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MTH302 Business Mathematics & Statistics Assignment No 2 Solution & Discussion Due Date:14-06-2012


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Please Discuss here about this assignment.Thanks

Assignment # 2

MTH302 (Spring 2012)

Total marks: 20

Lecture # 23 to 29

Due date:14-06-2012

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Question No. 1:

 

In a Rice Mill the weights (in Kilograms) of 50 packets are given below:

 

Observations

Frequency

10

4

15

5

20

9

40

16

60

8

80

5

100

3

Find the Mean and Mode. Also, find the Median using the Empirical Relationships of Mean, Median and Mode.

 

 

Question No. 2:

 

Suppose that the manager of a company wants to know if there exists any correlation between the sales of the product A and the product B. Suppose that he selects a random sample of the total sales of these products in  different days of the month. If you are a manager, then how will you calculate the correlation coefficient? (Use Manual Formula)

 

Days of

the month

Total sale of

the product A

Total sale of

the product B

1st day

15

16

4th day

14

13

7th day

10

11

10th day

18

19

15th day

15

13

18th day

20

22

22nd day

16

15

25th day

15

17

27th day

12

16

30th day

13

17

 

plz need solution

PLZ UPLOAD THE SOLUTION

mean=43.2

mode=40

median=40

correlation coefficient = 2.34

I found these answers i am sharing. can any one plz share the formulas and solution according to these answers?

ye Answers  hain yar

Mean=42.3

Median=16

Mode=the middle value of data= 5

ab in ka 

Empirical Relationships of Mean, Median and Mode.

Mean= 21.5

Mode=1033.59

Median=86.9

 

 

Correlation coefficient = 0.77

 Aj Raat ko 11 bje Salution update karon ga pray me to all

I will be waiting

thanx dear. but its not my answer somebody solved it for me and i had a BIG dout about the accuracy :( thats y demanded guidance from ppl like you :)

MARY PIYARE DOSTO AS SALAM ALIKUM.

ME NE AP SE WADA KIA THA K AJ ME ASSIGNMENT UPLOSD KRON GA TO LO ME NE ASSINGMENT UPDATE KR DIYA.

PRY FOR ME

Solution:

C-B

X

f

F *X

7.5-12.5

10

4

40

12.5-17.5

15

5

75

17.5-22.5

20

9

180

22.5-42.5

40

16

640

42.5-62.5

60

8

480

62.5-82.5

80

5

400

82.5-102.5

100

3

300

 

-

50

-

 

Mean=

Mean=42.3

Mode=

Mode=

Mode=24.83

We know that

Mode= 3medin-2Mean

Median=

Median=

Median=36.478

 

 

 

 

 

 

 

 

Answer 2

 

x

y

X*y

X2

Y2

15

16

240

225

256

14

13

182

196

169

10

11

110

100

121

18

19

342

324

361

15

13

195

225

169

20

22

440

400

484

16

15

240

256

225

15

17

255

225

289

12

16

192

144

256

13

17

221

169

289

∑ 148

∑ 159

∑ 2417

∑ 2264

∑ 2619

=∑XY-(∑X)(∑Y) ∕n

√ [∑x2-(∑X)2/n[∑Y2-(∑Y)2/n]

 =   2417-(148)(158)/10
√ [2264-(148)2/10][2619-(159)2/10

=   2417-2353.2
√ [2264-2190.4][2619-2528.1]

63.8
√ 74x90.9

         63.8
√6726.6

63.8         

81.8

= 0.7780

salaam bless u be happy 

Iss solution m jo class boundries bani h sahi h????   kia her boundry ka class interval diff ho sakta h??? plz guide me about...

i m waiting

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