DESCRIPTIVE

STATISTICS

Ms Nurazrin Jupri

SKEWNESS

Skewness measures the lack of symmetry in a data

distribution.

The skewed portion is the long and thin part of the curve.

A skewed distribution: the data are sparse at one end of

distribution but piled up at the other end.

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SKEWNESS IN RELATION TO

BETWEEN MEAN, MEDIAN & MODE

Mode : the highest point of the curve

Median : the middle value

Mean : located somewhere towards the tail of the

distribution

Affected by all values, including extreme values

Bell-shaped / normal distribution has NO SKEWNESS

Mean = Median = Mode

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MODE < MEDIAN < MEAN

Positively skewed

Skewed to the right

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MODE = MEDIAN = MEAN

Symmetrical

Zero-Skewness

Evenly or normally distributed

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MEAN < MEDIAN < MODE

Negatively skewed

Skewed to the left

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MEASURE OF SKEWNESS

To determine the difference between the mean and the mode of

the distribution

Mean – Mode = +ve distribution is right or positively

skewed

Mean – Mode = -ve distribution is left or negatively

skewed

Mean – Mode = 0 distribution is symmetrical

𝑷𝒆𝒂𝒓𝒔𝒐𝒏 𝒄𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 𝒔𝒌𝒆𝒘𝒏𝒆𝒔𝒔 =𝟑(𝒎𝒆𝒂𝒏 − 𝒎𝒆𝒅𝒊𝒂𝒏)

𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏

Ms Nurazrin Jupri

EXERCISE 1

1. What is the relationship between mean, median and

mode?

Find the mean median and mode of:

1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 6, 6, 7

• Mean is 4.

• Median is 4.

• Mode is 4.

Ms Nurazrin Jupri

EXERCISE 1 (CONT.)

What is the relationship between mean, median

and mode?

Find the mean, median and mode of:

0, 5, 10, 20, 40, 45, 45, 50, 50, 50, 60, 60, 60, 60, 60, 60, 70,

70, 70, 70, 70, 70, 70, 70

• The mean is 51.5.

• The median is 60.

• The mode is 70.

Ms Nurazrin Jupri

EXERCISE 1 (CONT.)

What is the relationship between mean, median

and mode?

• Find the mean, median, and mode of:

20, 20, 20, 20, 20, 20, 20, 20, 30, 30, 30, 30, 30, 30, 45, 45, 45, 50, 50, 60,

70, 90

• The mean is 36.1.

• The median is 30.

• The mode is 20.

Ms Nurazrin Jupri

QUARTILE

Normally used to describe positional values of large sets

of numerical data.

First quartile (Q1)

Second quartile (Q2)

Third quartile (Q3)

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FIRST QUARTILE (Q1)

Is a positional value where :

25% of the observations are smaller

75% of the observation are larger

Step 1: Find first quartile position

𝑸𝟏 =𝒏 + 𝟏

𝟒

Step 2: Arrange data

Step 3: Find first quartile value which correspond with first

quartile position.

Ms Nurazrin Jupri

THIRD QUARTILE (Q3)

Is a positional value where :

75% of the observations are smaller

25% of the observation are larger

Step 1: Find third quartile position

𝑸𝟑 =𝟑(𝒏 + 𝟏)

𝟒

Step 2: Arrange data

Step 3: Find third quartile value which correspond with

third quartile position.

Ms Nurazrin Jupri

EXAMPLE 1

The 3 year annual returns of 14 low-risk funds are given as

follows.

9.77 11.35 12.46 13.80 15.47 17.48 18.37

18.47 18.61 20.72 21.49 22.47 31.50 38.16

Find the first and third quartile.

𝑄1(𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛) =14 + 1

4= 3.75

Approximately, the forth position of data : 13.80

𝑄3(𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛) =3(14 + 1)

4= 11.25

Approximately, the eleventh position of data : 21.49

Ms Nurazrin Jupri

QUARTILES FOR GROUPED DATA

Step 1: Obtain the cumulative frequencies

Step 2: Identify the first and third quartile position by using

formula quartile position.

Step 3: Identify the first and third quartile classes.

• Quartile position

• Cumulative frequencies

Step 4: Find the first and third quartile by using formula.

𝑸𝟏 =𝒏

𝟒 𝑸𝟑 =

𝟑𝒏

𝟒

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QUARTILES FORMULA

First Quartile

Third Quartile

LCB = Lower class boundary

n = Number of observations

CF = Cumulative frequency before the quartile class

f = Frequency for quartile class

C = Class size

𝑸𝟏 = 𝑳𝑪𝑩𝟏 + (𝑪𝟏)

𝒏𝟒 − 𝑪𝑭𝟏

𝒇𝟏

𝑸𝟑 = 𝑳𝑪𝑩𝟑 + (𝑪𝟑)

𝟑𝒏𝟒 − 𝑪𝑭𝟑

𝒇𝟑

Ms Nurazrin Jupri

EXAMPLE 2

Table shows the distribution of test scores obtained by 42

students in Statistics class. Calculate Q1 and Q3.

Scores obtained Number of students

80-90 1

90-100 2

100-110 5

110-120 10

120-130 15

130-140 7

140-150 2

Total 42

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EXAMPLE 2 (CONT.)

Scores obtained Number of students Cumulative frequency

80-90 1 1

90-100 2 3

100-110 5 8

110-120 10 18

120-130 15 33

130-140 7 40

140-150 2 42

𝑸𝟑 =𝟑(𝟒𝟐)

𝟒= 𝟑𝟏. 𝟓 𝑸𝟏 =

𝟒𝟐

𝟒= 𝟏𝟎. 𝟓

Ms Nurazrin Jupri

EXAMPLE 2 (CONT.)

Scores obtained Number of students Cumulative frequency

80-90 1 1

90-100 2 3

100-110 5 8

110-120 10 18

120-130 15 33

130-140 7 40

140-150 2 42

𝑸𝟏 = 𝑳𝑪𝑩𝟏 + (𝑪𝟏)

𝒏𝟒 − 𝑪𝑭𝟏

𝒇𝟏

= 110 + 120 − 11010.5−8

10

= 110 + 2.5

= 112.50

𝑸𝟑 = 𝑳𝑪𝑩𝟑 + (𝑪𝟑)

𝟑𝒏𝟒 − 𝑪𝑭𝟑

𝒇𝟑

= 120 + 130 − 12031.5 − 18

15

= 120 + 9

= 129

Ms Nurazrin Jupri

INTERQUARTILE RANGE

The difference between the third and first quartiles in a set

of data.

One of dispersion measurement

𝑰𝒏𝒕𝒆𝒓𝒒𝒖𝒂𝒓𝒕𝒊𝒍𝒆 𝒓𝒂𝒏𝒈𝒆 = 𝑸𝟑 − 𝑸𝟏

Ms Nurazrin Jupri

SEMI-INTERQUARTILE RANGE

Known as Quartile Deviation

One of dispersion measurement

𝑸𝒖𝒂𝒓𝒕𝒊𝒍𝒆 𝑫𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏 = 𝑸𝟑 − 𝑸𝟏

𝟐

Ms Nurazrin Jupri

EXAMPLE 3

Refer to example 2 , find the interquartile and semi

interquartile range.

𝑰𝒏𝒕𝒆𝒓𝒒𝒖𝒂𝒓𝒕𝒊𝒍𝒆 𝒓𝒂𝒏𝒈𝒆 = 𝟏𝟐𝟗 − 𝟏𝟎𝟐. 𝟓 = 𝟐𝟔. 𝟓

𝑺𝒆𝒎𝒊 𝒊𝒏𝒕𝒆𝒓𝒒𝒖𝒂𝒓𝒕𝒊𝒍𝒆 𝒓𝒂𝒏𝒈𝒆 =𝟐𝟔. 𝟓

𝟐= 𝟏𝟑. 𝟐𝟓

Ms Nurazrin Jupri

COEFFICIENT OF VARIATION

Used while comparing distributions of different means and

variances

Gives the ratio of standard deviation to mean expressed

as percent.

𝑪𝑽 =𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏

𝒎𝒆𝒂𝒏× 𝟏𝟎𝟎

Ms Nurazrin Jupri

EXAMPLE 4

Typist Ani can type 40 words per minutes with standard

deviation of 5 while typist Jura can type 160 words per

minutes with standard deviation of 10. which typist is more

consistent in her work?

Standard deviation of Jura is twice than Ani

Ani can type four times the speed of Jura

𝑪𝑽𝑨𝒏𝒊 =𝟓

𝟒𝟎× 𝟏𝟎𝟎 = 𝟏𝟐. 𝟓%

𝑪𝑽𝑱𝒖𝒏𝒂 =𝟏𝟎

𝟏𝟔𝟎× 𝟏𝟎𝟎 = 𝟔. 𝟐𝟓%

It shows that the typing ability of typist Jura is more consistent than

typist Ani

Ms Nurazrin Jupri

EXERCISE 2

The investments of Karu and Kamal are given as below:

Whose investment is considered to be more consistent?

Karu Kamal

Profit (RM) 250 250

Standard Deviation 8.16 238.05

Ms Nurazrin Jupri