Elementary Statistics
Content
FAKULTI OF BUSINESS AND MANAGEMENT
SBST1303 ELEMENTARY STATISTICS
NO. MATRIKULASI
:
710624025377001
NO. KAD PENGNEALAN
:
710624025377
NO. TELEFON
:
019-4146812
E-MEL
:
[email protected]
PENSYARAH
:
LIM YAI FUNG
PUSAT PEMBELAJARAN
:
PPW KEDAH
QUESTION 1
A) I
AGE
Tally
23-26
IIII
Frequency(f)
4
RelativeFrequency 0.1
10
Relative
Frequency(%)
27-30
IIII I
31-34
IIII II
6
0.15
15
7
0.175
17.5
35-38
IIII
IIII
9
0.225
22.5
39-42
IIII
III
8
0.2
20
43-46
IIII
47-50
II
SUM
4
0.1
10
2
0.05
5
40
1
100
a) ii
Histogram of Frequency Distribution
10
9
8
7
Frequency
6
5
4
3
2
1
0
23-26
27-30
31-34
35-38
Age
39-42
43-46
47-50
a) iii
The Less-than or Equal Cumulative Distribution
Age
Frequency
(f)
23-26
27-30
31-34
35-38
39-42
43-46
47-50
Sum
4
6
7
9
8
4
2
∑ f = 40
Upper
Boundar
y
≤ 26.5
≤ 30.5
≤ 34.5
≤ 38.5
≤ 42.5
≤ 46.5
≤ 50.5
Cumulating
Process
Cumulative
Frequency
4
4+6
10 + 7
17 + 9
26 + 8
34 + 4
38 + 2
0
10
17
26
34
38
40
Cumulative
Frequency
(%)
0
25
42.5
65
85
95
100
Cumulative frequency of type less than
polygon for age of OUM graduates
45
40
Cumulative frequency (Age)
35
30
25
20
15
10
5
0
26.5
30.5
34.5
38.5
42.5
46.5
50.5
Upper boundary (OUM graduate)
b)
Cumulative frequency of type less than
polygon for age of OUM graduates
45
40
Cumulative frequency (Age)
35
Q3
30
25
Q2
20
15
25%
Q1
25%
10
5
25%
25%
0
26.5
30.5
34.5
38.5
42.5
46.5
50.5
Upper boundary (OUM graduate)
First quartile,r = 1
The data size = 40
Position = r / 4 (n + 1)
= 1 / 4 (40 + 1)
= 10.25
= 10 + 0.25
Q1 is at the position between tenth and eleventh,and it is 0.25 above the tenth
position.
Second quartile,r = 2
The data size = 40
Position = 2/4 (40+ 1)
= 20.5
= 20 + 0.5
Q2 is at the position between twentieth and twenty first,and it is 0.5 above the
twentieth position.
Third quartile, r = 3
The data size = 40
Position = 3/4( 40 + 1)
= 30.75
= 30 + 0.75
Q3 is at the position between thirtieth and thirty first,and it is 0.75 above the thirtieth
position.
b)
Mean Market A
µ1=23+21+24+23+24+23+22+22+23
9
µ =22.77
Mean Market B
µ2=20+20+21+22+22+23+21+21+23
9
µ =21.44
Mean Market A and B
µ1+µ2 = 22.77 + 21.44
9+9
= 2.45
Standard deviation Market A
= √(23-22.7)²+(21.22.77)²+(24-22.7)²+(23-22.77)²+(24-22.77)²+(23.77)²+(2222.77)²+(22-22.77)²+(23-22.77)²
= 0.9718
Standard devition Market B
= √(20-21.44)²+(20-21.44)²+(21-21.44)²+(22-21.44)²+(22-21.44)²+(23-21.44)²+(2121.44)²+(21-21.44)²+(23-21.44)²
= 1.13
Ii. Coefficient of variation Market A = Standard of variation/ Mean
= 0.97 / 22.77
= 0.04
Coefficient of variation Market B = Standard of variation/ Mean
= 1.13 / 21.44
= 0.05
QUESTION 2
a)
i. Mean = 5.01 + 4.98 + 5.02 + 5.00 + 4.99 + 5.00 + 4.97 + 5.02 + 5.00 + 5.01 + 5.00
12
=5
ii. Mode = 5.00
iii. Median = 4.97,4.98,4.99,5.00,5.00,5.00,5.00,5.01,5.01,5.01,5.02,5.02
= (12+1)/2=6.5th
Median = 5.00
iv. Variance = 0
v. Standard deviation
= √(5.01-5)² +( 4.98 -5)²+ (5.02 -5)²+ (5.00 -5)²+ (4.99-5)² + 5.00-5)² +( 4.97-5)² +
(5.02 -5)²+ (5.00-5)² +( 5.01-5)² + (5.00-5)² = 0.01505
SBST1303 ELEMENTARY STATISTICS
NO. MATRIKULASI
:
710624025377001
NO. KAD PENGNEALAN
:
710624025377
NO. TELEFON
:
019-4146812
E-MEL
:
[email protected]
PENSYARAH
:
LIM YAI FUNG
PUSAT PEMBELAJARAN
:
PPW KEDAH
QUESTION 1
A) I
AGE
Tally
23-26
IIII
Frequency(f)
4
RelativeFrequency 0.1
10
Relative
Frequency(%)
27-30
IIII I
31-34
IIII II
6
0.15
15
7
0.175
17.5
35-38
IIII
IIII
9
0.225
22.5
39-42
IIII
III
8
0.2
20
43-46
IIII
47-50
II
SUM
4
0.1
10
2
0.05
5
40
1
100
a) ii
Histogram of Frequency Distribution
10
9
8
7
Frequency
6
5
4
3
2
1
0
23-26
27-30
31-34
35-38
Age
39-42
43-46
47-50
a) iii
The Less-than or Equal Cumulative Distribution
Age
Frequency
(f)
23-26
27-30
31-34
35-38
39-42
43-46
47-50
Sum
4
6
7
9
8
4
2
∑ f = 40
Upper
Boundar
y
≤ 26.5
≤ 30.5
≤ 34.5
≤ 38.5
≤ 42.5
≤ 46.5
≤ 50.5
Cumulating
Process
Cumulative
Frequency
4
4+6
10 + 7
17 + 9
26 + 8
34 + 4
38 + 2
0
10
17
26
34
38
40
Cumulative
Frequency
(%)
0
25
42.5
65
85
95
100
Cumulative frequency of type less than
polygon for age of OUM graduates
45
40
Cumulative frequency (Age)
35
30
25
20
15
10
5
0
26.5
30.5
34.5
38.5
42.5
46.5
50.5
Upper boundary (OUM graduate)
b)
Cumulative frequency of type less than
polygon for age of OUM graduates
45
40
Cumulative frequency (Age)
35
Q3
30
25
Q2
20
15
25%
Q1
25%
10
5
25%
25%
0
26.5
30.5
34.5
38.5
42.5
46.5
50.5
Upper boundary (OUM graduate)
First quartile,r = 1
The data size = 40
Position = r / 4 (n + 1)
= 1 / 4 (40 + 1)
= 10.25
= 10 + 0.25
Q1 is at the position between tenth and eleventh,and it is 0.25 above the tenth
position.
Second quartile,r = 2
The data size = 40
Position = 2/4 (40+ 1)
= 20.5
= 20 + 0.5
Q2 is at the position between twentieth and twenty first,and it is 0.5 above the
twentieth position.
Third quartile, r = 3
The data size = 40
Position = 3/4( 40 + 1)
= 30.75
= 30 + 0.75
Q3 is at the position between thirtieth and thirty first,and it is 0.75 above the thirtieth
position.
b)
Mean Market A
µ1=23+21+24+23+24+23+22+22+23
9
µ =22.77
Mean Market B
µ2=20+20+21+22+22+23+21+21+23
9
µ =21.44
Mean Market A and B
µ1+µ2 = 22.77 + 21.44
9+9
= 2.45
Standard deviation Market A
= √(23-22.7)²+(21.22.77)²+(24-22.7)²+(23-22.77)²+(24-22.77)²+(23.77)²+(2222.77)²+(22-22.77)²+(23-22.77)²
= 0.9718
Standard devition Market B
= √(20-21.44)²+(20-21.44)²+(21-21.44)²+(22-21.44)²+(22-21.44)²+(23-21.44)²+(2121.44)²+(21-21.44)²+(23-21.44)²
= 1.13
Ii. Coefficient of variation Market A = Standard of variation/ Mean
= 0.97 / 22.77
= 0.04
Coefficient of variation Market B = Standard of variation/ Mean
= 1.13 / 21.44
= 0.05
QUESTION 2
a)
i. Mean = 5.01 + 4.98 + 5.02 + 5.00 + 4.99 + 5.00 + 4.97 + 5.02 + 5.00 + 5.01 + 5.00
12
=5
ii. Mode = 5.00
iii. Median = 4.97,4.98,4.99,5.00,5.00,5.00,5.00,5.01,5.01,5.01,5.02,5.02
= (12+1)/2=6.5th
Median = 5.00
iv. Variance = 0
v. Standard deviation
= √(5.01-5)² +( 4.98 -5)²+ (5.02 -5)²+ (5.00 -5)²+ (4.99-5)² + 5.00-5)² +( 4.97-5)² +
(5.02 -5)²+ (5.00-5)² +( 5.01-5)² + (5.00-5)² = 0.01505
