Linear Algebra Multiple Choice Questions on “Curve Fitting”.
1. Fit a straight line into the following data.
| x: |
0 |
1 |
2 |
3 |
4 |
5 |
| y: |
3 |
6 |
8 |
11 |
13 |
14 |
a) y=3.52+2.26x
b) y=3.52
c) y=2.26x
d) y=4+3x
Answer: a
Explanation: Here, N=6
Calculations of ∑x and ∑x2
| x |
y |
x2 |
xy |
|
| 0 |
3 |
0 |
0 |
|
| 1 |
6 |
1 |
6 |
|
| 2 |
8 |
4 |
16 |
|
| 3 |
11 |
9 |
33 |
|
| 4 |
13 |
16 |
52 |
|
| 5 |
14 |
25 |
70 |
|
| ∑x=15 |
∑y=55 |
∑x2=55 |
∑xy=177 |
|
We know that,
∑y=Na+b∑x
∑xy=a∑x+b∑x2
Substituting the values from the table into the equations-
55=(6)a+b(15) – (1)
177=(a)15+b(55) – (2)
Solving equations (1) and (2) simultaneously
a=3.52 and b=2.26
Thus the equation of the line is given by y=a+bx
Thus, the equation of the line is y=3.52+2.26x.
2. Fit a straight line y=a+bx into the given data:
(x,y):(5,12)(10,13)(15,14)(20,15)(25,16).
a) y=11
b) y=0.2x
c) y=11+0.2x
d) y=1.1+0.2x
Answer: c
Explanation: Here, N=5
Calculations of ∑x and ∑x2
| x |
y |
x2 |
xy |
|
| 5 |
12 |
25 |
60 |
|
| 10 |
13 |
100 |
130 |
|
| 15 |
14 |
225 |
210 |
|
| 20 |
15 |
400 |
300 |
|
| 25 |
16 |
625 |
400 |
|
| ∑x=75 |
∑y=70 |
∑x2=1375 |
∑xy=1100 |
|
We know that,
∑y=Na+b∑x
∑xy=a∑x+b∑x2
Substituting the values from the table into the equations-
70=(5)a+b(75) – (1)
1100=(a)75+b(1375) – (2)
Solving equations (1) and (2) simultaneously
a=11 and b=0.2
Thus the equation of the line is given by y=a+bx
Thus, the equation of the line is y=11+0.2x.
3. Fit a straight line y=a+bx into the given data by Actual Mean Method. What is the value of b?
| x: |
10 |
20 |
30 |
40 |
50 |
| y: |
22 |
23 |
27 |
28 |
30 |
a) 1.2
b) 0.15
c) 0.21
d) 0.8
Answer: c
Explanation: Solving by using Actual Mean Method
N=5
Mean of x=30 and y=26
| x |
y |
X=x-30 |
Y=y-26 |
X2 |
XY |
| 10 |
22 |
-20 |
-4 |
400 |
80 |
| 20 |
23 |
-10 |
-3 |
100 |
30 |
| 30 |
27 |
0 |
1 |
0 |
0 |
| 40 |
28 |
10 |
2 |
100 |
20 |
| 50 |
30 |
20 |
4 |
400 |
80 |
|
|
∑X=0 |
∑Y=0 |
∑X2=1000 |
∑XY=210 |
We know that,
∑y=Na+b∑x
∑xy=a∑x+b∑x2
Substituting the values from the table into the equations-
0=(5)a+b(0) – (1)
a=0
210=(a)0+b(1000) – (2)
b=0.21
Thus, the equation is
Y=a+Bx
Y=0+0.21X
Resubstituting X=x-30 and Y=y-26
y-26=0.21(x-30)
y=-4+0.21x
The equation of the line is given by y=a+bx
b=0.21 and a=-4.
4. Fit a straight line y=a+bx into the given data. Also estimate the production in the year 2000.
| Year(x): |
1966 |
1976 |
1986 |
1996 |
2006 |
| Production in lbs(y): |
10 |
12 |
13 |
16 |
17 |
a) 12.33
b) 14.96
c) 11.85
d) 18.67
Answer: b
Explanation: Solving by using Actual Mean Method
N=5
Mean of x=1986 and y=13
| x |
y |
X=x-1986 |
Y=y-13 |
X2 |
XY |
| 1966 |
10 |
-20 |
-3 |
400 |
60 |
| 1976 |
12 |
-10 |
-1 |
100 |
10 |
| 1986 |
13 |
0 |
0 |
0 |
0 |
| 1996 |
14 |
10 |
1 |
100 |
10 |
| 2006 |
16 |
20 |
3 |
400 |
60 |
|
|
∑X=0 |
∑Y=0 |
∑X2 =1000 |
∑XY=140 |
We know that,
∑y=Na+b∑x
∑xy=a∑x+b∑x2
Substituting the values from the table into the equations-
0=(5)a+b(0) – (1)
a=0
140=(a)0+b(1000) – (2)
b=0.14
Thus, the equation is
Y=a+Bx
Y=0+0.14X
Resubstituting X=x-1986 and Y=y-13
y-13=0.14(x-1986)
y=-265.04+0.14x
To find the production in the year 2000, substitute x = 2000.
y=-265.04+0.14(2000)
y=14.96.
5. Fit a straight line y=a+bx into the given data. What is the value of y when x=8 ?