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.

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