PROBABILITY
30.23 Mean = 4/π.Variance=2−(16/π^2 ). Probability thatXexceeds 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 atx=(r−1)/λ, and then use Stirling’s approx-
imation to find the maximum value.
30.29 Pr(kchicks hatching) =
∑∞
n=kPo(n, λ)Bin(n, p).
30.31 There is not much to choose between the schemes. In (a) the critical value of
the 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 governed
by a geometric distribution withp=q=^12 , and leads to an expected fine
of
∑∞
n=1^4 n(n−1)(1
2 )n. The sum can be evaluated bydifferentiating the result
∑∞
n=1pn=p/(1−p) with respect top, 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). The
marginal distributions aref(x)=3x(6x^2 − 8 x+ 3), and the same function ofy.
The covariance has the value− 3 /50, yielding a correlation of− 2 /3.
30.37 A=3/(24a^4 );μX=μY =5a/8;σ^2 X=σ^2 Y =73a^2 /960;E[XY]=3a^2 /8;
Cov[X, Y]=−a^2 /64.
30.39 (b) With the continuity correction Pr(xi≥15) = 0.0334. The probability that at
least three are 15 or greater is 7. 5 × 10 −^4.