SOLUTION: Let X be N(100, 10) and Y be N(10, 5). What is P(X − 4Y > 40) I think 4Y is N(40,20) but that gives me a negative value for the standard deviation of X-4Y.

Question 1171776: Let X be N(100, 10) and Y be N(10, 5). What is P(X − 4Y > 40)
I think 4Y is N(40,20) but that gives me a negative value for the standard deviation of X-4Y.
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Need to assume these are independent.
x is N~ (100,10) using variance
4y is N~(40, 80) because the constant*the variance is c^2*var(N) or 16*5=80
the distribution would be (60, 90), because when the two are combined, the variances are added, even when one distribution is subtracted from another. There is more variability when there is combining.
The sd is 9.486
z > (60-40)/9.486 or > 2.108
probability is 0.0175
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