Homework SourseAn individual who has automobile insurance from a certain co
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.
(a) Compute E(Y).
E(Y) = ?
(b) Suppose an individual with Y violations incurs a surcharge of $120Y2. Calculate the expected amount of the surcharge.
$ ?
| y | 0 | 1 | 2 | 3 |
| p(y) | 0.60 | 0.25 | 0.10 | 0.05 |
Solution
f= 1
fx = 0.6
Mean = fx / f = 0.6
Mean square = f x^2 / f = 1.1
Varriance = (Mean square) - (Mean)^2
Varriance = f x^2 - Mean^2 = 0.74
Stadard Dev= Var = 0.86
a. E(Y) = 0.6
b.
E(120Y^2) = 120 * E(Y^2)
E(Y^2) = Sum Y^2 * P(Y) = 0^2 (0.60) + 1^2 ( 0.25) + 2^2 ( 0.10) + 3^2 ( 0.05) = 1.1
therefore E(120Y^2) = 120* 1.1 = 132
| Values ( X ) | Frequency(f) | fx | ( X^2) | f x^2 |
| 0 | 0.6 | 0 | 0 | 0 |
| 1 | 0.25 | 0.25 | 1 | 0.25 |
| 2 | 0.1 | 0.2 | 4 | 0.4 |
| 3 | 0.05 | 0.15 | 9 | 0.45 |
