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Dear students,
The Assignment no.3 has been uploaded and it is graded. You have to solve only section C of Assignment 3.
Instructions about solving assignment must be strictly followed and provide assignment solution on MS Word file.
No assignment will be accepted through email so submit it within due date.
Section C
Question 1:
60 |
80 |
90 |
96 |
120 |
150 |
200 |
360 |
480 |
520 |
1060 |
1200 |
1450 |
2500 |
7200 |
The annual incomes (in thousands of $) of fifteen families is as
Calculate the
for the given data.
Question 2:
Find the correlation between candidates score in interview and written test for a random sample of 6 candidates, which is as
X |
2 |
4 |
5 |
6 |
8 |
11 |
Y |
18 |
12 |
10 |
8 |
7 |
5 |
Where X=Interview marks, Y=Marks of written test.
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Section A:
Question No 1:
Arrange the following set of data in ascending order and write down the frequency table showing the frequency of each term in front of it
Solution:
1,1,1,2,2,2,2,3,3,3,3,3,3,5,5,5,5,5,6,7,7,7,8,8,8,8,9,9,9,10
Item |
Frequency |
Percentage |
1 |
3 |
10.00% |
2 |
4 |
13.33% |
3 |
6 |
20.00% |
5 |
5 |
16.67% |
6 |
1 |
3.33% |
7 |
3 |
10.00% |
8 |
4 |
13.33% |
9 |
3 |
10.00% |
10 |
1 |
3.33% |
Grand Total |
30 |
100.00% |
Part 2: Find first second and third quartile of the following data
12, 18, 13,19,16,14,15,17,20.
Solution:
This is ungrouped data. First arrange the data in ascending order.
12,13,14,15,16,17,18,19,20
Now, find median or second quartile. 16
Median position: (n+1) /2 = (9+1)/2=10/2=5th position Counting median= 16
First quartile (Q1) First Quartile Position: (n+1)/4 =10/4=2.5th position So average of 2nd and 3rd positions values= (13+14)/2=13.5
Third quartile (Q3) Third Quartile Position: 3(n+1)/4 =30/4=7.5th position So average of 7th and 8th position values= (18+19)/2=18.5.
Question No 2:
Find the correlation coefficient r between variables X and Y using the following table (Do not use excel formulas to find correlation coefficient)
x |
4 |
3 |
5 |
7 |
y |
7 |
8 |
9 |
12 |
Solution:
Now from the given data
Hence the describe estimated regression line is
Y=a+bx
Section “B”:
Question No1.
In a Honda center sales man collected spare part prices $8, $2, $14, $13, $22, $18, $30, $25, $52, $46, $63 in one day. Find out the lower quartile, upper quartile and interquartile range for the above given data.
Solution:
By arranging the data we get values
2$,8$,13$,14$,18$,22$,25$,30$,46$,52$,63$
Median = ?
To find median we have to see if the data number is odd or even
If it is odd then it is calculated as
Median = n+1/2
As
N=11
Median= (11+1)/2=12/2=6^{th} value
22$
Lower Quartile = ?
Lower Quartile = n+1/4=(11+1)/4=12/4=3^{rd} value
Lower Quartile = 13$
Upper Quartile =?
Upper Quartile =3(n+1)/4=3(12)/4=36/4=9^{th} value
Upper Quartile = 46
Inter Quartile Range =?
Inter Quartile Range = Q3-Q1
Inter Quartile Range =46-13
Inter Quartile Range =33
Question No2.
Find the correlation coefficient r between variables X and Y using the following table (Do not use excel formulas to find correlation coefficient)
X |
8 |
9 |
10 |
11 |
Y |
10 |
5 |
12 |
11 |
ANSWER:
We have to find correlation coefficient r?
So, we have to find some more values
X |
Y |
XY |
||
8 |
10 |
64 |
100 |
50 |
9 |
5 |
81 |
25 |
45 |
10 |
12 |
100 |
144 |
120 |
11 |
11 |
121 |
121 |
121 |
=38 |
=38 |
=366 |
=390 |
=366 |
Putting value we get..
So the Correlation Coefficient is 0.41
Section “C”:
Question No 1:
60 |
80 |
90 |
96 |
120 |
150 |
200 |
360 |
480 |
520 |
1060 |
1200 |
1450 |
2500 |
7200 |
The annual incomes (in thousands of $) of fifteen families is as
Calculate the
G=(x1*x2*x3*……*xn) ^1/n
GM= (60*80*90*96*120*150*200*360*480*520*1060*1200*1450*2500*7200) ^1/15
=377.209 Ans
HM= n / (1/x1+1/x2+.....1/xn)
=15/(1/60+1/80+1/90+1/96+1/120+1/150+1/200+1/360+1/480+1/520+1/1060
+1/1200+1/1450+1/2500+1/7200)
= 186.3 Ans
Quartile divide data into 4 equal parts.
Position of Q1= (n+1)/4 = (15+1)/4 = 16/4 = 4
= 4th value = 120
Position of Q2 = 2(n+1)/4 = 2(15+1)/4 = 2(16)/4 = 32/4= 8
= 8^{th} value = 480
Position of Q3= 3(n+1)/4 = 3(15+1)/4 = 3(16)/4 = 48/4 = 12
= 12^{th} value = 1450
Question 2:
Find the correlation between candidates score in interview and written test for a random sample of 6 candidates, which is as
X |
2 |
4 |
5 |
6 |
8 |
11 |
Y |
18 |
12 |
10 |
8 |
7 |
5 |
Where X=Interview marks, Y=Marks of written test.
Solution:
Using formula
X |
Y |
XY |
||
2 |
18 |
4 |
324 |
36 |
4 |
12 |
16 |
144 |
48 |
5 |
10 |
25 |
100 |
50 |
6 |
8 |
36 |
64 |
48 |
8 |
7 |
64 |
49 |
56 |
11 |
5 |
121 |
25 |
55 |
Putting values in formula:
There is a negative correlation between interview marks and marks of the written test.
plzzzzzzzz send section D
FOR MEDIAN CALULATION FIRST ARRY THE DATA
2$,8$,13$,14$,18$,22$,25$,30$,46$,52$,63$
N=11 ,SO, ODD VAULE FOR MEDIAN =
Median= =
22$
FOR:
Lower Quartile = ?
Lower Quartile = = Lower Quartile = 13$
Upper Quartile = ?
Upper Quartile =3 =
Upper Quartile = 46$
Inter Quartile Range = ?
Inter Quartile Range = Q3-Q1
Inter Quartile Range =33$
2nd answer
So, find the vaule is that
X
Y
XY
8
10
64
100
50
9
5
81
25
45
10
12
100
144
120
11
11
121
121
121
=38
=38
=366
=390
=366
Putting value we get..
,
, , ,r=
is 0.41 Correlation Coefficient
SECTION D KAHA THA YEH KIA HAI ?
SECTION D ???????
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