6.43 a. From Ex. 6.41, Y has a normal distribution with mean u and variance oz/n.
b. For the given values, Y has a normal distribution with variance oz/n = 16/25. Thus,
the standard deviation is 4/5 so that
P(|Yu| s 1) =P(1 s 17: 1) =P(1.25 52: 1.25) = .7
6.54 a. The mgf for U = 2:1): is "v (t) = ear2" , which is recognized as the mgf for a
Poisson w/ mean 2: 7*:-
b. This is similar to 6.52. The distribution is binomial with m trials and p = $1115.
c. Following the same steps as in part (1 of Ex. 6.53, it
5.75 The marginal distributions for Y1 and Y2 are exponential with I3 = 1. And it was found in
Ex. 5.51 that Y1 and Y2 are independent. So:
a. E(Y1+ Y2) = E01) + E(Y2) = 2, V(Y1+ Y2) = V(Y1) + V(Y2) = 2.
b. Parl Y2 > 3) = P(Yl > 3 + Y2) =f fe'"aj/ldy2 =(1
a. The marginal probability mction is given in the table below.
b. No, evaluating binomial probabilities with n = 3, p = 1/3 yields the same result.
a. The marginal distribution of Y1 is hypergeometn'c with N = 9, n = 3, and r