Statistics Class 10

2/29/2012

Review

When playing roulette at the Bellagio casino in Las Vegas, a gambler is trying to decide whether to bet $5 on the number 13 or to bet $5 that the outcome is any one of these five possibilities: 0 or 00 or 1 or 2 or 3. From Example 8, we know that the expected value of the $5 bet for a single number is -26 . For the $5 bet that the outcome is 0 or 00 or ₵1 or 2 or 3, there is a probability of 5/38 of making a net profit of $30 and a 33/38 probability of losing $5.a. Find the expected value for the $5 bet that the outcome is 0 or 00 or 1 or 2 or 3.b. Which bet is better: A $5 bet on the number 13 or a $5 bet that the outcome is 0 or 00 or 1 or 2 or 3? Why?

Binomial Probability DistributionsA binomial probability distribution results from a procedure that meets all the following requirements.

Binomial Probability DistributionsA binomial probability distribution results from a procedure that meets all the following requirements.1. The procedure has a fixed number of trials.

Binomial Probability DistributionsA binomial probability distribution results from a procedure that meets all the following requirements.1. The procedure has a fixed number of trials.2. The trials must be independent. (The outcome of any individual trial doesn’t affect the probabilities in the other trials.)

Binomial Probability DistributionsA binomial probability distribution results from a procedure that meets all the following requirements.1. The procedure has a fixed number of trials.2. The trials must be independent. (The outcome of any individual trial doesn’t affect the probabilities in the other trials.)3. Each trial must have all outcomes classified into two categories (commonly referred to as success and failure).

Binomial Probability DistributionsA binomial probability distribution results from a procedure that meets all the following requirements.1. The procedure has a fixed number of trials.2. The trials must be independent. (The outcome of any individual trial doesn’t affect the probabilities in the other trials.)3. Each trial must have all outcomes classified into two categories (commonly referred to as success and failure).4. The probability of a success remains the same in all trials.

Binomial Probability DistributionsNote on IndependenceOften when selecting a sample we do so without replacement. This means that our events are dependent, and violate rule 2 of the binomial probability distribution. However we can use the 5% guideline for cumbersome calculations, and treat dependent events independent as long as the sample size is no more than 5% of the population size.

Binomial Probability DistributionsNotation for a Binomial Probability DistributionS and F (success and failure) denote the two possible categories of outcomes.

Binomial Probability DistributionsNotation for a Binomial Probability DistributionS and F (success and failure) denote the two possible categories of outcomes.• P(S)=p p=probability of success

Binomial Probability DistributionsNotation for a Binomial Probability DistributionS and F (success and failure) denote the two possible categories of outcomes.• P(S)=p p=probability of success• P(F)=q q=probability of failure

Binomial Probability DistributionsNotation for a Binomial Probability DistributionS and F (success and failure) denote the two possible categories of outcomes.• P(S)=p p=probability of success• P(F)=q q=probability of failure• n denotes the fixed number of trials

Binomial Probability DistributionsNotation for a Binomial Probability DistributionS and F (success and failure) denote the two possible categories of outcomes.• P(S)=p p=probability of success• P(F)=q q=probability of failure• n denotes the fixed number of trials• x denotes a specific number of successes in n trials

Binomial Probability DistributionsNotation for a Binomial Probability DistributionS and F (success and failure) denote the two possible categories of outcomes.• P(S)=p p=probability of success• P(F)=q q=probability of failure• n denotes the fixed number of trials• x denotes a specific number of successes in n trials• p denotes the probability of success in one of the n trials

Binomial Probability DistributionsNotation for a Binomial Probability DistributionS and F (success and failure) denote the two possible categories of outcomes.• P(S)=p p=probability of success• P(F)=q q=probability of failure• n denotes the fixed number of trials• x denotes a specific number of successes in n trials• p denotes the probability of success in one of the n trials• q denotes the probability of failure in one of the n trials

Binomial Probability DistributionsNotation for a Binomial Probability DistributionS and F (success and failure) denote the two possible categories of outcomes.• P(S)=p p=probability of success• P(F)=q q=probability of failure• n denotes the fixed number of trials• x denotes a specific number of successes in n trials• p denotes the probability of success in one of the n trials• q denotes the probability of failure in one of the n trials• P(x) denotes the probability of getting exactly x successes among the n trials

Binomial Probability DistributionsConsider an experiment in which 5 offspring peas are generated from 2 parents each having the green/yellow combination of genes for pod color. The probability of an offspring pea will have a green pod is . That is P(green pod) = 0.75. ¾Suppose we want to find the probability that exactly 3 of the 5 offspring peas have a green pod.a. Does this procedure result in a binomial distribution?b. If this procedure does result in a binomial distribution, identify the values of

Binomial Probability DistributionsConsider an experiment in which 5 offspring peas are generated from 2 parents each having the green/yellow combination of genes for pod color. The probability of an offspring pea will have a green pod is . That is ¾P(green pod) = 0.75. Suppose we want to find the probability that exactly 3 of the 5 offspring peas have a green pod.a. Does this procedure result in a binomial distribution?Yesb. If this procedure does result in a binomial distribution, identify the values of n=5, x=3, p=0.75, and q=0.25

Binomial Probability DistributionsDetermine whether the given procedure results in a binomial distribution. Surveying 12 jurors and recording whether there is a “no” response when they are asked if they have ever been convicted of a felony

Binomial Probability DistributionsDetermine whether the given procedure results in a binomial distribution. Surveying 12 jurors and recording whether there is a “no” response when they are asked if they have ever been convicted of a felonyBinomial

Binomial Probability Distributions Determine whether the given procedure results in a binomial distribution. Treating 50 smokers with Nicorette and recording whether there is a “yes” response when they are asked if they experience any mouth or throat soreness.

Binomial Probability Distributions Determine whether the given procedure results in a binomial distribution. Treating 50 smokers with Nicorette and recording whether there is a “yes” response when they are asked if they experience any mouth or throat soreness.Binomial

Binomial Probability Distributions Determine whether the given procedure results in a binomial distribution. Recording the number of children in 250 families

Binomial Probability Distributions Determine whether the given procedure results in a binomial distribution. Recording the number of children in 250 familiesNot binomial, there are more than two outcomes.

Binomial Probability DistributionsDetermine whether the given procedure results in a binomial distribution. Fifteen different Governors are randomly selected from the 50 Governors currently in office and the sex of each Governor is recorded.

Binomial Probability DistributionsDetermine whether the given procedure results in a binomial distribution. Fifteen different Governors are randomly selected from the 50 Governors currently in office and the sex of each Governor is recorded.Not binomial, not independent!

Binomial Probability DistributionsDetermine whether the given procedure results in a binomial distribution. Two hundred statistics students are randomly selected and each is asked if he or she owns a Ti-84 Plus Calculator.

Binomial Probability DistributionsDetermine whether the given procedure results in a binomial distribution. Two hundred statistics students are randomly selected and each is asked if he or she owns a Ti-84 Plus Calculator. No, but yes?!?!? We can use the 5% guideline for cumbersome calculations.

Binomial Probability DistributionsBinomial Probability FormulaIn a binomial Probability distribution, probabilities can be calculated by using the binomial probability formula.

Binomial Probability DistributionsBinomial Probability FormulaIn a binomial Probability distribution, probabilities can be calculated by using the binomial probability formula.First recall/learn: Factorial symbol (!) denotes the product of decreasing powers of positive whole numbers.

Binomial Probability DistributionsBinomial Probability FormulaIn a binomial Probability distribution, probabilities can be calculated by using the binomial probability formula.First recall/learn: Factorial symbol (!) denotes the product of decreasing powers of positive whole numbers. So and .

Binomial Probability DistributionsBinomial Probability FormulaIn a binomial Probability distribution, probabilities can be calculated by using the binomial probability formula.First recall/learn: Factorial symbol (!) denotes the product of decreasing powers of positive whole numbers. So and .

for wheren=number of trialsx=number of success among n trials(p=probability of success/q=probability of failure) in any one trial

Binomial Probability DistributionsThe probability of an offspring pea will have a green pod is . That is P(green pod) = 0.75. ¾Let’s use the binomial probability formula to find probability that exactly 3 of the 5 offspring peas have a green pod.

Binomial Probability DistributionsThe probability of an offspring pea will have a green pod is . That is P(green pod) = 0.75. ¾Let’s use the binomial probability formula to find probability that exactly 3 of the 5 offspring peas have a green pod. So n=5, x=3, p=0.75, and q=0.25

Binomial Probability DistributionsAssume that a procedure yields a binomial distribution with a trial repeated 14 times. Use Table A-1 to find the probability of 4 successes given the probability 0.60 of success on a single trial.

Binomial Probability DistributionsAssume that a procedure yields a binomial distribution with a trial repeated 5 times. Using the Binomial Probability formula find the probability of 2 successes given the probability .35 of success on a single trial.

Binomial Probability DistributionsThe brand name of McDonald’s has a 95% recognition rate. If a McDonald’s executive wants to verify that rate by beginning with a small sample of 15 randomly selected consumers, find the probability that exactly 13 of the 15 consumers recognize the McDonald’s brand name. Also find the probability that the number who recognize the brand name is not 13.

Homework!!!

• 5-3: 1-8,13, 33, 35, and 39.

for Binomial DistributionsFor a Binomial Distribution are given by the following formulas:

for Binomial DistributionsFor a Binomial Distribution are given by the following formulas:

for Binomial DistributionsFor a Binomial Distribution are given by the following formulas:

for Binomial DistributionsFor a Binomial Distribution are given by the following formulas:

for Binomial DistributionsFor a Binomial Distribution are given by the following formulas:

Now lets do example 1 and 2, then do problem 6 on the worksheet

for Binomial DistributionsUse the given values of n and p to find the mean μ and the standard deviation σ. Also, use the range rule of thumb to find the minimum usual value μ - 2σ and the maximum usual value μ + 2σ. 

for Binomial DistributionsUse the given values of n and p to find the mean μ and the standard deviation σ. Also, use the range rule of thumb to find the minimum usual value μ - 2σ and the maximum usual value μ + 2σ. Given: n = 60, p = 0.25 Mean: μ = np = (60)(.25) = 15 Standard deviation: σ = √(60 * .25 * .75) = 3.354 Min usual value: μ - 2σ = 15 – 2(3.354) = 8.292 Max usual value: μ + 2σ = 15 + 2(3.354) = 21.708

for Binomial DistributionsSeveral economics students are unprepared for a multiple-choice quiz with 25 questions, and all of their answers are guesses. Each question has five possible answers, and only one of them is correct. Find the mean and standard deviation for the number of correct answers for such students.Mean: μ = (25)(1/5) = 5Standard deviation: σ = √(25 *.2 *.8) = 2

for Binomial Distributions232 Mars, Inc., claims that 24% of its M&M plain candies are blue. A sample of 100 M&Ms is randomly selected. Find the mean and standard deviation for the numbers of blue M&Ms in such groups of 100.Mean: μ = np μ = (100)(.24) = 24 Standard deviation: σ = √(100 *.24 *.76) = 4.3

for Binomial DistributionsData Set 18 in Appendix B consists of a random sample of 100 M&Ms in which 27 are blue. Is this result unusual? Does it seem that the claimed rate of 24% is wrong?μ = 24 σ = 4.3The max usual values: 24 + 2(4.3) = 32.6 M&MsThe min usual values: 24 – 2(4.3) = 15.4 M&Ms

Homework!!!

• 5-3: 1-8,13, 33, 35, and 39.• 5-4: 1-12, 17, 19