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Question 1:                                                                                                                   Marks: 5+2=7

 

a)   Following are the heights (in centimeters) of 45 female high school studentsPrepare 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

 

 



Read more: STA301_assignment#2_SOLUTION - Virtual University of Pakistan http://vustudents.ning.com/forum/topics/sta301-assignment-2-solutio...

D= noooooo not fair i made it so please use my name by the way i posted it in the group also

Ðarker Than Black ma nay main discussion say iuthi hai ..islihay lihay ap above dekhu tu ap ki discussion ka link nichay a raha hai 

bacho wali assignment thi is pa larny wali kia bat thi wo b full solved nhi hy sara kam tu srudent ko karna hy har jga likha hy...............CALCULATE YOURSELF PLEASE

Fall2012_STA301_1_SOLUTION

See the attached file please 

Attachments:

aap ne is ki freq kaise bnai hai???????

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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