Blog: Mth302 Assignment No. 2 solution

MTH302 (Spring 2011)

Total marks: 40

Lecture # 23 to 29

Due date: 16-06-2011

DON’T MISS THESE IMPORTANT INSTRUCTIONS:

Upload assignment properly through LMS, (No Assignment will be accepted through email).

All students are directed to use the font and style of text as is used in this document.

This is an individual assignment, not a group assignment, so keep in mind that you are supposed to submit your own, self made & different assignment even if you discuss the questions with your class fellows. All similar assignments (even with some meaningless modifications) will be awarded zero marks and no excuse will be accepted. This is your responsibility to keep your assignment safe from others. Many solution files sent by students in assignment 1 are found to be copied and so awarded zero. You are therefore reminding here again.

All questions are graded and solve properly on word file.

Solve the assignment on MS word document and upload your word (.doc) files only. Do not solve the assignment on MS excel. If we get any assignment on MS excel or any format other than word file then it will not be graded.

Question 1: Marks=10

A shopkeeper wants to know whether there exists any correlation between sale of shampoo and sale of conditioner or not. Suppose that he selects a random sample of four brands out of all brands that he sale. The following information is obtained:

Brands

Sale of Shampoo

A

5

2

B

7

3

C

10

7

D

14

10

Question 2: Marks=10 

The weekly sale of a product A was recorded as below. Find the coefficient of variation of weekly sale for the Product A.

Product A

60

58

42

25

35

45

15

Question 3: Marks=10 

Find the Relative Frequency, Cumulative Frequency, and Percent Relative Frequency of the data given in the table below.

Class

Frequency

( f )

10 – 15

5

15 – 20

4

20 – 25

6

25 – 30

2

30 – 35

3

35 – 40

7

40 – 45

5

Question 4: Marks=10

The given data is 69, 93, 53, 45, 66, 71, 89, 95, 97, 103, 75

Determine the 5-point summary ( i.e. Smallest value, 1st Quartile (Q1), Median (Q2), 3rd Quartile (Q3), Largest value ).


Solution:

Q No. 2 Solution:

See Handouts:

60 58 42 25 35 45 15

Lecture Number 28 Page 204

Coefficient of variation is equal to the standard deviation divided by mean

multiply by 100.

Coefficient of variation=standard deviation/mean * 100

c.v=s/mean *100

Solve By yourself

Q No. 3 Solution

See Page # 170 Of handouts

Lecture Number 24

Relative frequency :

Relative Frequency of a class = Frequency of the class interval

---------------------------------------

Total Frequency

Cumulative Frequency:

If we add frequency of the second interval to the frequency of the first interval , then the

cumulative frequency for the second interval is obtained

Percent cumulative relative frequency:

This can be calculated same as cumulative frequency except now percent relative

frequency for each class interval is considered. The percent cumulative relative

frequency of the last class interval is 100% as all observations have been added.

Class 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35 35 – 40 40 – 45

Frequency (f) 5 4 6 2 3 7 5 32

Relative% Relative Frequency Frequency 0.1616 0.1212 0.1919 0.066 0.099 0.2222 0.1616 1100

Q No. 4:

See on Handouts

Lecture 26 Page Number 186

Quartiles divide data into 4 equal parts

Syntax

1st Quartile Q1=(n+1)/4

2nd Quartile Q2= 2(n+1)/4

3rd Quartile Q3= 3(n+1)/4

Lecture 27 Page Number 194

5-NUMBER SUMMARY:

5 number summary is:

• Smallest value

• 1st Quartile (Q1)

• Median (Q2)

• 3rd Quartile (Q3)

•Largest

Arrange the data in ascending order

45, 53, 66, 69, 71, 75, 89, 93, 95, 97, 103

Smallest Value = 45

1st Q= (n+1)/4

= (11+1)4

= (12)/4

=3

2nd Q= 2(n+1)/4

=2(11+1)/4

=2(12)/4

=2(3)

=6

3rd Q= 3(n+1)/4

=3(11+1)/4

=3(3)

=9

Largest Value= 103
:::::::::::::::::::::::::::::::::::::::

Class	Frequency f	Relative Frequency	% Relative Frequency
10-15	5	0.16	16
15-20	4	0.12	12
20-25	6	0.19	19
25-30	2	0.03	6
30-35	3	0.09	9
35-40	5	0.16	16
	32	1	100

Q No. 4

See on Handouts

Lecture 26 Page Number 186

Quartiles divide data into 4 equal parts

Syntax

1st Quartile Q1=(n+1)/4

2nd Quartile Q2= 2(n+1)/4

3rd Quartile Q3= 3(n+1)/4

Lecture 27 Page Number 194

5-NUMBER SUMMARY:

5 number summary is:

• Smallest value

• 1st Quartile (Q1)

• Median (Q2)

• 3rd Quartile (Q3)

•Largest value

Arrange the data in ascending order

45, 53, 66, 69, 71, 75, 89, 93, 95, 97, 103

Smallest Value = 45

1st

 Q= (n+1)/4

= (11+1)4

= (12)/4

=3

2nd

 Q= 2(n+1)/4

=2(11+1)/4

=2(12)/4

=2(3)

=6

Q2 Value: 75

3rd

 Q= 3(n+1)/4

=3(11+1)/4

=3(3)

=9

Largest Value= 103
::::::::::::::::::::::::::::::::::::::::::::::::::::::::::


Answer : NO (3)

Class Frequency Relative frequency % Relative frequency Cumulative
Frequency %Cumulative
Frequency
10-15 5 0.16 16 5 25
15-20 4 0.12 12 9 29
20-25 6 0.19 19 14 45
25-30 2 0.06 6 16 51
30-35 3 0.09 9 19 61
35-40 7 0.22 22 26 83
40-45 5 0.16 16 31 100
Total 32 1 100


Answer no. 4

First ordered array

45,53,66,69,71,75,89,93,95,97,103

Smallest value= 45

Largest value= 103

Q1= 1(n+1)/4

=11+1/4= 3

Q1= 3rd value =66

Q2= 2(n+1)/4

=2(11+1)/4

=6

Q2 = 6th value = 75

Q3 = 3(n+1)/4

. =3(11+1)/4

= 9

Q3 = 9th value = 95

Answer (2)
Mean = sum(A)/n = 280/ 7 = 40

CV= ?
SD = ?

A (X-MEAN) (X-MEAN)^2
60 -20 400
58 -18 324
42 -2 4
25 15 225
35 5 25
45 -5 25
15 25 625
Total=280 Total=1628

S.D = [{sum(x-mean)^2/n-1}^1/2]

S.D = (1628/6)^1/2

S.D = 16.47

C.V = S/mean*100 = 16.47 / 40*100 = 0.41 = 41.175



Ans(1)

Y X XY X^2 Y^2
5 2 10 4 25
7 3 21 9 49
10 7 70 49 100
14 10 140 100 196
Total=36 Total=22 Total=241 Total=162 Total=370

= 43/43.43

= 0.99

Very high correlation