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The Assignment no.3 has been uploaded and it is graded. You have to solve only section C of Assignment 3.
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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.
Tags:
mth302 assignment solution
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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