Suppose that , , and are independent random variables, each with a standard Normal (Z) distribution.

Question

Compute ( + < − ).

The normal distribution is a topic I have just started and I am not sure how to apply the CDF of the distribution to a problem such as this. I know, for example, how to calculate a normal distribution given a mean, standard deviation, and a condition. This problem, however, I cannot seem to find a starting point on so I was hoping for some guidance as to where to begin the problem.

A key fact is that linear combinations of independent normally distributed random variables are again normally distributed. You have given no context in the body of the Question to suggest whether you are aware of this circumstance, but it leads to the rather elegant solution already provided. — hardmath, Mar 10 '19 at 07:40
Ah ok, the solution makes much more sense now. Thank you for the link. — hmtkd, Mar 11 '19 at 18:52

J.G. · Accepted Answer · 2019-03-10T07:35:51.180

1

The trick is that $3^2+2^2+6^2=7^2$. Note that $3W+2X-6Y\sim N(0,\,49)$, so it's the same as asking for $P(V<-1)$ with the $N(0,\,1)$ variable $V:=\frac{3W+2X-6Y}{7}$ , which I'm sure you know approximates $0.16$.

edited Mar 10 '19 at 07:35

answered Mar 09 '19 at 22:49

J.G.

118,053

Could I ask how you were able to simplify $3W + 2X - 6Y ~ N(0,49)$ into $P(X < -1)$? That step confuses me. – hmtkd Mar 10 '19 at 06:45
@hmtkd Oh, one doesn't imply the other; I was just saying that an $N(0,,49)$ variable has the same probability of being $<-7$ as an $N(0,,1)$ variable such as $X$ does of being $<-1$. – J.G. Mar 10 '19 at 07:03

Suppose that , , and are independent random variables, each with a standard Normal (Z) distribution.

1 Answers1