statistics measures of central tendency (averages
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15.1
STATISTICS â MEASURES OF CENTRAL TENDENCY (AVERAGES) &
DISPERSION
DEFINITION OF CENTRAL TENDENCY / AVERAGES:
Central tendency (tending to the central value), which helps for finding performance and comparison
X ð
00-19 1 Minimum
20-39 3 Gradually increasing
40-59 7 Maximum
60-79 2 Gradually decreasing
80-99 1 Minimum
X - (Any variable: Height, Weight, Marks, Profits, Wages, and so on)
ð - Frequency, (Usually, repetitiveness, frequent happenings, number of times of occurrence)
List of Formula
Arithmetic Mean (ï¿œÌ ï¿œ) Geometric Mean (ð®ðŽ) Harmonic Mean (ð¯ðŽ)
Weighted Average
XÌ =â ð€ð
â ð€ G = (ð1
ð€1 à ð2ð€2 à ⊠à ðð
ð€ð)1
â ð€
Or ðº = ðŽðð¡ðððð (â ð€ ðððð
â ð€)
ð» =â ð€
âð€
ð
Combined Mean
xÌ =ð1xÌ 1 + ð2xÌ 2
ð1 + ð2
ðº = ðŽðð¡ðððð (ð1 log ðº1 + ð2 log ðº2
ð1 + ð2
) ð» =ð1 + ð2ð1
ð»1+
ð2
ð»2
15.2
Measures of Central Tendency (Averages)
Mean Partition Values: (Arrange the items in ascending order)
Mode (ðŽð) Arithmetic
(usual cases)
(Direct Method)
Geometric
(Comparisons
â ratios,
Proportions and %)
Harmonic
(Two units together
E.g. speed =
distance / time
Median (ðŽð) Fractiles (ðð)
Individual
XÌ =ð1 + ð2 ⊠ðð
ð
XÌ =â ðð
ðð=1
ð
xÌ =â ð
ð
GM = (ð1. ð2. ⊠ðð)1
ð
ðð ðºð = ðŽðð¡ðððð (â ððð ð
ð)
ð»ð =ð
â1
ð
If ânâ is odd:
ðð = (ð + 1
2)
ð¡â
ððð
(i.e. the middle obs)
If ânâ is even:
ðð =(
ð
2)
ð¡â+ (
ð
2+ 1)
ð¡â ððð
2
ð¹ð =ð(ð + 1)
ð¹
ðð = ððð ð¡ ð¢ð ð¢ðð
(ððððð¡ððð ððð¡âðð)
Discrete series
XÌ =â ðð
â ð =
â ðð
ð
= (ð1ð1 . ð2
ð2 . ⊠ðððð)
1
ð
ðð ðº = ðŽðð¡ðððð (â ð ððð ð
ð)
ð»ð =ð
âð
ð
ðð = ð ðð§ð ðð (ð¶ð >ð + 1
2) ð¹ð = ð ðð§ð ðð (ð¶ð >
ð(ð + 1)
ð¹)
Regular frequency
ðð = ððððð¡ððð ððð¡âðð
Irregular frequency
ðð = ðððð¢ðððð ððð¡âðð
Continuous / Grouped Frequency / (Interpolation Method)
XÌ =â ðð
â ð =
â ðð
ð
G = (ð1ð1 . ð2
ð2 . ⊠ðððð)
1
ð
ð¶ð ðº = ðŽðð¡ðððð (â ð ððð ð
ð)
ð»ð =ð
âð
ð
ðð = ð1 + (
ð
2â ðð
ðð¢ â ðð) à ð¶
ð¶ð ð +
ð
2â ð
ðà ð
ð¹ð = ð1 + (ð
ð
ð¹â ðð
ðð¢ â ðð) à ð¶
ð¶ð ð +ð
ð
ð¹â ð
ðà ð
ðð = ð1 + (ð0 â ðâ1
2ð0 â ðâ1 â ð1) à ð¶
Note:
1. Indirect / Shortcut / Assumed Mean (A) Method: Deviation Method (ð = ð â ðŽ): XÌ = ðŽ +â ð
ð & Step-Deviation Method (ð =
ðâðŽ
ð¶): xÌ = ðŽ +
â ð
ðà ð¶
2. Empirical relationship (thumb rule): If mode is ill-defined (ðð ððð ð ðð ððððððð¡ðððŠ ð ððð€ðð ððð ð¡ðððð¢ð¡ððð): XÌ â ðð = 3(XÌ â ðð) ðð ðð = 3ðð â 2XÌ
3. Fractiles: Quartiles (Q), Octiles (O), Deciles (D) and Percentiles (P)
15.3
Measures of Dispersion
Absolute Relative
(i) ððð§ð ð (ð) = ð¿ â ð ððšðððð¢ðð¢ðð§ð ðšð ð«ðð§ð ð(ð¶ð ð ) =
ð¿ â ð
ð¿ + ðà 100
(ii) ðð®ðð«ðð¢ð¥ð ððð¯ð¢ððð¢ðšð§ (ðð) =
ð3 â ð1
2
(Otherwise Semi inter quartile range)
ðð§ððð« ðªð®ðð«ðð¢ð¥ð ð«ðð§ð ð = ð3 â ð1
Coefficient of Quartile Deviation (Co QD)
ð¶ð ðð· =ð3 â ð1
ð3 + ð1
à 100
(iii) Mean Deviation (MD) about A, (ðš = XÌ , ðð , ðð) Coefficient of Mean Deviation (ð¶ð ðð·ðŽ)
ðð§ðð¢ð¯ð¢ðð®ðð¥: Mð·ðŽ =
1
ðâ|ð¥ â ðŽ| ð¶ð ðð·ðŽ =
ðð·ðŽ
ðŽÃ 100
ðð¢ð¬ðð«ððð: Mð·ðŽ =
1
ðâ ð|ð¥ â ðŽ|
ððšð§ðð¢ð§ð®ðšð®ð¬: ðð·ðŽ =
1
ðâ ð|ð â ðŽ|
(iv) Standard Deviation (s) Coefficient of Variation (CV)
ðð§ðð¢ð¯ð¢ðð®ðð¥: ð = ââ(ð â XÌ )2
ð ððâ
â ð2
ðâ XÌ 2
ð¶ð =ð
XÌ Ã 100
ðð¢ð¬ðð«ððð: ð = ââ ð(ð â XÌ )2
ð ððâ
â ðð2
ðâ XÌ 2
ððšð§ðð¢ð§ð®ðšð®ð¬: ð = ââ ð(ð â XÌ )2
ð ðð â
â ðð2
ðâ XÌ 2
ðððððððð = ð 2
Shortcut:
ð = ââ ðð2
ðâ (
â ðð
ð)
2
ðâððð ð = ð â ðŽ (ððð ððððð£ððð¢ðð ððð ððð ðððð¡ð) & ð =ð â ðŽ
ð¶ ððð ðððð¡ððð¢ðð¢ð
Comparison
Absolute Measure Relative Measure
1 Dependent of unit Independent of unit
2 Not considered for comparison considered for comparison
3 Not much difficult compared to Relative measure Difficult to compute and comprehend.
15.4
INDIVIDUAL OBSERVATIONS
Question 1: From the Individual Observations: 3, 6, 48 & 24, find out the following
Measures of Averages Measures of Dispersion
Arithmetic Mean Absolute Measure Relative Measure
Geometric Mean Range Coefficient of Range
Harmonic Mean Quartile Deviation Coefficient of Quartile Deviation
Median Mean Deviation Coefficient of Mean Deviation
Fractiles (ð1, ð3, ð6, ð·7 & ð75) Standard Deviation / Variation Coefficient of Variation
Mode
Answer:
Measures of Averages
Mean Formula Calculation Answer
AM XÌ =
â ð
ð
3 + 6 + 24 + 48
4
81
4
20.25
GM GM = (ð1 à ð2 à ⊠à ð)1
ð (3 Ã 6 Ã 24 Ã 48)1
4 (34. 44)1
4 12
HM ð»ð =ð
â1
ð
4
1
3+
1
6+
1
24+
1
48
4 Ã 48
16 + 8 + 2 + 1=
192
27
7.11
Note:
ð¿ 3 6 24 48
ð¥ðšð ð¿ 0.4771 0.7782 1.3802 1.6812
â log ð 4.3167
Formula Calculation Answer
GM ðŽðð¡ðððð (
â log ð
ð) ðŽðð¡ðððð (
4.3167
4)
11.94
Positional Average
Formula Calculations Answer
ðð ð ðð§ð ðð (ð + 1
2)
ð¡â
ððð ððð§ð ðð 2.5ð¡â ððð
6 + 0.5(24 â 6) 15 2ðð ððð + 0.5 (3ðð ððð â 2ðð ððð )
ð1 ð ðð§ð ðð (ð + 1
2)
ð¡â
ððð ððð§ð ðð 1.25ð¡â ððð
3 + 0.25(6 â 3) 3.75 1ð ð¡ ððð + 0.25 (2ðð ððð â 1ð ð¡ ððð )
ð3 ð ðð§ð ðð (3(ð + 1)
4)
ð¡â
ððð
ððð§ð ðð 3.75ð¡â ððð
3ðð ððð + 0.75 (4ð¡â ððð â 3ðð ððð )
24 + 0.75 (48 â 24)
ðð
ð75 ð ðð§ð ðð (75(ð + 1)
100)
ð¡â
ððð
ð6 ð ðð§ð ðð (6(ð + 1)
8)
ð¡â
ððð
ðµððð: ð3 = ð6 = ð75
15.5
ð·7 ð ðð§ð ðð (7(ð + 1)
10)
ð¡â
ððð ððð§ð ðð 3.5ð¡â ððð
24 + 0.5(48 â 24) ðð 3ðð ððð + 0.5 (4ð¡â ððð â 3ðð ððð )
Mode
Mode is ill-defined (Since all the observation has equal appearance)
Hence, the empirical relation is used to arrive ðð
Formula Calculations Answer
ðð ðððð â ðððð = 3(ðððð â ðððððð) 20.25 â ðððð = 3(20.25 â 15) 4.5
Measures of Dispersion (Absolute and Relative)
Formula Calculation Answer
1 Range (R) ð¿ â ð 48 â 3 45
Co â efficient of Range ð¿ â ð
ð¿ + ð =
48â3
48+3 0.8823
2 Quartile Deviation (ðžð«) ð3 â ð1
2
42 â 3.75
2
19.125
Coefficient of Quartile Deviation ð3 â ð1
ð3 + ð1
42 â 3.75
42 + 3.75
0.84
3 Mean Deviation (ðð·XÌ ) 1
ðâ|ð â XÌ |
63
4
15.75
Co â efficient of MD ðð·XÌ
ðððð
15.75
20.25
0.778
4 Standard Deviation (ð) â
â(ð â XÌ )2
ð â
1284.75
5
17.921
Or
â
âð2
ðâ (
âð
ð)
2
â2925
4â (
81
4)
2
17.921
ððð (ð) ð2 17.9212 321.16
ð¶ðððððððððð¡ ðð ð£ððððð¡ððð, ð£ðð(ð¥) ð
XÌ Ã 100
17.921
20.25Ã 100
88.49%
Working note
ð¿ |ð¿ â ï¿œÌ ï¿œ| ð¿ â ð (ð¿ â ï¿œÌ ï¿œ)ð ð¿ð
3 17.25 17.25 297.5625 9
6 14.25 14.25 203.0625 36
24 3.75 -3.75 14.0625 576
48 27.25 -27.75 770.0625 2304
Total 63 1284.75 2925
15.6
Question 2: Find Median, ðžð, ðžð,ð¶ð, ð«ð, ð·ðð for the observations: 1, 3, 6, 24, 48.
Answer:
Positional Average
Formula Calculations Answer
ðð ð ðð§ð ðð (
ð + 1
2)
ð¡â
ððð ððð§ð ðð 3ðð ððð 6 + 0.5(24 â 6) 6
ð1 ð ðð§ð ðð (
ð + 1
4)
ð¡â
ððð ððð§ð ðð 1.5ð¡â ððð 1 + 0.5(3 â 1) 2
1ð ð¡ ððð + 0.5 (2ðð ððð â 1ð ð¡ ððð )
ð3 ð ðð§ð ðð (
3(ð + 1)
4)
ð¡â
ððð
ððð§ð ðð 4.5ð¡â ððð
4ð¡â ððð + 0.5 (5ð¡â ððð â 4ð¡â ððð )
24 + 0.5 (48 â 24)
36 ð75
ð ðð§ð ðð (75(ð + 1)
100)
ð¡â
ððð
ð6 ð ðð§ð ðð (
6(ð + 1)
8)
ð¡â
ððð
ðµððð: ð3 = ð6 = ð75
ð·7 ð ðð§ð ðð (
7(ð + 1)
10)
ð¡â
ððð ððð§ð ðð 4.2ð¡â ððð 24 + 0.2(48 â 24) ðð. ð
4ð¡â ððð + 0.2 (5ð¡â ððð â 4ð¡â ððð )
Question 3: Discrete Frequency Distribution
x 10 11 12 13 14 15 16 17 18 19
f 8 15 20 100 98 95 90 75 50 30
Answer:
Measures of Averages
Formula Calculation Answer
1 Arithmetic Mean(xÌ ) âðð
ð
8727
581 15.02
2 Geometric Mean(ðºð) Antilog (â ð log ð
ð) Antilog (
682.4203
581) 14.95
3 Harmonic Mean (ð»ð) ð
âð
ð
581
39.25 14.802
Working Note:
ð¿ ð ðð¿ ð¥ðšð ð¿ ð ð¥ðšð ð¿ ð
ð¿
10 8 80 1.0000 8.0000 0.800
11 15 165 1.0414 15.6210 1.360
12 20 240 1.0792 21.5840 1.670
13 100 1300 1.1139 111.3900 7.690
15.7
14 98 1372 1.1461 112.3178 7.000
15 95 1425 1.1761 111.7295 6.330
16 90 1440 1.2041 108.3690 5.625
17 75 1275 1.2304 92.2800 4.411
18 50 900 1.2553 62.7650 2.780
19 30 570 1.2788 38.364 1.578
Total 581 8727 682.4203 39.25
Positional Average
Formula Calculations Answer Working Notes
ðð ð ðð§ð ðð (ð + 1
2)
ð¡â
ððð ððð§ð ðð 291ð ð¡ ððð
(ð. ð. ðð > 291) 15
ð¿ ð ðð
10 8 8
11 15 23
12 20 43
13 100 143
14 98 241
15 95 336
16 90 426
17 75 501
18 50 551
19 30 581
ð1 ð ðð§ð ðð (1(ð + 1)
4)
ð¡â
ððð ððð§ð ðð 145.5ð¡â ððð
(ð. ð. ðð > 145.5) 14
ð3 ð ðð§ð ðð (3(ð + 1)
4)
ð¡â
ððð
ððð§ð ðð 436.5ð¡â ððð
(ð. ð. ðð > 436.5)
17
ð75 ð ðð§ð ðð (75(ð + 1)
100)
ð¡â
ððð
ð6 ð ðð§ð ðð (6(ð + 1)
8)
ð¡â
ððð
ðµððð: ð3 = ð6 = ð75
ð·7 ð ðð§ð ðð (7(ð + 1)
10)
ð¡â
ððð ððð§ð ðð 407.4ð¡â ððð
ðð (ð. ð. ðð > 407.4)
Mode: Since there is a sudden increase in frequency from 20 to 100, we obtain mode by Grouping
Table
Grouping Table The highest frequency total in each of the six
columns of the grouping table is identified and
analyzed (Tally marks)
Total
Tally
Mark
(1) (2) (3) (4) (5) (6)
ð¿ ð (1) (2) (3) (4) (5) (6)
10 8 23
43
0
11 15 35
135
0
12 20 120
218
0
13 100 198
293
| | | 3
14 98 193
283
| | | | 4
15 95 185
260
| | | | 4
16 90 165
215
| | 2
17 75 125
155
| 1
18 50 80
0
19 30 0
15.8
Explanation to column
(ð) Original Frequency
(ð) grouping in âtwoâs
(ð) Leaving the first and grouping the
rest in âtwoâsâ
(ð) grouping in âthreeâsâ
(ð) Leaving the first and grouping in
âthreeâsâ
(ð) Leaving the first & second and
grouping in âthreeâsâ
Mode
Mode is ill-defined or bi-modal
(Since â14â and â15â occur equal number of times)
Hence, the empirical relation is used to arrive ðð
ðð ðððð â ðððð = 3(ðððð â ðððððð)
15.02 â ðððð = 3(15.02 â 15)
14.96
Points to Ponder:
Under Location Method, Mode = 13 (as the highest frequency is 100)
Under Grouping Method, Mode is ill- defined.
But, Under Empirical Relationship, Mode = 14.96, which brings the issues an accuracy
Measures of Dispersion
Formula Calculation Answer
1 Range (R) ð¿ â ð 19 â 10 10
Co â efficient of Range ð¿ â ð
ð¿ + ð
19 â 10
19 + 10 0.31
2 Quartile Deviation (ðžð«) ð3 â ð1
2
17 â 14
2 1.5
Coefficient of Quartile Deviation ð3 â ð1
ð3 + ð1
17 â 14
17 + 14 0.0967
3 Mean Deviation (ðð·XÌ ) 1
ðâ|ð â XÌ |
969.82
58.1 1.669
Co â efficient of MD ðð·XÌ
ðððð
1.669
15.02 0.111133
4 Standard Deviation (ð) ââ ð(ð â XÌ )2
ð â
2204.7628
581 3.80
ððð (ð) ð 2 3.802 14.44
ð¶ðððððððððð¡ ðð ð£ððððð¡ððð, ð£ðð(ð) ð
xÌ Ã 100
3.80
15.02Ã 100 25.29%
15.9
Working Note:
for MD For SD
ð¿ ð |ð¿ â ï¿œÌ ï¿œ| ð|ð¿ â ð| (ð¿ â ï¿œÌ ï¿œ) ð(ð¿ â ï¿œÌ ï¿œ)ð
10 8 5 .02 40.16 -5 .02 201.6032
11 15 4.02 60.30 -4.02 242.4060
12 20 3.02 68.40 -3.02 182.4080
13 100 2.02 202.00 -2.02 81.6080
14 98 1.02 99.96 -1.02 101.9592
15 95 0.02 1.90 -0.02 0.0380
16 90 0.98 88.20 0.98 86.4360
17 75 1.98 143.50 1.98 294.0300
18 50 2.98 1.49 2.98 444.0200
19 36 3.98 119.40 3.98 570.2544
â 581 969.82 2204.7628
Question 4: Continuous Frequency Distribution:
Marks 01-10 11-20 21-30 31-40 41-50 51-60 61-70 71-80 81-90 91-100
Number of Students 3 7 13 17 12 10 8 8 6 6
Also verify the empirical relation
Answer:
Measures of Averages
Formula Calculation Answer
A.M. (Direct Method) â ðð
ð
4375
90
48.61
A.M. (Short cut -Method) ðŽ + â ðð
ðà ð (A=45.5) 45.5 +
28
90Ã 10
48.61
ðŽ + â ðð
ðà ð (A = 55.5) 55.5 +
â620
90Ã 10
48.61
Geometric Mean, GM ðŽðð¡ðððð (
â ð log ð
ð) ðŽðð¡ðððð
150.5439
90
47.07
Harmonic Mean, HM ð
âð
ð
90
2.7905
32.25
15.10
Working Note:
Marks
(Class boundaries)
ð ð ðð ð =
ð â ðð. ð
ðð
ðð ð¥ðšð ð ð ð¥ðšð ð ð
ð
0.5 â 10.5 5.5 3 16.5 -4 -12 0.7404 2.2212 0.5454
10.5 â 20.5 15.5 7 108.5 -3 -21 1.903 13.3210 0.4516
20.5 â 30.5 25.5 13 331.5 -2 -26 1.4065 18.2845 0.5098
30.5 â 40. 5 35.5 17 603.5 -1 -17 1.5502 26.3534 0.4789
40.5 â 50.5 45.5 12 546.0 0 0 1.6580 19.8960 0.2637
50.5 â 60.5 55.5 10 555.0 1 10 1.7443 17.4430 0.1801
60.5 â 70.5 65.5 8 524.0 2 16 1.8162 14.5296 0.1221
70.5 â 80.5 75.5 8 604.0 3 24 1.8779 15.0232 0.1060
80.5 â 90.5 85.5 6 513.0 4 24 1.9320 11.5920 0.0701
90.5 â 100.5 95.5 6 573.0 5 30 1.9800 11.8800 0.0628
Total 90 4375.0 28 150.5439 2.7905
Positional Average and Mode
Formula Calculation Answer
Working Note
ðð ð +
ð
2â ð
ðà ð 40.5 +
45 â 40
12Ã 10 44.67
ð¿ ð ðð
0.5â10.5 3 3
10.5â20.5 7 10
20.5â30.5 13 23
30.5â40. 5 17 40
40.5â50.5 12 52
50.5â60.5 10 62
60.5â70.5 8 70
70.5â80.5 8 78
80.5â90.5 6 84
90.5â100.5 6 90
ð1 ð +
1ð
4â ð
ðà ð 20.5 +
22.5 â 10
13Ã 10 30.12
ð3 ð +
3ð
4â ð
ðà ð
60.5 +67.5 â 62
8Ã 10 67.38 ð6 ð +
6ð
8â ð
ðà ð
ð75 ð +
75ð
100â ð
ðà ð
ð3 = ð6 = ð75 = 67.38
ð·7 ð +
7ð
10â ð
ðà ð 60.5 +
63 â 62
8 Ã 10 61.75
ðð ð1 + (ð0 â ðâ1
2ð0 â ðâ1 â ð1
) à ð¶ 30.5 + (17 â 13
2 Ã 17 â 13 â 12) Ã 10 34.94
ðŽð ððððð ðð (ðð. ð â ðð. ð), since 17 is the highest frequency
Graphical Method: Ogive Curves for Positional Average:
Marks Number
of Students
Less than ogive curve More than ogive curve
UCL < cf LCL >cf
15.11
0.5 â 10.5 3 10.5 3 0.5 90 (= âð)
10.5 â 20.5 7 20.5 10 10.5 87
20.5 â 30.5 13 30.5 23 20.5 80
30.5 â 40. 5 17 40. 5 40 30.5 67
40.5 â 50.5 12 50.5 52 40.5 50
50.5 â 60.5 10 60.5 62 50.5 38
60.5 â 70.5 8 70.5 70 60.5 28
70.5 â 80.5 8 80.5 78 70.5 20
80.5 â 90.5 6 90.5 84 80.5 12
90.5 â 100.5 6 100.5 90 (= âð) 90.5 6
Verification of Empirical relation:
Mean â Mode = 3 (Mean - Median)
(i.e.,) 48.61 â 34.94 = 3 (48.61 â 44.67)
13.67 = 3 ( 4.006)
13.67 = 12.18, which is not true
Graphical Method
ðð = 35 (ðºðððâðððð ððð¡âðð)
ð1 = 30, ð3 = 45 & ð3 = 67
15.12
Measures of Dispersion
Formula Calculation Answer
1 Range (R) ð¿ â ð 100 â 1 99
Other-way 100.5 â 0.5 100
Co â efficient of Range ð¿ â ð
ð¿ + ð
100 â 1
100 + 1 0.98
2 Quartile Deviation (ðžð«) ð3 â ð1
2
67.38 â 30.12
2 18.63
Coefficient of Quartile Deviation ð3 â ð1
ð3 + ð1
67.38 â 30.12
67.38 + 30.12 0.38
3 Mean Deviation (ðð·XÌ ) 1
ðâ|ð â XÌ |
1843.54
90 20.48
Co â efficient of MD ðð·XÌ
ðððð
20.48
48.61 0.42
4 Standard Deviation (ð) ââ ð(ð â XÌ )2
ð â
53128.89
90 24.29
ððð (ð) ð2 (24.29)2 590.49
ð¶ðððððððððð¡ ðð ð£ððððð¡ððð, ð£ðð(ð) ð
XÌ Ã 100
24.29
48.61Ã 100 25.29%
Working Notes
Marks
(Class boundaries) ð ð |ð â ð| f|ð â ï¿œÌ ï¿œ| (ð¿ â ð)ð f (ð¿ â ð)ð
0.5 â 10.5 5.5 3 43.11 129.33 1858.4701 5575.4163
10.5 â 20.5 15.5 7 33.11 231.77 1096.2721 7673.9047
20.5 â 30.5 25.5 13 23.11 300.43 534.0721 6942.9373
30.5 â 40. 5 35.5 17 13.11 222.87 171.8721 2921.8252
40.5 â 50.5 45.5 12 3.11 37.32 9.6721 116.0652
50.5 â 60.5 55.5 10 6.89 68.90 47.4733 474.7210
60.5 â 70.5 65.5 8 16.89 135.12 285.2721 2282.1768
70.5 â 80.5 75.5 8 26.89 215.12 723.0721 5784.5768
80.5 â 90.5 85.5 6 36.89 221.34 1360.8721 8165.2326
90.5 â 100.5 95.5 6 46.89 281.34 2198.6721 13192.0326
Total 90 1843.54 53128.889
15.13
PROPERTIES: (A) MEASURES OF AVERAGES / CENTRAL TENDENCY
Arithmetic Mean
Property 1: If all the observations assumed by a variable are constants, say k, then the AM is also k.
Illustration: Consider 2, 2, 2
Property Calculation Answer
XÌ =ð + ð + ⯠+ ð
ð= ð
2 + 2 + 2
3 2
Property 2: (a) The algebraic sum of deviations of a set of observations from their AM is zero. And
(b) the sum of the square of the deviation taken from the Mean (XÌ ) is always minimum compared to
the deviations taken from any other Assumed Mean (ðŽ)
Illustration: Consider (X): 2, 3, 4
Property Formula Calculation Answer
XÌ = â ð 2 + 3 + 4
3 3
(a) â(ð â XÌ ) = 0
â ð(ð â XÌ ) = 0 â(ð â XÌ ) (2 â 3) + (3 â 3) + (4 â 3) 0
(b) â(ð â XÌ )2 †â(ð â ðŽ)2
â(ð â XÌ )2 (2 â 3)2 + (3 â 3)2 + (4 â 3)2 2
â(ð â ðŽ)2
ðâððð ðŽ = 4
(2 â 4)2 + (3 â 4)2 + (4 â 4)2 5
Property 3: AM is affected due to a change of origin (+/â) and / or scale (Ã/÷)
i.e., If ðŠ = ð + ðð¥, then the AM of y is given by yÌ = ð + ðï¿œÌ ï¿œ (where a is change of origin and b is change
of scale)
Illustration: Consider (ð) = 2, 3, 4,
Formula Calculation Answer ï¿œÌ ï¿œ =â ð
ð= ð + ðð
1 ð¿ = ð, ð, ð, ï¿œÌ ï¿œ =â ð¿
ð
ð + ð + ð
ð 3
2 ð = 4, 5, 6, ï¿œÌ ï¿œ =â ð
ð
4 + 5 + 6
3 5
Change of Origin (ð = 2)
ðµðððð ð = ð + 2 yÌ = ð + ðï¿œÌ ï¿œ 2 + 1 Ã3 5
3 ð = 0, 1, 2, ï¿œÌ ï¿œ =â ð
ð
0 + 1 + 2
3 1
Change of Origin (ð = â2)
ðµðððð ð = ð â 2 yÌ = ð + ðï¿œÌ ï¿œ â2 + 1 Ã3 1
4 ð = 4, 6, 8, ï¿œÌ ï¿œ =â ð
ð
4 + 6 + 8
3 6
Change of Scale (ð = 2)
ðµðððð ð = ð à 2 yÌ = ð + ðï¿œÌ ï¿œ 0 + 2 Ã3 6
15.14
5 ð = 1, 1.5, 2, ï¿œÌ ï¿œ =â ð
ð
1 + 1.5 + 2
3 1.5
Change of Scale (ð =1
2)
ðµðððð ð = ð Ã1
2 yÌ = ð + ðï¿œÌ ï¿œ 0 +
1
2Ã3 1.5
6 ð = 7, 9, 11, ï¿œÌ ï¿œ =â ð
ð
7 + 9 + 11
3 9 Change of Origin and
change of scale
(ð = 3)&(ð = 2) ðµðððð ð = 3 + 2 à ð yÌ = ð + ðï¿œÌ ï¿œ 3 + 2 Ã3 9
Property 4: If there are two groups containing ð1 and ð2 observations and ï¿œÌ ï¿œ1 and ï¿œÌ ï¿œ2 as the respective
arithmetic means, then the combined AM is given by (ï¿œÌ ï¿œ12) =ð1ï¿œÌ ï¿œÌ 1+ð2ï¿œÌ ï¿œÌ 2
ð1+ð2
Illustration Combined mean Calculation Answer
Group 1 Group II
ð1 = 5 ð2 = 15
ï¿œÌ ï¿œ1 = 9 ï¿œÌ ï¿œ2 = 5
ï¿œÌ ï¿œ12 =ð1ï¿œÌ ï¿œ1 + ð2ï¿œÌ ï¿œ2
ð1 + ð2
(5 Ã 9) + (15 Ã 5)
5 + 15 6
Points to Ponder:
1 In the case of ânâ number of groups, Combined mean (ï¿œÌ ï¿œ1âŠð) =âððï¿œÌ ï¿œÌ ð
âðð
2 If sizes of the group are same, then the combined Mean is the average of the group means
Explanation: If ð1= ð2 =n, then ï¿œÌ ï¿œ1+2 =ðï¿œÌ ï¿œÌ 1+ðï¿œÌ ï¿œÌ 2
ð+ð=
ð(ï¿œÌ ï¿œÌ 1+ï¿œÌ ï¿œÌ 2)
2ð=
ï¿œÌ ï¿œÌ 1+ ï¿œÌ ï¿œÌ 2
2
Illustration
Formula Calculation Answer
1 ð1 = 2, 3, 4, ï¿œÌ ï¿œ1 =â ð1
ð
2 + 3 + 4
3 3
2 ð2 = 4, 5, 6, ï¿œÌ ï¿œ2 =â ð2
ð
4 + 5 + 6
3 5
3 ï¿œÌ ï¿œ1+2 =ðï¿œÌ ï¿œ1 + ðï¿œÌ ï¿œ2
ð + ð
3 Ã 3 + 3 Ã 5
3 + 3 4
ï¿œÌ ï¿œ1+2 =ð(ï¿œÌ ï¿œ1 + ï¿œÌ ï¿œ2)
2ð
3(3 + 5)
2 Ã 3 4
ï¿œÌ ï¿œ1+2 =ï¿œÌ ï¿œ1 + ï¿œÌ ï¿œ2
2
3 + 5
2 4
3 If the averages are same, then the combined mean is the average itself
Explanation: If ï¿œÌ ï¿œ1 = ï¿œÌ ï¿œ2 = ï¿œÌ ï¿œ12
ï¿œÌ ï¿œ12 =ð1ï¿œÌ ï¿œ + ð2ï¿œÌ ï¿œ
ð1 + ð2
=ï¿œÌ ï¿œ(ð1 + ð2)
ð1 + ð2
Illustration
Formula Calculation Answer
1 ð1 = 2, 3, 4, ï¿œÌ ï¿œ1 =â ð1
ð
2 + 3 + 4
3 3
2 ð2 = 4, 2, ï¿œÌ ï¿œ2 =â ð2
ð
4 + 2
2 3
15.15
3 ï¿œÌ ï¿œ1+2 =ðï¿œÌ ï¿œ1 + ðï¿œÌ ï¿œ2
ð + ð
3 Ã 3 + 2 Ã 3
3 + 2 3
ï¿œÌ ï¿œ(ð1 + ð2)
ð1 + ð2
3(3 + 2)
2 + 3 3
ï¿œÌ ï¿œ1+2 = ï¿œÌ ï¿œ1 = ï¿œÌ ï¿œ2 3
Geometric Mean
Property 1: Transformation in terms of log function
ðºð = ðŽðð¡ðððð (1
ðâ ððð ð¥) ðð ððð ðºð =
1
ðâ ððð ð¥
Property 2: If all the observations assumed by a variable are constants, say ð > 0, then the GM of the
observations is also K.
Property Illustration Calculation Answer
(ð à ð à ⊠.à ð)1 ðâ = ð Consider: 2, 2, 2 = (2 à 2 à 2)1 3â 2
Property 3: GM of the product of two variables is the product of their GMâs.
Property 4: GM of the ratio of two variables is the ratio of the GMâs of the two variables.
Illustration Formula Calculation Answer
ð = 3, 6, 12 GM = (ð1 à ð2 à ⊠à ð)1
ð (3 Ã 6 Ã 12)1 3â 6
ð = 1, 2, 4 (1 Ã 2 Ã 4)1 3â 2
ð = 3, 12, 48 (3 Ã 12 Ã 48)1 3â 12
Property 3 Being ð = ð Ã ð GMð = GMð Ã GMð 6 Ã 2 12
ð = 3
1,6
2,12
4 (3 Ã 3 Ã 3)1 3â 3
Property 4 Being ð =ð
ð ðºðð§ =
GMð
GMð
6
2 3
Harmonic Mean:
Property 1: If all the observations taken by a variable are constants, say k, then the HM of the
observations is also k.
Property Illustration Calculation Answer
ð1
ð+
1
ð+ ⊠. +
1
ð
= ð ð = 2, 2, 2 31
2+
1
2+
1
2
2
Property 2: If there are two groups containing ðð and ðð observations and ð¿ð and ð¿ð as the
respective Harmonic Means, then the combined HM is given by (ï¿œÌ ï¿œðð) = ðð+ððððï¿œÌ ï¿œð
+ ððï¿œÌ ï¿œð
Illustration Combined H.M. Calculation Answer
Group 1 Group II ï¿œÌ ï¿œðð =ð1 + ð2ð1
ï¿œÌ ï¿œÌ 1+
ð2
ï¿œÌ ï¿œÌ 2
15 + 1015
3+
10
2
3.125
15.16
ð1 = 15 ð2 = 10
ï¿œÌ ï¿œ1 = 3 ï¿œÌ ï¿œ2 = 2
Median:
Property 1: If x and y are two variables, to be related by ð = ð + ðð for any two constants a and b,
then the median of y is given by ððð= ð + ðððð
(i.e., Median is affected due to a change of origin (+/â) and / or scale (Ã/÷))
Illustration: Consider (ð) = 2, 3, 4,
Formula Calculation Answer ððŽð= ð + ðð¿ðŽð
1 ð = 2, 3, 4, ï¿œÌ ï¿œðð= (
ð + 1
2)
ð¡â
ððð (3 + 1
2)
ð¡â
ððð 3
2 ð = 4, 5, 6, ï¿œÌ ï¿œðð= (
ð + 1
2)
ð¡â
ððð (3 + 1
2)
ð¡â
ððð 5 Change of Origin (ð = 2)
ðµðððð ð = ð + 2 ððð= ð + ðððð
2 + 1 Ã3 5
3 ð = 0, 1, 2, ï¿œÌ ï¿œðð= (
ð + 1
2)
ð¡â
ððð (3 + 1
2)
ð¡â
ððð 1 Change of Origin (ð = â2)
ðµðððð ð = ð â 2 ððð= ð + ðððð
â2 + 1 Ã3 1
4 ð = 4, 6, 8, ï¿œÌ ï¿œðð= (
ð + 1
2)
ð¡â
ððð (3 + 1
2)
ð¡â
ððð 6 Change of Scale (ð = 2)
ðµðððð ð = ð à 2 ððð= ð + ðððð
0 + 2 Ã3 6
5 ð = 1, 1.5, 2, ï¿œÌ ï¿œðð= (
ð + 1
2)
ð¡â
ððð (3 + 1
2)
ð¡â
ððð 1.5
Change of Scale (ð =1
2)
ðµðððð ð = ð Ã1
2 ððð
= ð + ðððð 0 +
1
2Ã3 1.5
6 ð = 7, 9, 11, ï¿œÌ ï¿œðð= (
ð + 1
2)
ð¡â
ððð (3 + 1
2)
ð¡â
ððð 9 Change of Origin and
change of scale
(ð = 3)&(ð = 2) ðµðððð ð = 3 + 2 à ð ððð= ð + ðððð
3 + 2 Ã3 9
Property 2: For a set of observations, the sum of absolute deviations is minimum when the deviations
are taken from the median.
Illustration: Consider (X): 0.5, 3, 4
Calculation Answer Property
ðð = (ð + 1
2)
ð¡â
ððð (3 + 1
2)
ð¡â
ððð 3
ï¿œÌ ï¿œ =â ð
ð
0.5 + 3 + 4
3 2.5
(a) â |ð â XÌ | |0.5 â 2.5| + |3 â 2.5| + |4 â 2.5| 4
(ð) < (ð)
(b) â |ð â ðð| |0.5 â 3| + |3 â 3| + |4 â 3| 3.5
15.17
Mode:
Property 1: If ð = ð + ðð, then ððð= ð + ðððð
(i.e., Mode is affected due to a change of origin (+/â) and / or scale (Ã/÷))
Illustration: Consider (ð) = 2, 3, 3, 4
Formula Calculation Answer ððŽð= ð + ðð¿ðŽð
1 ð = 2, 3, 3, 4, Most usual 3
2 ð = 4, 5, 5, 6, 5 Change of Origin (ð = 2)
ðµðððð ð = ð + 2 ððð= ð + ðððð
2 + 1 Ã3 5
3 ð = 0, 1, 1, 2, Most usual 1 Change of Origin (ð = â2)
ðµðððð ð = ð â 2 ððð= ð + ðððð
â2 + 1 Ã3 1
4 ð = 4, 6, 6, 8, Most usual 6 Change of Scale (ð = 2)
ðµðððð ð = ð à 2 ððð= ð + ðððð
0 + 2 Ã3 6
5 ð = 1, 1.5, 1.5, 2, Most usual 1.5
Change of Scale (ð =1
2)
ðµðððð ð = ð Ã1
2 ððð
= ð + ðððð 0 +
1
2Ã3 1.5
6 ð = 7, 9, 9, 11, Most usual 9 Change of Origin and
change of scale
(ð = 3)&(ð = 2) ðµðððð ð = 3 + 2 à ð ððð
= ð + ðððð 3 + 2 Ã3 9
(B) MEASURES OF DISPERSION: PROPERTY
Property Measure / Explanation
1 All the observations assumed by a variable are constant,
then measure of dispersion = 0
Range (R) = 0
Mean Deviation (MD) = 0
Standard Deviation (s) = 0
Illustration: Consider (ð¿): 2, 2, 2
Formula Calculation Answer
ï¿œÌ ï¿œ =â ð¿
ð
2 + 2 + 2
3 2
Range = L â S 2 â 2
0 ðð·XÌ =
1
ðâ|ð â XÌ |
|2 â 2| + |2 â 2| + |2 â 2|
3
ðð· = ââ(ð â XÌ )2
ð ââ
(ð â 2)2
3
2 Affected due to change of Scale, but not of origin ð ðŠ = 0 + |ð| à ð ð¥
ðð·yÌ = 0 + |ð| Ã MDxÌ
ð ðŠ = 0 + |ð| à ð ð¥Ì
3 Mean deviation takes its minimum value ðð·ðð=
1
ðâ|ð â ðð| is minimum
15.18
when A = Median
4 Combined SD ð 12 = â
ð1ð12 + ð2ð2
2 + ð1ð12 + ð2ð2
2
ð1 + ð2
where ð1 = ï¿œÌ ï¿œ1 â ï¿œÌ ï¿œ12 and ð2 = ï¿œÌ ï¿œ2 â ï¿œÌ ï¿œ12
Note: If ï¿œÌ ï¿œ1 = ï¿œÌ ï¿œ2 , then ï¿œÌ ï¿œ1 = ï¿œÌ ï¿œ2 = ï¿œÌ ï¿œ12
Then ð1 = 0 & ð2 = 0
⎠ð 12 = âð1ð1
2 + ð2ð22
ð1 + ð2
Illustration Calculation Answer
Group I Group II
ð1 = 5 ð2 = 15
ï¿œÌ ï¿œ1 = 9 ï¿œÌ ï¿œ2 = 5
ð 1 = 0.8 ð 2 = 0.5
ï¿œÌ ï¿œ12 = 6
ð 12 = â5 Ã (0.8)2 + (15 Ã (0.5)2) + (5 Ã 32) + (15 Ã (â1)2)
5 + 15
ð1 = ï¿œÌ ï¿œ1 - ï¿œÌ ï¿œ12 = 9 â 6 = 3
ð2 = ï¿œÌ ï¿œ2 â ï¿œÌ ï¿œ12 = 5 â 6 = â1
1.83
Problem for SD under Change of scale and origin
Formula Calculation Answer ï¿œÌ ï¿œ =â ð
ð= ð + ðð
1 ð¿ = ð, ð, ð, ï¿œÌ ï¿œ =â ð¿
ð
ð + ð + ð
ð 3
ð ð = ð¿ â ð 4 â 2 2
ðð·xÌ =â|ð â XÌ |
ð
â|ð â 3|
3
2
3
ð ð = ââ(ð â XÌ )2
ð â
â(ð â 3)2
3 0.82
2 ð = 4, 5, 6, ï¿œÌ ï¿œ =â ð
ð
4 + 5 + 6
3 5
Change of Origin (ð = 2)
ðµðððð ð = ð + 2 yÌ = ð + ðï¿œÌ ï¿œ 2 + 1 Ã3 5
ð ð = ð¿ â ð 6â4 2
ðð·YÌ =â|ð â YÌ |
ð
â|ð â 5|
3
2
3
ð ð = ââ(ð â YÌ )2
ð â
â(ð â 3)2
3 0.82
3 ð = 0, 1, 2, ï¿œÌ ï¿œ =â ð
ð
0 + 1 + 2
3 1
Change of Origin (ð = â2)
ðµðððð ð = ð â 2 yÌ = ð + ðï¿œÌ ï¿œ â2 + 1 Ã3 1
ð ð = ð¿ â ð 2 â 0 2
ðð·YÌ =â|ð â YÌ |
ð
â|ð â 1|
3
2
3
15.19
ð ð = ââ(ð â YÌ )2
ð â
â(ð â 1)2
3 0.82
4 ð = 4, 6, 8, ï¿œÌ ï¿œ =â ð
ð
4 + 6 + 8
3 6
Change of Scale (ð = 2)
ðµðððð ð = ð à 2 yÌ = ð + ðï¿œÌ ï¿œ 0 + 2 Ã3 6
ð ð = ð¿ â ð 8 â 4 4
ðð·YÌ =â|ð â YÌ |
ð
â|ð â 6|
3
4
3
ð ð = ââ(ð â YÌ )2
ð â
â(ð â 6)2
3 1.64
5 ð = 1, 1.5, 2, ï¿œÌ ï¿œ =â ð
ð
1 + 1.5 + 2
3 1.5
Change of Scale (ð =1
2)
ðµðððð ð = ð Ã1
2 yÌ = ð + ðï¿œÌ ï¿œ 0 +
1
2Ã3 1.5
ð ð = ð¿ â ð 2 â 1 1
ðð·YÌ =â|ð â YÌ |
ð
â|ð â 1.5|
3
1
3
ð ð = ââ(ð â YÌ )2
ð â
â(ð â 1.5)2
3 0.41
6 ð = 7, 9, 11, ï¿œÌ ï¿œ =â ð
ð
7 + 9 + 11
3 9 Change of Origin and
change of scale
(ð = 3)&(ð = 2) ðµðððð ð = 3 + 2 à ð yÌ = ð + ðï¿œÌ ï¿œ 3 + 2 Ã3 9
ð ð = ð¿ â ð 11 â 7 4
ðð·xÌ =â|ð â XÌ |
ð
â|ð â 9|
3
4
3
ð ð = ââ(ð â XÌ )2
ð â
â(ð â 9)2
3 0.41
Coefficient of Variation (CV): ð¶ð =ð
ï¿œÌ ï¿œÌ Ã 100
Illustration Calculation Comparison
Group 1 Group II
ï¿œÌ ï¿œ1 = 9 ï¿œÌ ï¿œ2 = 5
ð 1 = 0.8 ð 2 = 0.5
ï¿œÌ ï¿œ12 = 6
ð¶ð(ðŒ) =0.8
9 Ã 100 = 8.88%
ð¶ð(ðŒðŒ) =0.5
5Ã 100 = 10%
ð¶ð(ðŒ) = 8.88% < ð¶ð(ðŒðŒ) = 10%
More Stable
More Consistent
Less Variable
Less Dispersed
Less Stable
Less Consistent
More Variable
More Dispersed
EXTRA PROBLEMS
Comparison between Arithmetic Mean and Geometric Mean
Question 1: Find the average rate of return.
15.20
Year 1 2 3
Rate of Return (r %) 10% 60% 20%
Answer: The average rate of return
Formula Calculation Answer
GM G = (ð1 à ð2 à ⊠à ðð)1
ð (1.10 Ã 1.60 Ã 1.20)1 3â 1.283 ðð 128.3% ðð 28.3%
AM XÌ =
â ð
ð
1.10 + 1.60 + 1.20
3
1.3 ðð 130% ðð 30%
which is not possible
Comparison between Arithmetic Mean and Harmonic Mean
Question 2: An aeroplane covered a distance of 800 miles with four different speeds of 100, 200, 300
and 400 m/p.h for the first, second, third and fourth quarter of the distance. Find the average speed in
miles per hour.
Answer: The average speed is given by the H.M. of the given set of data.
Formula Calculation Answer
H M ð»ð =ð
â1
ð
41
100+
1
200+
1
300+
1
400
192 m/p.h
AM XÌ =
â ð
ð
100 + 200 + 300 + 400
4
250 m/p.h,
which is not true
Combined Mean
Question 3: Two groups of students reported mean weights of 160 kg and 150 kg respectively. Find
out, when the weight of both the groups together be 155 kg?
Answer:
Given Data Formula Calculation Answer
Group I Group II
Number ð1 ð2
Mean (kg.) XÌ 1 = 160 XÌ 2 = 150
Combined Mean: XÌ 12 = 155kg
XÌ 12 =ð1XÌ 1 + ð2ï¿œÌ ï¿œ2
ð1 + ð2
155 =160ð1 + 150ð2
ð1 + ð2
155ð1 + 155ð2 = 160ð1 + 150ð2
ð1 = ð2
Question 4: Show that for any two numbers a and b, standard deviation is given by |ðâð|
2
Answer: For two numbers a and b, AM is given by XÌ =ð+ð
2
The variance is =â(ðð â XÌ )2
2
=(ð â
ð+ð
2)
2
+ (ð â ð+ð
2)
2
2=
(ðâð)
4
2
+ (ðâð)2
4
2=
(ð â ð)2
4 â¹ ð =
|ð â ð|
4
(The absolute sign is taken, as SD cannot be negative).
Question 5: Prove that for the first n natural numbers, ðð âð2â 1
12 .
Answer: for the first n natural numbers AM is given by
15.21
XÌ =1 + 2 + ⊠⊠⊠+ ð
ð=
ð(ð + 1)
2ð=
ð + 1
2
⎠ðð· = ââ ðð
2
ðâ XÌ 2 = â
12 + 22 + 32 ⊠⊠. . +ð2
ðâ (
ð + 1
2)
2
âð(ð + 1)(2ð + 1)
6ðâ (
ð + 1
2)
2
= â(ð + 1)(2ð + 1)
6â (
ð + 1
2)
2
â(ð + 1)(2ð + 1)
6â
ð + 1
2Ã
ð + 1
2= â(ð + 1) (
(2ð + 1)
6â
ð + 1
4)
â(ð + 1)(4ð + 2 â 3ð â 3)
12= â
ð2 â 1
12
Thus, SD of first n natural numbers is SD = âð2 â 1
12
15.22
COMPARISON BETWEEN MEASURES OF CENTRAL TENDENCY N
o
Mea
sure
s
Ari
thm
etic
Mea
n
Geo
met
ric
Mea
n
Har
mo
nic
Mea
n
Med
ian
Mo
de
Ran
ge
Qu
arti
le
Dev
iati
on
Mea
n
Dev
iati
on
Sta
nd
ard
Dev
iati
on
1 Well defined Yes Yes Yes Yes
No (when the
number of
observations is
small, then use
Empirical
Relationship)
Yes Yes A may be
XÌ , ðð, ðð Yes
2 Easy to calculate &
simple to understand Yes No No Yes
Location Method,
but not Grouping
method
Yes Yes Yes No
3 Based on all the
items Yes
Yes (but able
to find only
for Positive
Values)
Yes
(ONLY
positive
values
and no
â0â)
No No No No Yes Yes
4
capable of further
mathematical
treatment
Yes
Yes (Useful
for
calculation of
Index
Numbers)
Yes
Yes (but only in
Mean Deviation,
no combined
Median)
No
No (But in case
of Quality
control and stock
market
fluctuations)
No
No (Useful for
Economists and
Businessmen and
in public reports)
Yes
5 Good basis for
comparison Yes Not much Yes
6 Necessary for
arrange of data No No No Yes No ------Not on Discussion-----
7 Affected by extreme
values Yes
Yes (Not
much
compared to
Yes No No Yes No Less than SD Yes
15.23
AM)
8
Not Precise â Mis-
leading impressions
(E.g. Average
number of persons is
1.5 which is not
possible)
No
No No
Yes (except
when Median
lies in between
two values)
Yes (except on
continuous series) ------Not on Discussion-----
9 Location (Inspection)
Method No No No
Yes (on
arrangement) Yes ------Not on Discussion-----
10 Graphical Method Yes (using Ogive
Curves) ------Not on Discussion-----
11
Calculated in the
case of open end
class intervals
No No No Yes Yes No Yes Based on âAâ No
12
Affected by
sampling
fluctuations
No
(least) No No Yes Yes Yes Yes Yes
Less
affected
13
Affected by Change
of origin Yes Yes Yes Yes Yes No No No No
Affected by Change
of Scale Yes Yes Yes Yes Yes Yes Yes Yes Yes
15.24
Explanations to Formulae:
1. Geometric Mean
Logarithmic formulae of Geometric Mean
Individual Observation Discrete Continuous
GM = âð¥1 à ð¥2 à ⊠.à ð¥ðð
log ðº. ð = log âð¥1 à ð¥2 à ⊠.à ð¥ðð
= 1
ðlog(ð¥1 à ð¥2 Ã. . .à ð¥ð)
= 1
ð(log ð¥1 + log ð¥2 + ⊠. log ð¥ð)
= 1
ðâ log ð¥
GM = Anti log (1
ðâ log ð¥)
GM = âð¥1ð1 à ð¥2
ð2 à ⊠. ð¥ððð
ð
log ðº. ð = log âð¥1ð1 à ð¥2
ð2 à ⊠. ð¥ððð
ð
= 1
ð[(log ð¥1
ð1 à ð¥2ð2 à ⊠. ð¥ð
ðð)]
= 1
ð[log ð¥1
ð1 + log ð¥2ð2 + ⊠. log ð¥ð
ðð]
= 1
ð[ð1 log ð¥1 + ð2 log ð¥2 + ⯠ðð log ð¥ð]
= 1
ðâ ð log ð¥
GM = Antilog 1
ðâ ð log ð¥
GM = â
ð1ð1 Ã ð2
ð2 Ã
⊠.à ðððð
ð
log ðº. ð = log â
ð1ð1 Ã ð2
ð2 Ã
⊠à ðððð
ð
= 1
ð[(log ð1
ð1 à ð2ð2 à ⊠à ðð
ðð)]
= 1
ð[log ð1
ð1 + log ð2ð2 + ⊠. log ðð
ðð]
= 1
ð[ð1 log ð1 + ð2 log ð2 + ⯠ðð log ðð]
= 1
ðâ ð log ð
GM = Antilog
1
ðâ ð log ð
15.25
Standard Deviation:
ð = ââ(ð â XÌ )2
ð
â(ð â XÌ )2 = â[ð2 â 2ðXÌ + XÌ 2]
â(ð â XÌ )2 = â ð2 â â(2ðXÌ ) + â XÌ 2
â(ð â XÌ )2 = â ð2 â 2XÌ â ð + ðXÌ 2
â(ð â XÌ )2 = â ð2 â 2â ð
ðâ ð + ð.
â ð
ð.â ð
ð
â(ð â XÌ )2 = â ð2 â 2(â ð)2
ð+
(â ð)2
ð
â(ð â XÌ )2 = â ð2 â 2(â ð)2
ð+
(â ð)2
ð
â(ð â XÌ )2 = â ð2 â(â ð)2
ð(2 â 1)
â(ð â XÌ )2
ð=
â ð2 â(â ð)2
ð
ð
â(ð â XÌ )2
ð=
ð â ð2â(â ð)2
ð
ð=
â ð2
ðâ (
â ð
ð)
2
=â ð2
ðâ XÌ 2
15.26
Graphical Method
Weighted Average:
1. Calculate goodwill using weighted average method:
Profit 20,000 10,000 (7000)
Weight 3 2 1
Missing Frequency:
1. Given N = 581 and Mean = 15. Find the missing frequencies.
x 10 11 12 13 14 15 16 17 18 19
f 8 15 x 100 98 95 y 75 50 30
2. Given Mean = 47, Median = 45, Mode = 35 and N= 90. Find the missing frequencies.
Marks 01-10 11-20 21-30 31-40 41-50 51-60 61-70 71-80 81-90 91-100
Number of Students 3 7 x 17 12 y 8 8 6 6
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