Question

An individual who has automobile insurance from a certain company is randomly selected. Let Y be the number of moving violations for which the individual was cited during the last 3 years. The pmf of Y is the following. y 0 1 2 3 p(y) 0.60 0.20 0.15 0.05 (a) Compute E(Y). E(Y) = (b) Suppose an individual with Y violations incurs a surcharge of $80Y2. Calculate the expected amount of the surcharge. $

Answer 1

Answer:

a) $E(Y) =\sum_{i=1}^n y_i P(Y=y_i)$

$E(Y) = 0*0.6 +1*0.2 +2*0.15 +3*0.15= 0.95$

b) $E(80Y^2) = 80 E(Y^2)$

$E(Y^2) =\sum_{i=1}^n y^2_i P(Y=y_i)$

$E(Y) = 0^2*0.6 +1^2*0.2 +2^2*0.15 +3^2*0.15= 2.15$

$E(80Y^2) = 80 E(Y^2)= 80*2.15 =172$

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:

$E(Y) =\sum_{i=1}^n y_i P(Y=y_i)$

And if we replace the values given we have:

$E(Y) = 0*0.6 +1*0.2 +2*0.15 +3*0.15= 0.95$

Part b

For this case we have defined a new random variable $80Y^2$ representing a subcharge, and we want to find the expected amount for this random variable, using properties of expected value we have:

$E(80Y^2) = 80 E(Y^2)$

And we can find $E(Y^2)$ on this way:

$E(Y^2) =\sum_{i=1}^n y^2_i P(Y=y_i)$

And if we replace the values given we have:

$E(Y) = 0^2*0.6 +1^2*0.2 +2^2*0.15 +3^2*0.15= 2.15$

And then replacing we got:

$E(80Y^2) = 80 E(Y^2)= 80*2.15 =172$