Statistics Lesson 1.pdf

12/04/2022
1
Khartoum International Developed School
(KIDS)
Statistics
Lesson 1
Measures of Central Tendency
1
Dr. Kamal Ramadan;KIDS; 2022
Dr. Kamal Ramadan;KIDS; 2022 2
 Definition:
Statistic is the set of methods and scientific
theorems which deal with collecting of
numerical data then depicted, described, and
analyzed these data and use the results in
predicting, reporting and investigation.
 Description Measures are:
1. Central Tendency
2. Dispersion
3. Skewness
4. Normality

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2
 The Functions of Statistic:
1. Collecting of numerical data about
specific case study.
2. Data Dispersion
3. Analysis the data
4. Use the results in predicting, reporting
and investigation
 Central Tendency:
Definition: It is the tendency of different
values to concentrate at a model value
representing the values in the distribution.
 Central Tendency Measures:
The most important measures are:
 The Arithmetic Mean
 The Median
 The Mode.

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3
 The Arithmetic Mean
 It is the sum of all values divided by their
numbers.
n
x
x
x
x n





2
1
x
x
Or using sigma:



n
i
i
n
x
x
1
 If the values are: x1, x2, . . . . .xn then the
arithmetic man is given by:
Where xi are the values and n their number:
x
Properties of Arithmetic Mean:
1. Has a clear definition and meaning
2. Easy to calculate
3. It is affected by all numbers in the data
 Disadvantages of Arithmetic Mean
1. Misleading when data contains extreme
values.
2. Cannot be calculated directly from group
frequency tables.

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4
x
Example 1:
Find the arithmetic mean for the following
values: 2, 0, 4, 10, 4, 6, 9,5
Solution
n
x
x
x
x n





2
1
8
40
8
5
9
6
4
10
4
0
2









x
5

x
12/04/2022 8
Example 2:
If the arithmetic mean for the values: 5, -3, 4,
8, y, 6 equal 7 find the value of y.
Solution
n
x
x
x
x n





2
1
6
20
6
6
8
4
3
5
7
y
y 







6
20
7
y

 y

 20
42
22

y

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5
Problem 1
 If the mean daily expenditure for a student
1200 USD, find the expenditure of this
student for the whole month (30 days).
Ans.: 36000 USD
Problem 2
 A group of students examined in math, if the
total sum of their marks 304, and the
arithmetic mean for their marks equal 38
find the number of student.
Ans.: 8
Arithmetic Mean for Two groups of Data
 If the number of values for first group is n1
and its mean is m1 and if the number of
values for the second group is n2 and its
mean is m2 then the arithmetic mean for the
two group when merged is:
2
1
2
2
1
1
n
n
n
m
n
m
x





This rule can be extended for 3 or more
groups

12/04/2022
6
Example 3:
Two classes 25 & 15 students are set for math
exam. If the arithmetic mean for their marks
are 60 & 80 respectively find the arithmetic
mean when the two classes merged together.
Solution
Given data: n1=25, n2=15, m1=60 , m2=80
2
1
2
2
1
1
n
n
n
m
n
m
x





5
.
67
40
2700
15
25
15
80
25
60







x
Problem 3:
A group consists of 10 boys and 15 girls, if the
mean height of the boys equal 1.6 m and for
girls equal 1.4 m, find the mean value for the
whole group.
Ans.: 1.48

12/04/2022
7
Arithmetic Mean for Frequency Tables:
 This type of tables gives the values (xr) and
the corresponding frequency (fr) in form of
columns and rows.
 The arithmetic mean can be calculated from
the rule:




 n
r
r
n
r
r
r
f
x
f
x
1
1
Where:
xr the numerical values number r (x1, x2,...)
fr the frequency of occurrence of the value xr
Example 4:
frequency table.
Solution
x 4 8 12 6 2
f 2 5 6 4 3
X F F .x
4 2 8
8 5 40
12 6 72
6 4 24
2 3 6
Sum 20 150




 n
r
r
n
r
r
r
f
x
f
x
1
1
From table:
f=20 & fx=150
5
.
7
20
150


x

12/04/2022
8
Arithmetic Mean for Class Frequency Tables:
This type of tables gives the values (xr) in
form of class [xi – xf] instead of actual values,
and gives the corresponding frequency (fr) for
the class. (in form of columns or rows)
The arithmetic mean can be calculated from
the rule:




 n
r
r
n
r
r
r
f
m
f
x
1
1
Where:
mr the class mid value [xi+xf/2]
fr the frequency & n is the number of classes
Example 5:
class frequency table.
Solution
class 0 - 10 10 - 20 20- 30 30 - 40 40- 50
frequency 3 7 8 6 1
class m F f.m
0 - 10 5 3 15
10 - 20 15 7 105
20 - 30 25 8 200
30 - 40 35 6 210
40 - 50 45 1 45
Sum 25 575




 n
r
r
n
r
r
r
f
m
f
x
1
1
From table:
f=25 & fm=575
23
25
575


x

12/04/2022
9
Assumed-Mean Method for Class Frequency
Tables:
This method greatly simplify the calculation
of the AM in previous method.
Here we choose the mid-value which has the
higher frequency as assumed value, denoted
by (w)
Then we find the deviation (D) between mid-
values and assumed mean (w), i.e.:
D=m-w
The arithmetic mean can be calculated from
the rule:





 n
r
r
n
r
r
f
D
f
w
x
1
1
Where:
w: is the assumed mi- value [with higher f]
fr the frequency & n is the number of classes
D: is the deviation [m – w]
Example 6:
class frequency table using assumed mean.
class 0 - 10 10 - 20 20- 30 30 - 40 40- 50
frequency 3 7 8 6 1

12/04/2022
10
Solution
class 0 - 10 10 - 20 20- 30 30 - 40 40- 50
frequency 3 7 8 6 1
class m D=m-w F fD
0 - 10 5 -20 3 -60
10 - 20 15 -10 7 -70
20 - 30 25 0 8 0
30 - 40 35 10 6 60
40 - 50 45 20 1 20
Sum 25 -50
From table:
choose w=25
then
f=25 & fD=-50
23
25
50
25 



x





 n
r
r
n
r
r
f
D
f
w
x
1
1
End of Lesson 1
Thank You
20

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