Mathematical Methods for Physics and Engineering - Matematica.NET

PROBABILITY30.23 Mean = 4/π. Variance = 2 − (16/π 2 ). Probability that X exceeds its mean=1− (2/π)sin −1 (2/π) =0.561.30.25 Consider, separately, 0, 1 and ≥ 2errorsonapage.30.27 Show that the maximum occurs at x =(r − 1)/λ, and then use Stirling’s approximationto find the maximum value.30.29 Pr(k chicks hatching) = ∑ ∞n=kPo(n, λ) Bin(n, p).30.31 There is not much to choose between the schemes. In (a) the critical value ofthe standard variable is −2.5 and the average fine would be 15.5 euros. For(b) the corresponding figures are −1.0 and 15.9 euros. Scheme (c) is governedby a geometric distribution with p = q = 1 , and leads to an expected fine2of ∑ ∞n=1 4n(n − 1)( 1 2 )n . The sum can be evaluated by differentiating the result∑ ∞n=1 pn = p/(1 − p) with respect to p, and gives the expected fine as 16 euros.30.33 (a) [12!(0.5) 6 (0.3) 3 (0.2) 3 ]/(6!3!3!)=0.0624.30.35 You will need to establish the normalisation constant for the distribution (36),the common mean value (3/5) and the common standard deviation (3/10). Themarginal distributions are f(x) =3x(6x 2 − 8x + 3), and the same function of y.The covariance has the value −3/50, yielding a correlation of −2/3.30.37 A = 3/(24a 4 ); µ X = µ Y = 5a/8; σX 2 = σ2 Y = 73a2 /960; E[XY ] = 3a 2 /8;Cov[X,Y ]=−a 2 /64.30.39 (b) With the continuity correction Pr(x i ≥ 15) = 0.0334. The probability that atleast three are 15 or greater is 7.5 × 10 −4 .1220