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Q No.1 (a)

 

 

Classes                        F          cumulative Frequency                     

 

 

148-152                       6                                  6                                 

 

153 -  157                    11                                17                               

 

158-    162                   14                                31

 

163 -      167                9                                  40

 

168-     172                  3                                  43

 

173     - 177                 2                                  45

 

                                    45

 

Q No.1 (a)

 

 

Classes                        F                       Relative frequency

 

 

148 - 152                     6                                  6/45 =0.1333                          

 

153   - 157                   11                                11/45= 0.244                                      

 

158 -   162                   14                                14/45= 0.311

 

163   -      167              9                                  9/45=    0.2

 

168 -    172                  3                                  3/45=   0.066

 

173     - 177                 2                                  2/45= 0.044

 

                                    45

 

Q No.1 (b)

 

(i)                 Ratio

(ii)               Nominal

(iii)             Ordinal

(iv)             Ratio

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Replies to This Discussion

Q NO.2 (a)

 

Find the Median

 

Class boundary                                               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=                   160                             

h=                     5

N/2=                95

c=                    38

 

Median   =                  L + h/f   (n/2 –c)

 

              =                    160+ 5/58(95- 38)

 

              =                    160 +0.0862(57)

 

Median   =                   164.9138        

 


P23            =       L + h/f (23n/100    -c)

 

Find the 23rd percentile

 

L= 160                    ,      h=5       ,               f= 58.                c=38

                

            =          160    + 5/58 (4370/100 -38)

            =          160      + 0.0862(5.7)

 

            =        160.4913

 

 

idea of solution of g.m

Q NO.2 (B)

Concept of Geometric Mean

The geometric mean is well defined only for sets of real positive numbers. In Geometric mean is calculated by multiplying all the numbers and taking th nth root of the total.

The formula of Geometric mean is

 

log G =                    Σ log X

                                    ________________

 

                                                N

 

 

 

 

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   

 

X                                                         Log X

 

18                                                        1.2553

 

19                                                        1.2788

 

19                                                        1.2788

 

19                                                        1.2788

 

19                                                        1.2788

 

ly wee kakao enjoy karo

 Zain kazami

Sta 301

Assignment solution

Question 1:

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 question 1

Here

Smallest value(X0)=148

Largest  value(Xm)=176

Range =Xm-x0=176-148=28

Class interval=h=range/no of classes

                        =28/6=4.6

                        =5

Class frequency

Class limit

Frequency

Relative Frequency

Relative cumulative frequency

1

148-152

6

6/45=0.133

0.133=

2

152-157

11

11/45=0.244

0133+0.244=0.377

3

158-162

14

14/45=0.311

0.377+0.311=

4

163-167

9

9/45=0.2

0.688+0.2=0.888

5

1681-72

3

3/45=0.66

0.888+0.66=0.954

6

173-177

2

2/45=0.044

0.954+0.044=0.998

 

 

 

 

 

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. 

Ans. Interval 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.

Answer  ranking or ordinal scale

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

Nominal scale

iv)                The weight of the person is 68 kg.

Ans. Ratio scale

Answer question 2

Median

Class boundries

Frequency

Cumulative frequency

145-150

4

4

150-155

6

4+6=10

155-160

28

10+28=38

160-165

58

38+58=96

165-170

64

96+64=160

170-175

30

160+30=190

Total

190

190

 

Essential of the formuila of the median

L    lower class limit

H     is size of the class

F    frequency

C    previous cumulative frequency

n/2 =190/2=95

median=l+h/f(n/2-c)

             =160+5/58(95-38)

             =160+5/58(57)

             =160+285/58

       =160+4.91

       =164.91

Harmonic means

 

Class frequency

Frequency

Midpoint(x)

1/X

F* 1/X

145150

4

147.5

1/147.5=0.0068

4*0.0068=0.0272

150155

6

152.5

1/152.5=0.0066

6*0.0066=0.0396

155160

28

157.5

1/157.5=0.0064

28*0.0064=0.1792

160165

58

162.5

1/162.5=0.0062

58*0.0062=0.3596

165170

64

167.5

1/167.5=0.0060

64*0.0060=0.384

170175

30

172.5

1/172.5=0.0058

30*0.0058=0.1740

Total

 

 

0.378

1.1636

 

 

Harmonic Mean= ∑f/∑{f*1/x)

Harmonic Mean=190/1.1635

Harmonic Mean=163.3

 

23rd  percentile=L+h/f(23(n)/100-c)

23rd  percentile=160+5/58{(23(190)/100-(38)}

23rd  percentile=160+5/58{(4370)/100-(38)}

23rd  percentile=160+5/58{(43.70-(38)}

23rd  percentile=160+5/58{(5.7)}

23rd  percentile=160+28.5/58

23rd  percentile=160+0.4913

23rd  percentile=160.49

Geometric Mean

                              nth positive root of n positive given value.We find geometric by this formula

Geometric Mean= (x1*x2*x3………….xn)1/n

where(xi>0)

G M=antilog{∑logX/n}

G M=(18*19*19*19*19*20*20*20*20*20*21*21*21*21*21*22*23*24*27*30*36)1/20

G M=antilog{∑logX/n}

x

logx

18

1.26

19

1.28

19

1.28

19

1.28

19

1.28

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

1.44

30

1.48

36

1.66

 

Geometric means = Antilog{26.82/20}

                              =Anti log 1.341

                              =

 

 

 

 

 

 

 

 

 

Zain kazami

Sta 301

Assignment solution

Question 1:

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 question 1

Here

Smallest value(X0)=148

Largest  value(Xm)=176

Range =Xm-x0=176-148=28

Class interval=h=range/no of classes

                        =28/6=4.6

                        =5

Class frequency

Class limit

Frequency

Relative Frequency

Relative cumulative frequency

1

148-152

6

6/45=0.133

0.133=

2

152-157

11

11/45=0.244

0133+0.244=0.377

3

158-162

14

14/45=0.311

0.377+0.311=

4

163-167

9

9/45=0.2

0.688+0.2=0.888

5

1681-72

3

3/45=0.66

0.888+0.66=0.954

6

173-177

2

2/45=0.044

0.954+0.044=0.998

 

 

 

 

 

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. 

Ans. Interval 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.

Answer  ranking or ordinal scale

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

Nominal scale

iv)                The weight of the person is 68 kg.

Ans. Ratio scale

Answer question 2

Median

Class boundries

Frequency

Cumulative frequency

145-150

4

4

150-155

6

4+6=10

155-160

28

10+28=38

160-165

58

38+58=96

165-170

64

96+64=160

170-175

30

160+30=190

Total

190

190

 

Essential of the formuila of the median

L    lower class limit

H     is size of the class

F    frequency

C    previous cumulative frequency

n/2 =190/2=95

median=l+h/f(n/2-c)

             =160+5/58(95-38)

             =160+5/58(57)

             =160+285/58

       =160+4.91

       =164.91

Harmonic means

 

Class frequency

Frequency

Midpoint(x)

1/X

F* 1/X

145150

4

147.5

1/147.5=0.0068

4*0.0068=0.0272

150155

6

152.5

1/152.5=0.0066

6*0.0066=0.0396

155160

28

157.5

1/157.5=0.0064

28*0.0064=0.1792

160165

58

162.5

1/162.5=0.0062

58*0.0062=0.3596

165170

64

167.5

1/167.5=0.0060

64*0.0060=0.384

170175

30

172.5

1/172.5=0.0058

30*0.0058=0.1740

Total

 

 

0.378

1.1636

 

 

Harmonic Mean= ∑f/∑{f*1/x)

Harmonic Mean=190/1.1635

Harmonic Mean=163.3

 

23rd  percentile=L+h/f(23(n)/100-c)

23rd  percentile=160+5/58{(23(190)/100-(38)}

23rd  percentile=160+5/58{(4370)/100-(38)}

23rd  percentile=160+5/58{(43.70-(38)}

23rd  percentile=160+5/58{(5.7)}

23rd  percentile=160+28.5/58

23rd  percentile=160+0.4913

23rd  percentile=160.49

Geometric Mean

                              nth positive root of n positive given value.We find geometric by this formula

Geometric Mean= (x1*x2*x3………….xn)1/n

where(xi>0)

G M=antilog{∑logX/n}

G M=(18*19*19*19*19*20*20*20*20*20*21*21*21*21*21*22*23*24*27*30*36)1/20

G M=antilog{∑logX/n}

x

logx

18

1.26

19

1.28

19

1.28

19

1.28

19

1.28

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

1.44

30

1.48

36

1.66

 

Geometric means = Antilog{26.82/20}

                              =Anti log 1.341

                              =

 

 

 

 

 

 

 

 

 

Zain kazami

Sta 301

Assignment solution

Question 1:

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 question 1

Here

Smallest value(X0)=148

Largest  value(Xm)=176

Range =Xm-x0=176-148=28

Class interval=h=range/no of classes

                        =28/6=4.6

                        =5

Class frequency

Class limit

Frequency

Relative Frequency

Relative cumulative frequency

1

148-152

6

6/45=0.133

0.133=

2

152-157

11

11/45=0.244

0133+0.244=0.377

3

158-162

14

14/45=0.311

0.377+0.311=

4

163-167

9

9/45=0.2

0.688+0.2=0.888

5

1681-72

3

3/45=0.66

0.888+0.66=0.954

6

173-177

2

2/45=0.044

0.954+0.044=0.998

 

 

 

 

 

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. 

Ans. Interval 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.

Answer  ranking or ordinal scale

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

Nominal scale

iv)                The weight of the person is 68 kg.

Ans. Ratio scale

Answer question 2

Median

Class boundries

Frequency

Cumulative frequency

145-150

4

4

150-155

6

4+6=10

155-160

28

10+28=38

160-165

58

38+58=96

165-170

64

96+64=160

170-175

30

160+30=190

Total

190

190

 

Essential of the formuila of the median

L    lower class limit

H     is size of the class

F    frequency

C    previous cumulative frequency

n/2 =190/2=95

median=l+h/f(n/2-c)

             =160+5/58(95-38)

             =160+5/58(57)

             =160+285/58

       =160+4.91

       =164.91

Harmonic means

 

Class frequency

Frequency

Midpoint(x)

1/X

F* 1/X

145150

4

147.5

1/147.5=0.0068

4*0.0068=0.0272

150155

6

152.5

1/152.5=0.0066

6*0.0066=0.0396

155160

28

157.5

1/157.5=0.0064

28*0.0064=0.1792

160165

58

162.5

1/162.5=0.0062

58*0.0062=0.3596

165170

64

167.5

1/167.5=0.0060

64*0.0060=0.384

170175

30

172.5

1/172.5=0.0058

30*0.0058=0.1740

Total

 

 

0.378

1.1636

 

 

Harmonic Mean= ∑f/∑{f*1/x)

Harmonic Mean=190/1.1635

Harmonic Mean=163.3

 

23rd  percentile=L+h/f(23(n)/100-c)

23rd  percentile=160+5/58{(23(190)/100-(38)}

23rd  percentile=160+5/58{(4370)/100-(38)}

23rd  percentile=160+5/58{(43.70-(38)}

23rd  percentile=160+5/58{(5.7)}

23rd  percentile=160+28.5/58

23rd  percentile=160+0.4913

23rd  percentile=160.49

Geometric Mean

                              nth positive root of n positive given value.We find geometric by this formula

Geometric Mean= (x1*x2*x3………….xn)1/n

where(xi>0)

G M=antilog{∑logX/n}

G M=(18*19*19*19*19*20*20*20*20*20*21*21*21*21*21*22*23*24*27*30*36)1/20

G M=antilog{∑logX/n}

x

logx

18

1.26

19

1.28

19

1.28

19

1.28

19

1.28

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

1.44

30

1.48

36

1.66

 

Geometric means = Antilog{26.82/20}

                              =Anti log 1.341

                              =

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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