Binomial and Geometric Distributions - UHcathy/Math2311/Lectures... · Geometric Distribution The geometric distribution is the distribution produced by the random variable X deﬁned

Binomial and Geometric DistributionsSection 3.2 & 3.3

Cathy Poliak, [email protected]

Office hours: T Th 2:30 pm - 5:15 pm 620 PGH

Department of MathematicsUniversity of Houston

February 11, 2016

Cathy Poliak, Ph.D. [email protected] Office hours: T Th 2:30 pm - 5:15 pm 620 PGH (Department of Mathematics University of Houston )Section 3.2 & 3.3 February 11, 2016 1 / 23

Outline

1 Beginning Questions

2 Binomial Distribution

3 Geometric Distribution


Popper Set Up

Fill in all of the proper bubbles.

Use a #2 pencil.

This is popper number 04.


Popper Questions

Wallen Accounting Services specializes in tax preparation forindividual tax returns. Data collected from past records reveals that 9%of the returns prepared by Wallen have been selected for audit by theInternal Revenue Service (IRS).

1. What is the probability that a customer of Wallen will be selectedfor audit?

a. 0.09 b. 0.91 c. 1 d. 02. Today, Wallen has six new customers. Assume the chances of

these six customers being audited are independent. Let X = thenumber of customers out of the six will be audited. What is theprobability that none of the six will be selected for audit? That isP(1st is not audited "and" 2nd is not audited "and" 3rd is notaudited "and" 4th is not audited "and" 5th is not audited "and" 6this not audited).

a. 0 b. 1 c. 0.568 d. 0.432


Audit Example

Let X be the number out of 6 customers that will be audited.

The possible values of X are X = {0,1,2,3,4,5,6}

What is the probability that exactly one out of the 6 customers willbe audited?


Audit Example

From this example we can note a couple of things.1. We are only looking at six customers, n = 6.

2. We are assuming the chances of these six customers beingaudited are independent.

3. For each customer, they will either be selected for audit or not beselected for audit.

4. We have the same probability of a person being selected for audit,p = 0.09 for each customer.

This type of setting occurs quite frequently that we can put amathematical model to these types of probability.


Binomial Setting

The previous example falls into a Binomial Setting which followsthese 4 rules.

1. There are a fixed number n of observations.2. The n observations are all independent. That is, knowing the

result of one observation tells you nothing about the otherobservations.

3. Each observation falls into one of just two categories, we call“success” and “failure.”

4. The probability of a success p is the same for each observation.


Popper Questions

Do these scenarios have a binomial setting?3. Rolling a die 50 times.

a. Yes b. No4. Rolling a die 50 times and finding the number of times that 5

occurs.a. Yes b. No

5. Determining the number of heads up out of three flips.a. Yes b. No


Binomial Probability Distribution

The distribution of the count X of successes in the Binomialsetting has a Binomial probability distribution.

Where the parameters for a binomial probability distribution is:I n the number of observationsI p is the probability of a success on any one observation

The possible values of X are the whole numbers from 0 to n.

As an abbreviation we say, X ∼ B(n,p).

Binomial probabilities are calculated with the following formula:

P(X = k) =n Ckpk (1− p)n−k


Steps to fining the probability

1. Make sure the assumptions are met for the Binomial setting.

2. Describe the success.

3. Determine p, the probability of success.

4. Determine n, the number of trials.

5. Put the probability question in terms of X. That is, P(X = x) orP(X ≤ x) or P(X ≥ x) or P(X < x) or P(X > X ).

6. Determine the probability. We can do this through the formula,inour calculator or R.


Using the Binomial Formula

With the example of Wallen customers being selected for audit, n = 6and p = 0.09. What is the probability that exactly one of the six will beselected for audit?

1. We have independence.

2. Success is being audited.

3. p = 0.09

4. n = 6

5. We want P(X = 1). We have met the 4 conditions of the binomialsetting so we can use the formula: P(X = k) =n Ckpk (1− p)(n−k).

P(X = 1) =6 C1(0.09)1(0.91)5 = 0.337


Using R or TI-83\84

To find probabilities in R.I To find P(X =k) use dbinom(k,n,p).I From previous example P(X = 1), k = 1, n = 6, p = 0.09> dbinom(1,6,0.09)[1] 0.3369774

I To find P(X ≤ k) use pbinom(k,n,p).

To find probabilities in TI-83\ 84 use the DISTR button this willgive you a list of probability distributions.

I To find P(X = k) use binompdf(n,p,k).I To find P(X ≤ k) use binomcdf(n,p,k).


Example #2

A fair coin is flipped 30 times.1. What is the probability that the coin comes up heads exactly 12

times? P(X = 12), n = 30, p = 0.5> dbinom(12,30,0.5)[1] 0.08055309

2. What is the probability that the coin comes up heads less than 12times? P(X < 12) = P(X ≤ 11)> pbinom(11,30,0.5)[1] 0.1002442

3. What is the probability that the coin comes up heads more than 12times? P(X > 12) = 1− P(X ≤ 12)> 1-pbinom(12,30,0.5)[1] 0.8192027


Mean and Variance of a Binomial Distribution

If a count X has the Binomial distribution with number of observationsn and probability of success p, the mean and variance of X are

µX = E [X ] = npσ2

X = Var [X ] = np(1− p)

Then the standard deviation is the square root of the variance.


From text 3.2 # 17

Suppose it is known that 80% of the people exposed to the flu virus willcontract the flu. Out of a family of five exposed to the virus, what is theprobability that:

1. No one will contract the flu?

2. All will contract the flu?

3. Exactly two will get the flu?

4. At least two will get the flu?


Example Continued

Suppose it is known that 80% of the people exposed to the flu virus willcontract the flu. Suppose we have a family of five that were exposed tothe flu.

1. Let X = number of family members contacting the flu. Create theprobability distribution table of X .

2. Find the mean and variance of this distribution.


Popper Question

Fifty percent of married couples paid for their honeymoon themselves.You randomly select 10 married couples and ask each if they paid fortheir honeymoon themselves.

6. What is the probability that exactly 5 out of the 10 paid for thehoneymoon themselves.

a. 0.5 b. 0.00488 c. 0.0009 d. 0.24617. What is the expected value, µX of the number of married couples

that paid for their honeymoon themselves out of 10 couplesasked?

a. 5 b. 10 c. 0 d. 5.58. What is the standard deviation, σX , of the number of married

couples that paid for their honeymoon themselves out of the 10couples asked?

a. 5 b. 2.5 c. 1.5811 d. 6.25


Geometric Distribution

The geometric distribution is the distribution produced by therandom variable X defined to count the number of trials needed toobtain the first success.

Examples: Flipping a coin until you get a head, Rolling a die untilyou get a 5.


Conditions

A random variable X is geometric if the following conditions are met:1. Each observation falls into one of just two categories, "success" or

"failure."

2. The probability of success is the same for each observation.

3. The variable of interest is the number of trials required to obtainthe first success.


Formula for Geometric Distribution

The probability that the first success occurs on the nth trial is:

P(X = n) = (1− p)n−1p

Where p is the probability of success.The probability that it takes more than n trials to see the firstsuccess is:

P(X > n) = (1− p)n

R commands:P(X = n) = dgeom(n-1,p) andP(X > n) = 1− pgeom(n − 1,p)TI-83\84: P(X = n) = geometpdf(p,n) andP(X > n) = 1− geomtcdf(p,n)


Mean and Variance of a Geometric Distribution

If a count X has the Geometric distribution with probability of successp, the mean and variance of X are

µx = E [X ] =1p

σ2x = Var [X ] =

1− pp2

The standard deviation is the square root of the variance.


From Text section 3.3 #8

A quarterback completes 44% of his passes. We want to observe thisquarterback during one game to see how many passes he makesbefore completing one pass.

1. What is the probability that the quarterback throws 3 incompletepasses bfore he has a completion?

2. How many passes can the quarterback expect to throw before hecompletes a pass?

3. Determine the probability that it takes more than 5 attempts beforehe completes a pass.

4. What is the probability that he attempts more than 7 passesbefore he completes one?


From Text Section 3.3 #14

Newsweek in 1989 reported that 60% of young children have bloodlead levels that could impair their neurological development. Assuminga random sample from the population of all school children at risk, find:

1. The probability that at least 5 children out of 10 in a sample takenfrom a school may have a blood lead level that may impairdevelopment.

2. The probability you will need to test 10 children before finding achild with a blood lead level that may impair development.

3. The probability you will need to test no more than 10 childrenbefore finding a child with a blood lead level that may impairdevelopment.


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Binomial and Geometric Distributions - UHcathy/Math2311/Lectures... · Geometric Distribution The geometric distribution is the distribution produced by the random variable X deﬁned