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STA301_Solved_Assignment#01

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plz measurement waly ka ans bta dain plz plz plz plz

Jazak Allah

Ash Dear ummmah

Class Number

 

class limit

Frequency

Relative Frequency

(r.f)

Commulative frequency

(c.f)

Relative comulative frequency

r.f * c.f=r.cf

1

148 ­­- 152

6

6/45 = 0.133

6

0.789

2

153 - 157

11

11/45 = 0.244

17

4.148

3

158 - 162

14

14/45 = 0.311

31

9.641

4

163 - 167

9

9/45 = 0.2

40

8

5

168 - 172

3

3/45 = 0.066

43

2.838

6

173 - 177

2

2/45 = 0.044

45

1.98

 

 

Total = 45

 

 

 

y relative cumulative frequency kasy ai?

yar ap ne relative cumulative frequency galat nikali hai.   is k lie formula ye hai:  c.f/total frequency.


yaha total frequency 45 hai. for example: first relative frequency:6/45=0.133   second:17/45=0.377  third:31/45=0.688 and so on.

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.

 

Ans:

                  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 Number

 class limit

Frequency

Relative Frequency

Relative cumulative frequency

1

148 ­­-152

6

6/45 = 0.133

0.133

2

153 – 157

11

11/45 = 0.244

0.133+0.244 = 0.377

3

 158 – 162

14

14/45 = 0.311

0.377+0.311=0.688

4

163 – 167

9

9/45 = 0.2

0.688+0.2=0.888

5

 168 – 172

3

3/45 = 0.066

0.888+0.066=0.954

6

173 - 177

2

2/45 = 0.044

0.954+0.044=0.998

 

 

Total = 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. 

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.

iv) The weight of the person is 68 kg.

Question 2:

 

a)      Find the median, harmonic mean and 23rd percentile for the following frequency distribution.

 

Class Boundaries

Frequency

CF

X

1/x

F(1/x)

145 – 150

4

4

147.5

0.0067

0.0268

150 – 155

6

10

152.5

0.0065

0.039

155 – 160

28

38

157.5

0.0063

0.1764

160 – 165

58

96

162.5

0.0061

0.3538

165 – 170

64

160

167.5

0.0059

0.3776

170 – 175

30

190

172.5

0.0057

0.171

 

 

 

 

 

 

 

Median:

 

 


                                                         

                                                         Plz calculate it             

Harmonic mean:

 

 

23rd Percentile:

                                                          

 

 

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   

 

 

Ans:

 

Geometric mean G of a set of n positive values X1, X2…Xn is defined as the positive nth root of their product.

 

 

                                      

 

 

 

 

X

logX

18

1.2552

19

1.2787

19

1.2787

19

12.787

19

1.2787

20

1.3010

20

1.3010

20

1.3010

20

1.3010

20

1.3010

21

1.3222

21

1.3222

21

1.3222

21

1.3222

22

1.3424

23

1.3617

24

1.3802

27

1.4313

30

1.4771

36

1.5563

 

∑logX =

 

 

 

   

 

 

G = antilog 1.33564 = 21.65

U can also calculat ur result from this calculator on side

http://www.sengpielaudio.com/calculator-geommean.htm

U can also calculate Ur ANTILOG from online calculator on the side

http://ncalculators.com/number-conversion/anti-log-logarithm-calcul...

please pay attention the formula of harmonic mean is not correct because it is used in ungrouped data as we know that in the question we have provided group data so please resolve your question with correct mean to say with grouped formula

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