Binomial (n, p): p(i) = ( )p'(1 - p)"-, for i = 0,1,...,n Mean = np: variance = np (1 - p): mgf is M(t) = (pet + 1 - p) Geometric(p): p(i) = (1 - p)-1p, for i = 1,2, Mean = 1/p; variance = (1 - p/p2; mgf is M(t) = (pet) [1 -(1-p) et]
Poisson(2): p(i) = e = * for i= 0, 1, 2,..., Mean and variance = a; mgf is M(t) = exp {2(et – 1)} p"(1 - p)"-" for n= r, r + 1,... Mean = 1/p:
NegBinomial (rp): p(n) = variance =1 (1-P)/p2
Uniform (a, b): f(x)= Mean is (a+b)/2, variance is (b-a)/12
mgf is M(t) = [eth - eta] / [t(b-a)].
Normal (u,02): f(x)= -(-u)°/20°-00 mgf is M(t) = exp {ut + (c2t2)/2}
Exponential(a): F(a) = 1-e-a, a 2 0; f(x) = le-x if x > 0. Mean is 1/2, variance is 1/72. mgf is M(t) =M(-t)
Gamma (n,): f(x) = le-AX(2x)n-1 /(n-1)!, if x 2 0. Mean is n/A, variance is n/a2.
mgf is M(t) = [Ma – t)]".
Markov's Inequality: P/X2 a} SPATE
Chebyshev's Inequality: P{ |X – E(X)/2 k} sp, and P{]X – E(X)/ 2 ko}s
Cov(X, Y ) = E(XY)-E(X)E(Y); Var(x) = E(X2)-[E(X)]2
A video can be downloaded from the web and moved to s student's mobile device in 12 minutes, on average. If the standard deviation of the time required is 2 minutes, then give a bound on the probability that 9 to 15 minutes are needed.