a) X and Y are independent random variables such that E(X)=20, Var(X) = 4, E(Y)=8 and Var(Y) = 2. Find:
i) E(X+3Y) and ii) Var(X-3Y)
b) Sixty-four percent of all airplanes arriving at an airport are late. Using the binomial formula, find the probability that in a random sample of seven airplanes, exactly four will arrive late.
c) Twenty-eight percent of adults contribute to charitable agencies on a regular basis. Let x be the number of adults in a random sample of 18 adults contribute to charitable agencies on a regular basis. Find the mean and the standard deviation of the probability distribution of .
btw this aint my homework. for reference.
i) E(X+3Y) and ii) Var(X-3Y)
b) Sixty-four percent of all airplanes arriving at an airport are late. Using the binomial formula, find the probability that in a random sample of seven airplanes, exactly four will arrive late.
c) Twenty-eight percent of adults contribute to charitable agencies on a regular basis. Let x be the number of adults in a random sample of 18 adults contribute to charitable agencies on a regular basis. Find the mean and the standard deviation of the probability distribution of .
btw this aint my homework. for reference.
-
A) E[X+3Y] = 44
Var [X+3Y] = 10
E[X-3Y] = -4
Var [X-3Y] = 10
B)7C4 .64^4 x.36^3 = 0.274
C) N = 18 P = .28
Binomial to Normal transformations.
Mean = NP Var = NP(1-P)
Mean = 5.04 Var = 3.6288
SD = 1.905
So Mean = 5.04 SD = 1.905
Var [X+3Y] = 10
E[X-3Y] = -4
Var [X-3Y] = 10
B)7C4 .64^4 x.36^3 = 0.274
C) N = 18 P = .28
Binomial to Normal transformations.
Mean = NP Var = NP(1-P)
Mean = 5.04 Var = 3.6288
SD = 1.905
So Mean = 5.04 SD = 1.905