Respuesta :
Answer:
[tex] E(Y) = 3 + 0.5 *2.7 = 3+1.35 = 4.35[/tex]
[tex] Var (Y) = 0.5^2 Var (X) = 0.25 *(1.2) = 0.3[/tex]
And then we have the following parameters for the random variable Y
[tex] \mu_Y = 4.35 , \sigma^2_Y = 0.3[/tex]
And the best option is:e. mean = $4.35, variance = $ .30.
Step-by-step explanation:
For this case we know the following info " The toll on the bridge is $3.00 per car plus $ .50 per person in the car"
So then we can create the following linear function:
[tex] Y = 3 + 0.5 X[/tex]
Where Y represent the total amount of money that is collected from a car. And X represent the number of people in the car.
We have the parameters for the random variable X given by:
[tex] E(X) = \mu_X = 2.7 , \sigma^2_x = 1.2[/tex]
So then we can find the expected value for the random variable Y like this:
[tex] E(Y) = 3 + 0.5 E(X)[/tex]
And if we replace the expected value for X we got this:
[tex] E(Y) = 3 + 0.5 *2.7 = 3+1.35 = 4.35[/tex]
Now we can find the variance for the random variable Y like this:
[tex] Var (Y) = Var (3) + 0.5^2 Var(X) + 2 Cov(3, 0.5X)[/tex]
We know that for a number we don't have variance since is a constant and the covariance between a number and a random variable is 0 so then we have just this for the variance of Y:
[tex] Var (Y) = 0.5^2 Var (X) = 0.25 *(1.2) = 0.3[/tex]
And then we have the following parameters for the random variable Y
[tex] \mu_Y = 4.35 , \sigma^2_Y = 0.3[/tex]
And the best option is:e. mean = $4.35, variance = $ .30.
