Exercise 1.52 (* Mathematica 7.0.0 *)

(a) Standard deviation of N(2)

StandardDeviation[PoissonDistribution[15×2.]]

(b) P[N (0.5) ≤5]

CDF[PoissonDistribution[15×0.5],5]

(c) P[(1/6)<(Subscript[S, 3])≤(1/3)]=P[N ((1/3)) ≥3]-P[N ((1/6)) ≥3]

CDF[PoissonDistribution[15×(1./6)],2]-CDF[PoissonDistribution[15×(1./3)],2]

CDF[GammaDistribution[3,(1/15)],(1./3)]-CDF[GammaDistribution[3,(1/15)],(1./6)]

(d) P[N (1) ≤10]

CDF[PoissonDistribution[15×1.],10]

(e) P((Subscript[S, 3])≤0.1)
(* In Mathematica, the 2nd. parameter of the Gamma distribution nis the reciprocal of λ *)

CDF[GammaDistribution[2,(1/15)],0.1]

(f) smallest λ: P[N(1)≥5]≥0.95

Plot[1-CDF[PoissonDistribution[λ],4],{λ,0.001,15}]
FindRoot[1-CDF[PoissonDistribution[λ],4]==0.95,{λ,0.001}]