Evaluating the probability of a Distribution

Question

$X_1 ∼ N(µ = 2, σ = 2), X_2 ∼ N(µ = 1, σ = 4), X_3 ∼ N(µ = −4, σ= 3):$ $X_1, X_2,$ and $X_3$ be independent and $Y = (X_1 + 2X_2 + X_3)^2.$

Determine $P(Y > E(Y)).$

My solution: I got the value of $E(Y) = 77$ and $Var(Y) = 11858.$

So, this means $Y \sim N(77,11858).$ (Is this right?) and $P(Y> E(Y))$ becomes $P(Y>77).$

Answer 1

Comment: It seems you are now on the right track. Consider the following simulation, which should be sufficiently accurate to see if any results are wrong. With sample sizes of a million, most means, standard deviations, and probabilities should be accurate to 2 or 3 significant digits. (Variances have squared units and so would not be accurate to so many digits.) Because $Y$ is normal with mean $0,$ the distribution of $Y^2$ is a multiple of a chi-squared distribution.

set.seed(2021)
m = 10^6
x1 = rnorm(m, 2, 2)
x2 = rnorm(m, 1, 4)
x3 = rnorm(m, -4, 3)
s = x1 + 2*x2 + x3
mean(s); sd(s)
[1] 0.007333673   # aprx E(S)
[1] 8.775375      # aprx SD(S)
y = s^2
mu.y = mean(y);  mu.y
[1] 77.00718      # aprx E(Y)
sd(y)
[1] 108.9997      # aprx SD(Y)
mean(y > mu.y)
[1]  0.317496     # aprx P[Y > E(Y)]

par(mfrow=c(1,2))
 hist(s, prob=T, col="skyblue2")
  curve(dnorm(x, mean(s), sd(s)), add=T, col="orange")
 hist(y, prob=T, col="skyblue2")
  abline(v = mu.y, col="orange", lwd=2)
par(mfrow=c(1,1))