Respuesta :
Answer:
See explanation below.
Step-by-step explanation:
Notation
Let's define the following events:
B= A boy rarely or never wear seat belts
P(B) = 0.18
G= A girl rarely or never wear seat belts
P(G) =0.10
Solution to the problem
For this case we are interested on the following variable Y ="people who wear a seat belt" .And the possible values for Y are 0,1,2.
The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.
One way
Using the definition of random variable we can find the expected value like this:
[tex] \mu = \sum_{i=1}^n Y_i P(Y_i) [/tex]
And the variance can be calculates like this:
[tex] Var(Y) = \sigma^2 = E(Y^2) -[E(Y)]^2 [/tex]
Where [tex] E(Y^2) = \sum_{i=1}^n Y^2_i P(Y_i) [/tex]
And then we can find the deviation like this:
[tex] Sd(Y) = \sigma = \sqrt{E(Y^2) -[E(Y)]^2} [/tex]
Second way
Let B the random variable who represent if a boy rarely or never wear seat belts with possible values B=0,1 and G the random variable who represent if a girl rarely or never wear seat belts with possible values G=0,1. We can find the expected value like this:
[tex] E(B+G) = E(B) +E(G) [/tex]
From definition of expected value.
And the variance like this:
Var(B+G) = Var(B) +Var(G) + 2Cov(B,G)[/tex]
If B and G are independent then [tex] Cov(B,G) = 0[/tex]
And the deviation is just this:
[tex] Sd(B+G) = sqrt{Var(B) +Var(G) + 2Cov(B,G)}[/tex]
