Post on 30-Jun-2020
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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.
Ms Nurazrin Jupri
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
Ms Nurazrin Jupri
MODE < MEDIAN < MEAN
Positively skewed
Skewed to the right
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MODE = MEDIAN = MEAN
Symmetrical
Zero-Skewness
Evenly or normally distributed
Ms Nurazrin Jupri
MEAN < MEDIAN < MODE
Negatively skewed
Skewed to the left
Ms Nurazrin Jupri
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)
Ms Nurazrin Jupri
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
Ms Nurazrin Jupri
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
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
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