Chapter 9Section 7

Chapter 9 Section 7

Mathematical Expectation

Ch9.7 Mathematical Expectation

To review from last Friday, we had the case of a binomial distribution given by the formula:

where p(s): the probability of a success, what we would like to happen.

p(~s): the probability of a fail, the opposite of what we would like to happen.

n: the number of trials

x: the number of desired “successes”

nCx p(~s)n-x p(s)x


In class we had an example of flipping a coin 7 times. We were given the following probabilities:

P(Heads) = .7

P(Tails) = .3

The following Binomial Distribution was developed to find various probabilities of flipping a Head in these 7 trials:

P(x) =

At the end of the day, with the use of your TI-Calc, we were able to make a table with all the corresponding probabilities

7Cx (.3)7-x (.7)x


# of success (heads)

Probability

0 .0002

1 .0036

2 .025

3 .0972

4 .2269

5 .3177

6 .2471

7 .0824

Lets examine this data with

the use of a scatter plot. What

would this scatter plot look like?


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0 2 4 6 8

# of Heads

Pro

bab

ilit

y


Based off of the scatter plot, if you were to make a guess for the number of heads to appear in the 7 flips, around what range would you guess? Why?

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# of HeadsPr

obab

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How do we determine this “expected” value with an algebraic approach? Lets first review over averages. How would you find the average of the following numbers: 2, 2, 2, 5, 5, 6, 6, 10, 10, 10 ?

Average = ------------ so:

2 + 2 + 2 + 5 + 5 + 6 + 6 + 10 + 10 + 10 (3)(2) + (2)(5) + (2)(6) + (3)(10)

----------------------------------------------------- = ----------------------------------------

10 10

Total

Number


2 + 2 + 2 + 5 + 5 + 6 + 6 + 10 + 10 + 10 (3)(2) + (2)(5) + (2)(6) + (3)(10)

----------------------------------------------------- = ----------------------------------------

10 10

=

=(.3)(2) + (.2)(5) + (.2)(5) + (.3)(10)

3 2 2 3(2) (5) (6) (10)

10 10 10 10


=(.3)(2) + (.2)(5) + (.2)(5) + (.3)(10)

This is an interesting way to explain the way to find an average. We now have something called a “weighted average.” The “weights” are the probabilities of such numbers showing up. Ex: we could say that in our average, we have a 30% chance of having a 2, a 20% chance of having a 5, and so on.


This idea of a weighted average is how we find the “expected value” of probability distribution. In order to calculate the expected value, we multiply each corresponding value by its probability and then add them together.

P(a)*A + p(b)*B + p(c)*C + … and so on

p(a), p(b), and p(c) are the probabilities, and A, B, C are the values. Lets look at an example:


Looking at our case of the 7 coin flips, we had the table: to find the “expected value” of this distribution, we will multiply the probabilities of each event by the value (number of heads), and then add them together.

.0002(0) + (.0036)(1) + (.025)(2) + (.0972)(3) + (.2269)(4) + (.3177)(5) + (.2471)(6) + (.0824)(7) = 4.9.

Does this agree with what you mentioned for the scatter plot?

# of success (heads)

Probability

0 .0002

1 .0036

2 .025

3 .0972

4 .2269

5 .3177

6 .2471

7 .0824


0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0 2 4 6 8

# of Heads

Prob

abili

ty

The expected value of 4.9 very much so matches our scatter plot of an expected value.


The interpretation of an expected value is what we expect to be our average value per trial. Each individual trial will not directly follow this expected value, but the long run average of the trials will follow closely to our expected value.


Example: You decide to start up a carnival game where a person pays $5 to flip a coin 3 times. The person wins money as follows:

0 Heads = $0, 1 Head = $5, 2 Heads = $10, and 3 Heads = $15.

You have weighted the coin as follows: the p(H) = .3 p(T) = .7

What is the customers expected value of winning?

We will need to make a payoff table showing the probabilities and the corresponding payoffs, to first find the probabilities we need to look at the binomial distribution.


# of heads Probability Payout p(s)*payout

0 .343 -5 -1.715

1 .441 0 0

2 .189 5 .945

3 .027 10 .27

The payout column is what the customer would win/lose given each outcome of heads appearing. -5 for 0 heads because the customer had to pay 5 dollars to play.Adding up the last column of this chart gives us an expected value of -.5which means that with this game, on average, we would make (or the customer would lose) around .5 cents per game.

Chapter 9Section 7

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