Transcript Probability Distributions and Expected Value

Probability
Distributions and
Expected Value
• In previous chapters, our
emphasis was on the probability
of individual outcomes.
• This chapter develops models for
distributions that show the
probabilities for all possible
outcomes of an experiment.
Random Variable (X)
• Has a single value, x, for each
outcome of an experiment.
To show all the possible outcomes,
a chart is normally used.
Discrete variables take values that are
separate (or that can be “counted”)
Continuous variables have an infinite
number of possible outcomes (usually
measurements that can have an
unlimited decimal place)
For example:
• The number of phone calls made
by a salesperson
–Discrete (1,2,3,4,5…..)
For example:
• The length of time a person spends
on the phone
–Continuous (1 min, 1.23min …..)
Let a DRV (X) be the possible
outcomes when rolling a die
The probability distribution could
be written in a table
X
1
2
3
4
P(X) = x
1/6
1/6 1/6 1/6
5
6
1/6 1/6
This is a uniform distribution, because all
the probabilities are the same.
A graph would look like this…
1/6
12345 6
Remember the PD for the sums
generated by rolling 2 dice?
7
Expected Value: Informal
When rolling 2 dice, the sum
that is generated most
frequently is called the
expected value. (7)
This can also be calculated
mathematically.
Multiply each roll by it’s
probability of occuring…
E(sum) = 2P(sum = 2) + 3P(sum = 3) + …
+…12P(sum = 12)
= 2 X 1 / 36 + 3 X 2 / 36 …
= 252 / 36
=7
The expected value, E(X), is the
predicted average of all possible
outcomes.
It is equal to the sum of the products of
each outcome with its probability
Expected Value of a Discrete
Random Variable
The sum of the terms of the form (X)(P[X])
E(X) = x1P(X = x1) + x2P(X = x2) +
…+
xnP(X = xn)
n
=
xiP(X = xi)
i=1
Ex 1: Suppose you toss 3 coins.
a) What is the likelihood that you would
observe at least two heads?
b) What is the expected number of heads?
Represent the theoretical
probability distribution as a table.
The DRV, X, represents the
number of heads observed.
X
P(X) = x
0 hs
1
8
1 hs
3
8
2 hs
3
8
3 hs
1
8
a) P(X = 2) + P(X = 3) = 3 / 8 + 1 / 8
=1/2
b) The expected number of heads
= 0(1 / 8) + 1(3 / 8) + 2(3 / 8) + 3(1 / 8)
=3/2
For a game to be fair, E(X) must be
zero Consider a dice game
If you roll a 1 2 3 you win $1.00
If you roll a 4 5 6 you pay $1.00
Is this game fair?
E(X) = (1)(1/6) + (1)(1/6) + (1)(1/6)
+ (-1)(1/6) + (-1)(1/6) + (-1)(1/6)
Page 374
1, 2(ex 2), 3a,c,
4, 9, 11,12, 19