Chapter 16

Week 6, Monday

Random Variables“A numeric value that is based on the outcome of a random event”

Example 1: Let the random variable X be defined as the number rolled on a fair die.

X P[X]1 1/6 = 17%2 1/6 = 17%3 1/6 = 17%4 1/6 = 17%5 1/6 = 17%6 1/6 = 17%

P[X=2] = 17%

P[X>2] = 17%+17%+17%+17%=68%

P[X=2.5] = 0%

This is called a“Probability Distribution”for the random variable, X

Random Variables“A numeric value that is based on the outcome of a random event”

Example 2: Let the random variable X be defined as the number of heads shown in three flips of a fair coin.

X P[X]0 1/8 = 12.5%1 3/8 = 37.5%2 3/8 = 37.5%3 1/8 = 12.5%

P[X=2] = 37.5%

P[X<2] = 12.5%+37.5% = 50%

P[X=2.5] = 0%

Random Variables“A numeric value that is based on the outcome of a random event”

Example 3: Let the random variable X be defined as the payout for a one year life insurance policy of $1,000,000 for a college student.

Outcome Label X P[X]Policyholder Dies $1,000,000 .01%Policyholder Lives $0 99.99%

P[X=0] = 99.99%

P[X<1,000,001] = .01%+99.99% = 100%

P[X=3,000] = 0%

NOTE: These are called “discrete random variables”,because X can only take ona finite number of values.

Later we will talk about adifferent type of random variable

Random Variables“A numeric value that is based on the outcome of a random event”

Example 3: Let the random variable X be defined as the payout for a one year life insurance policy of $1,000,000 for a college student.

Outcome Label X P[X]Policyholder Dies $1,000,000 .01%Policyholder Lives $0 99.99%

Suppose this random variable describes the payout policy for any traditional student at the University of Akron.Question: If a company decides to issue this policy to a random traditional student, what can they expect to lose on average?

μ = ][xxP Called: “Expected Value”Sometimes represented by E[X]

Random Variables“A numeric value that is based on the outcome of a random event”

Example 3: Let the random variable X be defined as the payout for a one year life insurance policy of $1,000,000 for a college student.

Outcome Label X P[X] xP[x]Policyholder Dies $1,000,000 .01% $1,000Policyholder Lives $0 99.99% $0

μ = ][xxP = $100 + $0 = $100

“A company can expect to pay out $100 per policyholder”

Recall: μ is the symbol for the population mean. Why is this appropriate?

Random Variables“A numeric value that is based on the outcome of a random event”

Example 3: Let the random variable X be defined as the payout for a one year life insurance policy of $1,000,000 for a college student.

Outcome Label X P[X] xP[x]Policyholder Dies $1,000,000 .01% $1,000Policyholder Lives $0 99.99% $0

μ = ][xxP = $100 + $0 = $100

There is a similar calculation for population standard deviation:

σ2 = ][)( 2 xPx μ Sometimes represented by VAR[X]Of course: σ = sqrt(σ2)

(Sometimes represented by SD[X])

Distributions to MemorizeSo far: the distributions were given to you, and you were asked to calculated probabilities (like P[X<3]), expected values, and variances.

Alternatively, sometimes you will be given a word problem and asked to derive the proper distribution (both x and P[x]).

You are responsible for two discrete distributions:(1) Equal-likelihood Distributions(2) Binomial Distributions

Distributions to Memorize(1) Equal-likelihood Distributions: When each of the k values for x has an equal likelihood of being chosen, P[x] = 1/k

Example 1: Roll a fair die. Each of the 6 possibilities is equally likely

X P[X]1 1/6 = 17%2 1/6 = 17%3 1/6 = 17%4 1/6 = 17%5 1/6 = 17%6 1/6 = 17%

Distributions to Memorize(1) Equal-likelihood Distributions: When each of the k values for x has an equal likelihood of being chosen, P[x] = 1/k

Example 2: Consider a deck of cards. Label jack=11, queen=12, king=13, and ace=14. Choose a card from a deck. There is an equal likelihood of picking #2 through #14 (13 possibilities). So P[x]=1/13 for x=2,3,…,14

X P[X]

2 1/13 = 7.7%3 1/13 = 7.7%…. ….12 1/13 = 7.7%13 1/13 = 7.7%14 1/13 = 7.7%

Distributions to Memorize(2) Binomial Distributions: Suppose you have n trials that are independent of each other and have the same sample space. Define an event, E, in that sample space as being successful. Let P[E]=p. Define the random variable X as the number of successes in n trials. This random variable follows a specific formula and is called the binomial distribution.

Example 1: Flip a coin 3 times. Define X as the number of heads observed.

Heads

Trial 1Tails

Heads

Trial 2Tails

Heads

Trial 3Tails Independent because,

for example, the result of the first toss does

not affect the result of the second toss.

n=3

Same Sample SpaceDefine E = {heads}

Then P[E] = p = 50%

X = Number of heads in 3 trials

^^ which was what we wanted ^^

This follows a binomial distribution (n=3, p=.5)

Distributions to Memorize(2) Binomial Distributions: Suppose you have n trials that are independent of each other and have the same sample space. Define an event, E, in that sample space as being successful. Let P[E]=p. Define the random variable X as the number of successes in n trials. This random variable follows a specific formula and is called the binomial distribution.

Example 2: There is a popquiz today; it is multiple choice, but you didn’t study and have to guess all the answers. There are 3 questions and 4 choices for each question. Define X as the number of right answers.

Wrong Ans

Trial 1Right Ans

Trial 2 Trial 3Right Ans Right Ans

Wrong Ans Wrong Ans

n=3Independent because

guessing (in)correctly for one question does not

change your chances of guessing another answer.

Same Sample SpaceDefine E = {Right Ans}Then P[E] = p = 25%

X = Number of right answers out of 3^^ which was what we wanted ^^

This follows a binomial distribution (n=3, p=.25)

Distributions to Memorize(2) Binomial Distributions: Suppose you have n trials that are independent of each other and have the same sample space. Define an event, E, in that sample space as being successful. Let P[E]=p. Define the random variable X as the number of successes in n trials. This random variable follows a specific formula and is called the binomial distribution.

If you recognize a distribution as binomial (n,p), then the distribution follows a complicated formula:

X P[X]0 nC0p0(1-p)n-0

1 nC1p1(1-p)n-1

… …n-1 nCn-1pn-1(1-p)n-(n-1)

n nCnpn-0(1-p)n-n

DO NOT MEMORIZE THIS FORMULA!

You will be given the distribution table.

Your job is to:(1) Recognize a situation as “binomial” (n,p)(2) Understand what n and p actually mean(3) Know how to solve problems with the table(4) Memorize the mean (np) and variance (npq)