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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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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.

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.

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

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.

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.

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

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.

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.

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

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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