Assignment SBST1303

September 15, 2017 | Author: alizaza | Category: Categorical Variable, Statistical Analysis, Statistics, Mathematics
Share Embed Donate


Short Description

Statistic Assignment...

Description

FACULTY OF INFORMATION TECHNOLOGY AND MULTIMEDIA COMMUNICATION

SBST1303 ELEMENTARY STATISTIC

Name: Matrix No: NRIC No. : Telephone No. : Email Address:

Tutor:

SBST1303 (Elementary Statistic)

QUESTION 1 a. Construct a relative frequency distribution for the data in Table 1 Num. Of Children

Frequency (f)

Relative Frequency (%)

0 1 2 3 4 5 SUM

7 7 18 20 7 1 60

7/60*100 = 11.666 7/60*100 = 11.666 18/60*100 = 30 20/60*100 = 33.333 7/60*100 = 11.666 1/60*100 = 1.666 99.997

b. Describe the distribution of the data The table above is Frequency Distribution Table of qualitative variable married couples. The first row is the category of the variable; the second row is the frequency of each categorical value with the total of 60 of couples married. The third row shows the relative frequency of the class is the ratio of its frequency to the total frequency. And the fourth row shows the relative frequency expressed in percentage by multiplying 100% to each relative frequency.

c. Use the frequency distribution table to construct a bar chart 25

Frequency

20

15

10

5

0 0

1

2

3

4

5

Num Of Children

2|Page

SBST1303 (Elementary Statistic)

d. Calculate the percentage of married couples having I.

Two children

(18) X 100=

30%

60

II.

At least two children:No of children

Couples Married

2

18

3

20

4

7

5

1 _____ 46 _____

(46) X 100= 76.666% 60

3|Page

SBST1303 (Elementary Statistic)

QUESTION 2 a. Using a class width of 7 and a first class lower limit of 30, construct a frequency distribution table using the data in Table 2.

Cholesterol Level

Frequency

30-36 37-43 44-50 51-57 58-64 65-71 72-78 SUM

4 3 10 11 5 4 3 40

Working:The number of class: K

= 6.29 =6

Class width : 7 72- 32 6

= 6.83 ≈ 7

b. Determine the following I. upper and lower boundaries and the class mid point for the second class Lower Boundaries of second class = 37+36 2 = 36.5 Upper Boundaries of second class = 43+44 2 = 43.5 Mid point for the second class

= 36.5 + 43.5 2 = 40 4|Page

SBST1303 (Elementary Statistic)

II.

relative frequency of the fifth class

Cholesterol Level

Frequency

58-64 SUM

5 40

Relative Frequency 0.13

Relative Frequency (%) 13

Working:Relative frequency

= 5/ 40 = 0.125

Relative frequency (%)

= 0.13 x 100 = 13

III.

Range of the data

= 11- 3 =8

5|Page

SBST1303 (Elementary Statistic)

c. With reference to the data in Table 2, construct a cumulative frequency polygon on a graph paper.

Cumulative Frequency Polygon 45 40

40 37

Cumulative frequency

35

33

30

28

25 20 17

15 10 7

5

4

0

0 29.5

36.5

43.5

50.5

57.5

64.5

71.5

78.5

Upper boundary

Working:Cholesterol Level

Frequency

23-29 30-36 37-43 44-50 51-57 58-64 65-71 72-78

0 4 3 10 11 5 4 3

SUM

40

Upper Boundary
View more...

Comments

Copyright ©2017 KUPDF Inc.
SUPPORT KUPDF