Interpolation Practice Questions

Interpolation is the process of estimating an unknown value that lies between two known values.

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

Solution:

Given

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

Since, y = y_0 + (x - x_0) \dfrac{y_1 - y_0}{x_1 - x_0}

y = 5 + (3 − (−5)) \dfrac{7 − 5}{5 − (-5)}

   = 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

Example 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, 

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)}+....

f(x) = 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)}+....

       = 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