¾As per A.L. Bowley, Statistics is called as Science of ...

As per A.L. Bowley, Statistics is called asScience of counting and averagesScience of counting and averages.

As per Boddington, statistics is science ofestimates and probabilities

As per Wallis and Roberts statistics isAs per Wallis and Roberts, statistics isregarded as a body of methods of decisionmaking in the face of uncertaintymaking in the face of uncertainty.

As per Horace Secrist :: Statistics isd d th t f f t ff t dregarded as the aggregate of facts affected

to a marked extent by multiplicity of causes,i ll d ti t dnumerically expressed or estimated

according to a reasonable standard ofll t d i t tiaccuracy, collected in a systematic manner

for a predetermined purpose and placed inl ti t h threlation to each other.

A captain of a cricket team needs to knowA captain of a cricket team needs to knowhow many players in his team are batsmen,bowlers wicket keepers and all rounder ?bowlers, wicket-keepers and all rounder ?

A Head of a state, a head of a family needsyall information related to them which helpsthem to tackle the situations.

The information is expressed in numbers,which is called statistical datawhich is called statistical data

I) Ungrouped data (or) Individual items (or)Raw data

II) Grouped Data

a) Discrete frequency Distribution

b) Continuous Frequency Distributionb) Continuous Frequency Distribution

Marks obtained by 30 students in a test ofMarks obtained by 30 students in a test of Mathematics subject out of 50 marks is given below ::below ::

15 21 40 30 18 16 23 27 17 40

34 32 36 19 28 41 42 31 35 40

38 39 44 43 37 36 42 28 33 40

H d l ifi iHence needs classification

There are 150 families in a village. Number offamilies having no child, one child, two children etcis given belowNo. of Children No. of families (F)

00 78

01 2201 22

02 22

03 1603 16

04 04

05 0305 03

06 05

150

M kM k 00 1010 1010 2020 2020 3030 3030 4040 4040 5050 5050 6060Marks Marks ObtainedObtained

00--1010 1010--2020 2020--3030 3030--4040 4040--5050 5050--6060

No. of No. of 1212 1818 2727 2020 1717 66studentsstudents

0-10, 10-20 etc are called class intervals and 12 is called0 10, 10 20 etc are called class intervals and 12 is calledfrequency.

A frequency is a number which tells how many students haveq y ytaken marks between 0 and 10. Hence in 0-10, 10 isexcluded in first class.

Class interval

The range of variables that canThe range of variables that can be divided into subgroups or g psub-rangers is called class intervals or classesintervals or classes.

WIDTH OF THE CI :: The difference betweenlower and upper class limits is called width of CIppClass frequency: The number of observationscorresponding to a particular class is calledp g pfrequency. i.e. it is a number which tells howmany observations fall in a particular classy pTypes Of Class IntervalsExclusive typeyp

If upper limit of first Class Interval = lower limitof second class interval then they are calledof second class interval then they are calledoverlapping class limits

Class Class IntervalsIntervals

Frequency Frequency

00--1010 22

1010 2020 331010--2020 332020--3030 55

Example:: If an observation is 10 then it falls in 2nd Class Intervals and not in 1st class intervals2 Class Intervals and not in 1 class intervals.

Class Class IntervalsIntervals

Frequency Frequency IntervalsIntervals

00--99 55

1010 1919 111010--1919 11

2020--2929 33

Upper Limit of 1st class interval ≠ Lower Limit of 2nd

class intervals

The following are marks scored by30 t d t i th ti t t30 students in mathematics test.Prepare a frequency distributiontable taking 5 as width of the class.

64 60 55 51 62 44 65 50 43 47 48 60 56 48 55 60 54 51 55 52 53 46 45 53 60 51 52 57 41

Class IntervalsClass Intervals Tallies FrequenciesFrequenciesClass IntervalsClass Intervals Tallies FrequenciesFrequencies

4040--45 (Ex)45 (Ex) |||||| 334545--5050 |||| 555050--5555 |||| |||| 1010|||| ||||

5555--6060 |||| 556060 6565 |||| || 666060--6565 |||| || 666565--7070 || 11

64 60 55 51 62 44 65 50 43 4764 60 55 51 62 44 65 50 43 47

52 48 60 56 48 55 60 54 51 55

52 53 46 45 53 60 51 52 57 41

Class IntervalsClass Intervals Tallies FrequenciesFrequenciesClass IntervalsClass Intervals Tallies FrequenciesFrequencies

4040--45 (Ex)45 (Ex) |||||| 334545--5050 |||| 555050--5555 |||| |||| 1010|||| ||||

5555--6060 |||| 556060 6565 |||| || 666060--6565 |||| || 666565--7070 || 11

Arithmetic Mean is defined as the sum of all theArithmetic Mean is defined as the sum of all the observations divided by total number of observations

I i d d b b lIt is denoted by symbol xRaw data : (ungrouped data)

x = ∑x

n

X = Value taken by the variables

∑x = Summation of X∑x = Summation of X

n = Total number of observation

Student Student AA BB CC DD EE FF GG HHMarksMarks 6767 7676 8282 4444 6060 7171 5454 6666

i.e. The mean marks of 8 students of the class = 65

Calculation of mean becomesdifficult if the raw data consists oftoo many numbers It is better totoo many numbers. It is better togroup the raw data into discrete and

i i S hcontinuous series. So thatcalculation of mean becomessimpler.

X = ∑fxX = ∑fx

N

X = Value taken by the variable

X = MeanX = Mean

f = Frequency corresponding to X

N = ∑f (Total member of observation)

H i h (iH i h (i 5050 5252 5454 5555 5656Heights (in Heights (in inches)inches)

5050 5252 5454 5555 5656

No.of personsNo.of persons 22 22 33 22 11pp

Solutions :: Calculation of mean by direct methodHeightsHeights PersonsPersons fxfxgg

5050 22 100100

5252 22 1041045454 33 1621625555 22 1101105656 11 5656

N = 10N = 10 ∑∑fxfx= 532= 532

X = ∑fx = 532X ∑fx 532

N 10

= 53.2

Therefore, mean height of individuals = 53.2

SHORT CUT METHOD

The formula is given by

X = A + ∑fd

NN

Where, X = Variables

A = Assumed Mean

d = (X-A)d (X A)

N = ∑f

Steps :::

i) T k d Ai) Take an assumed mean A

ii)Generally, the value corresponding to) y, p ghighest frequency is taken as assumedmean, as it reduces calculation.,

C l l t b h t tCalculate mean by short-cut method, taking data from example

Heights (in Heights (in inches)inches)

5050 5252 5454 5555 5656

No.of personsNo.of persons 22 22 33 22 11

Calculation of mean using short cut method

A = 54 inches

Height (X)

No.of persons (f)

D= (X-A) fd

A = 54 inches

(X) persons (f)

50 2 -4 -852 2 2 452 2 -2 -454 3 0 055 2 1 255 2 1 256 1 2 2

N = 10 ∑fd = 4 - 12

= -8

X = A + ∑fd

N

= 52+ = 54 – 0 8 = 53 2-8 52+ 54 0.8 53.210

X = 53.2 Inches

i.e. Mean height of persons is 53.2 inches

Note :: We get the same answer for mean whenNote :: We get the same answer for mean whenwe use direct and short cut methods

The formula is given by

X = ∑fxX ∑fx

N

X = Mid point of the class interval

X = Mean

f = Frequency corresponding to Xf Frequency corresponding to X

N = ∑f

k dk dMarks ScoredMarks Scored 00--1010 1010--2020 2020--3030 3030--4040 4040--5050No. of studentsNo. of students 22 33 55 44 11

Solution :: Calculation of meanSolution :: Calculation of mean

MarksMarks No.of No.of Students (f)Students (f)

Mid point = Mid point = xx

fxfxStudents (f)Students (f) xx

00--1010 22 55 10101010--2020 33 1515 45451010 2020 33 1515 45452020--3030 55 2525 1251253030--4040 44 3535 1401404040--5050 11 4545 4545

N =15N =15 ∑fx = 365∑fx = 365

X = ∑fx

N

= 365 /15 = 24.33

x = 24.33 Marks

i.e. Mean marks scored by the students is 24.33

DEFINITION OF MEDIAN

Median refers to the middle most value in a distribution.

i.e. It divides the distribution into two halves.

Calculation of Median

a)Raw Dataa)Raw Data

To calculate median for a raw data, the following procedure is considered

Steps ::Steps ::

i) Arrange the data in ascending order

ii) Find n +1 value 2

i) Apply the formula M = Size or value of n + 12

th

Item.

Example :: Calculate the median for the following ratio

Marks Marks S dS d

2525 1818 1616 2222 2020ScoredScored

Solution :: Calculation of Median

Sl.No. Sl.No. 11 22 33 44 55

Marks scored inMarks scored in 1616 1818 2020 2222 2525Marks scored in Marks scored in ascending orderascending order

1616 1818 2020 2222 2525

Here n +1 = 5 +1 = 6 = 32 2 2

M = size of n + 1 itemth

d

2

= size of 3rd item = 20

Therefore Median marks is 20Therefore Median marks is 20

DISCRETE SERIESF di t i d t di iFor a discrete series data, median is

calculated as follows ::

i) Find cumulative frequencies

ii) Fi d N l h Nii)Find N value where N = ∑fx

iii)Apply the formula M = size of item. In2

2

N th

) pp yother words locate a value which is justmore than value, from the cumulative

2

N2 ,

frequency column.2

iv) Read the corresponding X value. This) p ggives the value of Median

Calculate median marks from the following dataMarks scoredMarks scored 1010 2020 3030 4040 5050No. of studentsNo. of students 1010 1515 1717 66 22

Solutions :: Calculation of MedianMarks scored (x)Marks scored (x) No. of students No. of students

(f)(f)Cumulative Cumulative fr ifr i(f)(f) frequenciesfrequencies

1010 1010 10102020 1515 25252020 1515 25253030 1717 42424040 66 48484040 66 48485050 22 5050

N = 50N = 50

Here, = =   50 = 25 N

22

M = size of item = size of 25th item = 30N

2

th2

2

i.e. in the cumulative frequency column,th l th 25 i 42 ththe value more than 25 is 42. thecorresponding X value is 30.

M = 30 Marks

or

Median Marks is 30

Mode:Mode:Mode is defined as the value which is 

t d i b f ti irepeated maximum number of times in a data.  It may be denoted by the letter    Zy y

RAW DATA

Marks Marks ScoredScored

1212 1515 2020 1010 3030 2020 1010 2020

Solution:

Here, 20 occurs three times and otherHere, 20 occurs three times and othernumbers do not occur so often.

Therefore, 20 is the mode

E l l l t di d dExample :: calculate mean, median and mode for the following ungrouped data

90 70 53 95 72 7090, 70, 53, 95, 72, 70

Mean = AM = average = 90 + 70 + 53 + 95 + 72 + 70Mean AM average 90 + 70 + 53 + 95 + 72 + 70

6

= 450

66

= 75

Median = 53, 70, 70, 72, 90, 92Arrange in ascending order

Mode :

Mode = 70 repeated score (crude mode)

( ) M d 3 di 2 M(or) Mode = 3 median - 2 Mean

= 3(71) - 2 (75)

= 63 (Real Mode)

EXAMPLE 2One foot ball team scored 1,3, 2, 6, 4,0,2, 5, 3, 3, 1, 2, 4, 6,3 2 1 4 3 5 3 1 goals in the matches if played in order3, 2, 1, 4, 3, 5, 3,1, goals in the matches if played in order.Prepare frequency distribution table and find mean, medianand made Values (x)Values (x) ff fxfx cfcf

00 11 00 1111 44 44 5522 44 88 9933 66 1818 151533 66 1818 151544 33 1212 181855 22 1010 202055 22 1010 202066 22 1212 2222

N = ∑f = 22 ∑fx = 64

Mean = ∑fx = 64 = 2.9N 22N

Median N = 22 = 1122

Mdn = 3

Mode max frequency is 6

Mode = 3