A difference of three normals

MediumRandom Variables Statistics~4m

Let $X, Y, Z \sim \mathcal N(12, 12)$ be i.i.d. (mean $12$ , variance $12$ ). The probability $\mathbb{P}[X - Y > Z]$ can be written as $\Phi(a)$ for the standard normal CDF $\Phi$ . Find $a$ .

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$W=X-Y-Z$ is normal with mean $12-12-12=-12$ and variance $12+12+12=36$ ( $\sigma=6$ ).
Standardize $\mathbb{P}[W>0]$ : the threshold is $2\sigma$ above the mean, so it equals $\Phi(-2)$ .

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A difference of three normals

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