STAT PPT 1 AVERAGES (useful in analysing)

WHAT IS AN AVERAGE?
• SINGLE VALUE
• USED TO REPRESENT ALL VALUES IN A DATA SERIES
• LIES IN BETWEEN THE EXTREME VALUES
• ALSO KNOWN AS MEASURE OF CENTRAL TENDENCY

WHY DETERMINE THE AVERAGE?
• DESCRIBE THE CHARACTERISTICS OF A SERIES
• DRAW STATISTICAL INFERENCES
• COMPARE TWO OR MORE SERIES
• FACILITATE IMPORTANT DECISIONS

IDEAL CHARACTERISTICS OF AN AVERAGE
• SIMPLE TO COMPUTE
• CLOSE REPRESENTATION
• OBJECTIVELY DETERMINED
• CAPABLE OF FURTHER ALGAEBRIC OPERATIONS

MATHEMATICAL AVERAGES
CALCULATED USING AN ALGAEBRIC FORMAULA
BASED ON ALL VALUES OF THE DATA SERIES

ARITHMETIC MEAN
• DEFINED AS SUM OF ALL OBSERVATIONS DIVIDED BY THE TOTAL NUMBER
OF OBSERVATIONS
• SIMPLEST FORM OF AVERAGE
• MOST POPULARLY USED
• CAN BE COMPUTED FOR INDIVIDUAL/DISCRETE AND CONTINUOUS SERIES

ROLL NO. MARKS
(X)
DEVIATIONS
(X-A)
1 5 -20
2 15 -10
3 25 0
4 35 10
5 45 20
6 55 30
TOTAL 180 30
MEAN CALCULATION USING DIRECT METHOD:
ҧ
𝑥 = ෍
𝑥
𝑛
= 180/6
= 30
MEAN CALCULATION USING ASSUMED MEAN
METHOD:
ҧ
𝑥 = 𝐴 + ൗ
σ 𝑑
𝑛
= 25 + 30/6
=25 + 5
=30
= A
EXAMPLE 1: INDIVIDUAL SERIES

DISCRETE SERIES
FEW DATA POINTS ALONG WITH FREQUENCY
• DIRECT METHOD
ҧ
𝑥 = ൗ
𝛴𝑓𝑥
𝛴𝑓
WHERE,
ҧ
𝑥 IS A.M.
σ 𝑓𝑋 IS SUM OF PRODUCT OF ALL
OBSERVATIONS WITH THE FREQUENCY
𝛴𝑓 IS THE TOTAL NUMBER OF
OBSERVATIONS
• ASSUMED MEAN METHOD
ҧ
𝑥 = 𝐴 + ൘
σ f𝑑
𝛴𝑓
WHERE,
ҧ
𝑥 IS A.M.
A IS THE ASSUMED MEAN VALUE
σ f𝑑 IS SUM OF THE PRODUCT OF
FREQUENCY AND DEVIATIONS FROM
THE ASSUMED MEAN VALUE
OBSERVATIONS

MARKS
(X)
FREQUEN
CY (f)
DEVIATIONS
d = (X-A)
fx fd
5 10 -20 50 -200
15 20 -10 300 -200
25 30 0 750 0
35 50 10 1750 500
45 40 20 1800 800
55 30 30 1650 900
180 6300 1800
MEAN CALCULATION USING DIRECT METHOD:
ҧ
𝑥 = ൗ
𝛴𝑓𝑥
𝛴𝑓
= 6300/180
= 35
MEAN CALCULATION USING ASSUMED MEAN
METHOD:
ҧ
𝑥 = 𝐴 + ൘
σ f𝑑
𝛴𝑓
= 25 + 1800/180
=25 + 10
=35
= A
EXAMPLE 2: DISCRETE SERIES

CONTINUOUS SERIES
CLASS INTERVAL AND FREQUENCY IS GIVEN
• DIRECT METHOD
ҧ
𝑥 = ൗ
𝛴𝑓𝑚
𝛴𝑓
WHERE,
ҧ
𝑥 IS A.M.
σ 𝑓𝑚 IS SUM OF PRODUCT OF
MIDPOINT OF CLASS INTERVAL WITH
THE FREQUENCY
OBSERVATIONS
• ASSUMED MEAN METHOD
ҧ
𝑥 = 𝐴 + ൘
σ f𝑑
𝛴𝑓
WHERE,
ҧ
𝑥 IS A.M.
σ f𝑑 IS SUM OF THE PRODUCT OF
FREQUENCY AND DEVIATIONS FROM
THE ASSUMED MEAN VALUE
OBSERVATIONS

• STEP DEVIATION METHOD
ҧ
𝑥 = 𝐴 + ൘
{σ f𝑑′
𝛴𝑓} 𝑥 𝑖
WHERE,
ҧ
𝑥 IS A.M.
σ f𝑑’ IS SUM OF THE PRODUCT OF FREQUENCY AND DEVIATIONS FROM THE
ASSUMED MEAN VALUE
𝛴𝑓 IS THE TOTAL NUMBER OF OBSERVATIONS
i IS THE CLASS SIZE
CONTINUOUS SERIES
CLASS INTERVAL AND FREQUENCY IS GIVEN

EXAMPLE 3
CLASS
INTERVAL
MIDPOINT
(m)
FREQUENCY
(f)
DEVIATIONS
d = (m-A)
d’ = (m-A)/i fm fd fd’
0-10 5 10 -20 -2 50 -200 -20
10-20 15 20 -10 -1 300 -200 -20
20-30 25 30 0 0 750 0 0
30-40 35 50 10 1 1750 500 50
40-50 45 40 20 2 1800 800 80
50-60 55 30 30 3 1650 900 90
180 6300 1800 180
= A

SOLUTION USING STEP DEVIATION
METHOD
ҧ
𝑥 = 𝐴 + ൘
{σ f𝑑′
𝛴𝑓} 𝑥 𝑖
= 25 + [180/180] x 10
= 35

OBSERVATIONS
• USING THE SAME DATA, WE GET SAME ANSWER WITH ALL THE
METHODS
• THIS IS THE OBJECTIVITY TRAIT OF THE METHOD
• TYPE OF SERIES HAS AN EFFECT ON CHOICE OF METHOD
• USE ANY METHOD UNLESS SPECIFIED
• IN CASE OF CONTINUOUS SERIES, STEP DEVIATION METHOD
IS MOST APPROPRIATE

PROBLEM 1
DAILY
EXPENSE
FREQUENCY
40-59 50
60-79 ?
80-99 500
100-119 ?
120- 139 50
FIND THE MISSING FREQUENCY FROM THE GIVEN DATA OF 1000 FAMILIES:
THE VALUE OF MEAN IS 87.5.

PROBLEM 1
CALCULATION OF MEAN
DAILY
EXPENSE
CLASS
INTERVAL
FREQUENCY MIDPOINT
(m)
d’ = (m – A)/i fd’
40-59 39.5 – 59.5 50 49.5 -2 - 100
60-79 59.5 – 79.5 X 69.5 -1 - X
80-99 79.5 – 99.5 500 A = 89.5 0 0
100-119 99.5 – 119.5 400 – X 109.5 1 400 - X
120- 139 119.5 – 139.5 50 129.5 2 100
TOTAL = 1000 400 – 2X

SOLUTION
ҧ
𝑥 = 𝐴 + ൘
{σ f𝑑′
𝛴𝑓} 𝑥 𝑖
87.5 = 89.5 + [400 -2X/1000] x 20
X = 250
400 – X = 150
DAILY
EXPENSE
FREQUENCY
40-59 50
60-79 250
80-99 500
100-119 150
120- 139 50

PROBLEM 2
THE SUM OF DEVIATIONS OF A CERTAIN NUMBER OF
OBSERVATIONS MEASURED FROM 4 IS 72 AND THE SUM OF
DEVIATIONS OF SAME OBSERVATIONS MEASURED FROM 7 IS -3.
FIND THE NUMBER OF OBSERVATIONS AND THEIR MEAN.

SOLUTION
THE SUM OF DEVIATIONS IS EXPLAINED AS:
෍ 𝑑 = ෍(𝑋 − 𝐴)
ACCORDINGLY, σ 𝑋 − 4 = 72 𝐴𝑁𝐷 σ 𝑋 − 7 = −3
LET n BE THE NUMBER OF OBSERVATIONS AND ҧ
𝑥 BE THE MEAN
OF THE SERIES
VALUE CAN BE DETERMINED BY FORMING THE EQUATION

SOLUTION
EQUATION 1
ҧ
𝑥 = 𝐴 + ൗ
σ 𝑑
𝑛
ҧ
𝑥 = 4 + ൗ
72
𝑛
n ҧ
𝑥 − 4n = 72
EQUATION 2
ҧ
𝑥 = 𝐴 + ൗ
σ 𝑑
𝑛
ҧ
𝑥 = 7 + ൗ
− 3
𝑛
n ҧ
𝑥 − 7n = - 3
SOLVING THE TWO EQUATIONS SIMULTANEOUSLY, n = 25 AND ത
𝐱 = 6.88

MATHEMATICAL PROPERTIES OF A.M.
• 𝛴 𝑥 − ഥ
𝑥 = 0
• 𝛴 𝑥 − ഥ
𝑥 2 IS MINIMUM
• IF EACH VALUE OF THE DATA SERIES IS CHANGED (+/ - / x / ÷)
BY A CONSTANT, THE MEAN VALUE UNDERGOES A SIMILAR
CHANGE
• CALCULATION OF COMBINED AVERAGE
COMBINED MEAN = N1X1 + N2X2/ N1 +N2

PROBLEM 3
THE AVERAGE MONTHLY WAGE OF ALL WORKERS IN A FACTORY IS Rs.
444. IF THE AVERAGE WAGES PAID TO MALE AND FEMALE ARE Rs. 480
AND Rs. 360 RESPECTIVELY, FIND THE PERCENTAGE OF MALE AND
FEMALE WORKERS IN THE FACTORY.
SOLUTION:
X1 = 480 ; X2 = 360; X12 = 444
LET THE NUMBER OF MALE AND FEMALE WORKERS BE N1 AND N2
RESPECTIVELY

SOLUTION
COMBINED MEAN = N1X1 + N2X2/ N1 +N2
444 = 480N1 + 360N2/ N1 +N2
36N1 - 84N2 = 0
N1/ N2 =84/36
N1: N2 =7 : 3
THUS, THE PERCENTAGE OF MALE AND FEMALE WORKERS IN
THE FACTORY IS 70% AND 30% RESPECTIVELY.

PROBLEM 4
THE MEAN MARKS OF 99 STUDENTS WAS FOUND TO BE 60.
LATER, IT IS DISCOVERED THAT A SCORE OF 35 WAS MISREAD AS
53, ANOTHER SCORE WAS TAKEN AS 63 INSTEAD OF 36 AND
SCORE OF 5 WAS NOT TAKEN INTO ACCOUNT. FIND THE
CORRECT MEAN.
n = 99 ; X = 60
ACTUAL WRONG DIFFERENCE CHANGE IN n
35 53 -18 --
36 63 - 27 --
5 0 + 5 + 1
TOTAL = - 40 +1

SOLUTION
• SUM OF ALL OBSERVATIONS = n X MEAN = 99 X 60 = 5940
• CHANGE IN n = 99 +1 = 100
• CORRECT MEAN = 5940 -40 / 100 = 59

WEIGHTED A.M.
• DIFFERENT WEIGHTS ( RELATIVE IMPORTANCE)
• ALL THE DATA IN A SERIES VALUES ARE ASSIGNED WEIGHTS
• USEFUL IN CONSTRUCTION OF INDEX NUMBERS;
STANDARDISED RATES LIKE BIRTH RATE

FORMULA AND CALCULATION
1.MULTIPLY THE NUMBERS IN THE GIVEN DATA SET BY THEIR
RESPECTIVE WEIGHTS
2.SUM UP THE PRODUCTS
3.DIVIDE THE SUM OF THE PRODUCTS OBTAINED BY THE
TOTAL OF THE WEIGHTS
ഥ
𝒙𝒘 =
σ 𝒘𝒙
𝜮𝒘

PROBLEM 5
• A student takes three 100-point exams in statistics and score 80, 80 and 95.
The last exam is much easier than the first two, so the teacher has assigned it
less weight. The weights for the three exams are:
• Exam 1: 40 % of grade obtained.
• What is the student’s final weighted average for statistics?

SOLUTION
SCORE WEIGHTS PRODUCT
80 0.4 32
80 0.4 32
95 0.2 19
1.0 83
ഥ
𝒙𝒘 =
σ 𝒘𝒙
𝜮𝒘
ഥ
𝒙𝒘 =
𝟖𝟑
𝟏
THUS, THE FINAL GRADE OF THE
STUDENT IS 83.

LIMITATIONS OF A.M.
• UNDULY AFFECTED BY EXTREME VALUES
• IN OPEN DISTRIBUTION A.M. MAY NOT BE THE APPROPRIATE
MEASURE
• NON-HOMOGENOUS DATA, AVERAGE MAY GIVE MISLEADING
CONCLUSION

GEOMETRIC MEAN
• DEFINED AS THE nth OF THE PRODUCT OF ALL THE n ITEMS IN
A SERIES
• USED TO MEASURE THE AVERAGE CHANGE IN A VARIABLE
OVERA PERIOD OF TIME
• G.M. = 𝑥1 ⋅ 𝑥2 −−−−−− ⋅ 𝑥𝑛

USE OF LOG AND ANTILOG
• TO SIMPLIFY THE CALCULATION OF G.M. WE USE LOG AND
ANTILOG
• INDIVIDUAL SERIES:
G.M. = A.L. (𝛴log X)/n
• DISCRETE SERIES:
G.M. = A.L. (𝛴𝑓x log X)/n
• CONTINUOUS SERIES:
G.M. = A.L. (𝛴 𝑓 x log m)/n

PROBLEM 6
THE ANNUAL RATE OF DEPRECIATION OF A MACHINE IS 20% IN
YEAR 1, 15% IN YEAR 2, 10% IN YEAR 3 AND 5% IN THE 4TH YEAR.
CALCULATE THE AVERAGE ANNUAL RATE OF DEPRECIATION.
YEAR RATE OF DEP BASE YEAR VALUE (X) log X
I 20 80 1.9031
II 15 85 1.9294
III 10 90 1.9542
IV 5 95 1.9777
7.7644

SOLUTION
INDIVIDUAL SERIES:
G.M. = A.L. (𝛴log X)/n
= A.L. ( 7.7644/ 4)
= A.L. (1.9411)
= 0.8732 OR 87.32%
AVERAGE ANNUAL DEPRECIATION = 100 – 87.32 = 12.68%

HARMONIC MEAN
• DEFINED AS RECIPROCAL OF THE A.M. OF THE RECIPROCALS OF THE ITEMS IN
A SERIES
• USEFUL TO ASSIGN LARGER WEIGHTS TO SMALLER VALUES
• INDIVIDUAL SERIES:
H.M. = n ÷ 𝛴(1/X)
• DISCRETE SERIES:
H.M. = 𝛴𝑓 ÷ (𝛴𝑓 /𝑋)
• CONTINUOUS SERIES:
H.M. = 𝛴𝑓 ÷ (𝛴𝑓 /m)

PROBLEM 7
IN A CERTAIN OFFICE A LETTER IS TYPED BY A IN 4 MINUTES.
THE SAME LETTER IS TYPED BY B, C, D IN 5, 6 AND 10 MINUTES
RESPECTIVELY. WHAT IS THE AVERAGE TIME TAKEN IN
COMPLETING A LETTER? HOW MANY LETTERS CAN BE TYPED IN
A WORKING DAY COMPRISING 8 HOURS?

SOLUTION
EMPLOYEE TIME TAKEN
(X)
(1/X)
A 4 0.250
B 5 0.200
C 6 0.167
D 10 0.100
0.717
(i)
H.M. = n ÷ 𝛴(1/X) ; n =4
= 4/0.717
= 5.58 MINUTES PER LETTER
(ii) NO. OF LETTERS TYPED IN AN 8
HOUR DAY
= (8 X 60) /5.58 = 86 LETTERS

RELATIONSHIP BETWEEN
A.M. ,G.M. AND H.M.
• FOR ANY SET OF UNEQUAL POSITIVE NUMBERS, THE FOLLOWING RELATIONSHIP
HOLDS:
A.M. > G.M. > H.M.
• IF ALL VALUES IN THE DATA SET ARE EQUAL, THEN THE THREE AVERAGES WOULD
ALSO BE EQUAL
• THUS, GEOMETRIC MEAN OF TWO NUMBERS IS EQUAL TO THE GEOMETRIC MEAN
OF THEIR ARITHMETIC AND HARMONIC MEAN.

POSITIONAL AVERAGES
REPRESENTS THE DATA SERIES
NOT BASED ON ALL ITEMS OF THE SERIES
CONSIDERED AS AVERAGE DUE TO SPECIFIC PROPERTY/POSITION

MATHEMATICAL vs. POSITIONAL AVERAGES
• THE POSITIONAL AVERAGES ARE NOT BASED ON ALL OBSERVATIONS BUT THE
MATHEMATICAL AVERAGES ARE BASED ON THE WHOLE DATA SET
• COMPUTATION OF MATHEMATICAL AVERAGES CALLS FOR APPLICATION OF WELL-
DEFINED FORMULAE WHILE POSITIONAL AVERAGES CAN BE OBTAINED BY MERE
INSPECTION (IN RAW DATA AND IN DISCRETE FREQUENCY DISTRIBUTIONS)
• WHILE POSITIONAL AVERAGES CAN BE OBTAINED BY USING GRAPHS (IN CASE OF
CONTINUOUS FREQUENCY DISTRIBUTIONS), THE MATHEMATICAL AVERAGES
NEED TO BE CALCULATED

• THE POSITIONAL AVERAGES CAN BE CALCULATED IN OPEN-ENDED
DISTRIBUTIONS WHILE THE MATHEMATICAL AVERAGES CANNOT BE
• ALGEBRAIC MANIPULATION OF POSITIONAL AVERAGES IS NOT POSSIBLE. ON THE
OTHER HAND, THE MATHEMATICAL AVERAGES ARE AMENABLE TO FURTHER
ALGEBRAIC TREATMENT AS THEY HAVE WELL-DEFINED MATHEMATICAL
PROPERTIES
• THE MATHEMATICAL AVERAGES CAN BE DISTINGUISHED AS BEING SIMPLE AND
WEIGHTED. THERE IS NO QUESTION OF ANY POSITIONAL AVERAGES BEING
‘WEIGHTED’.

• IN COMPARISON TO MATHEMATICAL AVERAGES, THE POSITIONAL AVERAGES ARE
AFFECTED TO A LARGER EXTENT BY FLUCTUATIONS OF SAMPLING
• SAMPLING FLUCTUATIONS MEAN FLUCTUATIONS IN THE VALUES OF AN AVERAGE
OBTAINED FROM DIFFERENT SAMPLES OF SAME SIZE TAKEN FROM A GIVEN
POPULATION

MEDIAN
• MIDDLE VALUE OF A DISTRIBUTION
• 50% PERCENT OF THE VALUES LIE BELOW THE MEDIAN AND
50% PERCENT VALUES LIES ABOVE IT
• MEDIAN SPLITS THE DATA SERIES INTO TWO EQUAL HALVES

FORMULA
• INDIVIDUAL SERIES ( n IS ODD)
(i) ARRANGE ALL THE ITEMS IN A SERIES
IN A LOGICAL ORDER
(ii) THE MEDIAN POSITION WILL BE
CALCULATED AS FOLLOWS:
MEDIAN POSITION = [(n +1)/2]TH ITEM
WHERE,
n IS THE TOTAL NUMBER OF OBSERVATIONS
THUS, MIDDLE VALUE AMONG THE
ARRANGED DATA SET IS THE VALUE OF
MEAN
• INDIVIDUAL SERIES (n IS EVEN)
(i) ARRANGE ALL THE ITEMS IN A SERIES
IN A LOGICAL ORDER
(ii) THE MEDIAN POSITION WILL BE
CALCULATED AS FOLLOWS:
MEDIAN POSITION = [(n +1)/2]TH ITEM
n IS THE TOTAL NUMBER OF OBSERVATIONS
MEDIAN VALUE = MEAN{[(n/2) + 1TH] AND
(n/2)TH ITEM]}

PROBLEM 8
• FIND THE MEDIAN OF THE DAILY WAGES OF 7 WORKERS:
1000, 1500, 800, 900 1600, 2000, 1400
ALSO, FIND THE MEDIAN IF ANOTHER WORKER WITH A DAILY
WAGE OF ₹ 984 IS ADDED TO THE ABOVE DISTRIBUTION.

SOLUTION
• ARRANGE ALL OBSERVATIONS IN A LOGICAL ORDER:
800, 900, 1000, 1400, 1500, 1600, 2000
• Me position = (n+1/2)th item = (7+1)/2 = 4th item i.e. Me= 1400

SOLUTION
• ARRANGE ALL OBSERVATIONS IN A LOGICAL ORDER:
800, 900, 984, 1000, 1400, 1500, 1600, 2000
• Me position = (n+1/2)th item = (8+1)/2 = 4.5th item
• Me = Mean of 4th and 5th item of the series
• Me = (1000 + 1400)/2 = 1200

DISCRETE FREQUENCY DISTRIBUTION
• MEDIAN IS GIVEN BY THE VALUE OF (n+1)/2th ITEM
• TO DETERMINE MEDIAN, THE ‘LESS THAN’ CUMULATIVE FREQUENCIES ARE
OBTAINED
• LOCATE THE VALUE OF THIS ITEM HAVING REFERENCE TO THE CUMULATIVE
FREQUENCIES COLUMN
• FOR THIS, WE START FROM THE TOP AND MOVE DOWNWARD AS LONG AS THE
VALUE (n+1)/2 EXCEEDS THE CUMULATIVE FREQUENCY VALUES, AND STOP ONCE
THIS VALUE IS LOCATED.

PROBLEM 9
FIND THE MEDIAN OF THE FOLLOWING DISTRIBTUION OF MARKS OF
STUDENTS:
MARKS 100 150 80 200 250 180
NUMBER OF STUDENTS 24 26 16 20 6 30
1. ARRANGE THE ABOVE DISTRIBUTION IN ASCENDING ORDER:
MARKS 80 100 150 180 200 250

SOLUTION
2. CALCULATE CUMULATIVE FREQUENCY:
MARKS 80 100 150 180 200 250
CUMULATIVE FREQUENCY 16 40 66 96 116 122
MEDIAN IS GIVEN BY THE VALUE OF (n+1)/2th ITEM
Me = (122 +1)/2 = 61.5th ITEM
61.5th ITEM IS 150

ALGEBRAIC METHOD
(USED IN CONTINUOUS DISTRIBUTION)
• TO OBTAIN THE VALUE OF MEDIAN FIND THE CUMULATIVE FREQUENCIES
• CALCULATE n/2 AND THEN, WITH REFERENCE TO THE CUMULATIVE FREQUENCIES
COLUMN, DETERMINE THE CLASS INTERVAL IN WHICH MEDIAN FALLS.
• THIS IS TERMED AS THE MEDIAN CLASS. THE FOLLOWING FORMULA IS USED:
𝑀𝑒 = 𝑙1 +
𝑛
2
− 𝑐. 𝑓.
𝑓
𝑋 𝑖

PROBLEM 10
THE FOLLOWING IS THE DISTRIBUTION OF A SAMPLE OF 200 COMPANIES
ACCORDING TO THE PROFITS EARNED BY THEM IN THE LAST YEAR.
DETERMINE MEDIAN PROFIT FROM THE DISTRIBUTION.
PROFITS (IN ₹ LAKH) 0-25 25-50 50-75 75-100
NUMBER OF COMPANIES 30 50 80 40

SOLUTION
1. COMPUTE THE CUMULATIVE FREQUENCY
PROFITS (IN ₹ LAKH) 0-25 25-50 50-75 75-100
NUMBER OF COMPANIES 30 50 80 40
CUMULATIVE FREQUENCY 30 80 160 200
2. DETERMINE THE MEDIAN CLASS AS THE CLASS INTERVAL IN WHICH THE (n/2)th
ITEM LIES: 50-75
3. 𝑀𝑒 = 𝑙1 +
𝑛
2
− 𝑐.𝑓.
𝑓
𝑋 𝑖 = 50 +
100 − 80
80
𝑋 25 = 56.25

PROPERTIES OF MEDIAN
• THE SUM OF ABSOLUTE DEVIATIONS FROM MEDIAN IS THE MINIMUM.
• THAT IS TO SAY, THE AGGREGATE ABSOLUTE DEVIATIONS FROM A POINT
OTHER THAN MEDIAN CANNOT BE SMALLER THAN THE CORRESPONDING
AGGREGATE FROM MEDIAN OF THE OBSERVATIONS.
• SYMBOLICALLY, Σ|X –Me| IS MINIMUM

OTHER POSITIONAL VALUES
• MEDIAN IS THE VALUE THAT DIVIDES A DISTRIBUTION INTO TWO EQUAL
PARTS
• THERE ARE OTHER MEASURES AS WELL WHICH ARE USED TO DIVIDE THE
DISTRIBUTION IN MORE THAN TWO PARTS. THESE MEASURES INCLUDING
MEDIAN ARE KNOWN AS PARTITION VALUES
• THE VALUES THAT DIVIDE A SERIES OR DISTRIBUTION INTO FOUR PARTS
ARE TERMED AS QUARTILES,
• MEASURES THAT DIVIDE A SERIES INTO 10 AND 100 PARTS ARE CALLED,
RESPECTIVELY, DECILES AND PERCENTILES.

OTHER POSITIONAL VALUES
• FOR A DISTRIBUTION, THERE ARE THREE QUARTILES, NINE DECILES AND
NINETY-NINE PERCENTILES
• IT MAY BE NOTED THAT THE VALES OF SECOND QUARTILE, FIFTH DECILE
AND 50TH PERCENTILE ARE IDENTICAL TO THE MEDIAN VALUE.
• THE FIRST AND THIRD QUARTILES OF A DISTRIBUTION ARE RESPECTIVELY
CALLED LOWER QUARTILE AND UPPER QUARTILE.
• THE PARTITION VALUES REFLECT HOW THE VALUES IN DATA ARE
DISTRIBUTED AND THEY HELP TO LOCATE THE RELATIVE POSITION OF A
CERTAIN VALUE IN A GIVEN SERIES/DISTRIBUTION.

CALCULATION OF POSITIONAL VALUES
• THE PRINCIPLES AND METHODOLOGY OF USED FOR COMPUTING VARIOUS
PARTITION VALUES IS SAME AS FOR COMPUTATION OF MEDIAN.
• LIKE IN CASE OF MEDIAN, THE INDIVIDUAL OBSERVATIONS ARE
ARRANGED IN THE ASCENDING ORDER OF MAGNITUDE.
• THEN n IS DIVIDED BY 4, 10 OR 100 ACCORDINGLY AS QUARTILE, DECILES
OR PERCENTILES ARE REQUIRED TO BE CALCULATED, AND MULTIPLIED BY
KWHERE KTH QUARTILE, DECILE OR PERCENTILE IS DESIRED.

CALCULATION OF POSITIONAL VALUES
• THE FORMULAE, IN GENERAL FORM, FOR OBTAINING VARIOUS PARTITION
VALUES ARE:
• QUARTILEk = 𝑙1 +
𝑘(𝑛)
4
− 𝑐.𝑓.
𝑓
𝑋 𝑖
• DECILEk= 𝑙1 +
𝑘(𝑛)
10
− 𝑐.𝑓.
𝑓
𝑋 𝑖
• PERCENTILEk= 𝑙1 +
𝑘(𝑛)
100
− 𝑐.𝑓.
𝑓
𝑋 𝑖

PROBLEM 11
DETERMINE THE MEDIAN FOR THE FOLLOWING DISTRIBUTION
OF WEEKLY INCOME IN AN OFFICE:
WEEKLY INCOME (IN ₹ ‘000)
(EQUAL TO OR MORE THAN)
12 11 10 8 6 4 3 2 1
NUMBER OF WORKERS 0 0 14 26 42 54 62 70 80
CALCULATE:
• NUMBER OF WORKERS WITH WEEKLY INCOME BETWEEN ₹ 2,400 AND ₹ 10,500
• MAXIMUM INCOME OF THE LOWEST 25% WORKERS

SOLUTION
REWRITE THE GIVEN TABLE IN THE FORM OF CLASS INTERVALS:
WEEKLY INCOME (IN ₹ ‘000) 0-1 1-2 2-3 3-4 4-6 6-8 8-10 10-11 11-12
CUMULATIVE FREQUENCY 0 10 18 26 38 54 66 80 80
MEDIAN CLASS WILL BE 6,000 – 8,000
Me = 𝑀𝑒 = 𝑙1 +
𝑛
2
− 𝑐.𝑓.
𝑓
𝑋 𝑖 = 6000 +
40 − 38
16
𝑋 2000 = ₹ 6,250

SOLUTION
TO DETERMINE NUMBER OF WORKERS BETWEEN ₹ 2,400 AND ₹ 10,500:
0-1 1-2 2-3 3-4 4-6 6-8 8-10 10-11 11-12
ADD THE FREQUENCY BETWEEN THE GIVEN LIMITS:
8 X (600/1000) + 8 + 12 + 16 + 12 + 14 X (500/1000) = 60 WORKERS APPROXIMATELY

SOLUTION
TO DETERMINE MAXIMUM INCOME OF THE LOWEST 25% WORKERS:
0-1 1-2 2-3 3-4 4-6 6-8 8-10 10-11 11-12
DETERMINE FIRST QUARTILE: (n/4)th ITEM = 80/4 = 20th ITEM LIES IN INTERVAL 3,000
– 4,000
Q1 = 𝑙1 +
𝑛
4
− 𝑐.𝑓.
𝑓
𝑋 𝑖 = 3000 +
20 − 18
8
𝑋 1000 = ₹ 3,250

PROBLEM 12
USE THE FOLLOWING DATA TO COMPUTE THE UPPER AND
LOWER QUARTILES, D2, P5 AND P90.
MARKS BELOW 10 10-20 20-40 40-60 60-80 ABOVE 80
FREQUENCY 8 10 22 25 10 5
STEP 1: COMPUTE CUMULATIVE FREQUENCY
FREQUENCY 8 10 22 25 10 5
CUMULATIVE
FRQUENCY
8 18 40 65 75 80

SOLUTION
STEP 2: ASCERTAIN THE CI FOR EACH POSITIONAL VALUE:
FREQUENCY 8 10 22 25 10 5
CUMULATIVE
FRQUENCY
8 18 40 65 75 80
P5 D2
Q1 Q3 P90

SOLUTION
• D2 = 18
• P5 = 5
• Q1 = 21.82
• Q3 = 56
• P90 = 74

MODE
• MODE IS THE VALUE IN THE DATA WHICH OCCURS MOST FREQUENTLY
• IT IS TYPICAL IN THE SENSE THAT IT IS THE MOST PROBABLE VALUE
• IN THE CONTEXT OF A CONTINUOUS FREQUENCY DISTRIBUTION, MODE IS
DEFINED AS THAT VALUE OF THE VARIABLE WHERE THERE IS HIGHEST
CONCENTRATION OF OBSERVATIONS
• IT IS THE VALUE OF THE VARIABLE UNDER CONSIDERATION CORRESPONDING TO
THE HIGHEST ORDINATE OF THE SMOOTH FREQUENCY CURVE

CALCULATION OF MODE
• IN AN INDIVIDUAL SERIES, THE VALUE OF MODE IS THE SIZE OF THE ITEM
REPEATING THE HIGHEST NUMBER OF TIMES
• IN THE FOLLOWING DATA SET:
10, 6, 9, 7, 10, 8, 11, 7, 10, AND 12
THE VALUE 10 APPEARS MORE FREQUENTLY THAN ANY OTHER VALUE
THEREFORE, MODE = 10

CALCULATION OF MODE
• IN CASE OF DISCRETE FREQUENCY DISTRIBUTIONS, THE VALUE OF
MODE IS THE VALUE CORRESPONDING TO WHICH THE FREQUENCY IS
HIGHEST
EXAMPLE,
MARKS 10 14 16 22
FREQUENCY 8 18 40 65
THE VALUE 22 HAS THE HIGHEST FREQUENCY
THEREFORE, MODE = 22 MARKS

CALCULATION OF MODE
• ALGAEBRIC METHOD OF CALCULATION OF MODE:
𝑀𝑂𝐷𝐸 = 𝑀𝑜 = 𝑙1 + (
𝑓1
−𝑓0
|2𝑓1
−𝑓0
−𝑓2
|
) x i
• HERE, l1 IS THE LOWER LIMIT OF THE MODAL CLASS,
f1 IS FREQUENCY OF THE MODAL CLASS,
f0 AND f2 ARE RESPECTIVE FREQUENCIES OF THE PRE-MODAL AND POST
MODAL CLASSES, AND
i = WIDTH OF THE MODAL CLASS.

PROBLEM 13
Weight 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90
Frequency 4 12 40 41 27 13 9 4
CALCULATE THE VALUE OF MODE FOR THE GIVEN DATA
STEP 1: DETERMINE THE MODAL CLASS:
SINCE THE FREQUENCY OF 2 CLASS INTERVALS IS VERY CLOSE, PREPARE THE
GROUPING AND CLASSIFICATION TABLES.

GROUPING TABLE
TO DETERMINE THE CI WITH HIGHEST FREQUENCY
Weight 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90
Frequency 4 12 40 41 27 13 9 4
2 VALUE 16 81 40 13
2 VALUE - 52 68 24 -
3 VALUE 56 81 - -
3 VALUE - 93 49 -
3 VALUE - - 108 26

CLASSIFICATION TABLE
TALLY MARKS UNDER THE CI WITH HIGHEST FREQUENCY
CI 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90
I (f) 1
II 1 1
III 1 1
IV 1 1 1
V 1 1 1
VI 1 1 1
TOTAL 0 1 3 6 3 1 0 0
MODAL CLASS

COMPUTATION OF MODE
• ALGAEBRIC METHOD OF CALCULATION OF MODE:
𝑀𝑂𝐷𝐸 = 𝑙1 + (
𝑓1
−𝑓0
2𝑓1
−𝑓0
−𝑓2
) x i
𝑀𝑜 = 40 + (
41 − 40
2𝑋 41 −40 −27
) x 10
= 40.67

EMPIRICAL METHOD
IN CASE A SERIES IS NOT UNIMODAL
• MODE CAN ALSO BE ESTIMATED BY USING MEAN AND
MEDIAN IN CASE A SERIES IS NOT UNIMODAL
• ACCORDING TO THE EMPIRICAL RULE,
MODE = 3MEDIAN – 2MEAN

CHOICE OF AVERAGE
• MOST COMMONLY USED IS ARITHMETIC MEAN
• G.M. AND H.M. TO BE USED WHERE THE LARGER WEIGHTS
ARE TO BE ASSIGNED TO LOWER VALUES AND VICE-VERSA
• MEDIAN TO BE USED IN CASE OF EXTREME VALUES IN A
SERIES AND OPEN ENDED DISTRIBUTION
• MODE IS SUITABLE WHERE THE PURPOSE IS TO DETERMINE
THE MOST TYPICAL VALUE

STAT PPT 1 AVERAGES (useful in analysing)

More Related Content

Similar to STAT PPT 1 AVERAGES (useful in analysing) (20)

Recently uploaded (20)

STAT PPT 1 AVERAGES (useful in analysing)