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A textbook of Computer Based Numerical and Statiscal Techniques part 20

A textbook of Computer Based Numerical and Statiscal Techniques part 20. By joining statistical analysis with computer-based numerical methods, this book bridges the gap between theory and practice with software-based examples, flow charts, and applications. Designed for engineering students as well as practicing engineers and scientists, the book has numerous examples with in-text solutions. | 176 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Sol. The difference table for the given date is as follows x y vy v2 y v3 y 3 4.8 3.6 4 8.4 6.1 2.5 0.5 5 14.5 9.1 3 0.5 6 23.6 12.6 3.5 0.5 7 36.2 16.6 4 0.5 8 52.8 21.1 4.5 9 73.9 To obtain first term we use Newton s forward interpolation formula Here a 3 h 1 x 1 u -2 Hence we have 1 y3 uAy3 u u-1 A2 y3 u u -1 u - 2 A3 y3 3 3 2 3 3 3 On putting the subsequent values we get 1 4.8 -2 x 3.6 2 -3 2.5 2 3 4 0.5 3.1 2 6 Similarly to obtain tenth term we use Newton s backward interpolation formula. So a nh 9 h 1 a nh uh 10 u 1 u u 1 V72 u u 1 u 2 3 f 10 y9 uVy9 v y9 3 ----------------V y9 73.9 21.1 4.5 0.5 100 3 Example 4. Find the value 0 an annuity at 5- given the following table Rate 4 41 2 5 51 2 6 Annuity Value 172.2903 162.888 153.7245 145.3375 137.6483 INTERPOLATION WITH EQUAL INTERVAL m Sol. Difference table Rate Annuity Values V V2 V3 V4 4 172.2903 -9.4014 41 2 162.888 -9.1644 0.237 0.537 5 153.7245 -8.387 0.774 -0.0762 -0.6132 51 2 145.3375 -7.6892 0.6978 6 137.6483 .3 43 . 1 x 5 a 6 n x 8 8 2 y - y 6 -1.25 Vy 6 14 25 25 V2y 6 H-25 25 75 v3y 6 . y 2 3 - 137.6483 -1.25 -7.6892 -1-25X- .25 0.6978 -1.25 -0.25 0.75 00 -1.25 -0.25 0.75 1.75 3 4 147.2251 Approx. Example 5. In an examination the number of candidates who obtained marks between certain limits are as follows Marks 0 -19 20 - 39 40 - 59 60 - 79 80 - 99 No. of candidates 41 62 65 50 17 Find no. of candidates who obtained fewer than 70 marks. Sol. First we form the difference table. Marks less than x No.of candidates y V V2 V3 V4 19 41 62 39 103 65 3 -18 59 168 50 -15 -18 0 79 218 17 -33 99 235 178 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Here we have h 20 a 99 u 70 - 99 20 -1.45 Now on applying Newton s backward difference formula we get 70 f a uVf a u u 1 V2 f a u u 2 V3 f a u u 1 u 2 u 3 V4 f a 235 -1.45X17 -1.45 -0.45 x -33 -1.45 -0.45 0.55 x -18 2 6 235 - 24.65 - 10.76625 - 1.076625 235 - 36.492875 198.507 s 198 . Total no. of .