MELJUN CORTES DATA TYPES Rm104tr-13

7/29/2019 MELJUN CORTES DATA TYPES Rm104tr-13

1/39

Lesson 13 - 1

Year 1

CS113/0401/v1

LESSON 13TYPES OF DATA

Qualitative

Not usually numeric No particular order

Examples:

Colour, Types of Materials

Quantitative Numeric

Ordered

Measurable

Continuous E.g. Length, Age, Weight

Discrete

E.g. Shoe size, Number ofpeople


2/39

Lesson 13 - 2

Year 1

CS113/0401/v1

First stage in making raw data

understandable

RAW DATA

Number of sheets of listing paper

used by each of 120 jobs

Not easily digested!

17

24 11

14

18

17

7

5

21

6 11 18 22 14 6 17

14

8

12132712 189

14

18 14

13

21

8

27

9

11

16 27 21 14 11 19 7

10

29

17121419 129

23

17 24

7

13

14

17

21

8

17 19 24 26 2 5 18

14

16

7162813 148

19

27 9

18

8

24

19

7

13

14 16 19 11 17 23 12

25

16

15102118 1411

9

14 28

20

12

16

10

8

9

11 22 10 17 9 18 12

24

8

716520 710

DATA TABULATION (1)


3/39

Lesson 13 - 3

Year 1

CS113/0401/v1

Category

(No of sheets

used)

Tally Frequency

0 - 115 - 261111 1111 1111 1111 1111 1

10 - 371111 1111 1111 1111 1111

1111 1111 11

15- 311111 1111 1111 1111 1111

1111 120 - 161111 1111 1111 1

25 - 91111 1111

120Total

Frequency distribution table

DATA TABULATION (2)

Tabulate in (discrete) categories


4/39

Lesson 13 - 4

Year 1

CS113/0401/v1

FREQUENCY DISTRIBUTION(1)

Raw data

Raw data are collected data

which have been organized

numerically

Array

An array is an arrangement of

raw numerical data in ascendingor descending order of

magnitude. The difference

between the largest and smallest

number is called the range of the

data


5/39

Lesson 13 - 5

Year 1

CS113/0401/v1

FREQUENCY DISTRIBUTION(2)

Frequency distribution

When summarizing a large

number of raw data it is often

useful to distribute the data into

classes or categories and to

determine the number of

individuals belonging to each

class, called the class frequency


6/39

Lesson 13 - 6

Year 1

CS113/0401/v1

EXAMPLE

A set of 100 students obtainedfrom an alphabetical listing of an

university record.

Their weights ranging from 60kgto 74kg are tabulated.


7/39

Lesson 13 - 7

Year 1

CS113/0401/v1

Mass ( kilograms) Number of Students

60 - 62

63 - 65

66 - 68

69 - 71

72 - 74

5

18

42

27

8

Total 100

EXAMPLE

The first class or category, for

example consists of masses from 60

to 62 kg and is indicated by the

symbol 60 - 62. Since 5 students

have masses belonging to this class,

the corresponding class frequency is

5.

Data organized and summarized in

the above frequency distribution are

often called grouped data


8/39

Lesson 13 - 8

Year 1

CS113/0401/v1

CLASS INTERVAL

A symbol defining a class such as60 - 62 is called a class interval.

The end numbers 60 and 62, are

called the class limits.

The smaller number 60 is the

lower class limit and the larger

number 62 is the upper class

limit.


9/39

Lesson 13 - 9

Year 1

CS113/0401/v1

CLASS MARK

A class mark is the midpoint ofthe class interval and is obtained

by adding the lower and upper

class limits and dividing by two

In the previous examples, the

class mark of the interval 60 - 62

is (60 + 62) / 2 = 61


10/39

Lesson 13 - 10

Year 1

CS113/0401/v1

MEDIAN (1)

The median of a set of numbers

arranged in order of magnitude isthe middle value or the arithmetic

mean of the two middle values.

Example 1

The set of numbers

3, 4, 4, 5, 6, 8, 8, 8, 10

For an odd number of data the

median occurs at position

(N + 1) / 2

= 10 / 2

= 5th position

Therefore the median = 6


11/39

Lesson 13 - 11

Year 1

CS113/0401/v1

MEDIAN (2)

Example 2 The set of numbers

5, 5, 7, 9, 11, 12, 15, 18

For even number of data themedian is the average of the two

middle values

The median= (Pos 4 + Pos 5) / 2

= (9 + 11) / 2

= 10


12/39

Lesson 13 - 12

Year 1

CS113/0401/v1

For grouped data the median,obtained by interpolation is given

by

MEDIAN = L1 + C

Where

L1 = lower class boundary of the

median class(I.e. the classcontaining the median).

N = number of items in the data

(I.e. total frequency)

median

- 1

N

2

MEDIAN (1)


13/39

Lesson 13 - 13

Year 1

CS113/0401/v1

MEDIAN (2)

1 = sum of frequenciesof all classes lower

than the median

class

median = frequency of median

class

c = size of median classinterval


14/39

Lesson 13 - 14

Year 1

CS113/0401/v1

MEDIAN OF A GROUPEDFREQUENCY DISTRIBUTION

Draw a Cumulative FrequencyDiagram

Search for the middle value on

the c axis and read off thecorresponding value on the x axis

This is the median


15/39

Lesson 13 - 15

Year 1

CS113/0401/v1

MEDIAN FROM A

CUMULATIVE FREQUENCYDIAGRAM


16/39

Lesson 13 - 16

Year 1

CS113/0401/v1

MODE (1)

The mode of a set of numbers is

that value which occurs with the

greatest frequency, I.e. it is the

most common value. The mode

may not exit, and even of it does

exists it may not be unique

Example

The set

2, 2, 5, 7, 9, 9, 9, 10, 11, 12, 18

has mode 9

Example

The set

3, 5, 8, 10, 12, 15, 16

has no mode


17/39

Lesson 13 - 17

Year 1

CS113/0401/v1

MODE (2)

Example

The set

2, 3, 4, 4, 4, 5, 5, 7, 7, 7, 9

has mode 4 and 7 and is

called bimodal

A distribution having only one

mode is called unimodal


18/39

Lesson 13 - 18

Year 1

CS113/0401/v1

MODE OF A FREQUENCYDISTRICUTION

Ungrouped data

Mode is the x value which has

the highest value of

Grouped data

Cant find mode, only the modal

class


19/39

Lesson 13 - 19

Year 1

CS113/0401/v1

x f

51 - 55

55 - 60

61 - 65

12

16

10

MODAL CLASS

55 - 60 is the modal class

We dont know x values before

grouping, so we cant find the

mode exactly

N.B.

Actual mode might not even be in

this class


20/39

Lesson 13 - 20

Year 1

CS113/0401/v1

In cases where grouped datawhere frequency curve has been

constructed to fit the data, the

mode will be the value (or values)

of x corresponding to the

maximum point (or points) on thecurve, From a frequency

distribution or histogram the

mode can be obtained from the

following formula,

Mode = L1 + ((

1 + 2

1* c

MODE (1)


21/39

Lesson 13 - 21

Year 1

CS113/0401/v1

Where

L1 = lower class boundary ofmodal class

(i.e. class containing the

mode).

1 = excess of modal frequency

over frequency of next lower

class

2 = excess of modal frequency

over frequency of the next

higher class

c = size of modal class interval

MODE (2)


22/39

Lesson 13 - 22

Year 1

CS113/0401/v1

GROUPED MODE FROMHISTOGRAM (1)

Can only ESTIMATE

Assume mode is in Modal Class


23/39

Lesson 13 - 23

Year 1

CS113/0401/v1

Calculation Mode Estimate

= 25 + 5 x

= 25 + 5 x

= 25 + 1.9

= 26.9

40

40 + 64

40

104

GROUPED MODE FROMHISTOGRAM (2)


24/39

Lesson 13 - 24

Year 1

CS113/0401/v1

X =

X1 + X2 + X3+ .. + Xn

N

=

n

i=1

Xi

N

ARITHMETIC MEAN (1)

The arithmetic mean or the meanof a set of N numbers X1, X2, X3,

..., Xn is donoted by X is defined

as


25/39

Lesson 13 - 25

Year 1

CS113/0401/v1

ARITHMETIC MEAN (2)

Eight numbers:

7, 21, 13, 17, 23, 18, 9, 20

Add them = 128

Divide by 8 = 16

This is the arithmetic mean

It is the the most common

definition of average

It only works with quantitative

data


26/39

Lesson 13 - 26

Year 1

CS113/0401/v1

X =

1X1 + 2X2+ .. + nXn

1 + 2 + . n

=

n

i=1

iXi

in

i=1

X

ARITHMETIC MEAN (3)

If the number X1, X2, X3, ..., Xnoccurs 1, 2, 3, ..., n times

respectively, the arithmetic mean

is


27/39

Lesson 13 - 27

Year 1

CS113/0401/v1

MEAN OF A FREQUENCY

DISTRIBUTION

Mean age = = 20.77

(rounded to nearest integer, 21)

2077100

Age (x) xFrequency ()

17

18

19

20

21

22

23

24

25

26

3

8

14

21

24

13

7

6

3

1

51

144

266

420

504

286

161

144

75

26

= 100 x = 2077


28/39

Lesson 13 - 28

Year 1

CS113/0401/v1

HISTOGRAMS (1)

Only used for quantitative data

Histogram is like a bar chart, but

with no gaps between bars and

calibrated horizontal axis

Order of bars depends on value

and on horizontal scale


29/39

Lesson 13 - 29

Year 1

CS113/0401/v1

HISTOGRAMS (2)


30/39

Lesson 13 - 30

Year 1

CS113/0401/v1

HISTOGRAMS (3)


31/39

Lesson 13 - 31

Year 1

CS113/0401/v1

AREA IN HISTOGRAMS


32/39

Lesson 13 - 32

Year 1

CS113/0401/v1

Line of Code No of Programs

100 -

150 -

125 -

39

51

42

24

12

3

325 - 349

300 -

21275 -

30250 -

200 -

175 -

225 -

12

6

CUMULATIVE FREQUENCYDIAGRAMS (1)

Table 1:


33/39

Lesson 13 - 33

Year 1

CS113/0401/v1

Line of Code(less than)

CumulativeFrequency

100

150

125

132

81

39

15

3

0

325

300

201275

171250

200

175

225

222

234

240350

CUMULATIVE FREQUENCYDIAGRAMS (2)

Table 2:


34/39

Lesson 13 - 34

Year 1

CS113/0401/v1

020

40

60

80

100120

140

160

180

200

220

240

0 50 100 150 200 250 300 350

Lines of code (less than)

CummulativeFrequency

CUMULATIVE FREQUENCYDIAGRAMS(3)

Cumulative

Frequency

Curve


35/39

Lesson 13 - 35

Year 1

CS113/0401/v1

n

i=1

(Xi - X) 2

N

STANDARD DEVIATION (1)

The Standard Deviation of a setof N numbers X1, X2, ..., Xn is

denoted by S.D. and is defined by

S.D. =

Where

X = Arithmetic Mean

N = Total Number of element in

the set


36/39

Lesson 13 - 36

Year 1

CS113/0401/v1

nj=1

[ j (Xj- X)2 ]

n

j=1

i Xi2

i

i Xi2-

i( )

S.D.

or

S.D. =

STANDARD DEVIATION (2)(GROUPED DATA)

If X1, X2, ..., Xn occurs withfrequencies 1, 2, ..., n

respectively, the standard

deviation can be written as


37/39

Lesson 13 - 37

Year 1

CS113/0401/v1

Question 6 c) NCC 1/93

On test the actual access times

for 50 hard disc drives weredistributed as follows:

Calculate the mean access time andthe standard deviation.

Time (ms)

No. of Drives

22.6

3

22.7

1

23.022.9

106

22.8 23.223.1

914 25

23.3


38/39

Lesson 13 - 38

Year 1

CS113/0401/v1

Alternative Question 6cx22.6

22.7

22.8

22.9

23.0

23.1

23.2

23.3

f fx fx2

1

3

6

10

14

9

5

2

22.6

68.1

136.8

229.0

322.0

207.9

116.0

46.6

510.76

1545.87

3119.04

5244.10

7406.00

4802.49

2691.20

1085.781149.0 26405.24 (1 mark for each total) 2

2[1] [1]

Mean = 1149

50

= 22.98 [1]

S.D =fx2f

( X )2

=26405.24

50(22.98)2

= 0.156

[1]


39/39

Year 1

The variance of a set of data isdefined as the square of the

standard deviation and is thus

given by (S.D.)

Variance =

i.e.

Variance = (S.D.)2

n

j=1

[ j (Xj- X)2 ]

n

j=1j

VARIANCE

MELJUN CORTES DATA TYPES Rm104tr-13

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