Transcript Chapter 9_8 Lesson - Saint Joseph High School

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: nCx  p(~s)n-x  p(s)x
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”
Ch9.7 Mathematical Expectation
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) = 7Cx  (.3)7-x  (.7)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
Ch9.7 Mathematical Expectation
# of success
(heads)
0
1
2
3
4
5
6
7
Probability
.0002
.0036
.025
.0972
.2269
.3177
.2471
.0824
Lets examine this data with
the use of a scatter plot. What
would this scatter plot look
like?
Ch9.7 Mathematical Expectation
0.35
0.3
Probability
0.25
0.2
0.15
0.1
0.05
0
0
2
4
# of Heads
6
8
Ch9.7 Mathematical Expectation
0.35
0.3
0.25
Probability
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?
0.2
0.15
0.1
0.05
0
0
2
4
# of Heads
6
8
Ch9.7 Mathematical Expectation
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 ?
Total
Number
Average = ------------
so:
2 + 2 + 2 + 5 + 5 + 6 + 6 + 10 + 10 + 10
(3)(2) + (2)(5) + (2)(6) + (3)(10)
----------------------------------------------------- = ---------------------------------------10
10
Ch9.7 Mathematical Expectation
2 + 2 + 2 + 5 + 5 + 6 + 6 + 10 + 10 + 10
(3)(2) + (2)(5) + (2)(6) + (3)(10)
----------------------------------------------------- = ---------------------------------------10
10
=
3
2
2
3
(2)  (5)  (6)  (10)
10
10
10
10
=(.3)(2) + (.2)(5) + (.2)(5) + (.3)(10)
Ch9.7 Mathematical Expectation
=(.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.
Ch9.7 Mathematical Expectation
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:
Ch9.7 Mathematical Expectation
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)
0
1
2
3
4
5
6
7
Probability
.0002
.0036
.025
.0972
.2269
.3177
.2471
.0824
Ch9.7 Mathematical Expectation
The expected value of
4.9 very much so
matches our scatter
plot of an expected
value.
0.35
0.3
Probability
0.25
0.2
0.15
0.1
0.05
0
0
2
4
# of Heads
6
8
Ch9.7 Mathematical Expectation
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.
Ch9.7 Mathematical Expectation
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.
Ch9.7 Mathematical Expectation
# of heads Probability
0
.343
1
.441
2
.189
3
.027
Payout
-5
0
5
10
p(s)*payout
-1.715
0
.945
.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 -.5
which means that with this game, on average, we would make (or the
customer would lose) around .5 cents per game.