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Overview

This chapter will deal with the construction of

probability distributions

by combining the methods of Chapter 2 with the those of Chapter 4.

Probability Distributions will describe what will probably happen instead of what actually did happen.

Combining Descriptive Statistics Methods and Probabilities to Form a Theoretical Model of Behavior

5

4

Definitions to Form a Theoretical Model of Behavior

- Random Variable
a variable (typically represented by x) that has a single numerical value, determined by chance, for each outcome of a procedure

- Probability Distribution
a graph, table, or formula that gives the probability for each value of the random variable

Probability Distribution to Form a Theoretical Model of BehaviorNumber of Girls Among Fourteen Newborn Babies

x

P(x)

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

0.000

0.001

0.006

0.022

0.061

0.122

0.183

0.209

0.183

0.122

0.061

0.022

0.006

0.001

0.000

Probability Histogram to Form a Theoretical Model of Behavior

Definitions to Form a Theoretical Model of Behavior

- Discrete random variable
has either a finite number of values or countable number of values, where ‘countable’ refers to the fact that there might be infinitely many values, but they result from a counting process.

- Continuous random variable
has infinitely many values, and those values can be associated with measurements on a continuous scale with no gaps or interruptions.

Requirements for Probability Distribution to Form a Theoretical Model of Behavior

P(x) = 1

where x assumes all possible values

0 P(x) 1

for every value of x

Mean, Variance and Standard Deviation of a Probability Distribution

µ = [x•P(x)]

2= [(x - µ)2 • P(x)]

2=[x2 • P(x)] - µ2(shortcut)

=[x 2 • P(x)] - µ2

Roundoff Rule for Distributionµ, 2, and

Round results by carrying one more decimal place than the number of decimal places used for the random variable x. If the values of xare integers, round µ, 2, and to one decimal place.

E = Distribution [x • P(x)]

Event

Win

Lose

x

$499

- $1

P(x)

0.001

0.999

x • P(x)

0.499

- 0.999

E = -$.50

Definitions Distribution

- Binomial Probability Distribution
1. The experiment must have a fixed number of trials.

2. The trials must be independent. (The outcome of any individual trial doesn’t affect the probabilities in the other trials.)

3. Each trial must have all outcomes classified into two categories.

4. The probabilities must remain constantfor each trial.

For Distributionn = 15 and p = 0.10

Table B

Binomial Probability Distribution

P(x)

x

P(x)

n x

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

0.206

0.343

0.267

0.129

0.043

0.010

0.002

0.0+

0.0+

0.0+

0.0+

0.0+

0.0+

0.0+

0.0+

0.0+

0.206

0.343

0.267

0.129

0.043

0.010

0.002

0.000

0.000

0.000

0.000

0.000

0.000

0.000

0.000

0.000

15

Example: DistributionUsing Table B for n = 5 and p = 0.90, find the following: a) The probability of exactly 3 successesb) The probability of at least 3 successes

a) P(3) = 0.073

b) P(at least 3) = P(3 or 4 or 5)

= P(3) or P(4) or P(5)

= 0.073 + 0.328 + 0.590

= 0.991

Example Distribution: Find the probability of getting exactly 3 correct responses among 5 different requests from AT&T directory assistance. Assume in general, AT&T is correct 90% of the time.

This is a binomial experiment where:

n = 5

x = 3

p = 0.90

q = 0.10

Using the binomial probability chart to solve:

P(3)= 0.0.0729

Notation for Binomial Probability Distributions Distribution

n = fixed number of trials

x = specific number of successes in n trials

p = probability of success in one of n trials

q = probability of failure in one of n trials

(q = 1 - p)

P(x) = probability of getting exactly x success among n trials

Be sure that x and p both refer to the same category being called a success.

Binomial Probability Formula Distribution

n!

P(x) = • px•qn-x

(n - x )! x!

Probability of x successes among n trials for any one particular order

Number of

outcomes with exactly x successes among n trials

Method 1 Distribution

Binomial Probability Formulan!

- P(x) = • px• qn-x

(n - x )! x!

- P(x) = nCx• px•qn-x

for calculators with nCr key, where r = x

Example Distribution: Find the probability of getting exactly 3 correct responses among 5 different requests from AT&T directory assistance. Assume in general, AT&T is correct 90% of the time.

This is a binomial experiment where:

n = 5

x = 3

p = 0.90

q = 0.10

Using the binomial probability formula to solve:

P(3) = 5C3• 0.9 • 01 = 0.0.0729

2

3

Example: DistributionFind the mean and standard deviation for the number of girls in groups of 14 births.

- We previously discovered that this scenario could be considered a binomial experiment where:
- n = 14
- p = 0.5
- q = 0.5
- Using the binomial distribution formulas:
µ = (14)(0.5) = 7 girls

= (14)(0.5)(0.5) = 1.9 girls (rounded)

Example: DistributionDetermine whether 68 girls among 100 babies could easily occur by chance.

- For this binomial distribution,
- µ = 50 girls
= 5 girls

- µ + 2 = 50 + 2(5) = 60
- µ - 2 = 50 - 2(5) = 40
- The usual number girls among 100 births would be from 40 to 60. So 68 girls in 100 births is an unusual result.

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