Solve by elimination 3×+11y=4,-2×-6y=0
1 answer:
You might be interested in
The result for the given normal random variables are as follows;
a. E(3X + 2Y) = 18
b. V(3X + 2Y) = 77
c. P(3X + 2Y < 18) = 0.5
d. P(3X + 2Y < 28) = 0.8729
<h3>What is normal random variables?</h3>Any normally distributed random variable having mean = 0 and standard deviation = 1 is referred to as a standard normal random variable. The letter Z will always be used to represent it.
Now, according to the question;
The given normal random variables are;
E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8.
Part a.
Consider E(3X + 2Y)
Part b.
Consider V(3X + 2Y)
Part c.
Consider P(3X + 2Y < 18)
A normal random variable is also linear combination of two independent normal random variables.
Thus,
Part d.
Consider P(3X + 2Y < 28)
Therefore, the values for the given normal random variables are found.
To know more about the normal random variables, here
#SPJ4
The correct question is-
X and Y are independent, normal random variables with E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8. Determine the following:
a. E(3X + 2Y)
b. V(3X + 2Y)
c. P(3X + 2Y < 18)
d. P(3X + 2Y < 28)