Linear Algebra Questions and Answers – Curve Fitting

«
»

This set of Linear Algebra Multiple Choice Questions & Answers (MCQs) focuses 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
advertisement

a) y=3.52+2.26x
b) y=3.52
c) y=2.26x
d) y=4+3x
View Answer

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
View Answer

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.

advertisement

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
View Answer

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
View Answer

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
advertisement

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 ?

x: 1 2 3 4 5 6
y: 20 21 22 23 24 25

a) 45.2
b) 26
c) 28
d) 37
View Answer

Answer: b
Explanation: Here, N=6
Calculations of ∑x and ∑x2

x y x2 xy
1 20 1 20
2 21 4 42
3 22 9 66
4 23 16 92
5 24 25 125
6 25 36 216
∑x=21 ∑y=135 ∑x2=91 ∑xy=561

We know that,
∑y=Na+b∑x
∑xy=a∑x+b∑x2

Substituting the values from the table into the equations-
135=(6)a+b(21) – (1)
561=(a)21+b(91) – (2)

advertisement

Solving equations (1) and (2) simultaneously
a=4.8 and b=5.05
Thus the equation of the line is given by y=a+bx
Thus, the equation of the line is y=4.8+5.05x.
Putting x=8,
y=4.8+(5.05)×(8)
y=45.2.

Sanfoundry Global Education & Learning Series – Linear Algebra.

To practice all areas of Linear Algebra, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement
advertisement
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn