Answer:
Step-by-step explanation:
Here we have,
E(X) = 1, var(X) = 3, E(Y) = 0, var(Y) = 4
Since X and Y has normal distribution so W will also have normal distribution with mean
E(W) = E(05X-Y+6)
= 0.5E(X) - E(Y) +6
= 0.5* 1 -0+ 6
= 6.5
and variance
Var(W) = Var(0.5X-Y+6)
= 0.25Var(X)+Var(Y)
= 0.25 * 3 + 4
= 4.75
(b)
The z-score for W = 6 is
[tex]z=\frac{6-6.5}{\sqrt{4.75} } \\\\=-0.23[/tex]
The required probability is:
P(W>6) = P(z > -0.23)
= 0.5910
