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# i thing relative frequency ka jo ans ata ha usa round up krain ga

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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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.

 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

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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