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6/45=0.134

11/45=0.245

14/45= 0.312

9/45= 0.2

3/45=0.67

2/45=0.45

plz help me is correct or not?

and then r.c.f will be

0.134

0.379

0.691

0.891

1.561

1.606

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Replies to This Discussion

mine is

6/45 = 0.133

11/45 = 0.244

14/45 = 0.311

9/45 = 0.2

3/45 = 0.067

2/45 = 0.044

if wrong thn corect....

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

 

 

Median me :-

l = 160, h = 5, f = 58, n = 190, c = 38, it is correct according to lect. 7

Am I right?????

 

According to lectures Questn no 2 ma class boundries bhe change hon ge

 

yes your median values are correct but i'm too not sure about the value of that it should be c = 38 or c = 96 and the value of n will be 95 as n/2 = 95

no its wrong.........

0.133

0.24

0.311

0.2

0.06

0.04 is the correct r.f

check this

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