chapter 5 discrete probability distributions. random variable a numerical description of the result...

Chapter 5

Discrete Probability Distributions

Random Variable

A numerical description of the result of an experiment.

Discrete Random Variable

A random variable that can only assume some finite number of values.

Continuous Random Variable

A random variable that can assume any numerical value in an interval.

Discrete Probability Distribution

A list of the possible values of a discrete random variable and the associated probabilities.

Value Probability

1 1/6

2 1/6

3 1/6

4 1/6

5 1/6

6 1/6

Discrete Probability Function, f(x)

A function that provides the probability for each value of the discrete random variable

Required Conditions for a Discrete Probability Function

f(x) > 0 for all xSf(x) = 1

Discrete Uniform Probability Function

f(x) = 1/n

Where n is the number of values the random variable can assume.

Expected Value of a Random Variable

The average value of the variable over an infinite number of experiments.

E(x) = m = Sxf(x)

Example

Suppose your population consisted of 100 families with children where:

Number of Children

Number of Families

1 302 503 20

Example, cont.

Using the formula for the weighted mean, we could write:m = (Swixi)/(Swi)= [(1)(30)+(2)(50)+(3)(20)]/[30+50+20]= (1)(30/100)+(2)(50/100)+(3)(20/100)= (1)(.3)+(2)(.5)+(3)(.2)More generally: m = [S xi(wi/Swi)] = Sxf(x)

Variance of a Random Variable

The sum of the squared deviations from the mean weighted by the probabilities of a value occurring.

Var(x) = s2 = (S x-m)2f(x)

Standard Deviation of a Random Variable

xfx 2

Random Variable, Example

Assume a random variable can take on the following three values with the corresponding probabilities:x f(x)1 0.12 0.43 0.5

Find the expected value and variance of the random variable

Random Variable, Example

E(x) = (1)(.1) + (2)(.4) + (3)(.5) = 2.4

s2 = (1 - 2.4)2(.1) + (2 - 2.4) 2(.4) + (3 - 2.4) 2(.5) = (1.96)(.1) + (.16)(.4) + (.36)(.5)= .196 + .064 + .18= .44

Random Variable, Example

Assume the cost of participating in a “fifty-fifty” lottery is $5 and 100 people have purchased tickets. What is the expected value of purchasing a ticket?

E(x) = (.99)(-5) + (.01)(245) = -4.95 + 2.45 = -2.50

Random Variable, Example

Random variables.xlsx

Binomial Experiment

1. The experiment consists of a sequence of n identical trials.

2. Two outcomes are possible on each trial, one outcome is labeled success the other failure.

3. The probability of success, denoted by p, does not change from trial to trial.

4. The trials are independent.

Binomial Probability Function

xnx ppxnx

nxf

1

)!(!

!)(

Binomial Example

A die will be rolled three times. Success is defined as rolling a 1 or a 2. Failure is defined as rolling a 3 or higher. Assume we want to find the probability of having one success.

Given that definition:x = 1n = 3p = 1/3(1 - p) = 2/3

Binomial Example, Cont.

Success,1/3Success,1/3Success,1/3Failure, 2/3Failure, 2/3Success,1/3Failure, 2/3Failure, 2/3Success,1/3Success,1/3Failure, 2/3Failure, 2/3Success,1/3Failure, 2/3

Binomial Example, Cont.

Success,1/3Success,1/3Success,1/3Failure, 2/3Failure, 2/3Success,1/3Failure, 2/3Failure, 2/3Success,1/3Success,1/3Failure, 2/3Failure, 2/3Success,1/3Failure, 2/3

Binomial Example, Cont.

The portion of the binomial formula:px(1-p)(n-x)

represents the probability of going down one branch where there are x successes

In this case:(1/3)1(2/3)2

Binomial Example, Cont.

The other portion of the formula calculates how many ways we can have a given number of successes and failures.

n!x!(n-x)!

Getting x successes in n trials can be thought of as how many ways can x items be drawn from a group of n; a questions answered by the combinations formula.

Binomial Example, Cont.

In this case:S F F Position 1 is drawnF S F Position 2 is drawnF F S Position 3 is drawn

n! = 3! = 3! = 3x!(n-x)! 1!(3-1)! 2!

Expected Value and Variance for the Binomial Distribution

E(x) = m = npVar(x) = s2 = np(1 – p)

Practice

Assume the probability of getting a bad part (success) is 20 percent and the probability of getting a good part (failure) is 80 percent.

1. If we draw 6 parts and test them, what is probability of drawing 3 bad parts?

Practice, cont.

0.08192

0.5120.008123

4568.2.

!3!3

!6

2.12.)!36(!3

!6

1)!(!

!)(

3.3

363

xnx ppxnx

nxf

Practice, cont.

2. If we drawing at least 2 bad parts (given we draw 6)?

Practice, cont.

0.344640.2621440.3932161

0.262144110.327680.261

8.2.!6!0

!68.2.

!5!1

!61

2.12.)!06(!0

!62.12.

)!16(!1

!61

011)2(

6.05.1

060161

ffxf

Practice, cont.

3. What is the expected value of the distribution? The variance?

Practice, cont.

0.968.2.6)1(

48.)8(.6)(

pnp

npxE

Graded Homework

P. 199-200, #21, 23P. 209-210, #29,31