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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.

 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

 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)}

 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

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

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### Replies to This Discussion

aap ne 1st qs me array type data se class kaise bnai hai plz tell me like 148-152 ye wali??need method??

148 lower limit ha aur upper limit hum is traqiay say find kr sakaty hain.

upper limit = lower limit +(Class Interval - 1)

=148+(4)

=152

Ab  next upper calss k lia 152 main 5 add krain. aur is tra 5 add krtay jain sari classes pata chal jain gi.

Upper Class limit

148+5=153

153+5=158

158+5=163

163+5=168

168+5=173

Lower Class Limit

152+5=157

157+5=162

162+5=167

167+5=172

172+5=177

now

CLASS BOUNDRIES:

 148-152 153-157 158-162 163-167 168-172 173-177

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.

 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

 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)}

 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

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

Salam ! there is some confusion in Q2 part (a)...... where you find out the median and you put the values of l, h, f, n, and c are wrong. The correct values will be l = 160, h = 4, f = 58, n = 95, c = 96

actually here first we find the value of n that is n = 190 and than n / 2 = 190/2 = 95

Ater that all the values of l , h ,f, c  will come
after the calculations the answer of the median will be 163.93
Is it so ???

median value will be 159.7

bhai plzzz tel me Q 2 k B part harmonic mean mia 1/x   find kare ge mean X kis ko consider  kare ge

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

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