1. Probability Distribution
Chapter 5:
Probability Distribution

2. Types of Variables Chapter 1: Variable definition
A characteristic or attribute that can assume different values.
Chapter 5: Random variable
A variable whose value are determined by chance.

3. Random Variables Variables whose values are determined by chance.
Two Types of Variables
Discrete
Finite number of possible values
Continuous
Assumes all values between two values

4. Discrete Probability Distribution
Consists of the values a random variable can assume and the corresponding probabilities of the values. The probabilities are determined theoretically or by observations

5. What does this mean? Example:
Constructing a probability distribution for rolling a single die
Solution:
Sample Space: 1, 2, 3, 4, 5, 6
Probability: each has 1/6 of a chance

6. Construction a Probability Distribution
First, make a table
The Outcomes are placed on top
The probabilities are placed on the bottom
Outcome X 1 2 3 4 5 6
Probability
P(X) 1 6 1 6 1 6 1 6 1 6 1 6

7. Construction a Probability Distribution
Second, make a chart
P(X) 1 1 2 1 6 X 1 2 3 4 5 6 7

8. Rules of Probability Distribution
The sum of the probabilities of all the events in the sample space must equal 1
∑P(X)=1
Rule 2:
The probability of each event in the sample space must be between or equal to 0 and 1
0≤ P(X) ≤1

9. Practice
Page 258 #’s 1-25

10. Chapter 5 Section 2 Finding the Mean of Probability Distribution
Formula
μ= ∑X*P(X)
Mu(μ)= mean
∑ = sum of
X= outcomes
P(X)= probability of outcomes

11. Example of Probability Distribution Mean
Find the average number of spots that appear when a die is tossed.
Probability
P(X) 6 5 4 3 2 1
Outcome X

12. Example continued μ= ∑X*P(X)
μ= X •P(X ) + X •P(X ) + X •P(X ) + … + X •P(X )
μ= 1• • • + 4• • •
μ= 21 = 3.5
μ= 3.5* 1 1 2 2 3 3 n n 1 6 1 6 1 6 1 6 1 6 1 6 6
* Theoretically mean because there cannot be a 3.5 rolled with a die

13. Rounding Rule ONE DECIMAL to the RIGHT OF the GIVEN
The rounding rule for Mean, Standard Deviation, and Variance is:
to the RIGHT
OF the GIVEN

14. Formula for Variance of Probability Distribution
σ²= ∑[X²•P(X)] - μ²
σ = sigma = sum
Mu(μ)= mean
∑ = sum of
X= outcomes
P(X)= probability of outcomes

15. Variance of Probability Distribution
Outcome X 1 2 3 4 5 6
Probability
P(X) 1 1 1 1 1 1 6 6 6 6 6 6
σ²= ∑[X²•P(X)] - μ²
1²•1/6+2²•1/6+3²•1/6+4²•1/6+5²•1/6+6²•1/6
15.17
2.9 = σ²
- 3.5²
- μ²

16. Finding Standard Deviation of a Probability Distribution
Formula:
σ = √σ²
σ = √2.9
σ = 1.7

17. Assignment:
Page 267
#’s 1-10

18. Expected Value Example
One thousand tickets are sold at $1 each for a $350 TV. What is the expected value of the gain if you purchase one ticket?

19. Table Set Up for Expected Value
Win
Lose
Gain
$349
-$1
Probability 1
1,000
999
1,000
E(X) = 349 • (-1) • = 1
1,000
999
E(X) = -$0.65
This does not mean that you will lose $.65 if you participate. It means the that average lose of every person who plays will be $.65

20. Expected Value Formula μ= ∑X*P(X) E(X)= ∑X*P(X) E(X) = expected value
∑ = sum of
X= outcomes
P(X)= probability of outcomes
E(X)= ∑X*P(X)

21. Is it fair? How is a gambling game fair? Who does it favor? Example:
The expected value of the game is zero.
Who does it favor?
Expected value
Positive: The player
Negative: The house
Example:
Roulette: House wins $0.90 on every $1 bet
Craps: House wins $0.88 on every $1 bet

22. Your turn
One thousand tickets are sold at $1 each for four prizes ($100, $50, $25 and $10). After each prize drawing, the winning ticket is then returned to the pool of tickets. What is the expected value?

23. Table
Gain
Prob

24. Practice
Page # 268
12-18 even

25. Chapter 5 Section 3 The Binomial Distribution The Binomial Experiment
There must be a fixed number of Trials
Each Trial can have only two outcomes
Successful
Failure
Outcomes must be independent of each other
The probability must remain the same for each trial

26. Binomial Probability Formula
P(S) The symbol for the probability of success
P(F) The symbol for the probability of failure
p The numerical probability of success
q The numerical probability of failure
n The number of trials
X The number of successes in n trials

27. Example:
A coin is tossed 3 times. Find the probability of getting exactly two heads.
n = 3
X = 2
p = 1/2
q = 1/2

28. Page 277 # 4
A burglar alarm system has six fail-safe components. The probability of each failing is Find these probabilities.
Exactly three will fail
Fewer than two will fail
None will fail

29. Exactly three will fail
n = 6
X = 3
p = 0.05
q = 0.95
Use chart
0.002

30. Fewer than two will fail
X = 1 or 0
p = 0.05
q = 0.95 1 or
0.967
0.232 +
0.735

31. None will fail
n = 6
X = 0
p = 0.05
q = 0.95
0.735

32. Your turn
Page
#’s 11-12

33. Binomial Distribution
Mean Formula:
μ = n • p
Variance Formula:
σ²= n • p • q
Standard Deviation Formula:
σ=√n • p • q

34. Examples: No Examples today. I think you can handle it. Page 278

35. Multinomial Distribution

36. Page 284 Example 5-25
In a music store, a manager found that the probabilities that a person buys 0, 1, or 2 or more CDs are 0.3,.6, and .1 respectively. If 6 customers enter the store, find the probability that 1 won’t buy any CD’s, 3 will buy 1 CD, and 2 will buy 2 or more CDs.

37. Practice
Page 290 #’s 1-6

38. The Poisson Distribution
A discrete probability distribution that is useful when n is large and p is small and when the independent variables occur over a period of time.
Ex: area, volume, time

39. Formula The letter e is a constant approximately equal to 2.7183
Answers are rounded to four decimal places

40. Problem
If there are 200 typographical errors randomly distributed in a 500 page novel, find the probability that a given page contains exactly three errors.

41. Hypergeometric Distribution
Given a population two types
Ex: male and female
Success and failure
Without replacement
More accurate than Binomial Distribution

42. Formula a = population 1 b = population 2 n = total section
X = selection wanted

43. Example
Ten people apply for a job. Five have completed college and five have not. If a manager selects three applicants at random, find the probability that all three are graduates.
a = college graduates = 5
b = nongraduates = 5
n = total section = 3
X = selection wanted = 3

44. Page 287 Example 5-30 a = college graduates = 5 b = nongraduates = 5
n = total section = 3
X = selection wanted = 3

45. Practice
Page 291
#’s 17-21