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 the
arithmetic mean answer is comes same. So did not confuse and assume any value as
assumed 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 70
7 11 31 17 16 5 3
x f d= x-A f.d
10 7 10-40= -30 7 * -30= -210
20 11 20-40= -20 11 * -20= -220
30 31 30-40= -10 31 * -10= -310
40 17 40-40= 0 17 * 0= 0
50 16 50-40= 10 16 * 10= 160
60 5 60-40= 20 5 * 20= 100
70 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.d
10 7 10-30= -20 7 * -20= -140
20 11 20-30= -10 11 * -10= -110
30 31 30-30= 0 31 * 00= 0
40 17 40-30= 10 17 * 10= 170
50 16 50-30= 20 16 * 20= 320
60 5 60-30= 30 5 * 30= 150
70 3 70-30= 40 3 * 40= 120
N=90 fd = -140-110+0+170+320+150+120 = 510
fd 510
N 90
(Tip :- From above example, it observed that you will solve any taking assumed mean
from given example and solve it and answer comes as same.)
Ex (3) :-
C.I. f Xm
d=Xm
- A f.d
16-17 5 (16+17)/2= 33/2 = 16.5 16.5-10.5= 6 5 * 6 = 30
14-15 4 (14+15)/2= 29/2 = 14.5 14.5-10.5= 4 4 * 4 = 16
12-13 4 (12+13)/2= 25/2 = 12.5 12.5-10.5= 2 4 * 2 = 8
10-11 8 (10+11)/2= 21/2 = 10.5 10.5-10.5= 0 8 * 0 = 0
8-9 5 (8+9)/2= 17/2 = 8.5 8.5-10.5= -2 5 * -2 = -10
6-7 4 (6+7)/2= 13/2 = 6.5 6.5-10.5= -4 4 * -4 = -16
4-5 2 (4+5)/2= 9/2 = 4.5 4.5-10.5= -6 2 * -6 = -12
N=32 fd = 16
fd 16
N 32
OR
C.I. f Xm
d=Xm
- A f.d
16-17 5 (16+17)/2= 33/2 = 16.5 16.5-12.5= 4 5 * 4 = 20
14-15 4 (14+15)/2= 29/2 = 14.5 14.5-12.5= 2 4 * 2 = 8
12-13 4 (12+13)/2= 25/2 = 12.5 12.5-12.5= 0 4 * 0 = 0
10-11 8 (10+11)/2= 21/2 = 10.5 10.5-12.5= -2 8 * -2 = -16
8-9 5 (8+9)/2= 17/2 = 8.5 8.5-12.5= -4 5 * -4 = -20
6-7 4 (6+7)/2= 13/2 = 6.5 6.5-12.5= -6 4 * -6 = -24
4-5 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 single
common factor to reduce the figures to the minimum size. It is, in a way, a deliberate error
and it is compensated by multiplying the quotient. The adjusted or reduced value of “d ” in
now 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 and
multiplying)
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 fd’
90-99 2 94.5 (94.5-54.5)/10= 4 2*4= 8
80-89 9 84.5 (84.5-54.5)/10= 3 9*3= 27
70-79 12 74.5 (74.5-54.5)/10= 2 12*2= 24
60-69 17 64.5 (64.5-54.5)/10= 1 17*1= 17
50-59 20 54.5 (54.5-54.5)/10= 0 20*0= 0
40-49 9 44.5 (44.5-54.5)/10= -1 9*-1= -9
30-39 6 34.5 (34.5-54.5)/10= -2 6*-2= -12
20-29 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 30
10 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 ‘C’
A: Range 90-99, the number
start from 0 to 9, it means
0,1,2,3,4,5,6,7,8,9 = 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
40
80
OR
C.I. f Xm
d’= (Xm
-A)/c fd’
90-99 2 94.5 (94.5-44.5)/10= 5 2 * 5 = 10
80-89 9 84.5 (84.5-44.5)/10= 4 9 * 4 = 36
70-79 12 74.5 (74.5-44.5)/10= 3 12 * 3 = 36
60-69 17 64.5 (64.5-44.5)/10= 2 17 * 2 = 34
50-59 20 54.5 (54.5-44.5)/10= 1 20 * 1 = 20
40-49 9 44.5 (44.5-44.5)/10= 0 9 * 0 = 0
30-39 6 34.5 (34.5-44.5)/10= -1 6 * -1 = -6
20-29 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 fd’
42-46 4 44 (44-29)/5= 3 4 * 3 = 12
37-41 2 39 (39-29)/5= 2 2 * 2 = 4
32-36 7 34 (34-29)/5= 1 7 * 1 = 7
27-31 8 29 (29-29)/5= 0 8 * 0 = 0
22-26 4 24 (24-29)/5= -1 4 * -1 = -4
17-21 5 19 (19-29)/5= -2 5 * -2 = -10
12-16 3 14 (14-29)/5= -3 3 * -3 = -9
7-11 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 fd’
42-46 4 44 (44-24)/5= 4 4 * 4 = 16
37-41 2 39 (39-24)/5= 3 2 * 3 = 6
32-36 7 34 (34-24)/5= 2 7 * 2 = 14
27-31 8 29 (29-24)/5= 1 8 * 1 = 8
22-26 4 24 (24-24)/5= 0 4 * 0 = 0
17-21 5 19 (19-24)/5= -1 5 * -1 = -5
12-16 3 14 (14-24)/5= -2 3 * -2 = -6
7-11 2 9 (9-24)/5= -3 2 * -3 = -6
N=35 fd’ = 27
27
35
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 fd’
14-15 3 14.5 (14.5-8.5)/2= 3 3 * 3 = 9
12-13 8 12.5 (12.5-8.5)/2= 2 8 * 2 = 16
10-11 15 10.5 (10.5-8.5)/2= 1 15 * 1 = 15
8-9 20 8.5 (8.5-8.5)/2= 0 20 * 0 = 0
6-7 10 6.5 (6.5-8.5)/2= -1 10 * -1 = -10
4-5 4 4.5 (4.5-8.5)/2= -2 4 * -2 = -8
N=60 fd’ = 22
22
60
OR
C.I. f Xm
d’= (Xm
-A)/c fd’
14-15 3 14.5 (14.5-10.5)/2= 2 3 * 2 = 6
12-13 8 12.5 (12.5-10.5)/2= 1 8 * 1 = 8
10-11 15 10.5 (10.5-10.5)/2= 0 15 * 0 = 0
8-9 20 8.5 (8.5-10.5)/2= -1 20 * -1 = -20
6-7 10 6.5 (6.5-10.5)/2= -2 10 * -2 = -20
4-5 4 4.5 (4.5-10.5)/2= -3 4 * -3 = -12
N=60 fd’ = -38
(-38)
60
X = A + x c
fd’
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
2,3,4,5,6 = 5 intervals
value is 5
X = A + x c
fd’
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
4, 5 = 2 intervals
value is 2
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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 20
f: 3 7 (?) 20 8 5
Soln
:- Let us assume the missing frequency as ‘A’
x f fx
10 3 10*3 = 30
12 7 12*7 = 84
14 (A) 14*A = 14A
16 20 16*20 = 320
18 8 16*8 = 144
20 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 + 14A
655.75 + 15.25A = 678 + 14A
15.25A - 14A = 678 - 655.75
1.25A = 22.25
22.25
1.25
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 90 100 110
No.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 = 25
A + B = 25 - 13
A + B = 12, B = 12 - A
x f fx
50 1 50*1 = 50
60 3 60*3 = 180
70 (A) 70*A = 70 A
80 12-A 80*(12-A) = 960-80 A
90 6 90*6 = 540
100 2 100*2 = 200
110 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 - 5
A = 50/10 = 5 B = 7
A = 5
So, 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 (?) 125 128 130
No.of workers (f): 25 17 13 15 14 8 6 2
Soln
:- Let us assume the missing value a ‘A’.
x f fx
110 25 110*25 = 2750
112 17 112*17 = 1904
113 13 113*13= 1469
117 15 117*15 = 1755
A 14 A*14 = 14A
125 8 125*8 = 1000
128 6 128*6 = 768
130 2 130*2 = 260
N=100 fx = 9906 + 14A
So, missing value is 120.
Ex (7):- You are given the following incomplete information and its mean 25. find out the
missing frequencies.
Class interval (x): 0-10 10-20 20-30 30-40 40-50 Total
Frequencies (f): 5 - 15 - 5 = 45
Soln
:- Assume that the frequencies are ‘A’ and ‘B’. Total frequencies are 45.
So,
5+A+15+B+5= 45
A + B + 25 = 45
A + B = 45 - 25
A + B = 20
B = 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 = 50
A = 50/10 = 5
A = 5
79.60 =
79.60 =
X =
fx
N
9906 - 14A
100
115.86*100 = 9906 - 14A
11586 = 9906 - 14A
14A = 11586 - 9906
14A = 1680
1680
14
A = 120
115.86 =
A = = 120
x f Xm
fx
00-10 5 (10+0)/2=10/2=5 5 * 5 = 25
10-20 A (10+20)/2=30/2=15 A * 15 = 15A
20-30 15 (20+30)/2=50/2=25 15 * 25 = 375
30-40 20-A (30+40)/2=70/2=35 (20-A)*35 = 700-35 A
40-50 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 = 20
1125 = 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
fx
00-10 5 (10+0)/2=10/2=5 5 * 5 = 25
10-20 A (10+20)/2=30/2=15 A * 15 = 15A N=5+A+15+B+5
20-30 15 (20+30)/2=50/2=25 15 * 25 = 375 45 = 25 + A + B
30-40 B (30+40)/2=70/2=35 B*35 = 35 B 45 - 25 = A + B
40-50 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) 20A = 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 method
and 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 Total
No of Student (f): 5 18 - - 8 = 100
Soln
:- Assume that the frequencies are ‘A’ and ‘B’. Total frequencies are 100.
C.I. f Xm
fx
60-62 5 (60+62)/2=122/2=61 5 * 61 = 305
63-65 18 (63+65)/2=128/2=64 18 * 64 = 1152 N=5+18+A+B+8
66-68 A (66+68)/2=134/2=67 A * 67 = 67A 100 = 31 + A + B
69-71 B (69+71)/2=140/2=70 B * 70 = 70B 100 - 31 = A + B
72-74 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) 3A = 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 Total
Frequency (f): 2 4 5 10 6 9 7 = 43
Soln
:-
C.I. f Xm
d’= (Xm
-A)/c fd’
60-69 2 64.5 (64.5-34.5)/10= 3 2 * 3 = 6
50-59 4 54.5 (54.5-34.5)/10= 2 4 * 2 = 8
40-49 5 44.5 (44.5-34.5)/10= 1 5 * 1 = 5
30-39 10 34.5 (34.5-34.5)/10= 0 10 * 0 = 0
20-29 6 24.5 (24.5-34.5)/10= -1 6 * -1 = -6
10-19 9 14.5 (14.5-34.5)/10= -2 9 * -2 = -18
0-9 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 =
0,1,2,3,4,5,6,7,8,9 = 10 intervals
value is 10


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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 fd’
10-19 5 14.5 (14.5-54.5)/10= -4 5 * -4 = -20
20-29 6 24.5 (24.5-54.5)/10= -3 6 * -3 = -18
30-39 7 34.5 (34.5-54.5)/10= -2 7 * -2 = -14
40-49 8 44.5 (44.5-54.5)/10= -1 8 * -1 = -8
50-59 11 54.5 (54.5-54.5)/10= 0 11 * 0 = 0
60-69 6 64.5 (64.5-54.5)/10= 1 6 * 1 = 6
70-79 4 74.5 (74.5-54.5)/10= 2 4 * 2 = 8
80-89 2 84.5 (84.5-54.5)/10= 3 2 * 3 = 6
90-99 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 1-20
1 3 9 20 8 5 4
C.I. f Xm
d’= (Xm
-A)/c fd’
1-20 4 10.5 (10.5-70.5)/20= -3 4 * -3 = -12
21-40 5 30.5 (30.5-70.5)/20= -2 5 * -2 = -10
41-60 8 50.5 (50.5-70.5)/20= -1 8 * -1 = -8
61-80 20 70.5 (70.5-70.5)/20= 0 20 * 0 = 0
81-100 9 90.5 (90.5-70.5)/20= 1 9 * 1 = 9
101-120 3 110.5 (110.5-70.5)/20= 2 3 * 2 = 6
121-140 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 methods
a) 86, 70, 96, 93, 97, 94
Ans:- Short cut Method
x direct method x d=x - A Assumed Mean (A) = 93
70 70 70-93 = -23 d
86 x 86 86-93 = -7 N
93 N 93 93-93 = 0 -22
94 536 94 94-93 = 1 6
96 6 96 96-93 = 3 = 93 - 3.6667
97 X = 89.33 97 97-93 = 4 X = 89.33
x = 536 d = 22
(Tip:- Observe the above example, by using both the method i.e. “Direct” as will as
“Short cut” method, the answer will be same i.e. 89.33 , So student can be solve the
example any method.)
b) x: 5 10 15 20 25 30
f: 4 5 9 15 10 7
Ans:-
Direct Method Short cut Method
x f fx x f d= x-A f.d
5 4 5*4=20 5 4 5-20=-15 4* -15= -60
10 5 10*5=50 10 5 10-20=-10 5* -10= -50
15 9 15*9=135 15 9 15-20=-5 9* -5= -45
20 15 20*15=300 20 15 20-20=0 15* 0=0
25 10 25*10=250 25 10 25-20=5 10* 5=50
30 7 30*7=210 30 7 30-20=10 7* 10=70
N=f =50 fx = 965 N=50 fd = -35
fx fd
N N
965 -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 as
“Short cut” method, the answer will be same i.e. 19.3 , So student can be solve the
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 1
Ans:-
Above example solve by both the methods i.e. direct as will as short cut method.
X = A + x c
fd’
N
X = 34.5 + x 10 = 34.5 + (-0.6046 * 10) = 34.5 - 6.0465 =28.4534 = 28.45
X = A + x c
fd’
N
X = 54.5 + x 10 = 54.5 + (-0.72 * 10) = 54.5 - 7.2 = 47.30
X = A + x c
fd’
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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C.I. f Xm
f . Xm
x f Xm
d=Xm
- A f.d
80-84 2 82 164 80-84 2 82 20 2*20=40
75-79 0 77 0 75-79 0 77 15 0*15=0
70-74 10 72 720 70-74 10 72 10 10*10=100
65-69 17 67 1139 65-69 17 67 5 17*5=85
60-64 24 62 1488 60-64 24 62 0 24*0=0
55-59 19 57 1083 55-59 19 57 -5 19 * -5= -95
50-54 7 52 364 50-54 7 52 -10 7 * -10= -70
45-49 4 47 188 45-49 4 47 -15 4 * -15= -60
40-44 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 compare
the both method. If any doubt to how the values are calculate, then see the before example
to check.)
fXm
fd
N N
5188 (-20)
84 84
X = 61.76 X = 62 - 0.2380 = 61.7619
X = 61.76 X = 61.76
ByUsing Step Deviation Method
C.I. f Xm
d’= (Xm
-A)/c fd’
80-84 2 82 (82-62)/5 = 4 2 * 4 = 8
75-79 0 77 (77-62)/5 = 3 0 * 3 = 0
70-74 10 72 (72-62)/5 = 2 10 * 2 = 20
65-69 17 67 (67-62)/5 = 1 17 * 1 = 17
60-64 24 62 (62-62)/5 = 0 24 * 0 = 0
55-59 19 57 (57-62)/5 = -1 19 * -1 = -19
50-54 7 52 (52-62)/5 = -2 7 * -2 = -14
45-49 4 47 (47-62)/5 = -3 4 * -3 = -12
40-44 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 Step
Deviation” method, the answer will be same i.e. 61.76 , So student can be solve the
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 4
Ans:-
C.I. f Xm
f . Xm
x f Xm
d=Xm
- A f.d
18-19 3 18.5 55.5 18-19 3 18.5 6 3*6 = 18
16-17 8 16.5 132 16-17 8 16.5 4 8*4 = 32
14-15 15 14.5 217.5 14-15 15 14.5 2 15*2 = 30
12-13 20 12.5 250 12-13 20 12.5 0 20*0 = 0
10-11 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 compare
the both method. If any doubt to how the values are calculate, then see the before example
to check.)
fXm
fd
N N
794 44
60 60
X = 13.2333 X = 12.5 + 0.7333 = 13.2333
X = 13.23 X = 13.23
ByUsing Step Deviation Method
C.I f Xm
d’= (Xm
-A)/c fd’
18-19 3 18.5 (18.5-14.5)/2 = 2 3 * 2 = 6
16-17 8 16.5 (16.5-14.5)/2 = 1 8 * 1 = 8
14-15 15 14.5 (14.5-14.5)/2 = 0 15 * 0 = 0
12-13 20 12.5 (12.5-14.5)/2 = -1 20 * -1 = -20
10-11 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 Step
Deviation” method, the answer will be same i.e. 13.23 , So student can be solve the
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 c
fd’
N
X = A + x c
fd’
N
Formula
Formula

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Ex (13): - From the follwoing information relating to a certain industry consisting of 32
firms, calculate the average salary paid in the whole industry.
Income Group: 50-75 75-100 100-150 150-200 200-300
No. of Firms: 30 26 24 24 28
Average No. of
Workers : 5 8 5.5 6.5 1
Income
Group
C.I. f Xm
f . Xm
50-75 30*5 160 (50+75)/2=125/2=62.5 160*62.5 = 10000
75-100 26*8 208 (75+100)/2=175/2=87.5 208*87.5 = 18200
100-150 24*5.5 132 (100+150)/2=250/2=125 132*125 = 16500
150-200 24*6.5 130 (150+200)/2=350/2=175 130*175 = 22750
200-300 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 of
equal importance. This may not always be the case. When items vary in importance they
must be assigned some weights in proportion to their importance. The value of each item
is then multiplied by its weight. The products are summated and divided by the number of
weights, and not by the number of items. the quotient is the weighted arithmetic mean or
average.
Symbolically,
WX
W
‘Weight’ are numbers or percentages which stand for the relative importance of
items. 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 June
Price per Ton (Rs) 42.50 51.25 50.00 52.00 44.25 54.00
Tons Purchased 25 30 40 50 10 45
Ans :-
Prices Per Tons
Ton (Rs) Purchased
C.I. W Wx
42.50 25 42.50 * 25 = 1062.50
51.25 30 51.25 * 30 = 1537.50
50.00 40 50.00 * 40 = 2000.00
52.00 50 52.00 * 50 = 2600.00
44.25 10 44.25 * 10 = 0442.50
54.00 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 Y
A 1 70 80
B 2 65 64
C 3 58 56
D 4 63 60
Find the Weighted average marks of each candidate.
Ans:-
Marks by
Subjects Weights(W) X Y WX WY
A 1 70 80 1 * 70 = 70 1 * 80 = 80
B 2 65 64 2 * 65 = 130 2 * 64 = 128
C 3 58 56 3 * 58 = 174 3 * 56 = 168
D 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 arithmetic
mean for Y Candidate is 61.60
X = = = 113.1458
XW
= , WX = X * W , Where W = Weight
Wx
W
10072.50
200
Xw
= 50.3625
Xw
= 50.36
Xw
=
Xw
=
Wx
W
626
10
Xw
= 62.60
Xw
= 62.60
Xw
=
Xw
=
Wx
W
616
10
Xw
= 61.60
Xw
= 61.60
Xw
=
Xw
=

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Median :- “Median” is the value of that item in a series which divides the array into two
equal parts, one condisting of all the values less than it and the other consisting of all the
values more than it. It means, when the values of variable are arranged in an array form (in
an ascending or descending order of their magnitude), the value of middle item of the
array is the median and it is denote as Me.
“Me is the size of th
item” where ‘Me’ represents the median and ‘N’ the number
of items
(A) Individual Observation
Ex (1) :- Find the median value of the following data
X: 7, 12, 37, 32, 17, 22, 27 (Note: Arrange the data in increasing
Ans:- (ascending) order)
7 12 17 22 27 32 37 X
1 2 3 4 5 6 7 N
N + 1
2
7 + 1 8
2 2
Median (Me) = 22
Ex (1) :- Find the median value of the following data
X: 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
2
6 + 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 79
2 2 2
Median (Me) = 39.5
(B) Discrete Series
In discrete series, the value are already in the form of array and the frequencies
are recorded against each value.
However, for determining the size of item, a separate column is to be prepared
for cumulative frequencies (cf). The median size is first located with reference to the
cumulative 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, 80
f : 2 10 17 24 19 7 4 1
Ans:- Arrange the data in an array form with serial numbers (ascending order).
x f cf
60 10 10
62 2 2+10 =12
66 17 17+12 = 29
67 4 4+29 = 33
72 7 7+33 = 40
75 19 19+40 = 59
78 24 24+59 = 83
80 1 1+83 = 84
N=84
In above example Me = 42.5th
item, it means that find such c.f. value which is greater than
find th
item i.e. c.f. value. Here, we find 42.5th
item and from our table value 59 is greater
than 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, 75
f : 4 5 6 7 4 3 2 3 1
Ans: - Arrange the data in an array form with serial numbers (ascending order).
x f cf
60 4 4
61 3 3+4 = 7
63 3 3+7 = 10
66 5 5+10 = 15
69 2 2+15 = 17
71 4 4+17 = 21
74 7 7+21 = 28
75 1 1+28 = 29
78 6 6+29 = 35
N=35
In above example Me = 18th
item, it means that find such c.f. value which is greater than
find th
item i.e. c.f. value. Here, we find 18th
item and from our table value 21 is greater
than 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
= th item
=  4 th item
= 22
Median (Me) = = = = 39.5
Median (Me) = th item
= th item
=  3.5 th item
N + 1
2

Median is the size of th
item
84 + 1
2
It lies in 59 c.f.
Against 59 c.f. the value is 75
So, Median Marks is 75
(Note: Median value is identified by locating
the respective c.f. in which item falls.)
N + 1
2
Me = )th
= 42.5 th
item
)thN + 1
2

Median is the size of th
item
35 + 1
2
It lies in 21 c.f.
Against 21 c.f. the value is 71
So, Median Marks is 71
(Note: Median value is identified by locating
the respective c.f. in which item falls.)
N + 1
2
Me = )th
= 18 th
item
)thN + 1
2


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(c) Continous Series
A 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 item
and median class is locaed accordingly with reference to the cumulative frequency, which
covers 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 = 38
37.5-41.5 1 36 + 1 = 37
33.5-37.5 1 35 + 1 = 36
29.5-33.5 0 35 + 0 = 35
25.5-29.5 5 30 + 5 = 35
21.5-25.5 4 26 + 4 = 30
17.5-21.5 8 18 + 8 = 26
13.5-17.5 7 11 + 7 = 18
9.5-13.5 9 2 + 9 = 11
5.5-9.5 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 = 18
Median of Class interval is 18
Ex (2) C.I. : 10-15 15-20 20-25 25-30 30-35 35-40
f: 6 18 9 10 4 3
Ans:- Above series is already exclusive class interval series.
C.I. f c.f.
10-15 6 6
15-20 18 6 + 18 = 24
20-25 9 9 + 24 = 33
25-30 10 10 + 33 = 43
30-35 4 4 + 44 = 47
35-40 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)
25 - 24 1
9 9
Me = 20 + 0.1111 x 5 = 20 + 0.5555 = 20.5555
Me = 20.56
Median of Class interval is 20.56
Ex Marks C.I. : 20-40 40-60 60-80 80-100 100-120 120-140 140-160
No. 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.f
20-40 4 Less than 40 4 More than 20 4+54=58
40-60 6 Less than 60 4+6=10 More than 40 6+48=54
140-160 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 interval
is in the form of Less than or More than.
Median (Me) = L + ( ) x c
N/2 - c.f.
f
Note: How to identified the exclusive class
interval 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 value
is 4, but when lower class substract from
upper class then it find the value 3 (9 - 6 = 3).
then this class is an inclusive.
So, convert an inclusive class into exclusive
class interval series. That is 5.5 - 9.5.
Then 9.5 - 5.5 = 4. It is correct class interval.
N/2 - c.f.
f
Me = 17.5 +  ) x 4 = 17.5 +  ) x 4
N/2 - c.f.
f
Note:
Lower interval (L1
) = 10
Upper interval (L2
) = 15
c = L2
- L1
c = 15 - 10
c = 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 70
frequency: 2 6 11 21 27 36 43
Ans:- 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 2
20 6 10 - 20 6-2=4 6
30 11 20 - 30 11-6=5 11
40 21 30 - 40 21-11=10 21
50 27 40 - 50 27-21=6 27
60 36 50 - 60 36-27=9 36
70 43 60 - 70 43-36=7 43
N=43
Here, N = 43 then N/2 = 43/2 = 21.5th
item.
 L = 40, f = 6, c.f. = 21 and c = 10
21.5 - 21 0.5
6 6
Me = 40 + 0.08333 x 10 = 40 + 0.8333 = 40.8333
Me = 40.83
Less than : 10 20 30 40 50 60 70
frequency: 50 47 41 29 21 7 2
Ans:- 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 50
20 47 20 - 30 47-41=6 47
30 41 30 - 40 41-29=12 41
40 29 40 - 50 29-21=8 29
50 21 50 - 60 21-7=14 21
60 7 60 - 70 7-2=5 7
70 2 70 - 80 2 2
N=50
Here, N = 50 then N/2 = 50/2 = 25th
item.
 L = 40, f = 8, c.f. = 21 and c = 10
25 - 21 4
8 8
Me = 40 + 0.5 x 10 = 40 + 5 = 45
Me = 45
Ex (5):- In a group of 1000 wage earners, the monthly wages of 4% are below Rs. 60 and
those 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 it
into the original frequencies.
In problem Monthly C.I. s.f. F(%) f
given wages (Rs.)
below 60 Less than 60 0 - 60 4% 4% 40
under 62.50 Less than 62.50 60 - 62.5 15% 15-4=11% 110
62.50 - 95 62.50 - 95 62.5 - 95 Diff 100-30=70% 700
over 95 More than 95 95 - 100 15% 15-5=10% 100
100 & over 100 & over 100 & over 5% 5% 50
Note :- 100 = 4 + 11 + Diff + 10 + 5 = Diff + 30
100 - 30 = Diff
Diff = 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 below
C.I. f c.f.
0-60 40 40
60-62.50 110 40+110=150
62.5-95 700 110+150=850
95-100 100 850+100=950
100 & over 50 950+50=1000
N=1000
After Converting


N/2 - c.f.
f
Me = 40 +  ) x 10 = 40 +  ) x 10
N/2 - c.f.
f
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 - 95
So, L = 62.5, f = 700, c.f. = 150 and c = 95 - 62.5 = 32.5
500-150 350
700 700
Me = 62.5 + 0.5 x 32.5 = 62.5 + 16.25 = 78.75
Me = 78.75
Ex (6):- 10 percent of the workers in a firm, employing a total of 1000 workers, earn
between Rs. 5 and 9.99, 30 percent between Rs. 10 and Rs. 14.99, 250 workers between
Rs. 15 and Rs. 19.99 and the rest Rs. 20 and above. What is the median wage?
Inclusive Class Interval
Wages Rs. (C.I.) No. of workers f
5 - 9.99 10% 100
10 - 14.99 30% 300
15 - 19.99 250 250
20 & above (Diff) (Diff)
N=1000
After converting, the Exclusive Class Interval is given below
C.I. f c.f.
4.995 - 9.995 100 100
9.995 - 14.995 300 100+300 = 400
14.995 - 19.995 250 250+400 = 650
19.995 & above 350 650+350 = 1000
N=1000
Here, 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 = 5
500-400 150
250 250
Me = 14.995 + 0.4 x 5 = 14.995 + 2 = 16.995 = 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 60-70
Frequencies: 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 50-60
20 - 30 8 2+8 = 10 L = 50, f = 15, c = 10 &
30 - 40 6 6+10 = 16 c.f. = 16 + A
40 - 50 A 16+A = 16 + A Now put the values in formula
50 - 60 15 15+16+A = 31+A
60 - 70 10 10+31+A = 41+A
N=41+A
41+A 41+A
Me = 2 , 50 = 2
15 15
41+A 41*10 + 10A
50 - 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 70-80
Frequencies: 12 30 --- 65 --- 25 18
Complete 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 12
20 - 30 30 30+12 = 42
30 - 40 A A+42 = 42 + A
40 - 50 65 65+42+A = 107 + A
50 - 60 B B+107+A = 107+A+B
60 - 70 25 25+107+A+B = 132+A+B
70-80 18 18+132+A+B = 150+A+B
N=229
N/2 - c.f.
f
Me = 62.5 + ) x 32.5 = 62.5 +  ) x 32.5
1000=100+300+250+Diff
1000=650+Diff
Diff=1000-650
Diff=350
Me =14.995+ ) x 5 = 14.995 +  ) x 5
N/2 - c.f.
f
( ) - ( 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+18
229 = 150 + A + B
229 - 150 = A + B
79 = A + B
A + 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 229
Me = 2 , 46 = 2
f 65
114.5 - (42+A) , 114.5 - 42 - A
65 65
72.5 - A , 72.5 * 10 - 10 A
65 65
6 * 65 = 725 - 10 A Put, A = 33.5 on equation A + B = 79
390 = 725 - 10 A 33.5 + B = 79
10A = 725 - 390 B = 79 - 33.5
10 A = 335 B = 45.5
A = 335 / 10
A = 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.
Height : 5.1-6.0 6.1-7.0 7.1-8.0 8.1-9.0 9.1-10 10.1-11 11.1-12
No.of Plants: 3 8 27 25 17 11 9
C.I. C.I. f c.f.
5.1-6.0 5.05-6.05 3 3
6.1-7.0 6.05-7.05 8 8+3 = 11
7.1-8.0 7.05-8.05 27 27+11 = 38
8.1-9.0 8.05-9.05 25 25+38 = 63
9.1-10.0 9.05-10.05 17 17+63 = 80
10.1-11.0 10.05-11.05 11 11+80 = 91
11.1-12.0 11.05-12.05 9 9+91 = 100
N=100
N / 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 = 8.53
C.I. : 90-100 80-90 70-80 60-70 50-60 40-50 30-40 20-30
f : 2 9 12 17 20 9 6 5
Ans:
C.I. f c.f.
90-100 2 2+78 = 80
80-90 9 9+69 = 78
70-80 12 12+57 = 69
60-70 17 17+40 = 57
50-60 20 20+20 = 40
40-50 9 9+11 = 20
30-40 6 6+5 = 11
20-30 5 5
N=80
C.I. : 35-40 30-35 25-30 20-25 15-20 10-15 5-10
f : 1 5 12 15 10 4 3
Ans:
C.I. f c.f.
35-40 1 1+49 = 50
30-35 5 5+44 = 49
25-30 12 12+32 = 44
20-25 15 15+17 = 32
15-20 10 10+7 = 17
10-15 4 4+3 = 7
5-10 3 3
N=50
(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 of
table 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
20
20
20
Me = 50 + 1 x 10
Me = 50 + 10
Me = 60
Me = 50 + ( ) x 10
Me = 50 + ( ) x 10
N / 2 = 50 / 2 = 25
 L = 20 , f = 15, c.f. = 17, c = 5
25-17
15
8
15
Me = 20 + 0.5333 x 5
Me = 20 + 2.6666
Me = 22.6666
Me = 20 + ( ) x 5
Me = 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 in
the 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) order
7, 8, 9, 12, 13, 13, 13, 15
The mode is ‘13’ as it occurs the highest number of times (i.e. 3 times) in the
series. So the mode is clearly defined.
When the number of items increase, we may convert the data into a descrete
form. 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) order
61, 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 descrete
form. 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 of
highest frequencies or grouping or inspection (maxium concentration of frequencies).
The same procdure is adopted as it is followed in case discret series. After locating the
modal class, we have to interpolate the value of the mode within the modal class by using
the 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 20-30
f : 2 9 12 17 20 9 6 5
Ans:
C.I. f c.f.
90-100 2 2+78 = 80
80-90 9 9+69 = 78
70-80 12 12+57 = 69
60-70 17 17+40 = 57
50-60 20 20+20 = 40
40-50 9 9+11 = 20
30-40 6 6+5 = 11
20-30 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 5-10
f : 1 5 12 15 10 4 3
Ans:
C.I. f c.f.
5-10 3 3
10-15 4 4+3 = 7
15-20 10 10+7 = 17
20-25 15 15+17 = 32
25-30 12 12+32 = 44
30-35 5 5+44 = 49
35-40 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 of
highest frequencies. i.e. 20 and modal class
is (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 of
highest frequencies. i.e. 15 and modal class
is (20-25)
 f0
= 10, f1
= 15, f2
= 12, L = 20, c = 5
f1
- f0
(f1
- f0
) - (f2
- f1
)
Z = L + x c
f0
f1
f2
f2
f1
f0
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 taken
into consideration. It is denoted symbolically by ‘R’.
(i) Range = Largest value - Smallest value
R = L - S
(ii) Coefficient of Range (R) = =
Ex (1):- Compute the range and the Coefficient of Range of the series, and state which one
is 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, 29
Ans: (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 - S
L+ S L+ S L+ S
17 - 13 21 - 9 29 - 1
17+ 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 the
entire distribution into four equal parts.
First Quartile (Lower Quartile) Q1
is the value below which there are one fourth of the items
and 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 the
items 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 not
at all included in the computation. Thus we are considering only the middle half portion of
the distribution. The range so obtained is divided by two as we are considering only half of
the data. Thus the Quartil deviation measures the difference between the value of Q1
and
Q3
. 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 X
1 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 - R
Largest 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
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
- Q1
Q3
+ Q1

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8 24
4 4
Q1
= ( 2 )th
item Q3
= ( 6 )th
item
Q1
= 3 Q3
= 27
Q3
- Q1
27 - 3 24
2 2 2
Quartile Deviation (Q.D.) = 12
Q3
- Q1
27 - 3 24
Q3
+ 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 X
1 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
item
Q1
= ( 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 + 1.5
Q1
= 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 23.25
Q3
+ Q1
31.5+ 8.25 39.75
(3):- 3, 10, 12, 5 Arrange increasing (ascending) order
Ans:- 3 5 10 12 X
1 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
item
Q1
= ( 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 + 1.5
Q1
= 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 8
Q3
+ 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 X
1 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
Q1
= ( )th
item , Q3
= ( )th
item
Coffiecient of Q.D. = = = = 0.8
Q.D. = = = = 11.625
Q.D. = , Q1
= ( )th
item , Q3
= ( )th
item
Q1
= ( )th
item , Q3
= ( )th
item
Q1
= ( )th
item , Q3
= ( )th
item
Q1
= ( )th
item , Q3
= ( )th
item
Q.D. = = = = 11.625
Q.D. = , Q1
= ( )th
item , Q3
= ( )th
item
Q1
= ( )th
item , Q3
= ( )th
item
Q1
= ( )th
item , Q3
= ( )th
item
Q1
= ( )th
item , Q3
= ( )th
item
Q.D. = , Q1
= ( )th
item , Q3
= ( )th
item
Q1
= ( )th
item , Q3
= ( )th
item
Q1
= ( )th
item , Q3
= ( )th
item
Q1
= ( )th
item , Q3
= ( )th
item

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Q1
= ( 3 )th
item , Q3
= ( 9 )th
item
Q1
= 10 , Q3
= 49
Q3
- Q1
49 - 10 39
2 2 2
Q3
- Q1
49 - 10 39
Q3
+ 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= 43
50-59 49.5-59.5 4 4+37 = 41
40-49 39.5-49.5 5 5+32 = 37
30-39 29.5-39.5 10 10+22 = 32
20-29 19.5-29.5 6 6+16 = 22
10-19 9.5-19.5 9 9+7 = 16
0-9 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 26.33
Q3
+ 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 2
12-16 11.5-16.5 3 3+2 = 5
17-21 16.5-21.5 5 5+5 = 10
22-26 21.5-26.5 4 4+10 = 14
27-31 26.5-31.5 8 8+14 = 22
32-36 31.5-36.5 7 7+22 = 29
37-41 36.5-41.5 2 2+29 = 31
42-46 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 14.29
Q3
+ Q1
34.54+20.25 54.79
Q.D. = = = = 19.5
After Converting


Quartile Deviation(QD) = Q1
= L + ( ) x c & Q3
= L + ( ) x c
3N/4-cf
f
N/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 c
3N/4-cf
f
N/4-cf
f
Q3
- 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 the
values are despersed about the mean or the median or the mode. It is found by the
averaging all the deviations from the central tendency. These deviations are taken into
computations with regard to negative sign (i.e. all the deviations assumed as possitive).
In aggregating the devations, the algebraic negative signs are not taken into
account. It means all the devations are treated as positive ignoring the negative sign.
Mean deviation or average deviation is denoted symbolically by the Greak small
alphabet ‘’ (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 tedious
job. So, to make the things more simplified, a shor-cut method or formula is used as
under:
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 than
average (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 average
fB : the sum of the frequencies smaller than average
Ex (1):- From the following variables find the Mean Deviation and Coefficient of Mean
Deviation from the mean. ====> 10, 18, 3, 9, 6, 4
Ans: Arrange the data in an ascending order to have the short cut method applicable.
x d=x-X
3 3-8.34= 5.34 x 50
4 4-8.34= 4.34 n 6
6 6-8.34= 2.34 d 24
9 9-8.34= 0.66 n 6
10 10-8.34= 1.66  4
18 18-8.34= 9.66 x 8.34
x=50,n=6 d = 24
x Put all these values in formula
3
4 B=3 xB = 3+4+6=13 xA -xB - (A - B) X
6 B n
n=6, X = 8.34 37 - 13 - (3 - 3) * 8.34
9 6
10 A=3 xA = 9+10+18=37 24 - ( 0 ) * 8.34
18 A 6
24
6
From above observation, the value of Mean Deviation  = 4, So student can solve the
problem both the formula, but use easy one.
Ex (2):- From the following variables find the Mean Deviation and Coefficient of Mean
Deviation from the mean. ====> 15, 4, 3, 10, 12, 5, 1, 2, 6, 12
x d=x-X
1 1-7= 6 x 70
2 2-7= 5 n 10
3 3-7= 4 d 42
4 4-7= 3 n 10
5 5-7= 2  4.2
6 6-7= 1 x 7
10 10-7= 3
12 12-7= 5
12 12-7= 5
15 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 80 100 120 140 160 180
No of Batsmen: 6 19 40 23 65 83 55 20 9
Compute the Mean Deviation from mode and its Coefficient.
x f d=x-z fd
20 6 20-120=100 6*100=600 From the defination of mode
40 19 40-120=80 19*80=1520 The highest frequency is 83
60 40 60-120=60 40*60=2400  z = 120
80 23 80-120=40 23*40=920 fd 9180
100 65 100-120=20 65*20=1300 n 320
120 83 120-120= 0 83*0= 0
140 55 140-120=20 55*20=1100
160 20 160-120=40 20*40=800  28.69
180 9 180-120=60 9*60=540 z 120
n=6 fd = 9180
x f fx
20 6 20*6=120 B=6+19+40+23+65=153
40 19 40*19=760 xB = 120+760+2400+1840+6500=11620
60 40 60*40=2400
80 23 80*23=1840 A=83+55+20+9=167
100 65 100*65=6500 xA = 9960+7700+3200+1620=22480
Put all these values in formula
120 83 120*83= 9960 xA -xB - (A - B) z
140 55 140*55=7700 n
160 20 160*20=3200 22480-11620-(167-153) * 120
180 9 180*9=1620 320
10860 - (14*120) 10860 - 1680
320 320
9180
320
From above observation, the value of Mean Deviation  = 28.69, So student can solve the
problem both the formula, but use easy one.
Ex:- Find the Mean Deviation from following data.
4):- 50-59 60-69 70-79 80-89 90-99 100-109 110-119 120-129 130-139
4 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.d
130-139 4 134.5 4*134.5=538 134.5-94.5=40 4*40=160
120-129 8 124.5 8*124.5=996 124.5-94.5=30 8*30=240
110-119 14 114.5 14*114.5=1603 114.5-94.5=20 14*20=280
100-109 16 104.5 16*104.5=1672 9828 104.5-94.5=10 16v10=160
90-99 20 94.5 20*94.5=1890 104 94.5-94.5=0 20*0=0
80-89 16 84.5 16*84.5=1352 =94.5 84.5-94.5=10 16*10=160
70-79 14 74.5 14*74.5=1043 74.5-94.5=20 14*20=280
60-69 8 64.5 8*64.5=516 64.5-94.5=30 8*30=240
50-59 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.d
16-17 5 16.5 5*16.5=82.5 16.5-11=5.5 5*5.5= 27.5
14-15 4 14.5 4*14.5=58 14.5-11=3.5 4*3.5= 14
12-13 4 12.5 4*12.5=50 352 12.5-11=1.5 4*1.5= 6
10-11 8 10.5 8*10.5=4 32 10.5-11=0.5 8*0.5= 4
8-9 5 8.5 5*8.5=42.5 =11 8.5-11=2.5 5*2= 12.5
6-7 4 6.5 4*6.5=26 6.5-11=4.5 4*4.5= 18
4-5 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 “Root
Mean Square Deviation”. It is a second moment of a dispersion. Since the sum of the
square of the deviation from the mean is minimum, the deviations are taken only from
mean (but not from median or mode).
Standard Deviation is the root-mean-square average of all the deviations from
the 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 value
instead 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 same
factor. The formula works as under.
d’2
d’ 2
n n
fd’2
fd’ 2
n n
c - Common factor
Ex (1):- From the following variables find the Standard Deviation and Coefficient of
variation(c.v.) ====> 43, 48, 60, 30, 58, 23
Ans: Arrange the data in an ascending order
x d=x-X d2
d=(x-A) d2
23 23-43.67= -20.67 (-20.67)2
=427.25 23-48= -25 (-25)2
= 625
30 30-43.67= -13.67 (-13.67)2
=186.87 30-48= -18 (-18)2
= 324
43 43-43.67= -0.67 (-0.67)2
=0.45 43-48= -5 (-5)2
= 25
48 48-43.67= 4.33 (4.33)2
=18.75 48-48= 0 (0)2
= 0
58 58-43.67= 14.33 (14.33)2
=205.35 58-48= 10 (10)2
= 100
60 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 come
same. 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 of
variation(c.v.) ====> 50, 26, 37, 35, 34
x d=x-X d2
d=(x-A) d2
26 26-36.40 = -10.4 (-10.4)2
=108.16 26-35= -9 (-9)2
= 81
34 34-36.40 = -2.4 (-2.4)2
=5.76 34-35= -1 (-1)2
= 1
35 35-36.40 = -1.4 (-1.4)2
=1.96 35-35= 0 (0)2
= 0
37 37-36.40 = 0.6 (0.6)2
=0.36 37-35= 2 (2)2
= 4
50 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 Measure
Coefficient of Variation = * 100 ----- Relative Measure
  = or ----- Absolute Measure
Coefficient 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 ten
innings. Find who is better scorer and who is more consistent.
NEKO: 101 22 0 36 82 45 7 13 65 14
DECO: 97 12 40 96 13 8 85 8 56 16
Ans: Arrange the data in an ascending order.
NECO DECO
x d=x - X d2
= d*d x d=x-X d2
= d*d
0 0-38.5= -38.5 (-38.5)2
=1482.25 8 8-43.1= -35.1 (-35.1)2
=1232.01
7 7-38.5= -31.5 (-31.5)2
=992.25 8 8-43.1= -35.1 (-35.1)2
=1232.01
13 13-38.5= -25.5 (-25.5)2
=650.25 12 12-43.1= -31.1 (-31.1)2
=967.21
14 14-38.5= -24.5 (-24.5)2
=600.25 13 13-43.1= -30.1 (-30.1)2
=906.01
22 22-38.5= -16.5 (-16.5)2
=272.25 16 16-43.1= -27.1 (-27.1)2
=734.41
36 36-38.5= -2.5 (-2.5)2
=6.25 40 40-43.1= -3.1 (-3.1)2
=9.61
45 45-38.5= 6.5 (6.5)2
=42.25 56 56-43.1= 12.9 (12.1)2
=166.41
65 65-38.5= 26.5 (26.5)2
=702.25 85 85-43.1= 41.9 (41.9)2
=1755.61
82 82-38.5= 43.5 (43.5)2
=1892.25 96 96-43.1= 52.9 (52.9)2
=2798.41
101 101-38.5= 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 of
variation 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 100
marks 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 66
Find who is better in studies and if consistency is the criterion for awarding a prize, who
should 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*d
44 44-60.9= -16.9 (-16.9)2
=285.61 48 48-61.5= -13.5 (-13.5)2
=182.25
48 48-60.9= -12.9 (-12.9)2
=166.41 51 51-61.5= -10.5 (-10.5)2
=110.25
51 51-60.9= -9.9 (-9.9)2
=98.01 54 54-61.5= -7.5 (-7.5)2
=56.25
52 52-60.9= -8.9 (-8.9)2
=79.21 57 57-61.5= -4.5 (-4.5)2
=20.25
54 54-60.9= -6.9 (-6.9)2
=47.61 60 60-61.5= -1.5 (-1.5)2
=2.25
60 60-60.9= -0.9 (-0.9)2
=0.81 63 63-61.5= 1.5 (1.5)2
=2.25
72 72-60.9= 11.1 (11.1)2
=123.21 66 66-61.5= 4.5 (4.5)2
=20.25
72 72-60.9= 11.1 (11.1)2
=123.21 69 69-61.5= 7.5 (7.5)2
=56.25
76 76-60.9= 15.1 (15.1)2
=228.01 72 72-61.5= 10.5 (10.5)2
=110.25
80 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 variance
is less.
Ex (5):- Prices of particular commodity in five yearsin two cities are given below.
Tests: 1 2 3 4 5
CityA : 20 22 19 23 16
City B : 10 20 18 12 15
Find 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*d
16 16-20= -4 (-4)2
=16 10 10-15 = -5 ( -5 )2
= 25
19 19-20= -1 (-1)2
= 1 12 12-15 = -3 ( -3 )2
= 9
20 20-20= 0 (0)2
= 0 15 15-15 = 0 ( 0 )2
= 0
22 22-20= 2 (2)2
= 4 18 18-15 = 3 ( 3 )2
= 9
23 23-20= 3 (3)2
= 9 20 20-15 = 5 ( 5 )2
= 25
x=100 d2
= 30 x=75 d2
= 68
x 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 than
the 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 4
Matches A: 27 9 8 4 5
B: 17 9 6 5 3
Find the team which is more consistent.
Ans: Match A
x f fx d=x-A fd fd2
0 27 0*27=0 -2 27* -2 = -54 -2 * -54 = 108
1 9 1*9=9 -1 9* -1 = -9 -1* -9 = 9
2 8 2*8=16 0 8* 0 = 0 0 * 0 = 0
3 4 3*4=12 1 4* 1 = 4 1 * 4 = 4
4 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 B
x f fx d=x-A fd fd2
0 17 0*17=0 -2 17* -2 = -34 -2 * -34 = 68
1 9 1*9=9 -1 9* -1 = -9 -1* -9 = 9
2 6 2*6=12 0 6* 0 = 0 0 * 0 = 0
3 5 3*5=15 1 5* 1 = 5 1 * 5 = 5
4 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 17
No.of Stu (boys): 12 15 15 5 3
(girls): 13 10 12 2 1
Ans: Boys
x f fx d=x-A fd fd2
13 12 13*12=156 -2 12* -2 = -24 -2 * -24 = 48
14 15 14*15=210 -1 15* -1 = -15 -1* -15 = 15
15 15 15*15=225 0 15* 0 = 0 0 * 0 = 0
16 5 16*5=80 1 5* 1 = 5 1 * 5 = 5
17 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
Girls
x f fx d=x-A fd fd2
13 13 13*13=169 -2 13* -2 = -26 -2 * -26 = 52
14 10 14*10=140 -1 10* -1 = -10 -1* -10 = 10
15 12 15*12=180 0 12* 0 = 0 0 * 0 = 0
16 2 16*2=32 1 2* 1 = 2 1 * 2 = 2
17 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 for
durability and got the followig results.
Durability in ‘00’ hr: 17-19 19-21 21-23 23-25
Company A: 100 160 260 80
Company B: 30 420 120 30
Which 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’ fd’2
f fd’ fd’2
17-19 18 -1 100 -1*100= -100 -1* -100=100 30 -1*30=-30 -1*-30=30
19-21 20 0 160 0*160= 0 0*0 = 0 420 0*420=0 0*0 = 0
21-23 22 1 260 1*260=260 1*260=260 120 1*120=120 1*120=120
23-25 24 2 80 2*80=160 2*160=320 30 2*30=60 2*60=120
n=600 fd’=320 fd’2
=680 n=600 fd’=150 fd’2
=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’ fd’2
fd’ 2

n n n X
Put the values from above table and A = 20, c = 2
320 150
600 600
= 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 2
600 600 600 600
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’ company
as the variation in B bulbs is less.
Ex (9):- A purchasing agent obtained samples of lamps from two suppliers ‘A’ and ‘B’ with
the following information.
Length of the Life hr: 500-700 700-900 900-1100 1100-1300
Supplier A: 10 16 30 8
Supplier B: 3 42 12 4
Which Supplier’s lamps are more uniform?
Ans: A Supplier B Supplier
CI Xm
d’ f fd’ fd’2
f fd’ fd’2
500-700 600 -1 10 -1*10= -10 -1* -10=10 3 -1*3=-3 -1*-3=3
700-900 800 0 16 0*16= 0 0*0 = 0 42 0*42=0 0*0 = 0
900-1100 1000 1 30 1*30=30 1*30=30 12 1*12=12 1*12=12
1100-1300 1200 2 8 2*8=16 2*16=32 4 2*4=8 2*8=16
n=64 fd’=36 fd’2
=72 n=61 fd’=17 fd’2
=31
500+700 1200 Xm
- 800 600-800 -200
2 2 200 200 200
Similarly, all Xm
and d’ values find from above table.
fd’ fd’2
fd’ 2

n n n X
Put the values from above table and A = 800, c = 200
36 17
64 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’ Supplier’s lamps are more Uniform as their variation is less than the ‘A’ Supplier’s
lamps.
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 586.00 648.00
Average monthly wage Rs. 52.50 Rs. 47.50
Variance of the distribution Rs. 100.00 Rs. 121.00
Xm
= = = 18 d’ = = = = -1
X = A + * c ,  = - ( ) * c , CV = * 100

XA
= 20 + * 2 XB
= 20 + * 2
A
= - ( ) * 2 B
= - ( ) * 2 
 
 
 
CVA
= * 100 = * 100 CVB
= * 100 = * 100
Xm
= = = 600 d’ = = = = -1
X = A + * c ,  = - ( ) * c , CV = * 100

XA
= 800 + * 200 XB
= 800 + * 200
A
= - ( ) * 200 B
= - ( ) * 200 
 
 
 
CVA
= * 100 = * 100 CVB
= * 100 = * 100

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i) Which firm pays out the larger amount as monthly wages? and
ii) In which firm is there greater variability in individual wage?
Ans:- Firm A Firm B
n = 586, X = 52.50, variance = 100 n = 648, X = 47.50, variance = 121
 
 = variance X  = variance X
10 11
52.50 47.50
 = 10 , CV = 0.190476 * 100  = 11 , CV = 0.23147 * 100
CV = 19.05% CV = 23.15%
X = , x = X * n X = , x = X * n
x = 52.50 * 586 x = 47.50 * 648
x = 30765 Rs. x = 30780 Rs.
Factory B pays out more wages than factory A, and at the same it has greater variability
also.
Ex (11):- A study of large number of workers revealed an average pulse rate of 81 beats er
minute and standard deviation 12.2 beats. Measurement of heights gave averages 66.9
inches and standard deviation 2.7 inches. Are the industrial workers more variable in
respect of pulse rate?
Ans:- Pulse Rate Height in inches
X = 81,  = 12.2 X = 66.9,  = 2.7
 12.2  2.7
X 81 X 66.9
CV = 0.15061 * 100 = 15.06% CV = 0.04035 * 100 = 4.035%
CV = 15.06% CV = 4.04%
Yes the industrial workers are more variable in pulse rate. Pulse rate variable is more
than the Height in inches.
Karl Person’s Coefficient of Skewness:-
Karl Person has stated a formula for relative measure of skewness. That is why, the
formula is known as “Karl Person’s Coefficient of Skewness”. It is based on the differ-
ence between the Mean and Mode of the distribution which is divided by the standard
deviation. It is denoted by symbolically by ‘Skp’.
X - z 3( X - Me)
 
X - Mean, Me - Median, z - Mode,  - Standard deviation
Ex (12):- Find out the Arithmetic Mean, Median, Quartile Deviation, Mean Deviation ,
Standard Deviation from following given data. 53, 56, 64, 54, 46, 60, 94.
X : 46 53 54 56 60 64 94
N : 1 2 3 4 5 6 7
1) x1
+x2
+x3
+ x4
+ x5
+ x6
x
N N
46 + 53 + 54 + 56 + 60 + 64 + 94 427
7 7
Arithmetic MeanX = 61
2) N +1 th item
7+1 th item
8 th item
2 2 2
Median ( Me ) = 56
3) Q3
- Q1
N + 1 th item
3( N + 1 ) th item
2 4 4
N + 1 th item
3( N+1) th item
4 4
7 + 1 th item
8 th item
3(7 + 1) th item
3 * 8 th item
24 th item
4 4 4 4 4
Q1
= ( 2 ) th item
= 53 Q3
= ( 5 ) th item
= 64
Q1
= 56 Q3
= 64
Q3
- Q1
64 - 53 11
2 2 2
4) Mean Deviation 5) Standard Deviation
x d=x-X d2
46 46-61= -15 (-15)2
= 225
53 53-61= -8 (-8)2
= 64
54 54-61= -7 (-7)2
= 49
56 56-61= -5 (-5)2
= 25
60 60-61= -1 (-1)2
= 1
64 64-61= 3 (3)2
= 9
94 94-61= 33 (33)2
= 1089
x=427 d = 72 d2
= 1462
Mean Deviation ( ) = 10.29 and Standard Deviation () = 14.45
(Tip:- All the Methods can solve from only one example, so student practise all the
types from only one example for better understand and easy to solve any methods.)
CV = * 100 CV = * 100
 
 = 100 CV = * 100  = 121 CV = * 100 
 
x
n
x
n
CV = * 100 = * 100 CV = * 100 = * 100
 
1) Skp = or 2) Skp =

X = =
X = = = 61
Median (Me) = ( ) = ( ) = ( ) = ( 4 )th item
= 56
Quartile Deivation = Q1
= ( ) & Q3
= ( )
Q1
= ( ) Q3
= ( )
Q1
= ( ) = ( ) Q1
= ( ) = ( ) = ( )
Q. D. = = = = 5.6
X = =
X = 61
 = =
 = 10.2857
 = 10.29
x 427
N 7
d 72
N 7
d2
1462
N 7  = =
 = 208.8571
 = 14.4519
 = 14.45


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Ex (13):- Find out the Mean, Median, Mode, Quartile Deviation, Coefficient of QD, Mean
Deviation, Standard Deviation and coefficient of variation from following given data.
C.I. : 10-12 13-15 16-18 19-21 22-24 25-27 28-30
f: 2 1 6 2 3 4 4
Ans:- Since there is an inclusive class interval distribution, we shall convert it into
exclusive class interval series.
C.I. C.I. Xm
f cf d’ f.d’ fd’2
10-12 9.5-12.5 11 2 2 -2 2* -2= -4 -2 * -4 = 8
13-15 12.5-15.5 14 1 1+2=3 -1 1* -1= -1 -1 * -1 = 1
16-18 15.5-18.5 17 6 6+3=9 0 6*0 = 0 0 * 0 = 0
19-21 18.5-21.5 20 2 2+9=11 1 2*1 = 2 1 * 2 = 2
22-24 21.5-24.5 23 3 3+11=14 2 3*2 = 6 2 * 6 = 12
25-27 24.5-27.5 26 4 4+14=18 3 4*3 = 12 3 * 12 = 36
28-30 27.5-30.5 29 4 4+18=22 4 4*4 = 16 4 * 16 = 64
N=22 fd’= 31 fd’2
= 123
10+12 21 c = 12.5 - 9.5 Xm
-A 11 - 17 -6
2 2 c = 3 c 3 3
from above method, we find the values of Xm
, d’ and c.
1) Mean
fd’ 31
n 22
X = 17 + 4.2272 = 21.2272 = 21.23
2) Median (Me) N / 2 = 22 / 2 = 11  L = 18.5 , cf = 9 , f = 2 , c = 3
( N / 2 - cf) 11 - 9 2
f 2 2
Me = 18.5 + 3 = 21.5
3) Mode (z) Mean ( X ) = 21.23 , Median (Me) = 21.5
z = 3 Me - 2 X = 3 * 21.5 - 2 * 21.23 = 64.5 - 42.46 = 22.04
Mode (z) = 22.04
4) Quartile Deviation (QD)
N/4 - cf 3N/4- cf
f f
N = 22, N 22 3N 3 x 22 66
4 4 4 4 4
 L = 15.5, cf = 3, f = 6, c = 3 L = 24.5,cf = 14, f = 4, c = 3
5.5 - 3 16.5 - 14
6 4
2.5 2.5
6 4
= 15.5 + 0.4166 x 3 = 24.5 + 0.625 x 3
= 15.5 + 1.25 = 24.5 + 1.875
= 16.75 = 26.375
Q1
= 16.75 Q3
= 26.38
Q3
- Q1
26.38-16.75 9.63
2 2 2
Q3
- Q1
26.38-16.75 9.63
Q3
+ Q1
26.38+16.75 43.13
5) Mean Deviation () fd’
n
Put the value on formula 31
22
Mean Deviation () = 4.23
6) Standard Deviation ( )
fd’ = 31 , fd’2
= 123 , n = 22 , c = 3
fd’2
d’ 2
123 31 2
n n 22 22
 = 5.5909 - (1.40909)2
x 3 = 5.5909 - 1.9855 x 3
 = 3.6054 x 3 = 1.8988 x 3 = 5.6963 = 5.70
Standard Deviation ( ) = 5.70
 5.70
X 21.23
Coefficient of Variance (CV) = 26.83%
(Tip:- All the Methods can solve from only one example, so student practise all the
types of problems from only one example for better understand and easy to solve any
methods.)
Xm
= = = 11 d’ = = = = -2
X = A + * c = 17 + * 3 = 17 + (1.40909) * 3

Me = L + x c = 18.5 + x 3 = 18.5 + x 3 = 18.5 + (1 x 3)

= Q1
= L + ( ) x c & Q3
= L + ( ) x c
3N/4-cf
f
N/4-cf
f
Q3
- Q1
2
 Q1
= L +   x c Q3
= L +   x c
= = 5.5 = = = 16.5
 Q1
= 15.5 +  x 3 Q3
= 24.5 +  x 3

 Q1
= 15.5 +  x 3 Q3
= 24.5 +   x 3
Q.D. = = = = 4.815
 = x c fd’ = 31 , n = 22 , c = 3
 = x 3 = 1.40909 x 3 = 4.2272 = 4.23
= - ( ) x c = - ( ) x 3 

CV = * 100 = * 100 = 0.2683 * 100 = 26.83%




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Karl Pearsons Coefficient of Correlation: ( r ):-
The formula is based on arithmetic mean and standard deviation and it is most
widely used. The formula indicates whether the correlation is positive or negative. The
answer lies between +1 and -1 (Perfect Positive and Negative correlation respectively).
Zero represents the absence of correlation.
Karl Pearson’s coefficient of correlation is also known as the “Product Moment
Coeficient”. It is denoted by ‘r’ which is the symbol of the degree of correlation between
the two variables. “Coefficient of Correlation” is the numberical measure of the amount
of correlation existing between the two variables X and Y, the subject and the relative
respectively. The variable which is used as the standard is called the subect, and the
variable which is compared with the subject is called the relative. The Coefficient is
calculated by “dividing the product of all the deviations of each pair of observations from
their respective means by the product of the standard deviations of the two variables
multiplied by the number of items”.
xy Where, x = ( x - X ) = dx
n xy y = ( y - Y ) = dy
dx - Deviation of ‘x’ values of variable from there X i.e. (x - X).
dy - Deviation of ‘x’ values of variable from there Y i.e. (y - Y).
x
- Standard Deviation of ‘X’ varable.
y
- Standard Deviation of ‘Y’ varable.
n - Number of items paired.
The formula may be presented as under.
dxdy
n xy
The above formula can be simplified mathmetically as under.
dxdy dxdy
dx2
dy2
dx2
dy2
n n
(Note: No need to compute Standard Deviation).
These above formula are quite conveniently applied, if the X and Y of the variables are
integer or whole numbers.
When the X and Y of the variables are fractional figures, the computation
will be a tedious job. Under such circumstances, we can take the deviations from the
assumed means for the two variables (the next nearest whole number to the
actual mean preferably). Then we have the “Short cut” formula as under.
dx ) ( dy )
n
dx)2
dy)2
n n
(Note: Though the formula seems to be lengthy, the calculations are amazingly simplified.
dx ) ( dy )
n
X - Actual Mean of x Y - Actual Mean of y
Ax - Assumed Mean of x Ay - Assumed Mean of y
dxdy
n
dxdy
covariance n
xy xy
Probable Error: It is a difference resulting due to taking samples from the mass or
population. It is not possible to consider the entire population (census method) in statis-
tical analysis and arrive true or actual results. So there lies error in the sampling result as
compared to the actual result obtained in the census method.
Probable Error is a measure (a single fractional figure) which when ‘added to’ and
‘substracted from’ , gives us the two limits, within these two limits, it is probable that all
the results or answers (coefficient of correlation) of the sample pairs, selected from the
same population, will fail. Thus the probable error is a statistical measure which provied
for two limits within which all the answers, obtained from different sample pairs of the
population, will fail.
It is based on the standard errors multiplied by the probable factor. It is obtained by the
formula:
P.E. = Probable factor x Standard Error = 0.6745
Reason for taking the factor 0.6745 is that in a normal distribution 50% of the obser-
vation lie in the range   0.6745  where  is the mean and  is the standard deviation.
Ex (1):- From the following table calculate the coefficient of correlation by Karl Pearson’s
method. Arithmetic mean of X and Y variables are 6 and 8 respectively. Comment on the
result through the Probeable Error.
X: 6 2 10 - 8
Y: 9 11 - 8 7
r =
r = (Note: Deviations are taken from actual mean)
r = = r =
n x
   
dxdy -
r =
dx2
- dy2
- 
(a) Refer.... = n ( X - Ax ) ( Y - Ay )
(b) Refer.... Covariance =
(c) Refer.... =
1 - r2
n

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Ans: Let us find out the missing figures in X and Y variables.
X variable Y variable
x 6+2+10+A+8 y 9+11+B+8+7
n 5 n 5
26 + A 35 + B
5 5
6 x 5 = 26 + A 8 x 5 = 35 + B
30 = 26 + A 40 = 35 + B
A = 30 - 26 = 4 B = 40 - 35 = 5
A = 4 B = 5
Computation of Coefficient of Correlation
x y dx=(x-6) dx2
dy=(y-8) dy2
dxdy
6 9 6-6= 0 (0)2
= 0 9-8= 1 (1)2
= 1 0 x 1 = 0
2 11 2-6= -4 (-4)2
= 16 11-8= 3 (3)2
= 9 -4 x 3 = -12
10 5 10-6= 4 (4)2
= 16 5-8= -3 (-3)2
= 9 4 x -3 = -12
4 8 4-6= -2 (-2)2
= 4 8-8= 0 (0)2
= 0 -2 x 0 = 0
8 7 8-6= 2 (2)2
= 4 7-8= -1 (-1)2
= 1 2 x -1 = -2
x=30 y=40 dx=0 dx2
=40 dy=0 dy2
=20 dxdy= -26
dxdy -26 -26 -26
dx2
x dy2
40 x 20 800 28.2842
r = -0.92 ( High Degree of Negative Correlation )
1 - r2
1 - (-0.9192)2
1 - 0.8449 0.1550
n 5 5 2.2360
P.E. = 0.6745 x 0.06932 = 0.046756 = 0.04676
Probable Error (P.E.) = 0.04676
Ex (2):- Calculate the coefficient of corelation from the following data and calculate its
Probeable Error.
Marks in
Statistics(X): 30, 60, 30, 66, 72, 24, 18, 12, 42, 06
Account(Y): 06, 36, 12, 48, 30, 06, 24, 36, 30, 12
Ans:- Computation of Coefficient of Correlation
x y (x-36)/6 dx2
(y-24)/6 dy2
dxdy
dx dy
30 6 (30-36)/6= -1 (-1)2
= 1 (6-24)/6= -3 (-3)2
= 9 -1 x -3 = 3
60 36 (60-36)/6= 4 (4)2
= 16 (36-24)/6= 2 (2)2
= 4 4 x 2 = 8
30 12 (30-36)/6= -1 (-1)2
= 1 (12-24)/6= -2 (-2)2
= 4 -1 x -2 = 2
x y (x-36)/6 dx2
(y-24)/6 dy2
dxdy
dx dy
66 48 (66-36)/6= 5 (5)2
= 25 (48-24)/6= 4 (4)2
= 16 5 x 4 = 20
72 30 (72-36)/6= 6 (6)2
= 36 (30-24)/6= 1 (1)2
= 1 6 x 1 = 6
24 06 (24-36)/6= -2 (-2)2
= 4 (6-24)/6= -3 (-3)2
= 9 -2 x -3 = 6
18 24 (18-36)/6= -3 (-3)2
= 9 (24-24)/6= 0 (0)2
= 0 -3 x 0 = 0
12 36 (12-36)/6= -4 (-4)2
= 16 (36-24)/6= 2 (2)2
= 4 -4 x -2 = 8
42 30 (42-36)/6= 1 (1)2
= 1 (30-24)/6= 1 (1)2
= 1 1 x -1 = 1
06 12 (6-36)/6= -5 (-5)2
= 25 (12-24)/6= -2 (-2)2
= 4 -5 x -2 = 10
x=360 y=240 dx=0 dx2
=134 dy=0 dy2
=52 dxdy= 48
x 360 y 240
n 10 n 10
dxdy 48 48 48
dx2
x dy2
134 x 52 6968 83.4745
r = 0.575 ( Low Degree of Positive Correlation )
1 - r2
1 - (0.575)2
1 - 0.3306 0.6693
n 10 10 3.1623
P.E. = 0.6745 x 0.21164 = 0.1428 = 0.1428
The two limits, within which all the ‘r’s of different samples fall are,
0.575 0.575 ‘r’ answers fall within 0.718 to 0.432
+ 0.143 - 0.143 (Positive Range)
0.718 0.432
Ex (3):- Calculate the coefficient of corelation between income and weight from the follow-
ing data.
Income (Rs) X: 100 200 300 400 500 600
Weight (lbs) Y: 120 130 140 150 160 170
Ans:- Computation of Coefficient of Correlation
x y (x-350)/50 dx2
(y-145)/5 dy2
dxdy
dx dy
100 120 (100-350)/50= -5 (-5)2
= 25 (120-145)/5= -5 (-5)2
= 25 -5 x -5 = 25
200 130 (200-350)/50= -3 (-3)2
= 9 (130-145)/5= -3 (-3)2
= 9 -3 x -3 = 9
300 140 (300-350)/50= -1 (-1)2
= 1 (140-145)/5= -1 (-1)2
= 1 -1 x -1 = 1
400 150 (400-350)/50= 1 (1)2
= 1 (150-145)/5= 1 (1)2
= 1 1 x 1 = 1
500 160 (500-350)/50= 3 (3)2
= 9 (160-145)/5= 3 (3)2
= 9 3 x 3 = 9
600 170 (600-350)/50= 5 (5)2
= 25 (170-145)/5= 5 (5)2
= 25 5 x 5 = 25
x=2100 y=870 dx=0 dx2
=70 dy=0 dy2
=70 dxdy= 70
X = = Y = =
6 = 8 =
r = = = = = -0.9192
  
P.E. = 0.6745 = 0.6745 = 0.6745 = 0.6745
  
X = = = 36 Y = = = 24
r = = = = = 0.575
  
P.E. = 0.6745 = 0.6745 = 0.6745 = 0.6745
  

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x 2100 y 870
n 6 n 6
The deviation are divided by common factor and short cut method adopted.
dxdy 70 70 70
dx2
x dy2
70 x 70 490 70
r = 1 ( There is perfect Positive Correlation )
Ex (4):- Calculate Karl Person’s coefficient of corelation between percentages of pass
and failure from the following data.
No. of Student: 800 600 900 700 500 400
No. of Passed: 480 300 450 560 450 300
Ans:- Let us convert the data into the number of students passed and failed.
No. of Student: 800 600 900 700 500 400
No. of Passed (X): 480 300 450 560 450 300
No. of Failed (Y): 320 300 450 140 050 100
Pass % (X) : 060 050 050 080 090 075
Fail % (Y): 040 050 050 020 010 025
how to find fail stud --> 800 - 480 = 320 (Like that find others)
how to find % --> (480 / 800) * 100 = 60% (320 / 800) * 100 = 40%
x y dx=(x-60) dx2
dy=(y-35) dy2
dxdy
60 40 (60-60)= 0 (0)2
= 0 (40-35)= 5 (5)2
= 25 0 x 5 = 0
50 50 (50-60)= -10 (-10)2
= 100 (50-35)= 15 (15)2
= 225 -10 x 15=-150
50 50 (50-60)= -10 (-10)2
= 100 (50-35)= 15 (15)2
= 225 -10 x 15=-150
80 20 (80-60)= 20 (20)2
= 400 (20-35)= -15 (-15)2
= 225 20 x -15=-300
90 10 (90-60)= 30 (30)2
= 900 (10-35)= -25 (-25)2
= 625 30 x -25=-750
75 25 (75-60)= 15 (15)2
= 225 (25-35)= -10 (-10)2
= 100 15 x -10=-150
x=405 y=195 dx=45 dx2
=1725 dy=-15 dy2
=1425 dxdy=-1500
x 405 y 195
n 6 n 6
(Note: Deviation are taken from the assumed means and the short cut formula is used.
Because actual means 67.5 and 32.5 are in fractions. 67.5 - 7.5 = 60 & 32.5 + 2.5 = 35)
Put dx = 45, dy = -15, dx2
= 1725, dy2
= 1425, dxdy = -1500, n = 6
( 45 ) ( -15 )
6
45)2
 -15)2
6 6
-675
6
2025 225
6 6
-1500 - ( -112.5) -1500 + 112.5
1725 - 337.5 1425 - 37.5 1387.5 1387.5
-1387.5 -1387.5
37.25 x 37.25 1387.5
r = -1 ( There is perfect Negative Correlation )
(Note: 67.5 - 2.5 = 65 & 32.5 + 7.5 = 40, put the assumed mean 65 & 40 and solve it, the
answer is come same.)
x y dx=(x-65) dx2
dy=(y-40) dy2
dxdy
60 40 (60-65)= -5 (5)2
= 25 (40-40)= 0 (0)2
= 0 -5 x 0 = 0
50 50 (50-65)= -15 (-15)2
= 225 (50-40)= 10 (10)2
= 100 -15 x 10=-150
50 50 (50-65)= -15 (-15)2
= 225 (50-40)= 10 (10)2
= 100 -15 x 10=-150
80 20 (80-65)= 15 (15)2
= 225 (20-40)= -20 (-20)2
= 400 15 x -20=-300
90 10 (90-65)= 25 (25)2
= 625 (10-40)= -30 (-300)2
= 900 15 x -30=-750
75 25 (75-65)= 10 (10)2
= 100 (25-40)= -15 (-15)2
= 225 10 x -15=-150
x=405 y=195 dx=15 dx2
=1425 dy=-45 dy2
=1725 dxdy=-1500
( 15 ) ( -45 )
6
15)2
 -45)2
6 6
-675
6
225 2025
6 6
-1500 - ( -112.5) -1500 + 112.5
1425 - 37.5 1725 - 337.5 1387.5 1387.5
-1387.5 -1387.5
37.25 x 37.25 1387.5
X = = = 350 Y = = = 145
r = = = = = 1
  
X = = = 67.5 Y = = = 32.5
dx ) ( dy )
n
dx)2
dy)2
n n
dxdy -
r =
dx2
- dy2
- 
-1500 -
r =
1725 - 1425 -
-1500 -
r =
1725 - 1425 -
r = =
r = = = -1
 
 
   
-1500 -
r =
1425 - 1725 -
-1500 -
r =
1425 - 1725 -
r = =
r = = = -1
 
 
   

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Ex (5):- Following are the results of B.Com. examination in a college. Compute coefficient
of correlation between age and sucess in the examination and interpret the result. calcu-
late its Probeable Error.
Age of Candidates: 20-21 21-22 22-23 23-24 24-25 25-26
Candi. Appeared: 120 100 70 40 10 5
Successful Cand: 72 55 35 18 4 1
Ans:- Let us obtain the Mid-value of age group and convert the successful candidates into
percentages.
how to find mid value --> (20+21) / 2 = 41 / 2 = 20.5 (Like that find others)
how to find % --> (72 / 120) * 100 = 60% (55 / 100) * 100 = 55%
Age M.V. (x-23.5) dx2
% of (y-50)/5 dy2
dxdy
C.I. x dx y dy
20-21 20.5 20.5-23.5=-3 9 60 (60-50)/5= 2 4 -3 x 2= -6
21-22 21.5 21.5-23.5=-2 4 55 (55-50)/5= 1 1 -2 x 1= -2
22-23 22.5 22.5-23.5=-1 1 50 (50-50)/5= 0 0 -1 x 0=0
23-24 23.5 23.5-23.5=0 0 45 (45-50)/5= -1 1 0 x -1= 0
24-25 24.5 24.5-23.5=1 1 40 (40-50)/5= -2 4 1 x -2= -2
25-26 25.5 25.5-23.5=2 4 20 (20-50)/5= -6 9 2 x -6= -12
dx=-3 dx2
=19 dy=-6 dy2
=46 dxdy=-22
We are taking assumed means. In Y varaible we are dividing the deviation by the common
factor. So a short cut formula is used.
( -3 ) ( -6 )
6
-3)2
-6)2
6 6
18
6
9 36
6 6
-22 - 3 -25
19 - 1.5 46 - 6 17.5 40
-25 -25
4.1833 x 6.3245 26.4575
r = -0.95 ( High Degree of Negative Correlation )
1 - r2
1 - (-0.95)2
1 - 0.9025 0.04975
n 6 6 2.44948
P.E. = 0.6745 x 0.02031 = 0.01369 = 0.014
Ex (6):- Find Pearsonian coefficient of correlation between average profits and average
advertisement expenditure per shop and interpret. calculate its Probeable Error.
No. of shops: 12 18 25 20 10
Total Profit (Rs): 7200 5400 10000 3000 1800
Total Advertisement Exp(Rs): 1200 3600 7500 1000 600
Ans:- Let us obtain the average profits and average advertisement expenses X & Y.
Profit Advert. Expenses
No. of shops No. of Shops
7200 1200
12 12
(Like above find out all the values of X and Y)
x y (x-320)/10 dx2
(y-140)/10 dy2
dxdy
dx dy
600 100 (600-320)/10=28 784 (100-140)/10= -4 16 28 x -4 = -112
300 200 (300-320)/10= -2 4 (200-140)/10= 6 36 -2 x 6=-12
400 300 (400-320)/10= 8 64 (300-140)/10= 16 256 8 x 16=128
150 50 (150-320)/10= -17 289 (50-140)/10= -9 81 -17 x -9=153
180 60 (180-320)/10= -14 196 (60-140)/10= -8 64 -14 x -8=112
x=1630 y=710 dx= 3 dx2
=1337 dy= 1 dy2
=453 dxdy=269
x 1630 y 710
n 5 n 5
(Note: Deviations are taken from the assumed means. i.e. 326-6 = 320 & 142-2 = 140)
( 3 ) ( 1 )
5
3)2
1)2
5 5
dx ) ( dy )
n
dx)2
dy)2
n n
dxdy -
r =
dx2
- dy2
- 
-25 -
r =
19 - 46 -
- 25 -
r =
19 - 46 -
r = =
r = = = -0.94491
 
 
   
P.E. = 0.6745 = 0.6745 = 0.6745 = 0.6745
  
Average Profit = & Average Avrt. Expen =
Average Profit = = 600 & Average A.E. = = 100
X = = = 326 Y = = = 142
dx ) ( dy )
n
dx)2
dy)2
n n
dxdy -
r =
dx2
- dy2
-
269 -
r =
1337 - 453 -
 
 

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