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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.

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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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.

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

 

 

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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