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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.

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Presentation on theme: "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."— Presentation transcript:

1 Chapter 16 Week 6, Monday

2 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. XP[X] 11/6 = 17% 2 3 4 5 6 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

3 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. XP[X] 01/8 = 12.5% 13/8 = 37.5% 2 31/8 = 12.5% P[X=2] = 37.5% P[X<2] = 12.5%+37.5% = 50% P[X=2.5] = 0%

4 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 LabelXP[X] Policyholder Dies$1,000,000.01% Policyholder Lives$099.99% 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 on a finite number of values. Later we will talk about a different type of random variable

5 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 LabelXP[X] Policyholder Dies$1,000,000.01% Policyholder Lives$099.99% 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 = Called: “Expected Value” Sometimes represented by E[X]

6 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 LabelXP[X]xP[x] Policyholder Dies$1,000,000.01%$1,000 Policyholder Lives$099.99%$0 μ == $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?

7 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 LabelXP[X]xP[x] Policyholder Dies$1,000,000.01%$1,000 Policyholder Lives$099.99%$0 μ == $100 + $0 = $100 There is a similar calculation for population standard deviation: σ 2 = μ Sometimes represented by VAR[X] Of course: σ = sqrt(σ 2 ) (Sometimes represented by SD[X])

8 Distributions to Memorize So 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

9 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 XP[X] 11/6 = 17% 2 3 4 5 6

10 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 XP[X] 21/13 = 7.7% 3 …. 121/13 = 7.7% 131/13 = 7.7% 141/13 = 7.7%

11 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 1 Tails Heads Trial 2 Tails Heads Trial 3 Tails Independent because, for example, the result of the first toss does not affect the result of the second toss. n=3 Same Sample Space Define 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)

12 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 1 Right Ans Trial 2Trial 3 Right Ans Wrong Ans n=3 Independent because guessing (in)correctly for one question does not change your chances of guessing another answer. Same Sample Space Define 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)

13 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: XP[X] 0 n C 0 p 0 (1-p) n-0 1 n C 1 p 1 (1-p) n-1 …… n-1 n C n-1 p n-1 (1-p) n-(n-1) n n C n p n-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)


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