Chapter 3
“Descriptive statistics"
First: Measures of Central Tendency:
Measure
Mean
Weighted
mean
Median
Mode
Midrange

 "104
Definition
"Sum of values, divided by total number of values"
"Sum of values multiply the weights divided by total of
the weights"
"The value of data is divided into two equal parts that has
been ordered"
"Most frequent data value"
"the sum of the lowest and highest values
In the data set, divided by 2"
Symbol
 , X
X
MD
None
MR
(1) The Mean:
First: The Mean for ungrouped Data:
n
Xi
X 1  X 2  ............ X n 
Sample mean.  X 
 i 1
n
n
Example (3-1)
The data represent the number of days off per year for a sample of individuals
Selected from nine different countries.
"20, 26, 40, 36, 23, 42, 35, 24, 30"
Find the mean.
Solution:
9
X
X
i 1
n
i

20  26  40  36  23  42  35  24  30
 30.7  31 days
9
Second ; The Mean for Grouped Data : "Using the frequency distribution"
Mean sample  X 
fX
n
m
Where x m = the midpoints
n = Sum Frequency  f
1
Example (3-2)
If we have the following data:
Classis limits
(Frequency)
61116 21 26 31 36- 41
Sum
f
1
2
3
5
4
3
2
n =  f = 20
Find: The mean?
Solution
Step 1: Find the midpoints of each class and enter them in column:
Xm =
lower boundary  Upper boundary 5.5  10.5
35.5  40.5

 8....
 38
2
2
2
Step 2 : Make a table as shown:
Xm
Frequency = f
f Xm
8
1
8
13
2
26
18
3
54
23
5
115
28
4
112
33
3
99
38
2
76
Sum
n =  f = 20
Step 3: Sample mean  X 
f X
f X
n
2
m

m
= 490
490
 24.5 miles
20
(2)The Weighted Mean:
X
w1X1  w2 X 2  ....... w n X n
w1  w2  ..........w2

 w Xi
w
Where wi  The weights
i
i
Example (3-3):
Find the Weighted Mean of following data:
Course
Weighted (w)
degree (X)
English
3
4 points
Psychology
3
2 points
Biology
4
3 points
Physical
2
1 point
Solution:
w Xi (3  4) 

X

w

i
i
(3  2)  (4  3)  (2  1)
 2.7
3 3 4  2
(3)The Median: "Ungrouped Data"
First : When the data is an odd number:
Example (3-4)
The number of rooms in the seven hotels in city:
(713, 300, 618, 595, 311, 401, 292)
. Find the median.
Solution
Step 1: Arrange the data in order:  (292, 300, 311, 401, 595, 618, 713)
Step 2 Select the middle value: (292, 300, 311 , 401 , 595, 618, 713)

median
 MD = 401
The median 
3
Second: When the data is an even number:
Example (3-5)
The number of tornadoes that have occurred in the United States over an
8-year period follows:
(684, 764, 656, 702, 856, 1133, 1132, 1303)
Find the median.
Solution
Step 1 : Arrange the data in order:
(656, 684, 702 , 764 , 856 , 1132, 1133, 1303)

median
Step 2:
MD 
764  856
 810
2
(4)The Mode :
First: The Mode for ungrouped Data:
Example (3-6)
Find the mode for the number of branches that eight banks have:
1 . (18, 14, 34.5, 10, 11.3, 10, 12.4, 10)
2 . (401, 344, 209, 201, 227, 353,228,405)
Solution:
(1) 
 The mode = 10
 (mode = 0)
(2)Do not say that the mode is zero 
Second: The Mode for grouped Data "Frequency distribution":
Example (3-7):
Find the mode for grouped data
(Class)
5.5
10.5
15.5
20.5-25.5
25.5
30.5
35.5- 40.5

(Frequency) f
1
2
3


5
4
3
2
n = 20
 The mode = class is between (20.5 – 25.5)
Solution: 
4
(5)The Midrange:
lowest value  highest value
2
Example ( 3-6):
MR 
Find the midrange. (2, 3, 6, 8, 4, 1) 
 Solution: MR 
1 8
 4.5
2
( H . W )  For Exercises ; (19)  p 120
Exercises:
(1) Write Measures of Central Tendency?
(2) Find: (a) the mean (b) the Median (c) the mode (d) the midrange:
In data following:
(1) High Temperatures
62
72
66 79
83
61
62
85 72
64
74 71
42
38
91 66
77
90
74
63 64
68
42
Solution 
b. 68
c. mode = 0 d. 32.25)
 ( a. 68.1
(3) Complete the table and Find: (a) the mean,
Classes limit
Boundaries
Classes
0.5–
3.5–
6.5–
9.5–
12.5–15.5
Sum
(b) the mode
(Frequency) F
12
11
4
2
1
n =30
Solution:  (a . 5 ) (b .between ( 0.5-3.5)
(4) Find the weighted mean price of three models of automobiles sold. The
number and price of each model sold are shown in this list.
Model
A
B
C
Solution:  $9866.67
Number
8
10
12
Price $
10,000
12,000
8,000
5
Measures of Variation:
Measure
Definition
Symbol(s)
Range
Distance between highest value and lowest value
R
Variance
Average of the squares of the distance that each
 2 , S2
value is from the mean"
Standard
Square root of the variance

,
deviation
Coefficient
Is the standard deviation divided by the mean"
of Variation
(1)Range:
The Range 
 R = highest value - lowest value
Example (1):
Are shown here. Find the range:
Staff
Salary
Owner
100,000
Manager
40,000
Sales representative 30,000
Workers
25,000
15,000
18,000
Solution:
R = 100,000 - 15,000=85,000.
(2)Variance and Standard deviation
First : Variance and Standard deviation" Population" :
Population variance is:


2
 (X - M)

N
2
Where; M  Population mean
N  Population size
    2
Population standard deviation is 
6
C.V
S
Second : variance and standard deviation Sample For "ungrouped Data"
Variance Sample 
 S
2
 (X - X)

n -1
2

n  X i2  (  X )2
n(n  1)
Where X = sample mean
n = sample size
Standard Deviation Sample 
 S 
S2
Example (2)
Find the sample variance and standard deviation for the data:
(11.2, 11.9, 12 , 12.8, 13.4, 14.3)
Solution:
Variance 
 S 
2
Step 1
n  X i2  ( X )2
n (n  1)

  X 2  (11.2 )2 + (11.9)2 + (12)2 + (12.8)2 + (13.4)2 + (14.3)2  958.94
Step 2 
  X = 11.2 +11.9 +12 +12.8 +13.4 +14.3 = 75.6
 ( X ) 2  (75.6) 2  5715.36
Step 3

 Variance S 2 
6 (958 .94 )  5715.36
38.28

 1.276
6(5)
30
Standard Deviation 
 S 
s 2  1.276 1.13
Third: Sample Variance and Standard Deviation for " grouped Data" :
Variance 
 S 
2
n  f . X m2  (  f . X m ) 2
n ( n  1)
Where : X m = midpoint
f = frequency
n = Sum Frequency  f
7
Example (3)
Find the variance, the standard deviation and rang for the frequency distribution
of the data following:
Class boundaries
Frequency F
5.5–10.5
1
10.5–15.5
2
15.5–20.5
3
20.5–25.5
5
25.5–30.5
4
30.5–35.5
3
35.5–40.5
2
n
20
Solution:
Variance 
 S 
2
n  f . X m2  (  f . X m ) 2
n (n  1)
Step 1: Make a table as shown, and find the midpoint of each class:
Midpoint (Xm) Frequency ( f )
8
13
18
23
28
33
38
1
2
3
5
4
3
2
n=20
f .X m
8
26
54
115
112
99
76
X m2
64
169
324
529
784
1089
1444
 f  Xm  490
f . X m2
64
338
972
2,645
3,136
3,267
2,888
 f . X m2  13,310
Step 2: Substitute in the formula and solve for to get the variance:
20 (13,310) - (490)2
266,200- 240,100
Variance 
 S 2 


20 (20 - 1)
20(19)

 S 2 
26,100
 68.7
380
 S  68 .7  8.3
Step 3: the standard deviation 
 R=40.5 -5.5= 35
Rang: R = highest value - lowest value 
8
Or 
 R=41 - 6 = 35
(4)Coefficient of Variation:

 C.V 

X
Where:  = standard deviation
X = sample mean
Example (4):
The mean of the number of sales of cars over a 3-month period is 87car, and the
standard deviation is 5. The mean of the commissions is $5225, and the
standard deviation is $773. Compare the variations of the two .
‫ متىسط العمىالت‬.‫ سيارات‬5 ‫ واالوحراف المعياري‬،‫ سيارة‬87 ‫ أشهر هى‬3 ‫متىسط عذد مبيعات السيارات لمذة‬
. ‫ قارن االختالفات بيه العمىالت و عذد مبيعات السيارات‬. $ 773 ‫ االوحراف المعياري هى‬،$ 5225 ‫هى‬
Solution:
The coefficients of variation are:
( number of sales of cars )  X  87
C.V 
(commissions)  X  5225

S 5

 C.V
X
S  773
5
 0,057  0.06
87
773
 0.148  0.15
5225
The commissions are more variation than the sales
Exercises :
(1)What is the relationship between the variance and the standard
deviation?
 the standard deviation = The square root of the variance
Solution 
C.V 
  2
S  S2
Or
(2) What are the symbols used to represent the population
  2 , 
variance and standard deviation? Solution 
(3) what are the symbols used to represent the sample
 s 2 , s
Variance and standard deviation? Solution 
(4) Find the range, variance, and standard deviation for data:
 (48 , 254.7 , 15.9)
(1)(7, 37, 3, 8, 48, 11, 6, 0, 10, 3) Solution 
(2)60, 20, 40, 40, 45, 12, 34, 51, 30, 70, 42, 31, 69, 32, 8, 18, 50
 (62 , 332.4 , 18.2)
Solution 
9
(5)Find the "Range -the Variance -Standard deviation -Coefficient of
Variation" for data following:
Class limit Frequency ( f )
13–19
20–26
27–33
34–40
41–47
48–54
Xm
f .X m
X m2
f . X m2
2
7
12
5
6
1
 f .X
n=
m
 f .X

2
m

(172 Chapter 3 Data Description) ‫ملخص عه الفصل الثالث ص‬
(6) Which score has the highest?
a. X = 10
S =4

 Solution = 0.4
b. X = 120
S = 12

 Solution = 0.10
c . X = 60
S =8

 Solution = 0.13
(c is highest)
(7) Exam Scores On a philosophy comprehensive exam, this
distribution was obtained from 25 students.
Class boundaries
Frequency (F)
40.5–45.5
3
45.5–50.5
8
50.5–55.5
10
55.5–60.5
3
60.5–65.5
1
n=25
Find :
(a) the mean,
(b) the mode
(e)Standard deviation
(c) Range
(d) the Variance
(f) Coefficient of Variation"
10