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Assignment No.1 (Course STA301)
Fall 2012 (Total Marks 15)

Deadline
Your Assignment must be uploaded/ submitted before or on
23:59 7th November, 2012
STUDENTS ARE STRICTLY DIRECTED TO SUBMIT THEIR ASSIGNMENT BEFORE OR BY DUE DATE. NO ASSIGNMNENT AFTER DUE DATE WILL BE ACCEPTED VIA E.MAIL).

Rules for Marking
It should be clear that your Assignment will not get any credit IF:
• The Assignment submitted, via email, after due date.
• The submitted Assignment is not found as MS Word document file.
• There will be unnecessary, extra or irrelevant material.
• The Statistical notations/symbols are not well-written i.e., without using MathType software.
• The Assignment will be copied from handouts, internet or from any other student’s file. Copied material (from handouts, any book or by any website) will be awarded ZERO MARKS. It is PLAGIARISM and an Academic Crime.
• The medium of the course is English. Assignment in Urdu or Roman languages will not be accepted.
• Assignment means Comprehensive yet precise accurate details about the given topic quoting different sources (books/articles/websites etc.). Do not rely only on handouts. You can take data/information from different authentic sources (like books, magazines, website etc) BUT express/organize all the collected material in YOUR OWN WORDS. Only then you will get good marks.


Objective(s) of this Assignment:

• The assignment is being uploaded to build up the students concepts regarding measurement scale and manipulation of data.

• This assignment will strengthen the basic idea about the concept of measure of central tendency.





Assignment No: 1 (Lessons 1-9)
Question 1: Marks:5+2=7

a) Following are the heights (in centimeters) of 45 female high school students. Prepare a frequency distribution of the data where the class width is 5, the number of classes is 6 and the smallest observation is 148. Also obtain the relative frequency and relative cumulative frequency.

170, 151, 154, 160, 158, 154, 171, 156, 160, 157, 160, 157, 148, 165, 158, 159, 155, 151, 152, 161, 156, 164, 156, 163, 174, 153, 170, 149, 166, 154, 166, 160, 160, 161, 154, 163, 164, 160, 148, 162, 167, 165, 158, 158, 176.


b) Which level of measurement (scale of measurement) is suitable for the following data in each example?

i) The lake Saif-ul-Muluk is 10,578 feet above from the sea level.
ii) If customers go to the shop for buying shoes and try different size of shoes. The sizes of shoes are 7B, 10D and 12E.
iii) The numbering on T-shirts of players in the cricket team.
iv) The weight of the person is 68 kg.

Question 2: Marks:5+3=8

a) Find the median, harmonic mean and 23rd percentile for the following frequency distribution.

Class Boundaries Frequency
145 – 150 4
150 – 155 6
155 – 160 28
160 – 165 58
165 – 170 64
170 – 175 30

b) Explain the concept of Geometric mean? Also find the geometric mean for the following data.

18, 19, 19, 19, 19, 20, 20, 20, 20, 20, 21, 21, 21, 21, 22, 23, 24, 27, 30, 36


-END-

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Replies to This Discussion

plz upload the soluion here...as soon as i will get the solution i will to upload here......

please upload the solution

2 b)

median=X childa =l+h/f(n/2-C.F)

is ma

l= the lower class boundary of the medain class(n/2)

h=the class intrval size of the median class

f = the class frequncy of tmedian class

n=sigma f=total frequncy

C.F=the cumulative frequenchy preceding the medin class cumulativ freq

c.b              f                     c.f

145-150        4                 4

150-155        6                  10

155-160       28                 38

160-165      58                    96

165-170     64                  160

170-175    30                  190

toal           190

 

median class or the modal class = (n/2)th=190/2=95 th (to see in c.f)

so the medain r modal class is 160-165         58                    96

by putng the valus in formula

medain = 160 +5/58(95-38)

=160+5/58(57)

=164.9138



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STA301_Solved_Assignment#01

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Question 1:                                                                                                                   Marks: 5+2=7

 

a)   Following are the heights (in centimeters) of 45 female high school students. Prepare a frequency distribution of the data where the class width is 5, the number of classes is 6 and the smallest observation is 148. Also obtain the relative frequency and relative cumulative frequency.

 

170, 151, 154, 160, 158, 154, 171, 156, 160, 157, 160, 157, 148, 165, 158, 159, 155, 151, 152, 161, 156, 164, 156, 163, 174, 153, 170, 149, 166, 154, 166, 160, 160, 161, 154, 163, 164, 160, 148, 162, 167, 165, 158, 158, 176.

ANSWER:

Classes

 (f)

r.f

r.c.f

148 - 152

6

6/45 = ans

0.133

153 - 157

11

11/45 =ans

0.377

158 - 162

14

14/45 = ans

0.688

163 - 167

9

9/45 = ans

0.888

168 - 172

3

3/45 =ans

0.955

173 - 177

2

2/45 = ans

0.999

 

45

 

 

 

 

 

 

 

 

 

 

b)     Which level of measurement (scale of measurement) is suitable for the following data in each example?

i)                    The lake Saif-ul-Muluk is 10,578 feet above from the sea level. (Ratio Scale)

ii)                  If customers go to the shop for buying shoes and try different size of shoes. The sizes of shoes are 7B, 10D and 12E. (Interval Scale)

iii)                The numbering on T-shirts of players in the cricket team.  (Ordinal Scale)

iv)                The weight of the person is 68 kg. (Nominal Scale)

Question 2:                                                                                                                   Marks: 5+3=8

a)      Find the median, harmonic mean and 23rd percentile for the following frequency distribution.

 

Class Boundaries

Frequency

145 – 150

4

150 – 155

6

155 – 160

28

160 – 165

58

165 – 170

64

170 – 175

30

Solution:

Boundaries

f

c.f

145 – 150

4

4

150 – 155

6

10

155 – 160

28

38

160 – 165

58

96

165 – 170

64

160

170 – 175

30

190

median=l + h/f(n/2− c)

l = 164,h = 5,f = 64,n = 190,c = 96

CALCULATE YOURSELF PLEASE

 

Class Boundaries

Frequency

X

1/X

F(1/X)

145 – 150

4

(145 + 150)/2

0.007

0.03

150 – 155

6

-

0.007

0.04

155 – 160

28

-

0.006

0.2

160 – 165

58

-

0.006

0.4

165 – 170

64

-

0.006

0.4

170 – 175

30

-

0.006

0.2

 

190

 

 

TOTAL=

H.M =n/{∑f(1/x)}

 

CALCULATE YOURSELF PLEASE

 

 

boundaries

f

c.f

145 – 150

4

4

150 – 155

6

10

155 – 160

28

38

160 – 165

58

96

165 – 170

64

160

170 – 175

30

190

P23 = l + h/f(23n/100 − c)

l=164,h = 5,f = 64,n = 190,c = 96

CALCULATE YOURSELF PLEASE

 

b)     Explain the concept of Geometric mean? Also find the geometric mean for the following data.

18, 19, 19, 19, 19, 20, 20, 20, 20, 20, 21, 21, 21, 21, 22, 23, 24, 27, 30, 36

Solution:

x

logx

18

1.26

19

1.27

19

1.27

19

1.27

19

1.27

20

1.30

20

1.30

20

1.30

20

1.30

20

1.30

21

1.32

21

1.32

21

1.32

21

1.32

22

1.34

23

1.36

24

1.38

27

1.43

30

1.48

36

1.56

 

19.46

 

g.m = log G = (∑ log X)/n

CALCULATE YOURSELF PLEASE

 

 

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