Let k as 5^th term of difference table

Third order ditferenece as constant
k-130 = 6
k = 136
5^th term = 136

Since 4 values of f(x) are given
\( \Delta^{4} y_{0} =0 \)
\((\mathrm{E}-1)^{4} \mathrm{y}_{0} =0 \)
\(\left(\mathrm{E}^{4}-4 \mathrm{E}^{3}+6 \mathrm{E}^{2}-4 \mathrm{E}+1\right) \mathrm{y}_{0}=0 \)
\(\mathrm{y}_{4}-4 \mathrm{y}_{3}+6 \mathrm{y}_{2}-4 \mathrm{y}_{1}+\mathrm{y}_{0}=0 \)
\(467-4(397)+6(369)-4 \mathrm{y}_{1}+443=0 \)
\(4 \mathrm{y}_{1} =1536 \)
\(\mathrm{y}_{1} =384 \)

5 values as given
\( \Delta^{5} y_{0}=0 \)
\((\mathrm{E}-1)^{5} \mathrm{y}_{0}=0 \)
\(\left(\mathrm{E}^{5}-5 \mathrm{E}^{3}+10 \mathrm{E}^{3}-10 \mathrm{E}^{2}+5 \mathrm{E}-1\right) \mathrm{y}_{0}=0 \)
\(\mathrm{y}_{5}-5 \mathrm{y}_{4}+10 \mathrm{y}_{3}-10 \mathrm{y}_{2}+5 \mathrm{y}_{1}-\mathrm{y}_{0}=0 \)
\(32-5(24)+10 \mathrm{y}_{3}-10(14)+5(11)-7=0 \)
\(10 \mathrm{y}_{3}=180 \)
\(\mathrm{y}_{3}=18\)

Since four values of f(x) are given
\( \Delta^{4} \mathrm{y}_{0}=0 \\ (\mathrm{E}-1)^{4} \mathrm{y}_{0}=0 \)
\( \left(E^{4}-4 E^{3}+6 E^{2}-4 E+1\right) y_{0}=0 \)
\(y_{4}-4 y_{3}+6 y_{2}-4 y_{1}+y_{0}=0 \)
\(32-4(14)+6 y_{2}-4(5)+2=0 \)
\(6 y_{2}=56+20-32-2=42 \)
\(y_{2}=7 \)

We can write the given data as

\( x_{0} =75, \mathrm{~h}=5 \)
\(x_{0}+\mathrm{nh} =82 \)
\(5 \mathrm{n} =7 \Rightarrow \mathrm{n}=\frac{7}{5}=1.4\)

 \( \mathrm{y} =y_{0}+\frac{\mathrm{n} \Delta \mathrm{y}_{0}}{1 !}+\frac{\mathrm{n}(\mathrm{n}-1)}{2 !} \Delta^{2} \mathrm{y}_{0}+\ldots . . \)
\( =2459+\frac{1.4}{1 !}(-141)+\frac{(1 .-1)(-4)}{2 !}(-397)+\frac{(1.4)(1.4-1)(1 .--2)}{3 !}(45.1 \)
\(=2459-617.4-111.6-25.592 \)
\(\mathrm{y} =1704.408\)

Since three values of f(x) are given
\(\Delta^{3} y_{0}=0\)
\( (E-1)^{3} y_{0} =0 \)
\(\left(E^{3}-3 E^{2}+3 E-1\right) y_{0} =0 \)
\(y_{3}-3 y_{2}+3 y_{1}-y_{0} =0 \)
\(157-3(126)+3 y_{1}-100 =0 \)
\(y_{1} =107\)

Given
x_o = 6, x₁ = 7, x₂ = 10, x₃ = 12
y_o = 13, y₁ = 14, y₂ = 15, y₃ = 17
Using Lagrange's formula,
y(11) = \(13\frac { (4)(1)(-1) }{ (-1)(-4)(-6) } +14\frac { (5)(1)(-1) }{ (1)(-3)(-5) } +5\frac { (5)(4)(-1) }{ (4)(3)(-2) } +17\frac { (5)(4)(1) }{ (6)(5)(2) } \)
= 2.1666 - 4.6666 + 12.5 + 5.6666
= 15.6666
∴ y(x) = 15.6666

By data, we have
x_o = 40, x₁ = 50, x₂ = 60, x₃ = 70
y_o = 31, y₁ = 73, y₂ = 124, y₃ = 159.
Using Lagrange's formula, we get
\(y={ y }_{ 0 }\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } +{ y }_{ 1 }\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })({ x }_{ 1 }-{ x }_{ 3 }) } { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } +{ y }_{ 3 }\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \)
∴ y(42) = 31\(\frac { (-8)(-18)(-28) }{ (-10)(-20)(-30) } +73\frac { (2)(-18)(-28) }{ (10)(-10)(-20) } +124\frac { (2)(-8)(-28) }{ (20)(10)(-10) } +59\frac { (2)(-8)(-28) }{ (30)(20)(10) } \)
= 20. 832 + 36. 792 - 27. 776 + 7.632
y = 37. 48

Since 1995 lies at the table of the table, use Newton's backward interpolation formula.
Also, xn + nh ⇒ x  2001 + n (10) = 1995
⇒ 10n = 1995 - 2001 ⇒ n = \(\frac{-6}{10}\) ⇒ n = -0.6
\(\Rightarrow { y }_{ n }+\frac { n }{ n! } \nabla { y }_{ n }+\frac { n(n+1) }{ 2! } { \nabla }^{ 2 }{ y }_{ n }+\frac { n(n+1)(n-2) }{ 3! } { \nabla }^{ 3 }{ (y }_{ n })\)
The difference table is

∴ \(y=101+\frac { (0.6) }{ 1! } (8)+\frac { (-0.6)(-0.6+1) }{ 2! } (-4)+\frac { (-0.6)(-0.6+1)(-0.6+2) }{ 3! } (-1)+\frac { (-0.6)(-0.6+1)(-0.6+2)(-0.6+3) }{ 4! } (-3)\)
= 101 - (0.6)8 + \(\frac { (-0.6)(0.4) }{ 2 } (-4)+\frac { (-0.6)(0.4)(1.4) }{ 6 } (-1)+\frac { (0.6)(0.4)(1.4)(2.4)(-3) }{ 24 } \)
= 96.8368
Hence, population for the year 1995 is 96.837 thousands.

Let us calculate the number of men whose wages is less than Rs. 35 by using Newton's forward interpolation formula
x_o + nh = x ⇒ 30 + n (10) = 35
⇒ 10n = 35 - 30 = 5
⇒ n = \(\frac{5}{10}\) = 0.5
The difference table is

\(\Rightarrow { y }_{ o }+\frac { n }{ n! } \triangle { y }_{ o }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ o }+\frac { n(n+1)(n-2) }{ 3! } { \triangle }^{ 3 }{ (y }_{ o })\)
\(y(35)=9+\frac { 0.5 }{ 1! } (30)+\frac { (0.5)(0.5-1) }{ 2 } (5)+\frac { (0.5)(0.5-1)(0.5-2) }{ 6 } (2)\)
= 9 + 15 - 0.6 + 0.1 = 24 (approximately)
∴ Number of men getting wages between Rs. 30 and Rs. 35 is y(35) - y(30) = 24 - 9 = 15.

Since 82 lies at the beginning of the table, we can use Newton's forward interpolation formula
\(\Rightarrow { y }_{ o }+\frac { n }{ n! } \triangle { y }_{ o }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ o }+\frac { n(n+1)(n-2) }{ 3! } { \triangle }^{ 3 }{ (y }_{ o })\)
Also x₀ + nh = 82 ⇒ 75 + n(5) = 82 ⇒ 5n = 82 - 75 = 7
⇒ n = \(\frac75\) = 1.4
The difference table is

\(y=2459+\frac { 1.4 }{ 1! } (-441)+\frac { (1.4)(1.4-1) }{ 2! } (-397)+\frac { (1.4)(1.4-1)(1.4-2) }{ 3! } (457)\)
= 2459 - 617.4 - 111.6 - 25.592
y = 1704. 408 when x = 82.

Since x = 0.2 lies at the beginning of the table, use Newton's foward interpolation formula
\(\Rightarrow { y }_{ o }+\frac { n }{ n! } \triangle { y }_{ o }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ o }+\frac { n(n+1)(n-2) }{ 3! } { \triangle }^{ 3 }{ (y }_{ o })\)
Here h = 1, x_o = 0, x = 0.2
⇒ x₀ + nh = 0.2 ⇒ 0 + n(1) = 0.2 ⇒ n = 0.2
The forward difference table is

 ∴ y = 176 +\(\frac { 0.2 }{ 1! } (9)+\frac { (0.2)(0.2-1) }{ 2! } (0)+\frac { (0.2)(0.2-1)(0.2-2) }{ 3! } (-1)+\frac { (0.2)(0.2-1)(0.2-2)(0.3-3) }{ 4! } (4)\)

= 176 + 1.8 - 0.048 - 0.1344 = 177.6176
ஃ Hence when x = 0.2, y = 177.6176.

From the graph, it is clear that when x = 27, the value of y is 24.8

From the graph, it is found that the population for 1986 was 24 thousands.