\(x_{0}=1, x_{1}=2, x_{2}=3, x_{3}=4, x_{4}=7, x=5\)
\(y_{0}=2, y_{1}=4, y_{2}=8, y_{3}=16, y_{4}=128\)
Applying in Lagrange's formula we get
\( =-0.666+6.4-24+42.6666+8.533 \\ =32.933 \)

\(x_{0}=0, x_{0}=1, x_{0}=3, x_{3}=4, x=2\)
\( =\frac{5(1)(-1)(-2)}{(-1)(-3)(-1)}-\frac{(2(2)(-1)(-2)}{(1)(-2)(-3)}- (50) \frac{2(1)(-2)}{3(2)(-1)}-\frac{4105 /(2)(1)(-1)}{4(3)(1)}\)
\(=-0.833 \div 4 \div 33.333-17.5=19\)

\( x_{0}=40, x_{1}=50, x_{2}=60, x_{3}=70, x=42 \)
\(y_{0}=31, y_{2}=73, y_{2}=124, y_{3}=159 \)
\( y=j_{6} \frac{\left(x-x_{2}\right)\left(x-x_{2}\right)\left(x-x_{1}\right)}{\left(x-x_{2}\right)\left(x_{3}-x_{2}\right)\left(x_{2}-x_{1}\right)}-x_{3} \frac{\left(x-x_{3}\right)\left(x-x_{2}\right)\left(x-x_{2}\right)}{\left(x_{2}-x_{0}\right)\left(x_{2}-x_{2}\right)\left(x_{2}-x_{0}\right)} -1=\frac{\left(x-x^{2}\right)\left(x-x_{1}\right)\left(x-x_{2}\right)}{\left(x_{2}-x_{2}\right)\left(x_{2}-x_{2}\right)\left(x_{2}-x_{2}\right)}-t_{2} \frac{\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{2}\right)}{\left(x_{2}-x_{0}\right)\left(x_{3}-x_{1}\right)\left(x_{3}-x_{2}\right)} \)
\(=\frac{31(-8)(-18)(-28)}{(-10)(-20)(-301}-\frac{73(2)(-18)(-28)}{(10)(-10)(-20)}+124 \frac{(2)(-8)(-28)}{(20)(-10)(-10)} -\frac{159(2)(-8)(-18)}{(301(20)(10)} \)
= 20.832 + 36.792 - 27.776 + 7.632
= 37.48

\(x_{\mathrm{n}}=60, \mathrm{~h}=5, x_{\mathrm{n}}+\mathrm{nh}=58\)
\(5 n=-2 \Rightarrow n=\frac{-2}{5}=-0.4\)

\( \frac{m(n-1)(n-2)(n-3)}{4 !} \Gamma^{i} y \) \( y=68.4 \mathrm{~s}+\frac{(-0.4)}{1 !}(-6)+\frac{(-0.4)(-0.4+1)}{2 !}(2.4) -\frac{(-0.4)(-0.4-1)(-0.4+2)}{3 !}(-1.16) -\frac{(-0.4)(-0.4-1)(-0.4-2)(-0.4-3)}{4 !}(0.68 ) \)
\(=68.48+24-0.3408+0.07424-0.028288\\ =70.5851\ \)
\( \therefore \text { If } x=58, y=70.59 \)

\(x_{\mathrm{n}}=2001, \mathrm{~h} \doteq 10, x_{\mathrm{n}}+\mathrm{nh}=1995\)
\(n=\frac{-6}{10}=-0.6\)

\( \mathrm{y}= y_{n}+\frac{n}{1 !} \nabla y_{n}+\frac{n(\mathrm{n}+1)}{2 !} \nabla^{2} y_{n}+ \frac{n(\mathrm{n}+1)(\mathrm{n}+2)}{3 !} \nabla^{3} y_{s}+\frac{n(n+1)(\mathrm{n}+2)(\mathrm{n}+3)}{4 !} \nabla^{4} y_{n} \)
\(\mathrm{y}= 101+\frac{(-0.6)}{1 !}(8)+\frac{(-0.6)(-0.6+1)}{2 !}(-4)+\frac{(-0.6)(-0.6+1)(-0.6+2)}{3 !}(-1) \ +\frac{(-0.6)(-0.6+1)(-0.6+2)(-0.6+3)}{4 !}(-3) \)
= 101-4.8+0.48+0.056+0.1008 
= 96.8368

\( x_{0} =30, \mathrm{~h}=10, x_{0}+\mathrm{nh}=35 \)
10n = 5
n = 0.5

\( y =y_{0}+\frac{n}{1 !} \Delta y+\frac{n(n-1)}{2 !} \Delta^{2} y+\frac{n(n-1)(n-2)}{3 !} \Delta^{3} y_{n} \)
\(=9+\frac{(0.5)(30)}{1 !}+\frac{(0.5)(0.5-1)}{2 !}(5)+\frac{(0.5)(0.5-1)(0.5-2)}{3 !}(2) \)
 = 9+15-0.6+0.1 
= 24 
Number of men getting wages between
30 and 35 is y(35) - y(30) = 24-9 = 15

\( x_{0} =60, \mathrm{~h}=20, x_{0}+\mathrm{nh}=90 \\ 20 \mathrm{n} =30 \Rightarrow \mathrm{n}=1.5 \)

\( y=y_{0}+\frac{n}{1 !} \Delta y_{6}+\frac{n(n-1)}{2 !} \Delta^{2} y_{0}+\frac{n(n-1)(n-2)}{3 !} \Delta^{3} y_{0} \) \(+\frac{n(n-1)(n-2)(n-3)}{4 !} \Delta^{4} y_{0} \)
\( y(90)= 250+\frac{(1.5)}{1 !}(250)+\frac{(1.5)(1.5-1)}{2 !}(-20) +\frac{(1.5)(1.5-1)(1.5-2)}{3 !}(-10)+\frac{(1.5)(1.5-1)(1.5-2)(1.5-3)}{4 !}(20)\)
= 250+180.7 .5+0.625+0.46875 
\( =423.59 \sim 424 \)

\( x_{0} =1.7, \mathrm{~h}=0.1, x=x_{0}+\mathrm{nh}=1.75 \)
\(\mathrm{n}(0.1) =1.75-1.7 \\ \mathrm{n} =\frac{0.05}{0.1}=0.5\)

\( \mathrm{y} =y_{0}+\frac{n}{1 !} \Delta y+\frac{n(\mathrm{n}-1)}{2 !} \Delta^{2} y_{0}+\frac{n(\mathrm{n}-1)(\mathrm{n}-2)}{3 !} \Delta^{\prime} y_{0} \)
\(=5.474+\frac{0.5}{1 !}(0.576)+\frac{(0.5)(0.5-1)}{2 !}(0.06) \ +\frac{(0.5)(0.5-1)(0.5-2)}{3 !}(0.007) \)
 = 5.474+0.288-0.0075+0.0004375 
 = 5.7549375
 

\(x_{0}=0, \mathrm{~h}=1, x_{0}+\mathrm{nh}=0.2, \mathrm{n}=0.2\)

\( y =1.6+\frac{0.2}{1 !}(9)+\frac{(0.2)(0.2-1)}{2 !}(0)+\frac{(0.2)(0.2-1)(0.2-2)}{3 !}(-1) +\frac{(0.2)(0.2-1)(0.2-2)(0.2-3)}{4 !}(4) \)
 = 176+1.8-0.048-0.1344 
 = 177.6176

Since 5 values of f(x) are given
\( \Delta^{5} y_{0} =0 \)
\((E-1)^{5} y_{0} =0 \)
\(\left(E^{3}-5 E^{4}+10 E^{3}-10 E^{2}+5 E-1\right) y_{0} =0 \)
\(y_{5}-5 y_{4}+10 y_{3}-10 y_{2}+5 y_{1}-y_{0} =0 \)
\(390-5 y_{4}+10(306)-10(260)+5 y_{1}-200 =0 \)
\(y_{1}-y_{4} =-130 \quad \ldots(1) \)
\(\Delta^{5} y_{1} =0 \)
\((E-1)^{5} y_{1} =0 \)
\(\left(E^{5}-5 \mathrm{E}^{4}+10 \mathrm{E}^{3}-10 \mathrm{E}^{2}+5 \mathrm{E}-1\right) \mathrm{y}_{1} =0 \)
\(\mathrm{y}_{6}-5 \mathrm{y}_{5}+10 \mathrm{y}_{4}-10 \mathrm{y}_{3}+5 \mathrm{y}_{2}-\mathrm{y}_{1} =0 \)
\(430-5(390)+10 \mathrm{y}_{4}-10(306)+5(260)-\mathrm{y}_{1} =0 \)
\(10 \mathrm{y}_{4}-\mathrm{y}_{1} =3280 \)...(2)
Solving (1) and (2) we get
y₁ = 220 tons and y₂ = 350 tons

Given x_o = 0, x₁ = 3, x₂ = 5, x₃ = 6, x₄ = 8
y_o = 276, y₁ = 460, y₂ = 414, y₃ = 343, y₄ = 110
Lagrange's formula is
y = \(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 })(x-{ x }_{ 4 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 })({ x }_{ 0 }-{ x }_{ 4 }) } { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 })(x-{ x }_{ 4 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })({ x }_{ 1 }-{ x }_{ 3 })({ x }_{ 1 }-{ x }_{ 4 }) } { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 })(x-{ x }_{ 4 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 })({ x }_{ 2 }-{ x }_{ 4 }) } { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 })(x-{ x }_{ 4 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 })({ x }_{ 3 }-{ x }_{ 4 }) } { y }_{ 3 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 })(x-{ x }_{ 4 }) }{ ({ x }_{ 4 }-{ x }_{ 0 })({ x }_{ 4 }-{ x }_{ 2 })({ x }_{ 4 }-{ x }_{ 3 })({ x }_{ 4 }-{ x }_{ 4 }) } { y }_{ 4 }\)
⇒ 276 \(\frac { (1)(-1)(-2)(-4) }{ (-3)(-5)(-6)(-8) } +460\frac { (4)(-1)(-2)(-4) }{ (3)(-2)(-3)(-5) } +414\frac { (4)(1)(-1)(-4) }{ (5)(2)(-1)(-3) } +343\frac { (4)(1)(-1)(-2) }{ (6)(3)(1)(-2) } +110\frac { (4)(1)(-1)(-2) }{ (8)(5)(3)(2) } \)
⇒ y = -3.066 + 163.555 + 441.6 - 152.44 + 3.666
⇒ y = 453.311.

Using Newton's backward interpolation formula,
we can find y when x = 58.
∴ x + nh = x ⇒ 60 + n(5) = 58
⇒ 5n = 58 - 60 = -2
⇒ n = \(\frac{-2}{5}\)= -0.4
and y(58) = \({ y }_{ n }+\frac { n }{ n! } \nabla { y }_{ n }+\frac { n(n+1) }{ 2! } { \nabla }^{ 2 }{ y }_{ n }+\frac { n(n+1)(n-2) }{ 3! } { \nabla }^{ 3 }{ (y }_{ n })+\) .....
The difference table is

\(y(58)=68.48+\frac { (-0.4) }{ 1! } (-6)+\frac { (0.4)(-0.4+1) }{ 2! } (2.84)+\frac { (-0.4)(-0.4+1)(-0.4+2) }{ 3! } (-1.16)+\frac { (-0.4)(-0.4+2)(-0.4+2)(-0.4+3) }{ 3! } (0.68)\)
=68.48 + (0.4)(6) + \(\frac { (-0.4)(0.6) }{ 2 } (2.84)+\frac { (-0.4)(0.6)(1.6) }{ 6 } (-1.16)+\frac { (-0.4)(0.6)(1.6)(2.6) }{ 24 } (0.68)\)
= 68.48 + 2.4 - 0.3408 + 0.07424 - 0.028288
= 70.5851052
⇒ y(58) = 70.59
∴ Hence, premium for a policy maluting at the age of 58 is 70.59.

Let us calculate the number of students whose height is less than 90 ern using Newton's forward interpolation formula.
x_o+ nh = x ⇒ 60 + n(20) = 90 ⇒ 20n = 30
⇒ n = \(\frac32\) = 1.5
\({ y }_{ x }={ y }_{ o }+\frac { n }{ n! } \triangle { y }_{ o }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ o }+\frac { n(n+1)(n-2) }{ 3! } { \triangle }^{ 3 }{ (y }_{ o })+.......\)
The difference table is as follows:

∴y(90) = 250 + (1.5)(120)\(\frac { (1.5)(1.5-1) }{ 2! } (20)+\frac { (1.5)(1.5-1)(1.5-2) }{ 3! } (-10)+\frac { (1.5)(1.5-1)(1.5-2)(1.56-3) }{ 4! } (20)\)
= 250 + 180 - 7.5 + 0.625 + 0.46875
= 423.59 = 424 (app)
ஃ Number of students whose height is between 80 cm and 90 cm is y(90) - y(80)
= 424 - 370
= 54.

Since e^1.75 lies at the beginning of the table, we can use Newton's forward interpolation formula
∴ x_o + nh = x ⇒ 1.7 + n(0.1) = 1.75
⇒ n(0.1) = 1.75 - 1.7 = 0.05
⇒ n = \(\frac{0.05}{0.1}\) = 0.5
\({ y }_{ x }={ y }_{ o }+\frac { n }{ n! } \triangle { y }_{ o }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ o }+\frac { n(n+1)(n-2) }{ 3! } { \triangle }^{ 3 }{ (y }_{ o })+.......\)
The difference table is

∴ \(y\left( { e }^{ 1.75 } \right) =5.74+\frac { 0.5 }{ 1! } (0.576)+\frac { (0.5)(0.5-1) }{ 2! } (0.06)+\frac { (0.5)(0.5-1)(0.5-2) }{ 3! } (0.007)\)
y(e^1.75) = 5.474 + 0.288 - 0.0075 + 0.0004375
= 5.7549375

Since five values of f(x) are given, we assume that polynomial is of degree four.
Hence, fifth order differences are zeros.
∴ ∆5(y_k) = 0
⇒ (E - 1)5y_k = 0   (1)
Putting k = 0 in (1) we get
(E - 1)5y_o = 0 ⇒ (E₅- 5E₄ + 10E₃ - 10E₂+ 5E - 1)y_o = 0
⇒ y₅ - 5y₄ + 10y₃ - 10y₂ + 5y₁ - y_o = 0
⇒ 390 - 5y₄ + 10y₃ - 10y₂ + 5y₁ - y_o = 0
⇒ y₁ - y₄ = -130 (2)
Putting k = 1 in (1) we get,
(E - 1)5y₁ = 0
⇒ (E₅ - 5E₄ + 10E₃ - 10P + 5E -1) y₁ = 0
⇒ y₆ - 5y₅ + 10y₄ - 10y₃ + 5y₂ - y₁ = 0
⇒ 430 - 5(390) + 10y₄ - 10(306) + 5(260) - y₁ = 0
⇒ 10y₄ - y₁ = 3280

⇒ 9y₄ = 3150
⇒ y₄ = 350
Substituting y₄ = 350 in (2) we get,
y₁ - 350 = - 130 ⇒ y₁ = -130 + 350
⇒ y₁ = 220
∴ The productions for 1962 and 1965 are 220 The productions for 1962 and 1965 are 220