Is a branch of mathematics that deals with collection, organization, analysis and interpretation of data.

Terminology – ◦ Midpoint – Average to represent a class

◦ Lower (or upper) limit – limit of class

◦ Lower (or upper) boundary – boundary of class

10 - 14 15 - 19 20 - 24 25 - 29

Lower limit

Upper limit Boundary

Mid Point

14 + 15

2= 14.5

25 + 29

2= 27

Discrete Data – Data that has specific value or specific no. of decimal places. For example : ◦ Male, Female ◦ Very Hot, Hot, ◦ 1, 2, 3 (Physical items : chairs, bottles, children) ◦ 1.5, 2.55, 6.37 (Currency)

Continuous Data – Sometime, it is called as measurement data. Data that can be any integer value, positive of negative value, or value with infinity number of decimal places. For example : ◦ 6.23cm, 9.3463cm ◦ 91.141 ⁰C, 1.2 ⁰C

There are a numbers of charts or diagrams that can be used to represent statistical data. For example:

Frequency diagrams

Statistical diagrams

Pie Charts

Bar charts

Histogram

Line graphs

Stem-and-leaf diagrams

Table that presents frequency of each category.

There are two ways to present data in frequency table; ungrouped data or grouped data

Ungrouped Data

Each data class consists of single-valued data and its frequency

For example : Scores achieve by 20 students in a contest.

Scores No of

Students

0 2

1 2

2 3

3 4

4 5

5 4

Total 20

Grouped Data

Each data class consists of data within a particular interval.

For example : Time taken by 20 students to complete a math question

Time (in min) No of Students

3 – 4 5

5 – 6 7

7 – 8 5

9 – 10 3

Total 20

Measure that shows the central location of the data. Basically there are three common measures: ◦ Mode - Data with highest frequency

◦ Median - The middle term, or [(n+1)/2]th term which divides the data in ascending or descending order into 2 equal parts

◦ Mean - Average of the data, i.e. summation of all data divided by no of data.

n

x x ,Mean

Given a set of data which consists of 5, 4, 4, 5, 7. Determine mode, mean and median.

Mode - Data with highest frequency are 4 and 5. Thus, this is a set of multimodal data.

Median - Arrange data in ascending order

(4, 4), 5, (5, 7).

term th

2

15

Mean – Sum of all data / no of data

55

25

5

75445

n

x x ,Mean

Table shows the no of children per family for 20 families in town. Determine mode, median and mean.

No of Children No of family

0 2

1 4

2 5

3 7

4 2

Total 20

Mode – 3 children

Median = 10.5th term , that is average of 10th and 11th term. For easier calculation, construct another column to indicate cumulative frequency (total frequency of previous classes).

No of Children, No of family

(frequency, f ) Cumulative frequency, F

0 2 2 1 4 6 2 5 11 3 7 18 4 2 20

Total 20

10.5th

term is

here!!

Mean

No of Children, x No of family

(frequency, f ) f x

0 2 0 2 = 0

1 4 1 4 = 4

2 5 10

3 7 21

4 2 8

Total 20 43

children 15.220

43

f

xf x ,Mean

Table shows distance jumped by 15 students in a long jump event. Determine

modal class and mode value

median class and median value

mean.

Distance (m) No of students

4.0 – 4.4 3

4.5 – 4.9 5

5.0 – 5.4 6

5.5 – 5.9 1

Total 15

modal class – class with highest frequency = class 5.0 – 5.4

mode value is given as:

m033.5

)5.0()16()56(

5695.4

cLMode21

1m

Lm = Lower boundary of modal class 1 = fm – fprev

2 = fm – fnext

fprev = frequency of class before modal class fnext = frequency of class after modal class fm = frequency of modal class c = class width

median class – class which [(n+1)/2]th term median falls into

Median value is given as:

Distance (m)

No of students

Cumulative frequency

4.0 – 4.4 3 3

4.5 – 4.9 5 8

5.0 – 5.4 6 14

5.5 – 5.9 1 15

Total 15

m9.4

5.05

35.745.4

cf

F2

n

LMedianm

previous

m

8th term

is here!!

Lm = Lower boundary of median class Fprev = cumulative frequency of class before median class n = total frequency fm = frequency of median class c = class width

Median Class

Mean

Distance (m) No of

students, f Mid point, x f x

4.0 – 4.4 3 4.23=12.6

4.5 – 4.9 5 4.75=23.5

5.0 – 5.4 6 5.2 31.2

5.5 – 5.9 1 5.7 5.7

Total 15 73

242

4404.

..

742

9454.

..

m867.415

73

f

xf x ,Mean

To measure how far the data is spread out from its average

Basically there are four common measures: ◦ Range

◦ Interquartile range

◦ Variance

◦ Standard deviation

The difference between highest & lowest

Example : Determine range for data set which consists of 5,5,2,3,6,7,9

Value

Lowest

Value

HighestRange

729Range

The difference between Q3 and Q1

Q1 and Q3 are 1st and 3rd quartile respectively.

2nd quartile is equivalent to median.

Quartile – divide data in order into 4 equal parts

cf

F4

n3

L3Q

cf

F4

n

L1Q

3Q

previous

3Q

1Q

previous

1Q

LQ1 (Q3) = Lower boundary of Q1 (or Q3) class Fprev = cumulative frequency of class before Q1(or Q3) class n = total frequency fQ1 (Q3) = frequency of Q1(or Q3) class c = class width

1Q3QRange ileInterquart

Variance is given as:

Standard deviation is given as:

f

)xx(x

f

fxVariance

22

2

iancevarDeviation dardtanS

Given a set of data which consists of 5, 4, 4, 7, 2, 3, 5, 8, 9, 11. Determine range, interquartile range, variance and standard deviation.

Answer :

Range = 11 – 2 = 9

Interquartile Range :

Arrange data in order

2, 3, 4, 4, 5, 5, 7, 8, 9, 11

Interquartile range = Q3 – Q1 = 8 – 4 = 4

Q3 Q1

x 2 3 4 4 5 5 7 8 9 11

x2 4 9 16 16 25 25 49 64 81 121

025.21.4iancevarDeviation Standard

1.410

410

10

1218164492525161694

xf

fx Variance

8.510

58

10

11987554432

f

xx,Mean

22

Table shows distance jumped by 25 students in a long jump event. Determine range, interquartile range, variance and standard deviation.

Distance (m) No of students

4.0 – 4.4 1

4.5 – 4.9 3

5.0 – 5.4 5

5.5 – 5.9 7

6.0 – 6.4 8

6.5 – 6.9 1

Total 25

Range :

Distance (m) Mid Point, x No of students

4.0 – 4.4 4.2 1

4.5 – 4.9 4.7 3

5.0 – 5.4 5.2 5

5.5 – 5.9 5.7 7

6.0 – 6.4 6.2 8

6.5 – 6.9 6.7 1

Total 25

5.22.47.6

Class First

of Midpoint

Class Last

of MidpointRange

Interquartile range = Q3 – Q1 :

Determine Q1 and Q3 Class – class contains

Distance (m) Mid Point, x No of

students Cumulative frequency

4.0 – 4.4 4.2 1 1

4.5 – 4.9 4.7 3 4

5.0 – 5.4 5.2 5 9

5.5 – 5.9 5.7 7 16

6.0 – 6.4 6.2 8 24

6.5 – 6.9 6.7 1 25

Total 25

ly.respective term th4

1)3(n and th

4

1n

Q1 class

Q3 class

Interquartile range = Q3 – Q1

= 6.122 – 5.175

= 0.947

175.5)5.0(5

44

25

95.41Q

122.6)5.0(8

164

)25(3

95.53Q

Distance (m)

Mid Point, x

No of students

fx x2 fx2

4.0 – 4.4 4.2 1 4.2 17.64 17.64

4.5 – 4.9 4.7 3 14.1 22.09 66.27

5.0 – 5.4 5.2 5 26 27.04 135.2

5.5 – 5.9 5.7 7 39.9 32.49 227.43

6.0 – 6.4 6.2 8 49.6 38.44 307.52

6.5 – 6.9 6.7 1 6.7 44.89 44.89

Total 25 140.5 798.95

0.61123736.0Deviation Standard

0.373662.525

95.798x

f

fxVariance

62.525

5.140

f

fxx,Mean

222

Below shows the height of female and male students. Determine the mean, mode, variance, interquartile range and standard deviation of their height

Height (cm) Female Height (cm) Male

150-154 150-154

155-159 155-159

160-164 160-164

165-169 165-169

170-174 170-174

175-179 175-179

Total Total

cf

F2

n

LMedian

cLMode

m

previous

m

21

1m

iancevarDeviation Standard

xf

fxVariance

f

fxx,Mean

22

cf

F4

n3

L3Q

cf

F4

n

L1Q

3Q

previous

3Q

1Q

previous

1Q