discrete random variables & binomial distributioncathy/math3339/lecture/lecture5...discrete...

Discrete Random Variables & Binomial DistributionSec 4.1 - 4.6

Cathy Poliak, [email protected] in Fleming 11c

Department of MathematicsUniversity of Houston

Lecture 5 - 3339

Cathy Poliak, Ph.D. [email protected] Office in Fleming 11c (Department of Mathematics University of Houston )Sec 4.1 - 4.6 Lecture 5 - 3339 1 / 42

Outline

1 Random Variables

2 Probability Distribution

3 Expected Value

4 Standard Deviation

5 Chebyshev’s inequality

6 Rules for Means and Variances

7 Bernoulli Random Variables

8 Binomial Distribution

9 Cumulative Distributions


Review Questions

For each random variable, determine if it is:

a. Discrete b. Continuous

1. The number of cars passing a busy intersection between 4:30 PMand 6:30 PM.

2. The weight of a fire fighter.

3. The amount of soda in a can of Pepsi.


Random Variables

A random variable is a variable whose value is a numericaloutcome of a random phenomenon.Examples

I X = the sum of two dice, X = 2,3,4, . . . ,12.I X = the number of customers that order a muffin in a coffee shop

between 7:00 am and 9:00 am, X = 0,1, . . ..I X = weight of a box of Lucky Charms, X ≥ 0.

Random variables can either be discrete or continuous.


Discrete random variables

Discrete random variables has either a finite number of valuesor a countable number of values, where countable refers to thefact that there might be infinitely many values, but they result froma counting process.Example of discrete random variable:

I The sum of two dice.I The number of customers who order a muffin in a coffee shop

between 7:00 am and 9:00 am.

The possible values for a discrete random variable has "gaps"between each value.


Continuous random variables

Continuous random variables are random variables that canassume values corresponding to any of the points contained inone or more intervals.Example of continuous random variable:

I The weight of a box of Lucky Charms.

We want to know how both of these types of random variables aredistributed.


Probability Distribution

The probability distribution of a random variable X tells us whatvalues X can take and how to assign probabilities to those values. Thisis the "ideal" distribution for a random variable. Requirements for aprobability distribution:

1. The sum of all the probabilities equal 1.2. The probabilities are between 0 and 1, including 0 and 1.

We will determine the three characteristics of a probability distribution:shape, center, and spread.


Discrete Probability Distribution

The probability distribution of a discrete random variable X liststhe possible values of X and their probabilities. This is called aprobability mass function, pmf or frequency function.A probability distribution table of X consists of all possiblevalues of a discrete random variable with their correspondingprobabilities.


Example

Suppose a family has 3 children. What is the sample space of allpossible ways for gender?

Now suppose we want to the probability of the number of girls in afamily of 3 children, X would be:

What is the probability that in a family of 3 children, they are all girls?


Family of three

Given the probability distribution table of X for the number of girls in afamily of 3 children. Find P(X > 2), P(X < 1) and P(1 < X ≤ 3)

X = number of girls 0 1 2 3P(X) 0.125 0.375 0.375 0.125


Probability Distribution

Suppose you want to play a carnival game that costs 8 dollars eachtime you play. If you win, you get $100. The probability of winning is

3100 . Give the probability distribution, pmf, of the amount that you, theplayer, stand to gain.


Determining Center of a Discrete Random Variable

Suppose that X is a discrete random variable whose distribution is

Values of X x1 x2 x3 · · · xkProbability p1 p2 p3 · · · pk

To find the mean of the random variable X , multiply each possiblevalue by its probability, then add all the products:

µX = E [X ] = x1p1 + x2p2 + x3p3 + · · ·+ xkpk .

This is also called the expected value E [X ].Note: The list here is not a list of observations but a list of all possibleoutcomes. So we are finding µ, the population mean not x̄ , the samplemean.


Determine the Expected Value

The following is a probability distribution.

X 0 1 2 3P(X ) 0.7 0.1 0.05 0.15

Find the expected value (mean) of this probability distribution.


Example 2

Let X=The number of traffic accidents daily in a small city. Thefollowing table is the probability distribution for X .

X Probability0 0.101 0.202 0.453 0.154 0.055 0.05


Determining the spread of a Discrete Random Variable

The variance of a discrete random variable X is

σ2X = (x1 − µX )2p1 + (x2 − µX )2p2 + · · ·+ (xk − µX )2pk

=k∑

i=1

(xi − µX )2pi

The standard deviation of X is the square root of the variance

σX =√σ2

X .


Easier Calculation for Variance and StandardDeviation from a Discrete Probability Distribution

VAR(X) = σ2x = E(X 2)− [E(X )]2

Where E(X 2) = x21 p1 + x2

2 p2 + x23 p3 + . . .+ x2

n pn.


Determine the Standard Deviation

The following is a probability distribution.

X 0 1 2 3P(X ) 0.7 0.1 0.05 0.15

Find the standard deviation of this probability distribution.


Finding Expected Value and Standard Deviation in R

> x=c(0,1,2,3)> px=c(.7,.1,.05,.15)> ex = sum(x*px)> ex[1] 0.65> ex2 = sum(x^2*px)> ex2[1] 1.65> vx = ex2-ex^2> vx[1] 1.2275> sx = sqrt(vx)> sx[1] 1.107926


Determine Standard Deviation of Number of Accidents

Determine the standard deviation of the number of accidents on agiven day. The expected value is E [X ] = 2.

X 0 1 2 3 4 5P(X ) 0.10 0.20 0.45 0.15 0.05 0.05


Chebyshev’s Inequality

Chebyshev’s inequality, places a universal restriction on theprobabilities of deviations of random variables from their means.

If X is a random variable with mean µ and standard deviation σand if k is a postivie constant, then

P(|X − µ| > kσ) ≤ 1k2

That is: For any random variable, if we are k standard deviationsaway from the mean, then no more than 1

k2 ∗ 100% is beyond kstandard deviations from the mean. Or at least (1− 1

k2 ) ∗ 100% iswithin k standard deviations from the mean.


Chebyshev’s Inequality Application

Let X=The number of traffic accidents daily in a small city. Thefollowing table is the probability distribution for X . With, µ = 2 andσ = 1.18.

X Probability0 0.101 0.202 0.453 0.154 0.055 0.05

Confirm Chebyshev’s Inequality for k = 3.


Example

Suppose that in the small city on a given day there was rain. So therewould be at least one accident, the probability distribution of thenumber of accidents will be:

Number of accidents 1 2 3 4 5 6Probability 0.10 0.20 0.45 0.15 0.05 0.05

Compute the mean number of accidents.

Compute the variance of the number of accidents.


Rule for Means and Variances

If X is any random variable and a is a fixed numbers that is addedto all of the values of the random variable thenthe mean increases by that number:

E [a + X ] = a + E [X ].

the variance remains the same:

σ2a+X = σ2

X .

In the example of a rainy day, we added 1 accident to each value,1 + X and the probabilities remained the same.

E [1 + X ] = 1 + E [X ] = 1 + 2 = 3

σ21+X = σ2

X = 1.4


Rule for Means and Variances

If X is any random variable and b is a fixed numbers that ismultiplied to all of the values of the random variable thenthe mean is changed by that multiplier:

E [bX ] = bE [X ].

the variance is also changed by the square of the multiplier:

σ2bX = b2σ2

X .


Rule Means and Variances

Suppose X is a random variable with mean E [X ] and variance σ2X , and

we define W as a new variable such that W = a + bX , where a and bare real numbers. We can find the mean and variance of W by:

E [W ] = E [a + bX ] = a + bE [X ]

σ2W = Var [W ] = Var [a + bX ] = b2Var [X ]

σW = SD[W ] =√

Var [W ] =√

b2Var [X ] = |b|(SD[X ])


Example

Suppose you have a distribution X , with mean = 22 and standarddeviation = 3. Define a new random variable Y = 4X + 1.

1. Find the variance of X .

2. Find the mean of Y .

3. Find the variance of Y .

4. Find the standard deviation of Y .Cathy Poliak, Ph.D. [email protected] Office in Fleming 11c (Department of Mathematics University of Houston )Sec 4.1 - 4.6 Lecture 5 - 3339 26 / 42

Introduction Question

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. 0

2. What is the probability that a customer is not selected for audit?a. 0.09 b. 0.91 c. 1 d. 0


Bernoulli Random Variable

A Bernoulli random variable has only two possible values, usuallydesignated as 1 and 0.

Suppose we look at the i th customer of Wallen Accountingservices. Let X be the random variable that indicates that thecustomer is selected for audit. Thus X = 1 is the customer isselected for audit, X = 0 if not selected for audit. Selected foraudit is the "success" and not selected for audit is "failure."

The probability of success is p and the probability of failure is1− p. e.g. a coin is flipped (heads or tails), someone is eitheraudited or not audited, p = 0.09, 1 - p = 0.91.


Probability Function for Bernoulli Variable

f (x) = P(x) =

p, if x = 11− p, if x = 00, if x 6= 0,1

A compact way of writing this is:

f (x) = P(x) = px (1− p)1−x


The Bernoulli Process

The Bernoulli process must possess the following properties:1. The experiment consists of n repeated trials.

2. Each trial in an outcome that may be classified as a success orfailure.

3. The probability of success, denoted by p, remains constant fromtrial to trial.

4. The repeated trials are independent.


Audit Example

Today, Wallen has six new customers. Assume the chances of thesesix customers being audited are independent. This is a sequence ofBernoulli trials. We are interested in calculation the probability ofobtaining a certain number of people being selected for audit. Let Xiindicate the i th customer being selected for audit.Let Y = X1 + X2 + X3 + X4 + X5 + X6. What does Y represent?

What is the probability that Y = 0?



What is the probability that Y = n where n = 0,1,2,3,4,5,6?Cathy Poliak, Ph.D. [email protected] Office in Fleming 11c (Department of Mathematics University of Houston )Sec 4.1 - 4.6 Lecture 5 - 3339 31 / 42

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


Stopping at an intersection

Suppose that only 25% of all drivers come to a complete stop at anintersection with a stop sign when not other cars are visible. What isthe probability that of the 20 randomly chosen drivers,

1. Exactly 6 will come to a complete stop?

2. No one will come to complete stop?

3. At least one will come to a complete stop?

4. At most 6 will come to a complete stop?


Using R

To find P(X =k) use dbinom(k,n,p).

From previous example P(X = 6), k = 6, n = 20, p = 0..25> dbinom(6,20,.25)[1] 0.1686093

To find P(X ≤ k) use pbinom(k,n,p).

From previous example P(X ≤ 6)

> pbinom(6,20,0.25)[1] 0.7857819

What is the probability that less than 6 will come to a completestop? What is the probability of at least 6?


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

4. What is the probability that the coin comes up heads between 9and 13 times, inclusive? P(9 ≤ X ≤ 13) = P(X ≤ 13)− P(X ≤ 8)

> pbinom(13,30,0.5)-pbinom(8,30,.5)[1] 0.28427


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 ] = np

σ2X = Var [X ] = np(1− p)

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


Example #3

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. Find the mean

2. Find the variance of this distribution.


Cumulative Distributions

Any quantitative random variable X has a cumulative distributionfunction defined as

FX (x) = P(X ≤ x)

for all real numbers x .

For discrete random variables the relationship between the probabilityfunction, pmf, and the cumulative distribution function is

FX (x) =∑xi≤x

fX (xi)

where x1, x2, . . . are the values of X .


Example of Cumulative Distribution

What is the probability that at most 2 people will contract the flu?

P(X ≤ 2) = F (2) = P(X = 0) + P(X = 1) + P(X = 2)

= sum(dbinom(0 : 2,5,0.8))

or in RStudio: F (2) = pbinom(2,5,0.8).


Cumulative Distribution Function Properties

Any cumulative distribution function F has the following properties:1. F is a non-decreasing function defined on the set of all real

numbers.

2. F is right-continuous. That is, for each a,F (a) = F (a+) = limx→a+ F (x).

3. limx→−∞ F (x) = 0; limx→+∞ F (X ) = 1.

4. P(a < X ≤ b) = Fx (b)− Fx (a) for all real a and b, a < b.

5. P(X > a) = 1− Fx (a).

6. P(X < b) = Fx (b−) = limx→b− Fx (x).

7. P(a < X < b) = Fx (b−)− Fx (a).

8. P(X = b) = Fx (b)− Fx (b−).Cathy Poliak, Ph.D. [email protected] Office in Fleming 11c (Department of Mathematics University of Houston )Sec 4.1 - 4.6 Lecture 5 - 3339 41 / 42

Plot of Cumulative Distribution

0 1 2 3 4 5

0.0

0.2

0.4

0.6

0.8

1.0

x

flu.p

rob


discrete random variables & binomial distributioncathy/math3339/lecture/lecture5...discrete...

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