Tiêu đề 2nd stats ch 04.pdf ✅

Chapter 4 Interpolation and Extrapolation One and Two Marks Questions and Answers Question 1. Define Interpolation and Extrapolation Answer: Interpolation is the technique of estimating the value dependent variable(Y) for any intermediate ) value of the independent variable(X). I Extrapolation is the technique of estimating the value of dependent variable (Y) any value of independent variable (X) which is outside the given series. Question 2. Mention different methods of interpolation. Answer: • Binomial expansion method and • Newton’s advancing difference method Question 3. What is interpolation? Answer: Interpolation is the technique of estimating the value dependent variable( Y) for any intermediate value of the independent variable(X). Question 4. What is meant by extrapolation? Answer: Extrapolation is the technique of estimating the value of dependent variable (Y) any value of independent variable (X) which is outside the given series. Question 5. Write down the assumptions of interpolation and extrapolation. Answer: • The assumptions made in interpolation and extrapolations are: • There are no sudden jumps in the values of dependent variable(Y) from one period to another(X). • The rate of change of figures (Y) from one period to another(X) is uniform.

Chapter 4 Interpolation and Extrapolation Example 1. Estimate the production for the year 2010 and 2012 from the following data. Solution: Let X and Y be year and production. Since the known values of ‘Y’ are four, the fourth order leading differences will be zero, i.e. (y- l)4 = 0 or Δ4 = 0 (y – 1 )4 = y4 – 4y3 + 6y2 – 4y1 + y0 = 0 (i) and the second equation can be obtained by increasing the suffixes of each term of ‘Y’ by one, keeping the. coefficients same; we get y5 – 4y4 + 6y3 – 4y2 + y1 = 0 ......(ii) From equation (i) 52 – 4y3 + 6 (28) – 4(18) + 8 = 0 52 – 4y3 + 168 -72 + 8 = 0 So, by simplifying, – 4y3 = – 156 ∴ y3 = 39 ie. Production for the year 2010 is 139 (tons). From (ii) y5 – 4y4 + 6y3 – 4y2 + y1 = 0; Here y3 = 39 y5 – 4 x 52 + 6 x (39) – 4 x 28 + 18 = 0 y5 – 208 + 234 – 112 + 18 = 0 ∴ y5 = 68 i.e. the production for the year 2012 is 68(tons) Example 2. Cost of living indices of a Banglore for some years are given below. Interpolate the missing index number for and Solution:
Chapter 4 Interpolation and Extrapolation Since the known values are 5, the fifth leading differences will be zero, i.e. Δ5 = 0 Δ5 = (y – 1 )5 = y5 – 5y4 +10y3 – 10y2 + 5y1 – y0 = 0 .............(i) And the second equation can be obtained by, increasing the suffixes of each term of’y’ by one, keeping the coefficients same; ie. (y- 1)5 = y6 – 5y5 + 10y4 – 10y3+ 5y2 – y, = 0 .......... (ii) determine the value of Y2 from equation (i) 162 – 5(142) +10 (128) – 10y2 + 5(112) – 100 = 0 by simplifying, -10y2 = -1192 y2 = 119.2 Hence, the missing Index number for 1990 is 119.2. From (ii) y6 – 5(162) + 10(142) – 10(128) + 5(119.2) – 112 = 0 Here y2 =119.2 y6 – 810 + 1420 – 1280 + 596- 112 = 0. ∴ y6 = 186 Hence the cost of living index number for the year 2010 is 186. Example 3. Estimate the number of children in the ages 16 and 22 years from the following data: Solution: Let X and Y be age and no. of children. values of y are given, then 5th leading difference will be zero. Δ5 0 = (y – 1 )5 = y5 – 5y4 + 10y3 – 10y2 + 5y1 – y0 = 0 .... (i) and the second equation can be obtained as, increasing the suffixes of each term of ‘Y’ by one, keeping the coefficients same; we get : Δ5 = (y – 1)5 = y6 – 5y5+ 10y4 – 10y3 + 5y2 – y1 =0 .......(ii) From equation (i) y5 – 5y4 + 10y3 – 10y2+ 5y1 – y0 =0 ie„ 98 – 5(88) + 10y3 – 10(65) + 5(55) – 50 = 0 98 – 440 + 10y3 – 650 + 275 – 50 = 0 So, by simplifying, 10y3 – 767 = 0