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# Probability Distributions and Expected Value - PowerPoint PPT Presentation

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).

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### Probability Distributions and Expected Value

Random Variable (X) individual outcomes.

• 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 individual outcomes.(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: individual outcomes.

• The number of phone calls made by a salesperson

• Discrete (1,2,3,4,5…..)

For example: individual outcomes.

• 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 individual outcomes.

The probability distribution could be written in a table

This is a uniform distribution, because all the probabilities are the same.

A graph would look like this… individual outcomes.

1

1/6

2

3

4

5

6

Expected Value: Informal individual outcomes.

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) + … individual outcomes.

+…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

n possible outcomes.

= xiP(X = xi)

i = 1

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)

Ex 1: Suppose you toss 3 coins. possible outcomes.

• What is the likelihood that you would observe at least two heads?

• What is the expected number of heads?

1 possible outcomes.

3

3

1

8

8

8

8

Represent the theoretical probability distribution as a table.The DRV, X, represents the number of heads observed.

X 0 hs 1 hs 2 hs 3 hs

P(X) = x

a) P(X = 2) + P(X = 3) = 3 / 8 + 1 / 8 possible outcomes.

= 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 possible outcomes.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 possible outcomes.

1, 2(ex 2), 3a,c, 4, 9, 11,12, 19