Let X1 and X2 be two independent normal random variables

Let X1 ~N(2, 3) and X2  ~N(2, 1) be two independent normal random
variables. Find the probability, P(2X1 + X2 > 8).

any help would be great

Use the rule for adding independent normal random variables, convert the result to a standard normal and use the tables.

Hi, im really unsure how to do this question, but this is what ive tried.

im adding the two normal random var. using a formula i found here (http://en.wikipedia.org/wiki/Sum_of_nor ... _variables)

\(\displaystyle \(X1 + X2 \sim N ( 2+2, 3+1) \)\)

\(\displaystyle \(= X1 + X2 \sim N ( 4, 4) \)\)

but im stuck here, how do i times X1 by 2 to show that this is > 8, or is what im doing wrong?

thanks for any help .

ive made some more progress, is this right?

\(\displaystyle 2X1 + X2\)

\(\displaystyle =X1 \sim N(4,12) + X2 \sim N(2,1)\)

\(\displaystyle X1 + X2 \sim N(6,13)\)

\(\displaystyle P(X > 8)\)

\(\displaystyle Z = \frac{X - 6}{13}\)

\(\displaystyle Z = \frac{2}{13} = 0.153\)

\(\displaystyle P(X < 0.153) = 0.5596\)

waxydock said:

ive made some more progress, is this right?

\(\displaystyle 2X1 + X2\)

\(\displaystyle =X1 \sim N(4,12) + X2 \sim N(2,1)\)

\(\displaystyle X1 + X2 \sim N(6,13)\)

\(\displaystyle P(X > 8)\)

\(\displaystyle Z = \frac{X - 6}{13}\)

\(\displaystyle Z = \frac{2}{13} = 0.153\)

\(\displaystyle P(X < 0.153) = 0.5596\)

Click to expand...

OK until the last step. Don't you want \(\displaystyle P(Z > 0.153) = 0.4392\)?