A textbook of Computer Based Numerical and Statiscal Techniques part 23 pot
A textbook of Computer Based Numerical and Statiscal Techniques part 23 pot
... 206 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 7. Apply Stirling’s formula to find a polynomial of degree four which takes the values of y as given below: x y −− 1234 5 11111 [Ans. 42 28 1 33 uu−+ ] 8. ... 1 – .25 = 0.75 214 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 7. Apply Bessel’s interpolation formula, show that tan160° = 0.2867, Given that:...
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A textbook of Computer Based Numerical and Statiscal Techniques part 7 potx
... 1.4036 0.10203 = Since, x 3 and x 4 are approximately the same upto four places of decimal, hence the required root of the given equation is 1.4036. 48 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Using ... between 1.6866 and 2. 50 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES and f(3) = (3) 3 – 2(3) – 5 = 16 Therefore, a root lies betw...
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A textbook of Computer Based Numerical and Statiscal Techniques part 28 potx
... = 0.86742. Ans. Example 4. A switching path between parallel railroad tracks is to be a cubic polynomial joining positions (0,0) and (4,2) and tangent to the lines y = 0 and y = 2 as shown in ... shown in the figure. Apply Hermite’s interpolation formula to obtain this polynomial Sol. Since tangents are parallel to X-axis, y′ = 0 in both the cases. ∴ We have the table of values....
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A textbook of Computer Based Numerical and Statiscal Techniques part 33 pot
... a table, but no mathematical relationship between them is known as is often the case with experimental data. 6.4 GENERAL QUADRATURE FORMULA Let ()yfx= be a function, where y takes the values ... problem can be solved by Newton’s divided difference formula, we are giving here, as an alternative, Lagrange’s method. Lagrange’s polynomial, in this case, is given by NUMERICAL DIFFERENTIAT...
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A textbook of Computer Based Numerical and Statiscal Techniques part 38 potx
... 1.3997 3 =+ ++ = 364 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES which is same as the predicted value. ∴ 60.30 1.3997 yy== and 6 1.6997 y ′ = The result can be put in the tabular form x 0.20 ... differential equation () , dy yfxy dx ′ == by this method we first obtain the approximate value of 1n y + by Predictor formula and then improve the value of 1n y + by me...
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A textbook of Computer Based Numerical and Statiscal Techniques part 39 potx
... value y i and the true value y(x i ) at any stage is known as the total error. The total error at any stage is comprised of truncation error and round-off error. The most important aspect of ... DIFFERENTIAL EQUATION 371 Inherent stability is determined by the mathematical formulations of the problem and is dependent on the eigen values of Jacobian Matrix of the differential...
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A textbook of Computer Based Numerical and Statiscal Techniques part 42 potx
... method of least squares a parabola of the type 2 yabx=+ . Sol. Error of estimate for i th point (,) ii xy is () 2 ii i eyabx =−− By the principle of least squares, the values of a and b are ... = ∑ x 2 300 = ∑ xY ∑ Substituting the values of x ∑ , etc. and calculated by means of above table in the normal equations. We get, 5.7536 = 5A + 34B and 40.8862 = 3 4A...
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A textbook of Computer Based Numerical and Statiscal Techniques part 43 pot
... n ≤≤ and 1}jn≤≤ be a bivariate distribution. If we plot the corresponding values of x and ,y taking the values of x along x-axis and the values of y along y-axis, we obtain a collection of ... [Ans. 2 13 22 yx=+ ] 410 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 3. Prove that arithmetic mean of the coefficient of regression is greater t...
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A textbook of Computer Based Numerical and Statiscal Techniques part 54 potx
... that the sailors are on the average taller than soldiers? [Ans. Highly significant] 22. The yield of wheat in a random sample of 1000 farms in a certain area has a S.D. of 192 kg. Another random ... population with mean height 162.5 cms and standard deviation 4.5 cms? [Ans. H 0 : Accepted] 13. A random sample of 200 measurements from a large population gave a mean value...
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A textbook of Computer Based Numerical and Statiscal Techniques part 55 potx
... E(c)= accd N ++ bgbg and E(d)= bdcd N ++ bgbg abab cdcd ac bd N + + ++ ∴ χ 2 = aEa Ea bEb Eb cEc Ec dEd Ed –––– af af af af af af af af 2222 +++ (2) a – E (a) =a – abac N ++ bgbg = a a b c d a ac ... the hypothesis that male and female births are equally probable ? Sol. Let us set up the null hypothesis that the data are consistent with the hypothesis of equal probability for...
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