Solved Examples on Interpolation Formula

Example 1: Interpolate the value of y at x = 3, if two of the ordered pairs are (−5, 5), (5, 7).

Solution:

Given

x = 3, x₀ = −5, x₁ = 5, y₀ = 5, y₁ = 7 

Since, [Tex]y = y_0 + (x – x_0) \dfrac{y_1 – y_0}{x_1 – x_0} [/Tex]

[Tex]y = 5 + (3 − (−5)) \dfrac{7 − 5}{5 − (-5)} [/Tex]

   = 5 + 8(.2)

=5 + 1.6

y = 6.6

Example 2: Interpolate the missing value in the following table:

Solution:

Since the known values of Y are 4, the fourth leading difference will be zero. Thus, 

△⁴₀ = (y − 1)⁴ = y⁴ − 4y³ + 6y² − 4y¹ + y⁰ = 0

Substituting the values of y’s we have,

212 − 4(157) + 6y₂ − 4y₁ + y₀ = 0 

6y₂ = 628 − 212 + 428 − 100

y₂ = 124

Examples 3: The table below depicts insurance premiums payable to different age groups. Find the premium payable at 17.

Solution:

Age	Premium(Y)	△1	△2	△3	△4
15 = x0	11.1 = y0
		+1.5 = △10
25 = x1	12.6 = y1		+0.2 = △20
		+1.7 = △11		−0.1 = △30
35 = x2	14.3 = y2		+0.1 = △21		+0.4 = △40
		+1.8 = △13		+0.3 = △31
45 = x3	16.1 = y3		+0.4 = △22
		+2.2 = △14
55 = x4	18.3 = y4

Now, x = 17 – 15/25 – 15 = 2/10 = 0.2

Hence, Y₁₇ = 11.1 + 0.3 − 0.016 − 0.0048 − 0.0134 = 11.4

Thus premium payable at 17 is 11.4

Example 4: Find the value of y when x = 5 from the following data:

Solution:

Applying Lagrange Formula, 

[Tex]f(x) = \dfrac{(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)} + \dfrac{(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)}+\dfrac{(x − x_0)(x − x_1)(x − x_2)(x − x_4)}{(x_3 − x_0)(x_3 − x_1)(x_3 − x_2)(x_3 − x_4)}+…. [/Tex]

f(x) = [Tex]1.\dfrac{(5−3)(5-4)(5-6)(5-7)}{(2−3)(4-3)(4-6)(4-7)}+5.\dfrac{(5−2)(5−4)(5−6)(5−7)}{(3−2)(3−4)(3−6)(3−7)}+13.\dfrac{(5−2)(5−3)(5−6)(5−7)}{(4−2)(4−3)(4−6)(4−7)}+…. [/Tex]

       = 0.1 − 2.5 + 1.3 + 30.5 − 12.5

      = 28.6

Example 5: Reconstruct the following data with the class intervals halved using interpolation:

Solution:

For estimating the marks in the class interval 0-5, 10-15, and 20-25, we will have to interpolate the values of less than 5, less than 15 and less than 25 respectively. 

Interpolation of less than 5

x = 5−10/10 = −0.5

y₅ = y0 + x.△10 + [x(x−1).△²]/2 = 25

Similarly, 

Interpolation of less than 15 = 80

Interpolation of less than 25 = 155

Thus, the reconstructed table is as follows:

Marks	Cumulative Frequency	Marks	Number of Students
Less than 5	25	0 – 5	22
Less than 10	50	5 – 10	25
Less than 15	80	10 – 15	30
Less than 20	115	15 – 20	35
Less than 25	155	20 – 25	40
Less than 30	200	25 – 30	45

Interpolation formula is a method to find new values of any function using the set of available values through interpolation. It is an important statistical tool used to calculate the value between two points on the curve of a function from the given points which also lie on the same curve.

In statistical analysis and interpretation, sometimes it is found that a given series happens to be incomplete rather than complete, i.e., some values in the series remain unknown. But to derive correct results, it becomes essential to find the missing or unknown values in the series. The statistical technique that is used to estimate the unknown values on the basis of available data is called interpolation.