Stats 110a, Spring 2002 Name:

Friday, April 26 ID:

Quiz 4

A casino is testing a new computerized gambling game. Players push a button, and a continuous random number between 0 and 3 (inclusive) is drawn. If the number is within the interval [0,1), they win $10. If the number is in the interval [1, 2) they win $1.00. If the number is in the interval [2,3), they lose $175.

The density function that determines the frequency with which the random numbers appear is given by

f(x) = (2/9)x for x in [0,3]

= 0 else

a) What's the expected value for someone playing this game?

Let X be the continuous RV that the computer uses to determine the payoff. Now let Y represent the payoff, i.e. the winnings. Then

Y = 10 if [0 <= X < 1]

Y = 1 if [1 <= X < 2]

Y = -1.75 if [2 <= X < 3]

So Y is a discrete RV, and we are asked to find E(Y) = sum ( y * P(Y = y))

= 10 * P(Y = 10) + 1 *P(Y =1) + -1.75 * P(Y = -1.75)

= 10*P(0 <= X < 1) + 1*P(1 <= X < 2) -1.75 *P(2 <= X < 3)

Now to get the probabilities, you can either sketch the pdf, note that its a triangle, and figure out the appropriate areas, or you can do an integral:

P(0 <= X < 1) = int ((2/9)x ds integrated from 0 to 1 is

(2/9) (1/2)x² from 0 to 1 = 1/9.

The others are

P(1 <= X <2) = 3/9

P(2 <=X<3) = 5/9.

Plug these in above, and you get

E(Y) = 0.47222 (So maybe this Casino will not be in business for long.)

b) What's the standard deviation of winnings?

Var(Y) = E(Y - .4722)² = (10 - .4722)² * (1/9) + (1 - .4722)² * (3/9) + (-1.75 - .4722)² *(5/9) = 12.92284

SD(Y) = sqrt(12.92284) = 3.594835