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Karnataka 2nd PUC Statistics Question Bank Chapter 4 Interpolation and Extrapolation
2nd PUC Statistics Interpolation and Extrapolation Exercise Problems
Question 1.
Define Interpolation.
Answer:
It is the technique of estimating the value of the dependent variable (Y) for any intermediate value of the independent variable (X)
Question 2.
Define extrapolation.
Answer:
It is the technique of estimating the value of dependent variable (Y) for any value of the independent variable (X) which is outside the range of the given series
Question 3.
Differentiate between interpolation & Extrapolation.
Answer:
Interpolation |
Extrapolation |
It is the process of calculating the value of the dependent variable for any intermediate value of the independent variable. | It is the process of calculating the value of the dependent variable for any value of the independent variable which is outside the range of the given series. |
Question 4.
Write down the assumptions of interpolation and extrapolation?
Answer:
Assumptions of Interpolation & Extrapolation:
- There are no sudden change in the values of dependent variable from one period to another
- There is a sort of uniformity in the rise or fall of the values of the dependent variable
- There will be no Consecutive missing values in the series
Question 5.
Mention different methods of interpolation.
Answer:
- Binomial expansion method
- Newton’s advancing difference method
Solve for x means find the value of x that would make the equation you see true.
Question 6.
Write down the conditions for application of Binomial expansion method of interpolation and extrapolation.
Answer:
1. The values should be in arithmetic progression, These should be a common difference between the values of the in depend variable.
2. The value of ‘x’ for which the value of ‘y’ is to be interpolated must be one of the values of x.
Question 7.
Write down the formula of Binomial expansion method for 4 values of y
Answer:
If known values are 4
(y – 1)^{4} = y_{4} – 4y_{3} + 6y_{2} – 4y_{1} + y_{o} = 0
Question 8.
Expand (y – 1)^{5}
Answer:.
(y – 1)^{5} = y_{5} – 5y_{4} + 10y_{2} – 10y_{2} + 5y_{1} – y_{0}= 0
Question 9.
Newton’s method of advancing differences.
Answer:
If the ‘x’ series is in descending order. Convert them in an ascending order and then apply Newton formula.
Question 10.
Write down Newton’s formula for interpolation
Answer:
Newton’s Formula
Question 11.
Write the formula to find the value of’x’ in finding the missing value of’y’ using Newton’s method of interpolation
Answer:
Question 12.
From the following data relating to production. Estimate the production for 2006 and 2010
Answer:
since known values are 4
∴ (y – 1) ^{4} = 0
y_{4} – 4y_{3} + 6y_{2} – 4y_{1} + y_{0} = 0 …… (1)
The second equation can be calculated by increasing the suffixes of each term of ‘y’ by one and let the coefficients same.
We get
y_{5} – 4y_{4} + 6y_{3} – 4y_{2} + y_{1} =0 …..(2)
Substitute y values in equation (1)
y_{4} – 4y_{3} + 6y_{2} – 4y_{1} + y_{0} = 0
→ 26 – 4(15) + 6(10) – 4y_{1} + 5 = 0
-4y_{1} + 31 = 0
y_{1} = \(\frac{-31}{-4}\) = 7.75
Production in 2006 is 7.7 5 tonnes
Now substitute y values in equation (2)
y_{5} – 4y_{4} + 6y_{3} – 4y_{2} + y_{1} = 0
y_{5} – 4(26) + 6(15) – 4(10) + 7.75 = 0
y_{5} – 104 + 90 – 40 + 7.75 = 0
y_{5} = 46.25
Production for the year 2010 is 46.25 tonnes
Question 13.
Estimate sales of a business concern for the year 2011
Answer:
Since Know the values are 4
∴ (y – 1)^{4} = 0
y_{4} – 4y_{1} + 6y_{2} – 4y_{1} + y_{0} = 0 …. (1)
The second equation can be calculated by increasing the suffixes of each term of ‘y’ by 1, and let the coefficients same.
We get
y_{5} – 4y_{4} + 6y_{3} – 4y_{2} + y_{1} = 0 ….. (2)
Substitute ‘y’ values in equation (1)
y_{4} – 4y_{3} + 6y_{2} – 4y_{1} + y_{0} = 0
65 – 4 (38) + 6 (25) – 4y_{1} + 13 = 0
65 – 152 + 150 – 4y_{1} + 13 = 0
76 – 4y_{1} = 0
y_{1} = 76/4
y_{1} =19
Now substitute ‘y’ values in equation (2)
y_{5} – 4y_{4 }+ 6y_{3} – 4y_{2} + y_{1} = 0
y_{5} – 4(65) + 6 (38) – 4(25) + 19 = 0
y_{5} = -260 + 228 – 100 + 19 = 0
y_{5} – 113 = 0
y_{5} = 113
Question 14.
Interpolate the business in 1982 and extrapolation for the following year 1984 from the following data
Answer:
Since known values are 5
∴ (y – 1)^{5} = 0
y_{5} – 5y_{4} + 10y_{3} – 10y_{2} + 5y_{1} – y_{0} = 0 …. (1)
The second equation can be calculated by increasing the suffixes of each term of ‘y’ by one and let the coefficient same.
The second equation will be
y_{6} – 5y_{5} + 10y_{4} – 10y_{3} + 5y_{2} – y_{1} = 0 … (2)
Substitute y values in equation (1)
780 – 5y_{4} + 10(365) – 10(235) + 5(150) – 80 = 0
-5y_{4} + 2750 = 0
y_{4} = \(\frac{2750}{5}\) ⇒ y_{4} = 550
The business in the year 1982 is 550 Lakhs Substitute y values in equation (2)
y_{6} – 5(780) + 10(550) – 10(365) + 5(235) – 150 = 0
y_{6} – 3900 + 5500 – 3650 + 1175 – 150 = 0
⇒ y_{6} – 1025 = 0 ⇒ y_{6} =1025
The business for the year 1984 is 1025 Lakhs.
Question 15.
Estimate the production for the years 2000 and 2010 with the help of the following table
Answer:
Since known values are 5
∴ (y – 1)^{5} = 0
y_{5} – 5y_{4} + 10y_{3} – 10y_{2} + 5y_{1} – y_{0} = 0 … (1)
The second equation can be calculated by increasing the suffixes of each term of ‘y’ by one and let the coefficient same.
The second equation will be
y_{6} – 5y_{5} + 10y_{4} – 10y_{3} + 5y_{2} – y_{1} = 0 …. (2)
Substitute y values in equation (1)
40 – 5y_{4} + 10(26) – 10(19) + 5(12) – 5 = 0
40 – 5y_{4} + 260 – 190 + 60 – 5 = 0
-5y_{4} + 165 = 0
y_{4} = \(\frac{165}{5}\) ⇒ y<_{4} = 33
Substitute y values in equation (2)
y_{6} – 5(40) + 10(33) – 10(26) + 5(19) – 12 = 0
y_{6} – 200 + 330 – 260 + 95 – 12 = 0
y_{6} = 47
Question 16.
Use Newton’s method to find the number of employees whose wages ₹ 600 per day
Answer:
⇒ 36 – 7.5 – 0.750 – 0.4375 – 0.32813
Y_{600} = 26.98 ⇒ Y_{600} = 27
27 employees earning ₹ 600.
Question 17.
The following table shows the expectation of life at different ages. Find the expectation of life at age 26.
Answer:
x = \(\frac{26-15}{5}=\frac{11}{5}\)
⇒ 30 – 22 – 1.32 – 0.176 – 0.1408
⇒ 26 years
The expectation of life at 26 years is 26 years.
Question 18.
The table shows the expectation of life at different ages. Find the expectation of life at the age 26.
Answer:
⇒ 30 – 22 – 1.32 – 0.176 – 0.1408 ⇒ 26.1632
Expectation of life at age 26 is 26.1632
Question 19.
Find the number of persons below the age of 70 years from the following data
Answer:
⇒ 333 + 400 – 46.875 – 13.4375 – 4.275
= 668.4297 ⇒ 668
There are 668 people below age of 70 years.
2nd PUC Statistics Interpolation and Extrapolation Pratical Assignments
Question 1.
Interpolate and extrapolate the production for the years 2000 and 2010 with the help of following table:
Answer:
Let x and y be the year and production
The number of known values of ‘y’ : n = 6. So, the Binomial expansion (y – 1)^{6} = 0.
i.e., y_{6} – 6y_{5} + 15y_{4} – 20y_{3} + 15y_{2} – 6y_{1} + y_{0} = 0 …(1)
33 – 6y_{5} + 15 (18) – 20(15) + 15(13) – 6 (11) + 10 =0
-6y_{5} + 142 = 0; – 6y_{5} = -142; y_{5} = \(\frac{142}{6}\) = 23.67
Jons is the production for the year
For extrapolation the suffixes of ‘y’ in equation (1) is increased by ‘1’
y_{7} – 6y_{6} + 15y_{5} – 20y_{4} + 15y_{3} – 6y_{2} + 6y_{1} =0
y_{7} – 6(33) + 15(23.67) – 20(18) + 15(15) – 6(13) + 11 = 0
y_{7} – 198 + 355.05 – 360 + 225 – 78 + 11 = 0
y_{7} – 44.95 = 0
∴ y_{7} = 44.95 Jons is the production for the year 2010.
Question 2.
Interpolate and extrapolate the production for the years 1989 and 1991 with the help of the following table.
Answer:
Let x and y be the year and production.
The number of known values of ‘y’ ; n = 4, the Binomial expansion (y – 1)^{4} = 0
i.e., y_{4} – 4y_{3} + 6y_{2} – 4y_{1} + y_{0} = 0 …(1)
135 – 4y_{3} + 6(126) – 4(122) +1 20 = 0
-4y_{3} + 523 = 0; y_{3} = \(\frac{523}{4}\) = 130.75
is the production for the year 1989.
For extrapolation the suffixes of y has increased by in equation (1)
y_{5} – 4y_{4} + 6y_{3} – 4y_{2} + y_{1} = 0
y_{5} – 4(135) + 6(130.75) – 4(126) + 122 = 0
y_{5} – 540 + 784.5 – 504 + 122 = 0
y_{5} – 137.5 = 0; y_{5} = 137.5
is the production for the year 1991.
Question 3.
Interpolate and extrapolate the production for the years 1982 and 1985 from the following data.
Answer:
Let x and y be the year and production
Number of known values of ‘y’ n = 5 and so, the Binomial expansion (y – 1)^{5} = 0
i.e., y_{5} – 5y_{4} + 10y_{3} – 10y_{2} + 5y_{1} – y_{0} = 0 …(1)
780 – 5(525) + 10y_{3} – 10 (150) + 5 (120) – 100 = 0
780 – 2625 + 10y_{3} – 1500 + 600 – 100 = 0
10y_{3} – 2845 = 0;
∴ y_{3} = \(\frac{2845}{10}\) = 284.5
Tons is the production for the year 1982.
For extrapolation suffixes of ‘y’ has increased by ‘1’ in equation (1). We get:
y_{6} – 5y_{5} + 10y_{4} – 10y_{3} + 5y_{2} – y_{1} =0.
y_{6} – 5(780) + 10 (525) y_{6} – 865 = 0 + 10(284.5) + 5(150) – 120 = 0
y_{6} – 3900 + 5250 – 2845 + 750 – 120 = 0;
∴ y_{6} = 865 Tons is the production for the year 1985
Question 4.
Estimation the production for the years 1965 and 1975 with the help of the following table.
Answer:
Let x and y be the year and production.
The number of known values of ‘y’ n = 4 and so, the Binomial expansion (y – 1)^{4} = 0
ie., y_{4} – 4y_{3} + 6y_{2} – 4y_{1} + y_{0} = 0 …(1)
210 – 4y_{3} + 6(150) – 4(120) + 100 = 0
210 – 4y_{3} + 900 – 480 + 100 = 0
-4y_{3} + 730 = 0: 4y_{3} = -730
∴ y_{3} = \(\frac { 730 }{ 4 }\) = 182.5
Tons is the production for the year 1965.
For extrapolation suffixes of y has increased by 1 in equation (1) we get:
y_{5} – 4y_{4} + 6y_{3} – 4y_{2} + y_{1} = 0
y_{5} – 4(210) + 6(182.5) – 4(150) + 120 = 0
y_{5} – 840 + 1095 – 600 + 120 = 0 .
y_{5} – 225 = 0
∴ y_{5} = 225 Tons is the production for the year 1975.
Question 5.
The anual sales of a company are given below. Interpolate and extrapolate the sales for the year 2002 and for the year 2005.
Answer:
Let x and y be the year and Sales
The number of known values of y n = 4 and so, the Binomial expansion (y – 1)^{4} = 0
i.e., y_{4} – 4y_{3} + 6y_{2} – 4y_{1} + y_{o} = 0 …(1)
282 – 4(238) + 6y_{2} – 4(163) +125 =0
282 – 952 + 6y_{2} – 652 + 125 = 0
6y_{2} – 1197 = 0,
∴ y_{2} = \(\frac { 1197 }{ 6 }\) = 199.5 is the sales for the year 2002.
For Extrapolation, the suffixes of ‘y’ has increased by T in equation (1) We get
y_{5} – 4y_{4} + 6y_{2} – 4y_{2} + y_{1} = 0
y_{5} – 4(282) + 6(238) – 4(199.5) + 163 = 0
y_{5} – 1128 + 1428 – 798 + 163 = 0
i.e., y_{5} – 335 = 0
∴ y_{5} = 335 is the sales for the year 2005.
Question 6.
From the following data, obtain the value of y when x = 9 by using Newton’s forward difference method
Answer:
Here the value of ‘X’ for which the value of ‘y’ to be interpolate is not in the range of ‘x’ so use, Newtons’ method.
The number of known values of ‘y’ n =5, so prepare leading difference table up to Δ^{4}.
Foot note: – 1 – (- 1) = 0
x = \(\frac{9-3}{4}\) = 1.5;
(x – 1) = 1.5 – 1 = 0.5; (x – 2) = 1.5 – 2 = (-0.5)
(x – 3) = (1.5 – 3) = (-1.5).
The Newtons equation of Interpolation is :
= 42 + 1.5 + 1.125 + 0.0625 + 0
∴ y_{0} = 44.6875
Question 7.
Calculate y when x is 12 from the following by interpolation method:
Answer:
The no of Known values of ‘y’ n = 5,, so preparing leading different table up to Δ^{4}
Foot note: – 1 – (- 3) = -1 + 3 = 2; 0 – 2 = -2.
x = \(\frac{12-10}{10}\) = 0.2;
(x -1) = 0.2 – 1 = -0.8; (x – 2) = 0.2 – 2 = (-1.8);
(x – 3) = (0.2 – 3) = (-2.8)
The Newtons equation of Interpolation is:
Y_{12} = 23 + 1.4 + 0.24 + 0.096 + 0.0672
∴ y_{12} = 24.8032
Question 8.
Following is data regarding annual net life insurance premium. Using Newtons method estimate the premium at the age of 26 years.
Answer:
Let x and y be the age and annual net premium.
The number of known values of ‘y’ n = 4, so prepare leading differences up to Δ^{3}
x = \(\frac{26-20}{5}\) = 1.2
(x – 1) = 1.2 – 1 = 0.2; (x – 2) = 1.2 – 2 = (-0.8);
The Newton’s equation of Interpolation is:
Y_{26} = 1426 + 186 + 40.2 – 0 = ₹ 1616.2
is the premium for the age 26 year.
Question 9.
Below are given the wages earned by workers per day in certain factory. Calculate the number of workers earning up to Rs. 750 per day.
Answer:
Let ‘x’ and ‘y’ be the wages per day and number of workers. The number of known values of y are n = 6, so prepare leading differences up to Δ^{5}.
x = \(\frac{750-500}{100}\) = 2.5;
(x – 1) = 2.5 – 1 = 1.5; (x – 2) = 2.5 – 2 = 0.5, (x – 3) = 2.5 – 3 = (-0.5), (x – 4) = 2.5 – 4 = (-1.5)
The Newton’s formula of Interpolations is :
Y_{750} = 50 + 250 + 93.75 + 0 + 1.1953 + 0
= 394.9453 = 395 Workers.
Question 10.
From the following data estimate the number of persons earning wages below Rs. 90 per day.
Answer:
Let x and y be the wages per day and no. of persons. Here class intervals are converted into less than / below type, because ‘y’ value to be interpolate is below 90 then prepare the leading difference table upto Δ^{4}, since there are n = 5 known values of y.
Here
x = \(\frac{90-40}{20}\) = 2.5
(x – 1) = 2.5 – 1 = 1.5; (x – 2) = 2.5 – 2 = 0.5,
(x – 3) = 2.5 – 3 = (-0.5)
The Newton’s formula of Interpolation Is:
= 500 + 600 – 75 – 6.25 – 1.5625
= 1017.1875 persons.