Respuesta :
Answer:
a) [tex] E(Y) =\sum_{i=1}^n y_i P(Y=y_i)[/tex]
[tex] E(Y) = 0*0.6 +1*0.2 +2*0.15 +3*0.15= 0.95[/tex]
b) [tex] E(80Y^2) = 80 E(Y^2)[/tex]
[tex] E(Y^2) =\sum_{i=1}^n y^2_i P(Y=y_i)[/tex]
[tex] E(Y) = 0^2*0.6 +1^2*0.2 +2^2*0.15 +3^2*0.15= 2.15[/tex]
[tex] E(80Y^2) = 80 E(Y^2)= 80*2.15 =172[/tex]
Step-by-step explanation:
Previous concepts
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.
The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).
Solution to the problem
For this case we have defined the following random variable Y="number of moving violations for which the individual was cited during the last 3 years. "
And we have the distribution for Y given:
y 0 1 2 3
P(y) 0.6 0.2 0.15 0.15
Part a
For this case the expected value is given by:
[tex] E(Y) =\sum_{i=1}^n y_i P(Y=y_i)[/tex]
And if we replace the values given we have:
[tex] E(Y) = 0*0.6 +1*0.2 +2*0.15 +3*0.15= 0.95[/tex]
Part b
For this case we have defined a new random variable [tex]80Y^2[/tex] representing a subcharge, and we want to find the expected amount for this random variable, using properties of expected value we have:
[tex] E(80Y^2) = 80 E(Y^2)[/tex]
And we can find [tex] E(Y^2)[/tex] on this way:
[tex] E(Y^2) =\sum_{i=1}^n y^2_i P(Y=y_i)[/tex]
And if we replace the values given we have:
[tex] E(Y) = 0^2*0.6 +1^2*0.2 +2^2*0.15 +3^2*0.15= 2.15[/tex]
And then replacing we got:
[tex] E(80Y^2) = 80 E(Y^2)= 80*2.15 =172[/tex]
