Linear Algebra Questions and Answers – Curve Fitting

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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
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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.

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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.

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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
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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
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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

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)

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.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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