Find $P(2\bar{X}-\bar{Z}>3\bar{Y})$

Question

Let $(X_{1},X_{2},X_{3},...,X_{8})\sim N(8,16)$ is independent random sample , $(Y_{1},Y_{2},Y_{3})\sim N(1,9)$ is independent random sample and $(Z_{1},Z_{2},Z_{3})\sim N(6,10)$ is independent random sample. Find $P(2\bar{X}-\bar{Z}>3\bar{Y})$.

Their sample means are (respectively) , $\bar{X}\sim N(8,\frac{16}{8})=N(8,2)$ , $\bar{Y}\sim N(1,\frac{9}{3})=N(1,3)$ and $\bar{Z}\sim N(6,\frac{10}{3})$

from here I can not calculate the probability. How can I express $P(2\bar{X}-\bar{Z}>3\bar{Y})$? Any idea will be appreciated.

Answer 1

Assuming the necessary independence conditions,

Consider that your probability is the same as the following

$$\mathbb{P}[2\overline{X}_8-\overline{Z}_3-3\overline{Y}_3>0]$$

and the rv $2\overline{X}_8-\overline{Z}_3-3\overline{Y}_3$ is still gaussian....moreover for independence:

$$W=2\overline{X}_8-\overline{Z}_3-3\overline{Y}_3\sim N\left(7;\frac{115}{3}\right)$$

thus

$$\mathbb{P}[W>0]=1-\Phi\left(\frac{-7}{\sqrt{115}}\sqrt{3}\right)\approx 87.09\%$$