Interpolation and Extrapolation Business Statistics way of answering under few methods

1. INTERPOLATION
AND
EXTRAPOLATION
Concept of Business statistics

2. Meaning and definitions
• Interpolation is the process or technique of finding the missing value within the given data.
Extrapolation is the process of finding the future value which lies outside the given set of data.
Definition:
According to W.M. Harper “interpolation consists in reading a value, which lies between two
extreme points. Extrapolation means reading that lies outside the two extreme points.”

3. Lets understand this concepts with examples
• Example 1:
• Example2: Find the sales value for 2021
X 1 2 3 4 5
Y 10 15 ------ 25 24
Year 2016 2017 2018 2019 2020
sales 100 120 130 125 145

4. The applications of the above techniques is based on the following
assumptions.
• 1. there are no sudden jumps, from one period to another period.
• 2. there is a uniformity in the rate of change in the values of variables.
• 3. there is a definite and stable relationships between the two variables.

5. Uses / significance of interpolation and extrapolation.
• The technique are helpful in filling the gaps that exist in statistical data.
• They are help full in obtaining the median and mode in continuous series.
• They are required when the data insufficient due to difficulties in collecting .
• They are use full in obtaining the future data also.

6. Methods of interpolation and extrapolation
• There are two methods of interpolation and extrapolation. They are
• Binominal Expansion method
• Newton’s method of advancing differences.
• Binominal Expansion method:
• This method is used for both technique and applicable only when following conditions are
fulfilled by the given or collected data.
• 1. The independent variables(x) advances by equal intervals like 2,4,6,8 or 3,6,9,12 and so on.
• 2. the value of (x), for which the value of (y) is to be interpolated, is one of the class limits of X
series.

7. There are two ways to adopt binominal expansion method.
1.By using Binominal Theorem.
2. BY using Pascal’s Triangle.
• steps to be followed in above method as follows.
Under Binominal Theorem Under Pascal’s triangle
Step 1. Find leading differences Find coefficients using Pascal’s triangle
Step2. Analysis of the Differences Formation of equations
Step3. finding the value. Find the values.

8. Illustration1.Solving the problem using Binominal Theorem when known values are Two.
X 1 2 3
Y 4 ---- 14
Leading difference
X Y ∆1
∆2
1 4 𝑦0 -4(𝑦1 − 𝑦0) = ∆0
1
2 -- 𝑦1 14(𝑦2 − 𝑦1)=∆1
1
18(∆1
1
−∆0
1
)
3 14 𝑦2

9. • Analysis of the difference
• ∆2 =0
• ∆1
1
−∆0
1
=0
• (𝑦2−𝑦1) − (𝑦1 − 𝑦0)=0
• 𝑦2 − 𝑦1 − 𝑦1+ 𝑦0 = 0
• 𝑦2 − 2𝑦1 + 𝑦0 = 0
• Finding the missing value
• 𝑦2 − 2𝑦1 + 𝑦0 = 0
• 14-2𝑦1+4=0
• 18-2𝑦1=0
• 18=2𝑦1
• 𝑦1=18/2=9

10. Illustration2.Solving the problem using Binominal Theorem when known values are
Three.
X 1 2 3
Y 4 ---- 14
Leading difference
X Y ∆1 ∆2 ∆3
1 4 𝑦0 4(𝑦1 − 𝑦0) = ∆0
1
2 8 𝑦1 -8(𝑦2 − 𝑦1)=∆1
1
-4(∆1
1
−∆0
1
)=∆0
2
34(∆1
2
-∆0
2
)=∆0
3
3 --- 𝑦2 22(𝑦3-𝑦2)=∆2
1
30(∆2
1
-∆1
1
)=∆1
2
4 22 𝑦3

11. • Analysis of the difference
• ∆3 =0
∆1
2
-∆0
2
= 0
(∆2
1
-∆1
1
)−(∆1
1
−∆0
1
)=0
((𝑦3- 𝑦2)- 𝑦2 − 𝑦1 ) − ( 𝑦2 − 𝑦1 -( 𝑦1 − 𝑦0))=0
• Y3-y2-y2+y1-y2+y1+y1+y0=0
• Y3-3y2+3y1-y0=0
• Finding the missing value
• 𝑦3-3𝑦2+3𝑦1-𝑦0=o
• 22-3𝑦2+3(8)-4=0
• 22-3𝑦2+24-4=0
• 22+24-4=3𝑦2
• 42=3𝑦2
• 𝑦2=42/3
• 𝑦2=14

12. Using above steps we can able to get following formulas and same
formulas can apply directly for finding values.

13. Under Pascal’s triangle
• Find coefficients using Pascal’s triangle
• Formation of equations
• 1. Record the ‘Y’ alphabet with the
subscripts up to 𝑦0.
• 2. place the signs alternatively starting
from the first which must be Plus.

14. Illustration 1:Find the sales value for the year 2020 from the data given below.
year Sales in 00units
2014 240
2015 300
2016 320
2017 360
2018 400
2019 450
• Workings:
• Here 6years sales values are given hence
we needs to prepare 7rows pascal’s
triangle.

15. • After preparation of triangle, assign ‘Y’
alphabet with subscripts to known values.
year Sales in
2014 240(𝑦0)
2015 300(𝑦1)
2016 320((𝑦2)
2017 360(𝑦3)
2018 400(𝑦4)
2019 450(𝑦5)
2020 -------(𝑦6)
Place the signs alternatively starts from plus sign.
+(𝑦6)-(𝑦5)+ (𝑦4)- (𝑦3)+ (𝑦2)- (𝑦1)+ (𝑦0)
Join the last rows triangle value to above and
frame the formula.
(𝑦6)-(6𝑦5)+(15𝑦4)-(20𝑦3)+ (15𝑦2)- (6𝑦1)+ (𝑦0)=0
Using above formula find the needed value.

16. • (𝑦6)-(6𝑦5)+(15𝑦4)-(20𝑦3)+ (15𝑦2)- (6𝑦1)+ (𝑦0)=0
• (𝑦6)-(6*450)+(15*400)-(20*360)+ (15*320)- (6*300)+ (240)=0
• 𝑦6-2700+6000-7200+4800-1800+240=0
• 𝑦6-11700+11040=0
• 𝑦6-660=0(shift -660 to RHS)
• 𝑦6=660
• So sales value for 2020 is 66000 units.

17. Illustration 2:Find the sales value for the year 2017 from the data given
below.
year Sales in 00units
2014 150(𝑦0)
2015 235(𝑦1)
2016 365((𝑦2)
2017 ----(𝑦3)
2018 525(𝑦4)
2019 780(𝑦5)
• Workings:
• Here 5years sales values are given hence
we needs to prepare 6rows pascal’s
triangle.

18. • Place the signs alternatively starts from plus sign.
• (𝑦5)- (𝑦4)+ (𝑦3)- (𝑦2)+(𝑦1)- (𝑦0)
• Join the last rows triangle value to above and frame the formula.
• (𝑦5)- (5𝑦4)+ (10𝑦3)- (10𝑦2)+(5𝑦1)- (𝑦0)=0
Using above formula find the needed value.
(𝑦5)- (5𝑦4)+ (10𝑦3)- (10𝑦2)+(5𝑦1)- (𝑦0)=0
780- (5*525)+ (10𝑦2)- (10*365)+(5*235)- (150)=0
780-2625+(10𝑦2)-3650+1175-150=0
10𝑦2 +1955-6425=0
10𝑦2 -4470=0
10𝑦2 =4470
𝑦2 =4470/10
𝑦2 =447

19. Newton’s method of Advancing differences.
• This method is used only in interpolation, more concerned
with the interpolation of figures of (y) between two values of
‘X’ variables.it is based on advancing differences. We have to
proceed with the obtaining of differences between the ‘Y’
values extending the process till only one difference remains in
the end.

20. Following are the steps to be followed in applying Newton’s
formula
• Step1. find the leading differences, until only one differences remain in the end.
• step2. find the ‘X’ value by using the following formula
• X= item of interpolation -- item of origin
Difference between Adjoining items
Step3. Apply the Newton’s formula
Y= 𝑦0+X∆0
1
+x(x-1) ∆0
2
+x(x-1)(x-2) ∆0
3
+ x(x-1)(x-2)(x-3) ∆0
4
--------------so on
1x2 1x2x3 1x2x3x4

21. X Y
10 50
20 150
30 300
40 500
50 700
60 800
Illustration 1: Below are the wages earned by workers per month in a certain factory
interpolate the number of workers earning ₹35.

22. X Y ∆1 ∆2 ∆3 ∆4
∆5
10 50 𝑦0 100 (𝑦1 − 𝑦0) = ∆0
1
20 150 𝑦1 150 (𝑦2 − 𝑦1)=∆1
1 50 (∆1
1
−∆0
1
)=∆0
2 0 (∆1
2
-∆0
2
)=∆0
3
-50(∆0
4
)
30 300 𝑦2 200 (𝑦3-𝑦2)=∆2
1 50 (∆2
1
-∆1
1
)=∆1
2 -50 (∆2
2
-∆1
2
)=∆1
3
-50(∆1
4
) 0(∆0
5
)
40 500 𝑦3 200 (𝑦4-𝑦3)=∆3
1 0 (∆3
1
-∆2
1
)=∆2
2 -100 (∆3
2
-∆2
2
)=∆2
3
50 700 𝑦4 100 (𝑦5-𝑦4)=∆4
1 -100 (∆4
1
-∆3
1
)=∆3
2
60 800 𝑦5

23. Step 2: X= item of interpolation -- item of origin
Difference between Adjoining items
X=35-10
10
X=25/10
X=2.5

24. Step3‘:
Y= 𝑦0+ X∆0
1
+x(x-1) ∆0
2
+x(x-1)(x-2) ∆0
3
+ x(x-1)(x-2)(x-3) ∆0
4
+ x(x-1)(x-2)(x-3)(x-4) ∆0
5
1x2 1x2x3 1x2x3x4 1X2X3X4X5
• 50+2.5(100)+2.5(2.5-1) (50)+ 2.5(2.5-1)(2.5-2)(0)+2.5(2.5-1)(2.5-2)(2.5-3)(-50)+2.5(2.5-1)(2.5-2)(2.5-3)(2.5-4)(0)
1X2 1x2x3 1x2x3x4 1x2x3x4x5
=50+250+93.75+0+1.953+0
= 395.70 or 396 workers

25. Illustration 2: Below are the wages earned by workers per month in a certain factory
interpolate the number of workers earning ₹25.
X Y
10 50
20 150
30 300
40 500
50 700
60 800

26. X Y ∆1
∆2
∆3
∆4
∆5
10 50 𝑦0 100 (𝑦1 − 𝑦0) = ∆0
1
20 150 𝑦1 150 (𝑦2 − 𝑦1)=∆1
1 50 (∆1
1
−∆0
1
)=∆0
2 0 (∆1
2
-∆0
2
)=∆0
3
-50(∆0
4
)
30 300 𝑦2 200 (𝑦3-𝑦2)=∆2
1 50 (∆2
1
-∆1
1
)=∆1
2 -50 (∆2
2
-∆1
2
)=∆1
3
-50(∆1
4
) 0(∆0
5
)
40 500 𝑦3 200 (𝑦4-𝑦3)=∆3
1 0 (∆3
1
-∆2
1
)=∆2
2 -100 (∆3
2
-∆2
2
)=∆2
3
50 700 𝑦4 100 (𝑦5-𝑦4)=∆4
1 -100 (∆4
1
-∆3
1
)=∆3
2
60 800 𝑦5

27. Step2: X= item of interpolation -- item of origin
Difference between Adjoining items
X=25-10
10
X=15/10
X=1.5

28. Step3:
Y= 𝑦0+ X∆0
1
+x(x-1) ∆0
2
+x(x-1)(x-2) ∆0
3
+ x(x-1)(x-2)(x-3) ∆0
4
+ x(x-1)(x-2)(x-3)(x-4) ∆0
5
1x2 1x2x3 1x2x3x4 1X2X3X4X5
50+1.5(100)+1.5(1.5-1) (50)+ 1.5(1.5-1)(1.5-2)(0)+1.5(1.5-1)(1.5-2)(1.5-3)(-50)+1.5(1.5-1)(1.5-2)(1.5-3)(1.5-4)(0)
1X2 1x2x3 1x2x3x4 1x2x3x4x5
=50+150+18.75+0-1.17+0
= 217.57 or 218 workers.