91Ó°ÊÓ

It is given that X, Y and Z are independent and identical variables that follow standard normal distribution.

Therefore,

\(\begin{array}{l}X \sim N\left( {0,1} \right)\\Y \sim N\left( {0,1} \right)\\Z \sim N\left( {0,1} \right)\end{array}\)

Therefore, the mean of X,Y and Z is 0 and variance is 1.

Let us define a new variable

\(W = 3X + 2Y - 6Z\)

The expectation is:

\(\begin{aligned}{}E\left( W \right) &= E\left( {3X + 2Y - 6Z} \right)\\ &= E\left( {3X} \right) + E\left( {2Y} \right) + E\left( { - 6Z} \right)\\& = 3E\left( X \right) + 2E\left( Y \right) - 6E\left( Z \right)\\& = 0\end{aligned}\)

Since the variables X, Y and Z are i.i.d., they have 0 covariance and hence their covariance term vanishes in the variance formula.

The variance is:

\(\begin{aligned}{}V\left( W \right) &= V\left( {3X + 2Y - 6Z} \right)\\ &= V\left( {3X} \right) + V\left( {2Y} \right) + V\left( { - 6Z} \right)\\ &= 9V\left( X \right) + 4V\left( Y \right) + 36V\left( Z \right)\\& = 49\end{aligned}\)

We have to find:

\(\begin{aligned}{}\Pr \left( {3X + 2Y < 6Z - 7} \right) &= \Pr \left( {3X + 2Y - 6Z < - 7} \right)\\ &= \Pr \left( {W < - 7} \right)\\ &= \Pr \left( {\frac{{W - 0}}{{\sqrt {49} }} < \frac{{ - 7 - 0}}{{\sqrt {49} }}} \right)\\& = \Pr \left( {Z < \frac{{ - 7}}{7}} \right)\\ &= \Pr \left( {Z < - 1} \right)\\ \approx 0.1587\end{aligned}\)

Hence the final answer is 0.1587.