Direct Series

8/22/2019 Direct Series

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Direct series

( B-a) direct method

Formula =

= = N

X = Value

F= Frequencies of Value

Steps:- F ( each item is multiplied by its frequency)

Sum total of FX

sum total of frequencies

ILLUSTRATION

Following is the weekly wage earning of 19 workers:

Wages 10 20 30 40 50

No. Of

Workers

4 5 3 2 5

Calculate the arithmetic mean using Direct Method.

Solution:

Wages (X) No. of Workers or frequency Multiple of the Value of X and

Frequency (FX)

10 4 4 x 10 = 40

20 5 5 x 20 = 100

30 3 3 x 30 = 90

40 2 2 x 40 = 80

50 5 50 x 5 = 250

= 560


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= 29.47

Mean wage earning of 19 workers = 29.47

(B-a) short cut Method

A= Assumed Mean

Sum of Frequencies.

Steps

Firstly take A ( Assumed mean from Values (X)

2. Secondly, find d (deviation)

= ( X-A)

3. Thirdly, fd = multiple of d and f

= (f x d)

4. Add up, separely the positive (+) and, negative (-) values, of all fd. Find out the difference

between the two to get

5. Move, to got , add up all frequencies

6. Lastly, put the values into formula.

Numerical


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ILLUSTRATION

Following are the wages of 19 workers:

Wages 10 20 30 40 50

No. of

Workers

4 5 3 2 5

Calculate arithmetic

(Assumed Average, A =30)

Wages (X) No. of Workers or

Frequency (F)

Deviation(d=X-A)

(A=30)

Multiple of Deviation

and Frequency (fd)

10 4 10-30= -20 4 x (-20)= -80

20 5 20-30= -10 5 x (-10) = -50

30 3 30-30 =0 3x0=0

40 2 40-30=10 2x10=20

50 5 50-30=20 5x20=100

=30+

=30-

= 30-0.53 = 29.47

Arithmetic Mean= 29.47

(B-c) step- Deviation Method

Formula =

X C


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D =

= d = X-A

A= assumed mean

C= Common difference

Numerical

ILLUSTRATION

Following are the wages of 19 workers:

Wages 10 20 30 40 50No. of

Workers

4 5 3 2 5

Calculate arithmetic mean using Step-deviation method

(Assumed Average, A =30)

Wages (X) No. of Workers

or Frequency (F)

Deviation(d=X-A)

(A=30)

Step-deviation

D=

(C=10)

Multiple of

Deviation and

Frequency (fd)

10 4 10-30= -20 -20/10 = 2 4x(-2)=-8

20 5 20-30= -10 -10/10=2 5 x (-1) = -5

30 3 30-30 =0 0/10=0 3x0=0

40 2 40-30=10 10/10=1 2x1=2

50 5 50-30=20 20/10=2 5x2=10


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A= 30, C=10 and

=

= -0.053

Putting these values in the formula,

30+

x10=30-0.053x10

= 30-0.53=29.47

Arithmetic Mean = 29.47

(C) Continuous series:-

(C-a) direct method

M= mid value = L1+L2/2

F= Frequency

N= No. of value

Numerical

ILLUSTRATION

The following table shows marks in English secured by students of class X in your school in

their examination. Calculate mean marks using Direct Method.

Marks 0-10 10-20 20-30 30-40 40-50

No. of

Students

20 24 40 36 20


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Solution:-

Marks Mid-

value{m=L1+L2}

No. of students or

Frequency (F)

Multiple of Mid-value

and Frequency (fm)

0-10 0+10/2=5 20 20x5=100

10-20 10+20/2=15 24 24x15=360

20-30 20+30/2=25 40 40x25=1,000

30-40 30+40/2=35 36 36x35=1,260

40-50 40+50=45 20 20x45=900

= 3,620/140=25.86 marks.

(c-b) short cut method

A= Assumed mean (from M)

D= M-A

F= Frequency

Steps

1.

Firstly calculate M= L1+L2/22. Take A from M d= M-A3. F x d = 4. Sum total of Frequency =

Numerical


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ILLUSTRATION

The following table shows marks secured by the students of a class in an examination in English.

Marks 0-10 10-20 20-30 30-40 40-50

No. of

students

20 24 40 36 20

Calculate mean marks using Short-cut Method.

SOLUTION:

( Assumed Average, A = 25)

Marks Mid-

value{m=L1+L2}/2

Frequency (f) Deviation (d=m-

A)

(A=25)

Multiple of

Deviation and

Frequency (fd)

0-10 0+10/2=5 20 5-25= -20 20x-20= -400

10-20 10+20/2=15 24 15-25 = -10 24x-10=-240

20-30 20+30/2=25 40 25-25=0 40x0=0

30-40 30+40/2=35 36 35-25=+10 36x+10=36040-50 40+50/2=45 20 45-25=+20 20x +20=400


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= 25+0.86=25.86

Mean marks = 25.86

(C-c) step-deviation Method

X C

D= d/c

C= common differences

Steps:

1. M= L1+L2/22. Take A from M3. D= M-A4. D= d/c5. Fd= f xd6. Sum total of frequency =

Numerical

ILLUSTRATION:

The following table show marks obtained by the students of a class in their test in English.

Marks 0-10 10-20 20-30 30-40 40-50

No. of

Students

20 24 40 36 20


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SOLUTION:

Marks Mid-

value(=L1+L2)/2

Frequency (f) Deviation

(d=m-A)

(A=25)

Step-

deviation(d)

{d= m-A}/C

(C=10)

Multiple of

step-

deviation and

frequency

(fd)

0-10 5 20 -20 -2 -40

10-20 15 24 -10 -1 -24

20-30 25 40 0 0 0

30-40 35 36 +10 +1 +36

40-50 45 20 +20 +2 +40

Mean,

X C

= 25+0.086X10

= 25+0.86=25.86

Key concept

Q1. Calculation of A.M in case of cumulative frequency series?

Sol. Firstly convert cumulative frequency into simply frequency distribution series than solve it,by Direct Method.

Q2. Calculation of A.M in a Mid-value series?

Sol. Directly solve the question, there is no need to convert mid value series into simple

frequency distribution series.


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Q3. In case of Inclusive Series?

Ans. Directly solve question by applying direct method. These is no need to convert it into

exclusive series.

Illustration

Marks in Statistics of the students of Class XI are given below. Find out arithmetic mean.

Marks No. of Students

Less than 10 5

Less than 20 17

Less than 30 31

Less than 40 41

Less than 50 49

Solution:

A Cumulative frequency distribution should first be converted into a Simple Frequency

Distribution, as under:

Conversion of a Cumulative Frequency Distribution into a simple Frequency Distribution

Marks No. of Studenta

0-10 5

10-20 17-5 = 12

20-30 31-17 = 14

30-40 41-31 = 10

40-50 49-41 = 8

Now mean value of the data is obtained using Direct Method as under:


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Calculation of mean

Marks Mid-

value(=L1+L2)/2

Frequency (f) Multiple of

step-

deviation and

frequency

(fm)

0-10 5 5 25

10-20 15 12 180

20-30 25 14 350

30-40 35 10 350

40-50 45 8 360

,65

Arithmetic mean = 25.82

Illustration

The following table shows marks in economics of the students of a class. calculate arithmetic

mean.

Marks No. of students

More than 0 30

More than 2 28

Mora than 4 24

More than 6 18

More than 8 10

Solution:


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Converting cumulative frequency distribution into a simple frequency distribution, we get the

following:

Marks No. of students

0-2 30-28 = 2

2-4 28-24 = 4

4-6 24-18 = 6

6-8 18-10 = 8

8-10 10-0 = 10

Arithmetic mean of this continuous series is estimated below, using Direct Method.

Calculation of arithmetic mean using direct method

Marks Mid-

value(=L1+L2)/2

Frequency (f) Multiple of

step-

deviation andfrequency

(fm)

0-2 1 2 2

2-4 3 4 12

4-6 5 6 30

6-8 7 8 56

8-10 9 10 90


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Mean marks = 6.33

Calculation of corrected arithmetic mean

Sometimes in the calculation of arithmetic mean some item or values are wrongly written.

Accordingly, the mean value goes wrong. But such mean value can be corrected by

() ( )

Illustration

Following table gives marks in statistics of the student of a class. find out mean marks.

Mid-

value

5 10 15 20 25 30 35 40

No. of

students

5 7 9 10 8 6 3 2

Solution

In this series, mid values are already given. The calculation of arithmetic mean involves the

same procedure as in case of exclusive series.

Calculation of Arithmetic mean.


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Mid- value(m) No. of students or frequency

(f)

Multiple of Mid-value and

Frequency (fm)

5 5 25

10 7 70

15 9 135

20 10 200

25 8 200

30 6 180

35 3 105

40 2 80

995

50

Arithmetic mean = 19.9 marks

Illustration

The following table shows monthly pocket expenses of the students of a class. find out the

average pocket expenses.

Pocket

Expenses

20-29 30-39 40-49 50-59 60-69

No. of

students

10 8 6 4 2

Solution


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Calculation of arithmetic mean of inclusive series is the same as of exclusive series.

Pocket

Expenses

Mid-value

(m = l1+l2/2)

Frequency (f) Deviation (d=

m-A)

(A = 44.5)

Step deviation

d = d/c

Fd

20-29 24.5 10 -20 -2 -20

30-39 34.5 8 -10 -1 -8

40-49 44.5 6 0 0 0

50-59 54.5 4 10 1 4

60-69 64.5 2 20 2 4

20

30

Arithmetic mean = Rs. 37.83

Calculation of missing value

If mean of the series is known, but one value is missing, the missing value may be found using

the following formula

1 2 n

illustration

suppose mean of a series of 5 itmes is 30. Four values are, 10,15,30 and 35 respectively. Find the

missing (5th value) of the series.

Solution

Assume 5th

value as X5.

Given X1 = 10, X2 = 15, X3 = 30, X4 = 35, X5 =?


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10 15 30 35 5

5

90 5

5

90 5

150 = 90 + X5

X5 = 15090 = 60

Thus, value of the 5th

item = 60

Calculation of Weighted Arithmetic mean

In simple AM, all items of a series are taken as of equal significance, but sometimes we may

give greater significance to some items and less to others. For example, A housewife may give

more significance to food, less to entertainment. So we can say that when different items of a

series are weighted according to their relative importance, the average of such series is called

weighted arithmetic average.

Formula

W = weight

X = Value

Numerical

Illustration

Calculate weighted mean of the following data.


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Marks(X) 81 76 74 58 70 73

Weight

(W)

2 3 6 7 3 7

Solution:

Calculation of weighted mean

Marks (X) Weight (w) WX

81 2 162

76 3 228

74 6 444

58 7 406

70 3 210

73 7 511

Weighted mean,

1961

28

Weighted mean = 70.04 marks.

Combined Arithmetic Mean.

Given the mean values of two or more parts of a series and the number of items in each part, one

can get combined Arithmetic mean or mean of the series as a whole

Formula

Here


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X1,2 = Combined arithmetic mean of parts 1 and 2 of a series

X1 = A.M of part 1 of the series.

X2 = A.M of part 2 of the series.

N1 = Number of items in part 1 of the series.

N2 = Number of items in part 2 of the series.

Numerical

Illustration

60 students of Section A of class XI, obtained 40 mean marks in statistics, 40 students of section

B obtained 35 mean marks in Statistics. Find out mean marks in Statistics for class XI as a

whole.

Solution

Given,

N1 = 60, X1 = 40, X2 = 35

We know

Thus, combined arithmetic mean = 38

ILLUSTRATION 1.


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In the following frequency distribution, the frequency of the class interval (40-50) is not known.

Find it, if the arithmetic mean of the distribution is 52:

Wages (Rs) 10-20 20-30 30-40 40-50 50-60 60-70 70-80

No. of Workers 5 3 4 ? 2 6 13

SOLUTION:

Wages (Rs) Midvalue (m) Frequency Fm

10-20 15 5 75

20-30 25 3 75

30-40 35 4 140

40-50 45 F 45f

50-60 55 2 110

60-70 65 6 390

70-80 75 13 975

= 33 + f fm = 1,756 + 45f

176545f

Or, 52 (33 + f) = 1,765 + 45f

1,716 + 52f = 1,765 + 45f

52f45f = 1,7651,716 => 7f = 49 => f = 7


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ILLUSTRATION 2.

if the arithmetic mean of the following series is 115.86; find the missing value.

Wages (Rs) 110 112 113 117 ? 125 128 130

No. of Workers 25 17 13 15 14 8 6 2

SOLUTION:

Let the missing value be X.

Wages (x) No. of Workers (f) Fx

110 25 2,750

112 17 1,904

113 13 1,469

117 15 1,755

X 14 14X

125 8 1,000

128 6 768

130 2 260

f = 100 fX = 9,906+ 14X

990614

100

115.86 x 100 = 9,906 + 14X

14X = 11,5869,906 = 1,680

1680

= 120


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Calculation of Arithmetic Mean for Diligent Series

Type of Series Direct Method Short Cut or Assumed

mean Method

Step Deviation

Method

1. IndividuateSeries

d

d

1. Discrete Series

Or

fd

d

2. ContinuousSeries

fd

d

# Define Merit & Demerit of Arithmetic Mean ?

Ans :- # Merit

1. Based on all items :- A.M is based on all the items in a series. It is, therefore arepresentative value of the different items.

2. Stability :- A.M is a of central tendency. This is because changes in the Simple of a serieshave minimum effect on the arithmetic average.

3. Basis of Comparison :- State and certain arithmetic mean can be easily used forcomparision.


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# Demerits :->

Effect of Extreme value :-> The main defect of a mean is that it eblected by extreme value of the

series. For example 1 jf Monthly income of 5 workers are

Workers Monthly Income (Rs)

A 6000

B 3500

C 2500

D 1500

E 1000

The average income of all the five workers would be

60003500250015001000

5

14500

5

Certainly this is not such an accurate mean of the monthly income of the monthly income of 5

workers.

2. Laughable Conclusions :- A.M sometime offers laughable conclusions. Jf there are 50students in class XI and 31 student in class XII, the average strength of these two classes.

Would come to

2= 30.5 Students. Which is very funny because these cannot be half students.

3. Mean value may not figure in the series at all :- The mean value which dose not figure inthe series at all .

Just time , the average of 2,3 and 7 is

3

Which is not these in the series.

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