Question

The number of people in a car that crosses a certain bridge is represented by the random variable X, which has a mean value
`\mu_X` = 2.7, and a variance
`\sigma^2_X` = 1.2. The toll on the bridge is $3.00 per car plus $ .50 per person in the car. The mean and variance of the total amount of money that is collected from a car that crosses the bridge are:a. mean = $1.35, variance = $ .30.b. mean = $8.60, variance = $ .30.c. mean = $8.60, variance = $ .60.d. mean = $4.35, variance = $3.30.e. mean = $4.35, variance = $ .30.

Answer 1

`E(Y) = 3 + 0.5 *2.7 = 3+1.35 = 4.35`

`Var (Y) = 0.5^2 Var (X) = 0.25 *(1.2) = 0.3`

And then we have the following parameters for the random variable Y

`\mu_Y = 4.35 , \sigma^2_Y = 0.3`

And the best option is:e. mean = $4.35, variance = $ .30.

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:

`Y = 3 + 0.5 X`

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:

`E(X) = \mu_X = 2.7 , \sigma^2_x = 1.2`

So then we can find the expected value for the random variable Y like this:

`E(Y) = 3 + 0.5 E(X)`

And if we replace the expected value for X we got this:

`E(Y) = 3 + 0.5 *2.7 = 3+1.35 = 4.35`

Now we can find the variance for the random variable Y like this:

`Var (Y) = Var (3) + 0.5^2 Var(X) + 2 Cov(3, 0.5X)`

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:

`Var (Y) = 0.5^2 Var (X) = 0.25 *(1.2) = 0.3`

And then we have the following parameters for the random variable Y

`\mu_Y = 4.35 , \sigma^2_Y = 0.3`

And the best option is:e. mean = $4.35, variance = $ .30.