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.