Expected Value
Central tendencies
 Mean
 Mode
 Median
 The most frequently used measure for
describing central tendency is the expected
value or average.
 Remember the values included in the
average are “weighted” by the frequency.
Random Variable A numerical variable
whose value depends on
the outcome of a chance
experiment
Discrete Probability Distribution
1) Gives the values associated with
each possible x value
2) Usually displayed in a table, but
can be displayed with a
histogram or formula
Discrete probability distributions
3)For every possible x value,
0 < P(x) < 1.
4) For all values of x,
S P(x) = 1.
Suppose you toss 3 coins & record
the number of heads.
The random variable X defined as ...
The number of heads tossed
List the values of the Random
variable.
X
0
1
List the corresponding
probabilities.
P(X)
.125
.375
2
.375
3
.125
Let x be the number of courses for which
a randomly selected student at a certain
university is registered.
X
1 2 3
P(X).02 .03 .09
4 5 6 7
? .40 .16 .05
Why
does
this
not
start
at
zero?
.25
P(x = 4) =
P(x < 4) = .14
P(x < 4) = .39
P(x > 5) = .61
What is the probability that the student
is registered for at least five courses?
Expected Value is the mean
weighted by its probability.
Mean: Average
Weighted Mean: each Random
variable X multiplied by its
probability
Formulas for mean
μ x = x i pi
Let x be the number of courses for which
a randomly selected student at a certain
university is registered.
X
1
2
3
4
5
6
7
P(X).02 .03 .09 .25 .40 .16 .05
What is the mean of this
distribution?
m=
1(.02)+2(.03)+3(.09)+4(.25)+5(.40)+6(.16)+7(.05)
Setting up an expected value problem
Each day a student puts $.50 in a candy
machine to buy a $.35 pack of gum. She
observes three possible outcome. 80% of the
time she gets the gum and $.15 change.
16% of the time she gets the gum and no
change. 4% of the time she gets the gum
and the machine returns her $.50. Over a
period of time, what is the cost of the gum?
 DO NOT take the mean of three possible
costs .35, .50 and 0. This = 28.3, and is
NOT the expected value. The average cost
(or expected value) takes into account that
they DO NOT occur equally often.
Set up a discrete probabilty
WE emphasize that the term EXPECTED
distribution
VALUE does not mean the value we expect
in an everyday sense. In this gum
example, the amount paid for the gum is
Cost
0.35
0.50 to pay 0
never(X)
$.36. You
would never expect
that, however, $.36 is what you would pay
P(x)
0.80 of purchases,
0.16 or long 0.04
with a large number
period of time. This is the Expected Value.
P(x) is the probability that the cost will occur.
E(x) = (0.35)(0.80) +(0.50)(0.16)+(0)(0.04) = 0.36
Here’s a game:
If a player rolls two dice
and gets aAsum
of 2is or
12,
fair game
one where
the cost to play EQUALS
he wins $20.theIf
he
gets
a
expected value!
0 $5. 5The cost
20
7,X he wins
to
P(X)
7/9
1/6
1/18
roll the dice one time is
$3. Is this game fair?
NO, since m = $1.944 which is less
than it cost to play ($3).
Another way to look at the same
problem
In this case a fair
game is
where
theinto
expected
We can incorporateone
the cost
of play
the
This is
not a fair
game when wevalue
do the=0!
probability
distribution.
game.
X
(0-3)
(5-3)
P(X)
7/9
1/6
(20 – 3)
E(x) = (-3)(7/9) + (2)(1/6)+(17)(1/18) = -1.06
1/18
Your turn
 An Auto repair shop’s records show that
fifteen percent of the cars serviced need
minor repairs averaging $100. Sixty-five
percent need moderate repairs averaging
$550, and twenty percent need major repairs
averaging $1100.
 What is the expected cost of the repair of a
car selected at random?
Your Turn
 A game involves throwing a pair of dice. The
player will win, in dollars, the sum of the
numbers thrown. How much should the
organization charge to enter the game if they
want to break even?
 If they want to average $1.50 per person
profit.
 Gambling affords a familiar illustration of the
notion of an expected value. Consider
Roulette. After bets are placed, the croupier
spins the wheel and declares one of thirty –
eight numbers, 00, 0, 1, 2….36 to be the
winner. We will assume that each of these
thirty-eight numbers is equally likely.
Suppose we bet $1 on an odd number
occurring.
18 9
Px(1) = P( X = 1) = =
38 19
20 10
Px(1) = P( X = 1) =
=
38 19
9
 10 
$1   $1  = $0.053 = 5cents
 19 
 19 
Winning a dollar
Losing a dollar
Expected value