# 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

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

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

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

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

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

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.

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