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

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

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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

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?

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

 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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