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STA-301 Solution of Assignment No. 2...Learn Statistics Concepts With Me


Assignment No: 2 (Lessons 10-15)

Question 1:                                                                                                                           Marks: 7

 

For the given cumulative frequency table of students of different age groups, calculate the coefficient of standard deviation and coefficient of variation.

 

Age in years

Cumulative frequency of students (cf)

5-8

3

9-12

15

13-16

24

17-20

51

21-24

57

25-28

60

 

Question 2:                                                                                                                           Marks: 8

 

From the following data of hours worked in a factory (x) and output units (y), determine the regression line of y on x, the linear correlation coefficient and interpret the result of correlation coefficient.

Hours (X)

91

102

83

93

89

72

82

85

79

Production (Y)

300

302

315

330

300

250

300

340

315

 

 Solution with concepts


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Here is complete Solution

Little Confusion in Ansari Brother Assignment at the end 2nd Question.

See my Assignment.

and also post me right or wrong?



 Assignment No.2 (Course STA301)

 Spring 2012 (Total Marks 15)

 

Assignment No: 2 (Lessons 10-15)

Question 1:                                                                                                                           Marks: 7

 

For the given cumulative frequency table of students of different age groups, calculate the coefficient of standard deviation and coefficient of variation.

 

Age in years

Cumulative frequency of students (cf)

5-8

3

9-12

15

13-16

24

17-20

51

21-24

57

25-28

60

 

 

Solution:

 

Age in Years

C.F

X

F

Fx

Fx2

 5 - 8

3

6.5

3

19.5

126.5

9 – 12

15

10.5

12

126

1323

13 – 16

24

14.5

9

130.5

1892.25

17 – 20

51

18.5

27

499.5

9240.75

21 – 24

57

22.5

6

135

3037.5

25 - 28

60

26.5

3

79.5

2106.75

Total

210

99

60

990

17726.75


            Variance =S2 = ∑fx2 / ∑f – (∑fx/∑f)2

                                    17726.75 / 60 – (990/60)2

                                    295.44583 – (16.5)2

                                    295.44583 – 272.25

                                    23.19583

            Standard Deviation = √S2 = √23.19583 = 4.816204938

 

            Coefficient of Standard Deviation = S/Mean

                                                            _

                                                Mean= X = 990/60 = 16.5

                                                            S = 4.81620

            So                                            4.816204938 / 16.5 = 0.291891208

                        Coefficient of Variation = 4.81620 / 16.5

                                                               = 0.291891208 * 100

                        Coefficient of Variation = 29.1891208                                             

 

Question 2:                                                                                                                           Marks: 8

 

From the following data of hours worked in a factory (x) and output units (y), determine the regression line of y on x, the linear correlation coefficient and interpret the result of correlation coefficient.

Hours (X)

91

102

83

93

89

72

82

85

79

Production (Y)

300

302

315

330

300

250

300

340

315

 

Solution:

X

Y

X.Y

X2

Y2

91

300

27300

8281

90000

102

302

30804

10404

91204

83

315

26145

6889

99225

93

330

30690

8649

108900

89

300

26700

7921

90000

72

250

18000

5184

62500

82

300

24600

6724

90000

85

340

28900

7225

115600

79

315

24885

6241

99225

776

2752

238024

67518

846654

 

                  ∑X =     776

                  ∑Y =     2752

                  ∑X.Y = 238024

                  ∑X2   = 67518

                  ∑Y2   = 846654

 

Regression line Y on X

                  Byx  =   n∑xy – (∑x) (∑y) / n∑ x2 – (∑x)2

                           = 9(238024) – (776)(2752) / 9 (67518) – (776)2

                           = 9(238024) – (2135552) / 9 (67518) – (602176)

                           = 2142216 – 2135552 / 607662 – 602176

                          = 6664 / 5486

                           = 1.2147284

                              _      _

                      A  = y -  bx

                           = ∑y / n – b (∑x/n)

                           = 2752 /9 – 1.2147284 (776/9)

                           = 305.77777 – 1.2147284 (86.22222)

                           = 201.0411907

                              _

                  Y= a+bx

                    = 201.0411904 + 1.2147284

 

Linear Coefficient of Correlation:

 

                      ∑xy – (∑x)(∑y)/n

                r = --------------------------------------------

                     √ [∑x2 – (∑x)2 / n ] [∑y2 – (∑y)2 / n]

 

                      (238024) – (776)(2752) /9

                  =  -----------------------------------------------------

                      √ [67518 – (776)2 / 9] [846654 – (2752)2 / 9]

 

                      238024 – 237283.5556

                   = -------------------------------------------------------

                     √ (67518 – 66908.4444) (846654 – 841500.4444)

                      740.4444

                   = --------------------------

                      √ (609.5556) (5153.556)

                       740.4444

                   = -------------------

                      √3141378.92

                       740.444

                   = ------------------

                     1772.393557

                   = 0.417765003

 

dear answer same to hy mera or apka m ne round kr k likha hy bs

yeah formula

Byx  =   n∑xy – (∑x) (∑y) / n∑ x2 – (∑x)2

 

kahan se mila handpout mae to nahi

u have used Persons co-efficient correlation formula

STA301 Assignment No 2 Solution Spring 2012

One more idea solution 

Attachments:

Don't you think that we have to make class boundaries in Q#1 plzz tell me fast 

In  first question i think class boundaries banani hai ? 

maryum aap nay first question kar liya hai ?

yyes

hahahah nahi dear there is no need to make class boundaries becz mid points class interval or c.b se same hi aty hain so no need to make it 

ok thankx

please if any one can send me teh Stat301 handouts

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