Central Tendency & Dispersion

MEASURES OF CENTRAL
TENDENCY & DISPERSION

BirinderSingh,AssistantProfessor,PCTE
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INTRODUCTION
Measures of central tendency are statistical
measures which describe the position of a
distribution
They are also called statistics of location, and
are the complement of statistics of dispersion,
which provide information concerning the
variance or distribution of observations.
In the univariate context, the mean, median and
mode are the most commonly used measures of
central tendency.
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CENTRAL TENDENCY OR AVERAGE
 An average is a single value which represents the
whole set of figures and all other individual items
concentrate around it.
 It is neither the lowest value in the series nor the
highest it lies somewhere between these two extremes.
 The average represents all the measurements made on
a group, and gives a concise description of the group as
a whole.
 When two are more groups are measured, the central
tendency provides the basis of comparison between
them.
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DEFINITION
Simpson and Kafka defined it as “ A measure of
central tendency is a typical value around which other
figures congregate”
Waugh has expressed “An average stand for the whole
group of which it forms a part yet represents the whole”.
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FUNCTIONS OF AVERAGE
 Brief Description
 Helpful in Comparison
 Helpful in formulation of policies
 Basis of Statistical Analysis
 Representation of the Universe
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CHARACTERISTICS OF A GOOD AVERAGE
 Easy to understand
 Simplified
 Uniquely defined
 Represent the whole group or data
 Not affected by extreme values
 Capable of further algebraic treatment
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TYPES OF AVERAGES
Types of
Averages
Mathematical
Averages
Arithmetic
Mean (AM)
Geometric
Mean (GM)
Harmonic
Mean (HM)
Positional
Averages
Median (M) Mode (Z)
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ARITHMETIC MEAN
 Most popular & widely used measure
 It is defined as the value which is obtained by
adding all the items of a series and dividing this
total by the number of items.
 It is also referred as mean and is denoted as 𝑋
where 𝑋 =
𝐸𝑋
𝑁
 It is of two types:
 Simple Arithmetic Mean
 Weighted Arithmetic Mean
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METHODS TO SOLVE INDIVIDUAL SERIES
 Direct Method
𝑋 =
𝐸𝑋
𝑁
;
𝑋 = Arithmetic Mean, EX = Sum of observations
N = No. of observations
 Shortcut Method
𝑋 = A +
𝐸𝑑𝑥
𝑁
;
A = Assumed Mean, dx = X-A
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PRACTICE PROBLEMS
Q1: The pocket allowances of 10 students are as:
15, 20, 30, 22, 25, 18, 40, 50, 55, 65.
Calculate the A.M. of their allowances using
both methods.
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METHODS TO SOLVE DISCRETE SERIES
 Direct Method
𝑋 =
Ʃ𝑓𝑋
𝑁
;
𝑋 = Arithmetic Mean, f = frequency, N = Ʃf
 Shortcut Method
𝑋 = A +
Ʃ𝑓𝑑𝑥
𝑁
;
A = Assumed Mean, dx = X-A, N = Ʃf
.
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PRACTICE PROBLEMS
Q1: Calculate Arithmetic Mean from the following
data using both methods:
Wages 10 20 30 40 50
No. of Workers 4 5 3 2 5
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METHODS TO SOLVE CONTINUOUS SERIES
 Direct Method
𝑋 =
Ʃ𝑓𝑋
𝑁
;
𝑋 = Arithmetic Mean, f = frequency, N = Ʃf
 Shortcut Method
𝑋 = A +
Ʃ𝑓𝑑𝑥
𝑁
;
A = Assumed Mean, 𝑑𝑥 = X-A, N = Ʃf
 . Step Deviation Method
𝑋 = A +
Ʃ𝑓𝑑𝑥′
𝑁
x 𝑖 ;
A = Assumed Mean, 𝑑𝑥′ =
𝑋 −𝐴
𝑖
, N = Ʃf,
i = common factor
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PRACTICE PROBLEMS
Q1: Compute AM by Step Deviation Method:
Ans: 25.85
(Inclusive Series)
Ans: 37.83
Marks 0-10 10-20 20-30 30-40 40-50
No. of students 20 24 40 36 20
Size 20-29 30-39 40-49 50-59 60-69
Frequency 10 8 6 4 2
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16
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PRACTICE PROBLEMS
(Cumulative Frequency Series)
Ans: 25.81
(Cumulative Frequency Series)
Ans: 6.33
Marks
Less
than 10
Less
than 20
Less
than 30
Less
than 40
Less
than 50
Size
More
than 0
More
than 2
More
than 4
More
than 6
More
than 8
Frequency 30 28 24 18 10
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PRACTICE PROBLEMS
Q5: Find the missing frequency if mean is 52:
Ans: 7
Q6: Find the missing value if mean of the data is
115.86:
Ans: 120
Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80
No. of
students
5 3 4 ? 2 6 13
Wages 110 112 113 117 ? 125 128 130
No. of workers 25 17 13 15 14 8 6 2
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PRACTICE PROBLEMS
Q7: The sum of the deviations of a certain number of items
measured from 2.5 is 50 and from 3.5 is -50. Find number of
items and mean. Ans: 100, 3
Q8: The mean height of 25 male workers is 61 cms and the
mean height of 35 female workers is 58 cms. Find the
combined mean height of 60 workers in the factory.
(Combined AM) Ans: 59.25
Q9: The mean wage of 100 workers in a factory, running in
two shifts of 60 and 40 workers respectively is Rs. 38. The
mean wage of 60 workers working in the morning shift is
Rs. 40. Find the mean wage of 40 workers working in the
evening shift.
(Combined AM) Ans: 35
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PRACTICE PROBLEMS – TYPICAL EXAMPLES
Q10: The mean monthly salary paid to all
employees in a certain company was Rs. 600. The
mean monthly salaries paid to male & female
employees were Rs. 620 and Rs. 520 respectively.
Find the percentage of male to female employees in
the company.
Q11: A bookseller has 150 books of Physics &
Chemistry whose average price is Rs. 40 per book.
Average price of Physics book is Rs. 43 and that of
Chemistry is Rs. 35. Find the number of books of
each subject with the book seller.
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PRACTICE PROBLEMS – TYPICAL EXAMPLES
Q12: The mean of 100 items is 80. By mistake one item
is misread as 92 instead of 29. Find the correct mean.
(Ans: 79.37)
Q13: The mean of 5 observations is 7. Later on it was
found that two observations 4 and 8 were wrongly taken
instead of 5 and 9. Find the correct mean.
(Ans: 7.4)
Q14: The average daily price of a share of a company
from Monday to Friday was Rs. 130. If the highest &
lowest price during the week were Rs. 200 and Rs. 100
respectively, find the average daily price when the
highest & lowest price are not included. (Ans: 116.67)
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DISADVANTAGES OF MEAN:
• It is affected by extreme values.
• It cannot be calculated for open end
classes.
• It cannot be located graphically
• It gives misleading conclusions.
• It has upward bias.
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GEOMETRIC MEAN (GM)
 It is defined as the nth root of the product of all
the n values of the variable.
 GM = 𝑋1. 𝑋2. 𝑋3. … . . 𝑋𝑛
𝑛
where
𝑋1. 𝑋2. 𝑋3. … . . etc. are the various values of the series
n = number of items
 If there are two items in a series say 2 & 8, then their
GM = 2 𝑥 8
2
= 16
2
= 4
 If there are three items in a series say 2 ,4 & 8, then
their GM = 2 𝑥 4 𝑥 8
3
= 64
3
= 4
 If, the number of items in a series is very large, it
would be difficult to calculate the geometric mean.
 For calculating GM, log values are used.
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GEOMETRIC MEAN – USING LOGARITHMIC
CALCULATION
 GM = (𝑋1. 𝑋2. 𝑋3. … . . 𝑋𝑛)1/ 𝑛
Taking log on both sides
Log GM =
1
𝑛
(log𝑋1 + 𝑙𝑜𝑔𝑋2 + 𝑙𝑜𝑔𝑋3 + … . . +𝑙𝑜𝑔𝑋𝑛)
Log GM =
Ʃ 𝑙𝑜𝑔𝑋
𝑛
Taking Antilog on both sides
GM = Antilog (
Ʃ 𝑙𝑜𝑔𝑋
𝑛
)
 The values of logs and antilog can be obtained from the
log table and antilog tables.
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CALCULATION OF GM – INDIVIDUAL
SERIES
 Find the logarithms of the given values.
 Find the sum total of logs, i.e. Ʃ log X.
 Divide Ʃ log X by the number of items (N)
 Calculate Antilog of the value
 The result will be the Geometric Mean
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PRACTICE PROBLEMS – INDIVIDUAL
SERIES
Q1: Calculate Geometric Mean of the following
series:
i. 180, 190, 240, 386, 492, 662
ii. 2574, 475, 75, 5, 0.8, 0.08, 0.005, 0.0009
iii. 0.9841, 0.3154, 0.0252, 0.0068, 0.0200, 0.0002,
0.5444, 0.4010
Answers:
i. 317.9
ii. 1.841
iii. 0.0511
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CALCULATION OF GM – DISCRETE SERIES
& CONTINUOUS SERIES
 Find the logarithms of the given values i.e. log X
 Multiply the frequency with the corresponding
log values
 Find the sum total of logs, i.e. Ʃ f log X.
 Divide Ʃ f log X by the total number of frequencies
(N = Ʃf)
 Calculate Antilog of the value
 The result will be the Geometric Mean
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PRACTICE PROBLEMS – DISCRETE &
CONTINUOUS SERIES
Q4: Calculate Geometric Mean of the following:
(Ans: 8.82)
Q5: Calculate Geometric Mean of the following:
(Ans: 19.27)
X 6 7 8 9 10 11 12
F 8 12 18 26 16 12 8
Marks 0-10 10-20 20-30 30-40 40-50
Students 3 4 6 3 2
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COMBINED GEOMETRIC MEAN
 If G1 & G2 are the Geometric Means of two
groups having N1 & N2 items, then the combined
GM is given by the following formula:
G12 = Antilog (
𝑁1
log 𝐺1
+𝑁2
log 𝐺2
𝑁1
+𝑁2
)
Q6: The GM of a sample of 10 items was found to
be 20 and that of a sample of 20 items was found to
be 15. Find the combined GM. (Ans: 16.51)
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WEIGHTED GEOMETRIC MEAN
 If G1 & G2 are the Geometric Means of two
groups having W1, W2, …… weights, then the
weighted GM is given by the following formula:
G12 = Antilog (
𝑊1
log 𝐺1
+𝑊2
log 𝐺2
+ …..
𝑊1
+𝑊2
+ ……..
)
Q6: Calculate weighted GM: (Ans: 119.3)
Items Index No. Weights
Food 120 7
Rent 110 5
Clothing 125 3
Fuel 105 2
Others 140 3
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AVERAGE RATE OF GROWTH OF
POPULATION
 GM is also used to compute the average annual
percent increase in population, prices when the
values of the variables at the beginning of the first
and at the end of the nth period are given. The
average annual percent increase may be computed by
applying the formula:
Pn = P0 (1+r)n where
r = the average rate of growth
n = Number of years
P0 = Value at the beginning of the period
Pn = Value at the end of the period
Formula: r = Antilog (
log 𝑃 𝑛 −log 𝑃0
𝑛
) – 1
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PRACTICE PROBLEMS – AVERAGE RATE OF
GROWTH OF POPULATION
Q7: The population of a country has increased from
84 million in 1983 to 108 million in 1993. Find the
annual rate of growth of population. (Ans: 2.6%)
Q8: The population of a town was 10000. At first, it
increased at the rate of 3% p.a. for the first three
years and then it decreased at the rate of 2% p.a.
for the next 2 years. What will be the population of
the town after 5 years? (Ans: 10494)
Q9: The population of a city was 1,00,000 in 1980
and 1,44,000 in 1990. Estimate the population at
the middle of 1980-1990. (Ans: 120000)
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HARMONIC MEAN (H.M.)
 It is a mathematical average.
 It is based on the reciprocal of items.
 It is defined as the reciprocal of the artihmetic
average of the reciprocal of the values of irs
various items.
 HM =
𝑁
Σ (
1
𝑋
)
 It is useful in finding averages involving speed,
time, price and ratios.
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CALCULATION OF HARMONIC MEAN –
INDIVIDUAL SERIES
 Find out the reciprocals of the values of the series
 Add the values of the reciprocals to get Σ(
1
𝑥
)
 Divide the number of items by the sum total of
reciprocals.
 The final value is the Harmonic Mean.
Q1: Calculate HM of the following:
2, 4, 7, 12, 19 Ans: 4.86
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CALCULATION OF HARMONIC MEAN –
DISCRETE & CONTINUOUS SERIES
 Divide each frequency by the respective values of
the variable.
 Obtain the total Σ(
𝑓
𝑥
)
 Divide the number of items by the Σ(
𝑓
𝑥
)
 The final value is the Harmonic Mean.
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PRACTICE PROBLEMS – H.M.
Q2: The following table gives the marks obtained
by students in a class. Calculate the H.M.
Ans: 23.24
Q3: Calculate the H.M. for the following:
Ans: 20.48
Marks 18 21 30 45
No. of students 6 12 9 2
Marks 0-10 10-20 20-30 30-40 40-50
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WEIGHTED HARMONIC MEAN
 Weighted H.M. =
Σ𝑊
Σ(
𝑊
𝑋
)
Q4: Find the weighted H.M. of the items 4, 7, 12,
19, 25 with weights 1, 2, 1, 1, 1 respectively.
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APPLICATIONS OF HARMONIC MEAN
Q5: An aeroplane covers the four sides of a square field at
speeds of 1000, 2000, 3000 and 4000 km/hr respectively.
What is its average speed? Ans: 1920
km/hr
Q6: A cyclist covers first three kms at an average speed of 8
km/hr, another 2 kms ar 3 km/hr and the last two kms at 2
km/hr. Find the average speed and verify your answer.
Ans: 3.42 km/hr
Q7: Typist A can type a letter in 5 minutes, B in 10 minutes
& C in 15 minutes. What is the average number of letters
typed per hour per typist? Ans: 7.33
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RELATIONSHIP BETWEEN
A.M., G.M. & H.M.
 G.M. = 𝐴. 𝑀. 𝑥 𝐻. 𝑀.
 When all the values of the series differ in size,
A.M. > G.M. > H.M.
 When all the values of the series are equal,
A.M. = G.M. = H.M.
Q8: If AM of two numbers is 10 and their GM is 8, find the
two numbers and HM. Ans: 16, 4, 6.4
Q9: Using the values 2, 4 and 8, verify that
A.M. > G.M. > H.M.
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Median is a central value of the distribution, or the value
which divides the distribution in equal parts, each part containing
equal number of items. Thus it is the central value of the variable,
when the values are arranged in order of magnitude.
Connor has defined as “ The median is that value of the variable
which divides the group into two equal parts, one part comprising
of all values greater, and the other, all values less than median”
MEDIAN
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CALCULATION OF MEDIAN – DISCRETE
SERIES
 Arrange the data in ascending or descending order.
 Calculate the cumulative frequencies.
 Apply the formula : M = Size of
𝑵+𝟏
𝟐
𝒕𝒉 𝒊𝒕𝒆𝒎
 Now locate
𝑁+1
2
th items in the cumulative frequency
column. The value to be selected which is equal to or higher
than
𝑁+1
2
value.
 Median is the value of the variable corresponding to the
selected value.
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PRACTICE PROBLEMS – DISCRETE SERIES
Q1: Calculate the median from the following data:
X: 10 12 14 16 18 20 22
f: 2 5 12 20 10 7 3
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CALCULATION OF MEDIAN – CONTINUOUS
SERIES
 Arrange the data in ascending order
 Calculate the cumulative frequencies.
 Find out the median size by using the formula: N/2
 Determine the median class in which median lies
 Apply the formula : M = l1 +
𝑵
𝟐
−𝒄.𝒇.
𝒇
x i; where
l1 = lower limit of the median class
c.f. = cumulative frequency of the class preceeding the median class
f = frequency of the median class
i = size of the class interval of the median class
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PRACTICE PROBLEMS – CONTINUOUS SERIES
Q1: Calculate the median from the following data:
(Ans: 17.5)
Q2: Calculate Median: (Ans: 36.25)
X: 0-5 5-10 10-15 15-20 20-25 25-30 30-35 35-40 40-45
f: 6 12 17 30 10 10 8 5 2
Value Frequency
Less than 10 4
Less than 20 16
Less than 30 40
Less than 40 76
Less than 50 96
Less than 60 112
Less than 70 120
Less than 80 125
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Q3: Calculate the median from the following data: (Inclusive)
(Ans: 25)
Q4: Calculate Median: (Unequal Class Interval)
(Ans: 32.67)
Q5: Calculate Median (Descending Class Intervals)
(Ans: 18.125)
X: 1--10 11-20 21-30 31-40 41-50
f: 4 12 20 9 5
Size 10-15 15-17.5 17.5-20 20-30 30-35 35-40 40 +
f: 10 15 17 25 28 30 40
Marks 30-35 25-30 20-25 15-20 10-15 5-10 0-5
Student
s
4 8 12 16 10 6 4
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Q6: Calculate Median: (Mid Value Series)
(Ans: 45.385)
Q7: Find the missing frequencies if N = 100 & M = 30.
(Ans: 15, 10)
Q8: Find the missing frequencies if N = 229 & M = 46.
(Ans: 34*, 45)
X: 5 15 25 35 45 55 65 75
f: 15 7 11 10 13 8 20 16
Size 0-10 10-20 20-30 30-40 40-50 50-60
f: 10 ? 25 30 ? 10
Size 10-20 20-30 30-40 40-50 50-60 60-70 70-80
f: 12 30 ? 65 ? 25 18
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PRACTICE PROBLEMS
Q9: The median of the few number of observations
is 48.6. The four items having values are 35, 36.5,
49.1, 50 were added to given series, what will be
the new median? (Ans: 48.6)
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MERITS OF MEDIAN
 Median can be calculated in all distributions.
 Median can be understood even by common people.
 Median can be ascertained even with the extreme items.
 It can be located graphically
 It is the most appropriate average in case of open ended
classes
 It is the most suitable average in dealing with qualitative
facts such as beauty, intelligence, honesty etc.
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DISADVANTAGES OF MEDIAN
 It is not based on all the values.
 It is not capable of further mathematical treatment like
arithmetic mean
 It is affected fluctuation of sampling.
 In case of even no. of values it may not the value from the data.
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QUARTILES, DECILES & PERCENTILES
 Just as median divides the series into two equal
parts, there are other measures which divide the
series
 Quartiles: These divide a series into 4 equal parts.
For any series, there are three quartiles called Q1
(lower – 25%), Q2 (median – 50%), Q3 (upper – 75%)
 Deciles: These divide a series into 10 equal parts.
For any series, there are 9 deciles denoted by D1, D2,
….. D9.
 Percentiles: These divide a series into 100 equal
parts. For any series, there are 99 percentiles
denoted by P1, P2, ….. P99.
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CALCULATION OF QUARTILES, DECILES &
PERCENTILES
For Individual &
Discrete Series
For Continuous
Series
Formula to be used in
Continuous Series
Q1 = Size of
𝑁+1
4
th item Q1 = Size of
𝑁
4
th item
Q1 = l1 +
𝑵
𝟒
−𝒄.𝒇.
𝒇
x i
Q3 = Size of
3(𝑁+1)
4
th item Q3 = Size of
3𝑁
4
th item
Q3 = l1 +
𝟑𝑵
𝟒
−𝒄.𝒇.
𝒇
x i
D1 = Size of
𝑁+1
10
th item D1 = Size of
𝑁
10
th item
D1= l1 +
𝑵
𝟏𝟎
−𝒄.𝒇.
𝒇
x i
D9 = Size of
9(𝑁+1)
10
th item D9 = Size of
9𝑁
10
th item
D9 = l1 +
𝟗𝑵
𝟏𝟎
−𝒄.𝒇.
𝒇
x i
P1 = Size of
𝑁+1
100
th item P1 = Size of
𝑁
100
th item
P1 = l1 +
𝑵
𝟏𝟎𝟎
−𝒄.𝒇.
𝒇
x i
P99 = Size of
99(𝑁+1)
100
th
item
P99 = Size of
99𝑁
100
th item
P99 = l1 +
𝟗𝟗𝑵
𝟏𝟎𝟎
−𝒄.𝒇.
𝒇
x i
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PRACTICE PROBLEMS
Q1: From the following data, calculate Q1, Q3, D5, P25
21, 15, 40, 30, 26, 45, 50, 54, 60, 65, 70
Ans: 26, 60, 45, 26
Q2: From the following data, calculate Q1, Q3, D6, P85
Ans: 13,15,14,16
Q3: Calculate Q1, Q3, D8, P56
Ans: 14.7, 34, 36.3, 26.4
X 10 11 12 13 14 15 16 17 18
f 3 4 5 12 10 7 5 2 1
Wages 0-10 10-20 20-30 30-40 40-50
No. of Workers 22 38 46 35 19
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PRACTICE PROBLEMS – TYPICAL
Q4: The first & third quartiles of the following data
are given to be 25 marks and 50 marks respectively
out of the given data below:
Find out the missing
frequencies when N = 72
Ans: 12, 11, 3
Marks Frequency
0-10 4
10-20 8
20-30 ?
30-40 19
40-50 ?
50-60 10
60-70 5
70-80 ?
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MODE
 Mode is the most frequent value or score in the distribution.
 It is defined as that value of the item in a series.
 It is denoted by the capital letter Z.
 It is the highest point of the frequencies distribution curve.
 Croxton and Cowden : defined it as “the mode of a
distribution is the value at the point armed with the item tend to
most heavily concentrated. It may be regarded as the most
typical of a series of value”
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CALCULATION OF MODE – INDIVIDUAL
SERIES
 Inspection Method: In this method, simply identify
the value that occurs most frequently in a series.
Q1: Find the mode from the following data:
8, 10, 5, 8, 12, 7, 8, 9, 11, 7 &
7, 8, 10, 15, 10, 22, 20, 26, 20, 34, 20, 6, 10 &,
10, 15, 20, 25, 30
(Ans: 8, Bi Modal, No Mode)
 By changing the Individual Series into Discrete
Series: First convert the series into discrete, then
identify the value corresponding the highest frequency,
that value will be mode.
Q2: Find the mode from the following data:
8, 10, 5, 8, 12, 7, 8, 9, 11, 7
(Ans: 8)
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CALCULATION OF MODE – DISCRETE
SERIES
 Inspection Method: By inspecting, identify the
value whose frequency is maximum, is Mode.
Q3: Find the mode from the below data:
(Ans: 14000)
Q4: Find the mode from the below data:
(Ans: 7)
Income 11000 12000 13000 14000 15000 16000
No. of persons 2 4 7 10 4 3
Marks 5 6 7 8 9 10
No. of students 8 11 15 2 3 1
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SERIES
 Grouping Method: In some cases, it is possible
that value having the highest frequency may not
be the modal value.
 It happens where the difference between the
maximum frequency and the frequency
preceeding or succeeding it is very small and
items are heavily concentrated on either side.
 In such cases, mode can be determined by
grouping method.
 Here, modal value is determined by preparing
two tables – Grouping Table & Analysis Table
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SERIES
Q5: Find the mode using grouping method:
(Ans: 12)
Q6: Find the mode using grouping method:
(Ans: 40)
X 7 8 9 10 11 12 13 14 15 16 17
f 2 3 6 12 20 24 25 7 5 3 1
X 20 25 30 35 40 45 50 55
f 1 3 5 9 14 10 6 4
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CALCULATION OF MODE – CONTINUOUS
SERIES
 Firstly, modal class is determined by using
inspection method or grouping method, as similar
to discrete series
 After determining modal class, mode can be
found out using the following formula:
Z = 𝑙1 +
𝑓1
− 𝑓0
2𝑓1
− 𝑓0
− 𝑓2
𝑥 𝑖
𝑙1 = lower limit of the modal class
𝑓1 = frequency of the modal class
𝑓0 = frequency of the pre modal class
𝑓2 = frequency of the post modal class
𝑖 = size of the modal class
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CALCULATION OF MODE – CONTINUOUS
SERIES
 Few Important points to remember:
 If the first class is the modal class, the 𝑓0 is taken as
zero.
 If the last class is the modal class, the 𝑓2 is taken as
zero.
 If the modal value lies outside the modal class
(Failure of Formula), then the following formula is
used to calculate the mode:
Z = 𝑙1 +
𝑓2
𝑓0
+ 𝑓2
𝑥 𝑖
 If mode is ill defined (Bi-Modal Series), then we use
the formula
Z = 3M – 2𝑋
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PRACTICE PROBLEMS - MODE
Q7: Calculate Mode: Ans: 18
Q8: Calculate Mode: (Cumulative Frequency Series) Ans: 27.73
Wages 0-5 5-10 10-15 15-20 20-25 25-30 30-35
No. of workers 3 7 15 30 20 10 5
Marks between No. of students
10 and 15 4
10 and 20 12
10 and 25 30
10 and 30 60
10 and 35 80
10 and 40 90
10 and 45 95
10 and 50 97
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PRACTICE PROBLEMS - MODE
Q9: Calculate Mode: (Unequal Class Intervals)
Ans: 33.47X f
0-5 4
5-10 8
10-20 10
20-25 9
25-30 13
30-40 30
40-44 6
44-50 9
50-70 12
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PRACTICE PROBLEMS – MODE (TYPICAL)
Q10: Calculate mode: (Bi-Modal Series)
(Ans: 𝑋 = 49.51, M = 49.69, Z = 50.05)
Q11: Calculate mode: (Failure of Formula)
(Ans: Z = 48.89)
Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90
No. of students 4 6 20 32 33 17 8 2
Marks 25-
35
35-
45
45-
55
55-
65
65-
75
75-
85
85-
95
95-
105
105-
115
No. of
students
4 44 38 28 6 8 12 2 2
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PRACTICE PROBLEMS – MODE (TYPICAL)
Q12: Calculate mode: (First Class as Modal Class)
(Ans: Z = 160)
Q13: Calculate mode: (Last Class as Modal Class)
(Ans: Z = 163.58)
Marks 100-200 200-300 300-400 400-500 500-600
Marks 155-157 157-159 159-161 161-163 163-165
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PRACTICE PROBLEMS – MODE (MISSING
FREQUENCY)
Q14: The median and mode for the distribution are Rs.
25 and Rs. 24 respectively. Find the missing
frequencies.
(Ans: 25 & 24)
Q15: The median and mode of the following wage
distribution of 230 workers are known to be Rs. 33.5
and Rs. 34 respectively. Find the missing values:
(Ans: 60, 100 & 40)
Expenditure 0-10 10-20 20-30 30-40 40-50
No. of families 14 ? 27 ? 15
Wages 0-10 10-20 20-30 30-40 40-50 50-60 60-70
No. of workers 4 16 ? ? ? 6 4
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ADVANTAGES OF MODE :
• Mode is readily comprehensible and
easily calculated
• It is the best representative of data
• It is not at all affected by extreme
value.
• The value of mode can also be
determined graphically.
• It is usually an actual value of an
important part of the series.
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DISADVANTAGES OF MODE :
 It is not based on all observations.
 It is not capable of further
mathematical manipulation.
 Mode is affected to a great extent by
sampling fluctuations.
 Choice of grouping has great
influence on the value of mode.
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CONCLUSION
• A measure of central tendency is a measure
that tells us where the middle of a bunch of
data lies.
• Mean is the most common measure of
central tendency. It is simply the sum of the
numbers divided by the number of numbers
in a set of data. This is also known as
average.
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• Median is the number present in the middle when the
numbers in a set of data are arranged in ascending or
descending order. If the number of numbers in a data
set is even, then the median is the mean of the two
middle numbers.
• Mode is the value that occurs most frequently in a set
of data.
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Central Tendency & Dispersion

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