3. (20 points) Let X and Y be two independent random variables with distribution
ID: 2979009 • Letter: 3
Question
3. (20 points) Let X and Y be two independent random variables with distributions respectively as X- Normal(4, 25) and Y- Normal(2, 9) (a) (2 points) Calculate P(1 < X <=7). (b) (5 points) Define the random variable V = X -5Y . Calculate P(V > 0). (c) (6 points) Let Y1, Y2, ..., Y5 be five independent random draws from the distribution of Y. Define S5 =sum of all Yi on i =1 to i=5 and W = X -S5. Calculate P(W > 0). Explain why or why not you get a different answer from part (b). (d) (7 points) Now, let X1, X2,..., Xn be n independent random draws from the distribution of X. Define the random sum Sn = X1 + X2 + ... + Xn = sum of all Xi on i=1 to i=n What is the (exact) distribution of Sn? Specify its mean and variance. Calculate the smallest value of n such that P(Sn > 0) > 0.999.Explanation / Answer
Assume that Z is the standard normal distribution N(0,1):
a)
P(1<X<7) = P((1-4)/5 < Z < (7-4)/5) = P(-0.6 < Z < 0.6) =
2P(Z<0.6) - 1 = 2*0.7257-1 = 0.4514
b)
E[V] = E[X]-5E[Y] = 4-5*2 = -6
Var[V] = Var[X]+25Var[V] = 25 + 25*9 = 250 -> STD[V] = sqrt(250) = 15.8
-> P(V > 0) = P(Z > (0-(-6))/15.8) = P(Z > 0.38) = 1 -0.6480 = 0.352
c)
E[W] = 4-5*2 = -6
Var[W] = 25+9+9+9+9+9 = 70 -> STD[W] = sqrt(70) = 8.37
-> P(W > 0) = P(Z > (0-(-6)/8.37) = P(Z > 0.72) = 1-0.7642 = 0.2358
In part b we get one sample and multiply it by 5, but in part (c) we take 5 different samples, so the standard deviation in this case will be less than part (b).
d)
Sn is a normal distribution with mean = 4n and variance = 25n : N(4n,25n)
variance = 25n -> standard deviation = 5sqrt(n)
P(Sn > 0) =
P(Z > (0 - 4n)/5sqrt(n)) =
P(Z > -0.8sqrt(n)) > 0.999 -> P(Z < 0.8sqrt(n)) > 0.999
(from standard normal table) -> 0.8sqrt(n) > 3.08
-> n > (3.08/0.8)^2
-> smallest value for n = 15
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