We will use Newton’s backward interpolation formula to find the polynomial.
\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\frac { n }{ n! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\frac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+..\)

To find y in terms of x
\(\therefore\) x_n + nh = x, x_n = 7, h = 1 \(\Rightarrow\) n = x − 7
\({ y }_{ (x) }=29+(x-7)(7)+\frac { (x-7)(x-6) }{ 2 } (1)\)
= \(29+7x-49+\frac { 1 }{ 2 } \left( { x }^{ 2 }-13x+42 \right) \)
= \(\frac { 1 }{ 2 } \left[ { x }^{ 2 }+x+2 \right] \)

Since the required value is at the end of the table, apply backward interpolation formula

.\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\frac { n }{ n! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\frac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+...\)
To find y at x = 7.5
\(\therefore\) xn + nh = 7.5, xn = 8, h = 1 \(\Rightarrow\) n = –0.5
\({ y }_{ (x=7.5) }=512+\frac { -0.5 }{ 1! } 169+\frac { 0.5(-0.5+10 }{ 2! } 42+\frac { -0.5(-0.5+1)(-0.5+2) }{ 3! } 6\)
= 421.88

\(\Delta \)\(\left[ \frac { 5x+12 }{ { x }^{ 2 }+5x+6 } \right] \)
By Partial fraction method
\(\frac { 5x+12 }{ { x }^{ 2 }+5x+6 } =\frac { A }{ x+3 } +\frac { B }{ x+2 } \)
\(A=\frac { 5x+12 }{ x+12 } [x=-3]=\frac { -15+12 }{ -1 } =\frac { -3 }{ -1 } =-3\)
\(B=\frac { 5x+12 }{ x+3 } \)[x = -2] \(=\frac { 2 }{ 1 } =2\)
\(\frac { 5x+12 }{ { x }^{ 2 }+5x+6 } = \left[ \frac { 3 }{ x+3 } +\frac { 2 }{ x+2 } \right] \)
\(\Delta \frac { 5x+12 }{ { x }^{ 2 }+5x+6 } =\Delta \left[ \frac { 3 }{ x+3 } +\frac { 2 }{ x+2 } \right] \)
\(=\left[ \frac { 3 }{ x+1+3 } -\frac { 3 }{ x+3 } \right] +\left\{ \frac { 2 }{ x+1+2 } -\frac { 2 }{ x+2 } \right\} \)
\(=3\left[ \frac { 1 }{ x+4 } -\frac { 1 }{ x+3 } \right] +2\left[ \frac { 1 }{ x+3 } -\frac { 1 }{ x+2 } \right] \)
\(=\left[ \frac { -3 }{ (x+4)(x+3) } -\frac { 2 }{ (x+3)(2+3) } \right] \)
\(=\frac { -5x-14 }{ (x+2)(x+3)(x+4) } \)

Let x be the weight and y be the number of students.
Difference table of cumulative frequencies are given below

Let us calculate the number of students whose weight is below 70. For this we use forward difference formula
\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\frac { n }{ n! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\frac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+....\)
To find y at x = 70
\(\therefore\) x₀+nh = 70, x₀ = 40, h = 20
40+n(20) = 70 \(\Rightarrow\) n = 1.5
\({ y }_{ \left( x=70 \right) }=250+1.5\left( 120 \right) +\frac { \left( 1.5 \right) \left( 0.5 \right) }{ 2! } \left( -20 \right) +\frac { \left( 1.5 \right) \left( 0.5 \right) \left( -0.5 \right) }{ 3! } \left( -10 \right) +\frac { \left( 1.5 \right) \left( 0.5 \right) \left( -0.5 \right) \left( -1.5 \right) }{ 4! } \left( 20 \right) \)
= 250 + 180 - 7.5 + 0.625 + 0.46875
= 423.59
\(\cong \) 424.
Number of students whose weight is between
60 and 70 = y(70)−y(60) = 424−370 = 54

Since we apply the forward interpolation formula,last four values of f(x) are taken into consideration (Take the values from x = 3).
The forward interpolation formula is
\({ y }_{ (x={ x }_{ 0 }+nh) }={ y }_{ 0 }+\frac { n }{ 1! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{0 }+\frac { n(n-1)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+...\)
x₀ + nh = 3.2, x₀ = 3,y = 1
\(\therefore n=\frac { 1 }{ 5 } \)
The difference table is

y_(x=3.2) = \(16+\cfrac { 1 }{ 5 } (4)+\cfrac { \frac { 1 }{ 5 } \left( \frac { -4 }{ 5 } \right) }{ 2 } \left( 0 \right) +\cfrac { \frac { 1 }{ 5 } \left( \frac { -4 }{ 5 } \right) \left( \frac { -9 }{ 5 } \right) }{ 6 } \times 10\)
= 16+0.8+0+0.48
= 17.28