assignment
TRANSCRIPT
FACULTY OF SCIENCE AND TECHNOLOGY
SEMESTER JANUARY 2012
SBST1303
ELEMENTARY STATISTIC
MATRICULATION NO : 800717126061001
IDENTITY CARD NO. : 800717-12-6061
TELEPHONE NO. : 014-3343554
E-MAIL : [email protected]
LEARNING CENTRE : OUM, Petaling Jaya
SBST1303
1
TABLE OF CONTENTS
Contents
1. a Frequency Distribution Table 2
Page
1. b(i) Histogram of Frequency Distribution Table For The Pulse Rates of Males 3
1. b(ii) Histogram of Frequency Distribution Table For The Pulse Rates of Females 4
1. c(i) Cumulative Frequency For The Pulse Rates of Males 5
1. c(ii) Cumulative Frequency For The Pulse Rates of Females 6
2. a Find Mean, Mode, Median and Standard Deviation 7
2. b Pearson’s Coefficient of Skewness 13
ATTACHMENT 14
SBST1303
2
1.a
Table 1.1: Frequency Distribution Table for the Pulse Rate of Males.
Frequency Distribution Table
Class Frequency (f) Cumulative Frequency
50 – 59 6 6 60 – 69 17 23 70 – 79 8 31 80 – 89 8 39 90 – 99 1 40
Table 1.2: Frequency Distribution Table for the Pulse Rate of Females.
Class Frequency (f) Cumulative Frequency 60 – 69 12 12 70 – 79 13 25 80 – 89 10 35 90 – 99 3 38
100 – 109 2 40
Class
1.b (i)
Freq
uenc
y
Figure 1.1: Histogram Of Frequency Distribution Table For The Pulse Rates Of Males
2
10
12
14
4
18
16
313940
70 - 7980 - 8990 - 99
881
623
50 - 5960 - 69
617
CFFrequencyClass
90 - 990
17
6
8 8
1
50 - 59 60 - 69 70 - 79 80 - 89
8
6
1.b (ii)
Class
Freq
uenc
y
2
10
12
14
4
18
16
353840
Figure 1.2: Histogram Of Frequency Distribution Table For The Pulse Rates Of Females
CFFrequencyClass
100 - 1090
60 - 69 70 - 79 80 - 89 90 - 99
8
6
1225
1213
100 - 109
60 - 6970 - 7980 - 8990 - 99
1032
1213
10
32
44
12
8
4
40
36
32
28
24
Freq
uenc
y
20
16
0
Class
40 - 4950 - 5960 - 6970 - 79
23≤ 59.5≤ 69.5
617
1.c (i)
80 - 8990 - 99
Cumulative FrequencyFrequencyUpper
Boundary00≤ 49.56
Figure 1.3: Cumulative Frequency For The Pulse Rates Of Males
313940
≤ 79.5 881
≤ 89.5≤ 99.5
Upper Boundary (Pulse Rates)
99.549.5 59.5 69.5 79.5 89.5
44
12
8
4
40
36
32
28
24
Freq
uenc
y
20
16
0
Class
50 - 5960 - 6970 - 7980 - 89
1.c (ii)
90 - 99100 - 109
Figure 1.4: Cumulative Frequency For The Pulse Rates Of Females
3538
25≤ 69.5≤ 79.5
1213
12
Cumulative FrequencyFrequencyUpper
Boundary00≤ 59.5
40
≤ 89.5 1032
≤ 99.5≤ 109.5
99.5 109.5
Upper Boundary (Pulse Rates)
59.5 69.5 79.5 89.5
SBST1303
2. a
Find Mean, Mode, Median and Standard Deviation
Table 2.1: Lower Boundary, Class Mid-Point, Upper Boundary for the Pulse Rates of Males
MALES
Class Lower Boundary
Class Mid-Point (x)
Upper Boundary
Frequency (f)
50 – 59 49.5 54.5 59.5 6 60 – 69 59.5 64.5 69.5 17 70 – 79 69.5 74.5 79.5 8 80 – 89 79.5 84.5 89.5 8 90 – 99 89.5 94.5 99.5 1
i. Mean = location or position of data distribution
Formula
𝜇𝜇 = ∑ 𝑓𝑓𝑖𝑖 𝑥𝑥𝑖𝑖∑ 𝑓𝑓𝑖𝑖
= 6(54.5)+17(64.5)+8(74.5)+8(84.5)+1(94.5)6+17+8+8+1
= 69.75 ≈
70
SBST1303 ii. Mode = observation (or the number) which has the largest frequency
Formula
𝓍𝓍 � = 𝐿𝐿𝐵𝐵 + 𝐶𝐶 � ∆𝐵𝐵∆𝐵𝐵 + ∆𝐴𝐴
�
= 59.5 + 10 � 17−6(17−6)+(17−8)
�
= 59.5 + 10 � 1111+9
�
= 59.5 + (110/20)
=
iii. Median = summarise distribution in two equal parts of 50% each
65
Formula
𝑥𝑥� = 𝐿𝐿𝐵𝐵 + 𝐶𝐶�𝑛𝑛 +1
2 − 𝐹𝐹𝐵𝐵 �
𝑓𝑓𝑚𝑚
(a) Where (n + 1)/2 = (40 + 1)/2 = 20.5
(b) The median class is 60 – 69 with the following:
𝑓𝑓𝑚𝑚 = 17, 𝐿𝐿𝐵𝐵 = 59.5, 𝐶𝐶 = 10, 𝐹𝐹𝐵𝐵 = 6
= 59.5 + 10 (𝟐𝟐𝟐𝟐.𝟓𝟓 − 𝟔𝟔)𝟏𝟏𝟏𝟏
= 68.03 ≈ 68
SBST1303
iv. Standard Deviation
Formula
𝜎𝜎 = �∑ 𝑥𝑥𝑖𝑖
𝑛𝑛
2
− �∑ 𝑥𝑥𝑖𝑖
𝑛𝑛 �2
= �4545
− �405
�2
= √90.8 − 64
=
5.18
SBST1303
Table 2.2: Lower Boundary, Class Mid-Point, Upper Boundary for the Pulse Rates of Females
FEMALES
Class Lower Boundary
Class Mid-Point (x)
Upper Boundary
Frequency (f)
60 – 69 59.5 64.5 69.5 12 70 – 79 69.5 74.5 79.5 13 80 – 89 79.5 84.5 89.5 10 90 – 99 89.5 94.5 99.5 3
100 – 109 99.5 104.5 109.5 2
i. Mean = location or position of data distribution
Formula
𝜇𝜇 = ∑ 𝑓𝑓𝑖𝑖 𝑥𝑥𝑖𝑖∑ 𝑓𝑓𝑖𝑖
= 12(64.5)+13(74.5)+10(84.5)+3(94.5)+2(104.5)12+13+10+3+2
=
77
SBST1303 ii. Mode = observation (or the number) which has the largest frequency
Formula
𝓍𝓍 � = 𝐿𝐿𝐵𝐵 + 𝐶𝐶 � ∆𝐵𝐵∆𝐵𝐵 + ∆𝐴𝐴
�
= 69.5 + 10 � 13−12(13−12)+(13−10)
�
= 69.5 + 10 � 11+3
�
= 69.5 + (10/4)
=
iii. Median = summarise distribution in two equal parts of 50% each
72
Formula
𝑥𝑥� = 𝐿𝐿𝐵𝐵 + 𝐶𝐶�𝑛𝑛 +1
2 − 𝐹𝐹𝐵𝐵 �
𝑓𝑓𝑚𝑚
(c) Where (n + 1)/2 = (40 + 1)/2 = 20.5
(d) The median class is 70 – 79 with the following:
𝑓𝑓𝑚𝑚 = 13, 𝐿𝐿𝐵𝐵 = 69.5, 𝐶𝐶 = 10, 𝐹𝐹𝐵𝐵 = 12
= 69.5 + 10 (20.5 − 12)13
= 76.04 ≈ 76
SBST1303
iv. Standard Deviation
𝜎𝜎 = �∑ 𝑥𝑥𝑖𝑖
𝑛𝑛
2
− �∑ 𝑥𝑥𝑖𝑖
𝑛𝑛 �2
Formula
= �4265
− �405
�2
= √85.2 − 64
=
4.60
SBST1303
2. b
Pearson’s Coefficient of Skewness
𝑃𝑃𝐶𝐶𝑃𝑃 (1) = (𝑀𝑀𝑀𝑀𝑀𝑀𝑛𝑛 − 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 )𝑃𝑃𝑆𝑆𝑀𝑀𝑛𝑛𝑀𝑀𝑀𝑀𝑆𝑆𝑀𝑀 𝐷𝐷𝑀𝑀𝐷𝐷𝑖𝑖𝑀𝑀𝑆𝑆𝑖𝑖𝑀𝑀𝑛𝑛
= 𝑥𝑥̅ − 𝑥𝑥�𝑠𝑠
MALES
= (70−65)5.18
=
Value obtained show it positively skewed
0.97
𝑃𝑃𝐶𝐶𝑃𝑃 (1) = (𝑀𝑀𝑀𝑀𝑀𝑀𝑛𝑛 − 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 )𝑃𝑃𝑆𝑆𝑀𝑀𝑛𝑛𝑀𝑀𝑀𝑀𝑆𝑆𝑀𝑀 𝐷𝐷𝑀𝑀𝐷𝐷𝑖𝑖𝑀𝑀𝑆𝑆𝑖𝑖𝑀𝑀𝑛𝑛
= 𝑥𝑥̅ − 𝑥𝑥�𝑠𝑠
FEMALES
= (77−72)4.60
=
Value obtained show it positively skewed
1.07
14
ATTACHMENT
REFERENCES
Prof Dr Mohd Kidin Shahran (2011). Module SBST1303 Elementary Statistics. Open
University Malaysia.