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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

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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?

# 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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