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Probability Distributions and Expected Value

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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.

- 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)

- The number of phone calls made by a salesperson
- Discrete (1,2,3,4,5…..)

- The length of time a person spends on the phone
- Continuous (1 min, 1.23min …..)

The probability distribution could be written in a table

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

1

1/6

2

3

4

5

6

7

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

n

= xiP(X = xi)

i = 1

The sum of the terms of the form (X)(P[X])

E(X) = x1P(X = x1) + x2P(X = x2) + … + xnP(X = xn)

- What is the likelihood that you would observe at least two heads?
- What is the expected number of heads?

1

3

3

1

8

8

8

8

X 0 hs 1 hs 2 hs 3 hs

P(X) = x

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

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)

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