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
View Answer
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
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
View Answer
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
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) 27
b) 26
c) 28
d) 37
View Answer
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 | 120 |
| 6 | 25 | 36 | 150 |
| ∑x=21 | ∑y=135 | ∑x2=91 | ∑xy=490 |
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)
490=(a)21+b(91) – (2)
Solving equations (1) and (2) simultaneously
a=19 and b=1
Thus the equation of the line is given by y=a+bx
Thus, the equation of the line is y=19+x.
Putting x=8,
y=19+8
y=27
Sanfoundry Global Education & Learning Series – Linear Algebra.
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