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STA301 Statistics and Probability Assignment No 1 Solution & Discussion Due Date:07-11-2012

 

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   

 

 

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Replies to This Discussion

yeap assignment ban jaye gi............ 

aur ban gai hai :) chk on page 3

secnd qstion k a part ma ap ko medin ka formula dekan ha 

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

No it's wrong! mein ny upload key hain file page 3 pr chk karo.

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




Q2 part1 

median =164.9138

H.M.=164.862

P23=160.49

Q2 Partb

G.M.=21.66

match with yr answers and justify

Galat hai. CF k column mein 96+64 karna hai 160 banay ga aur nichy walaq 190.

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