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Statistics Class 10 2/29/2012.

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Presentation on theme: "Statistics Class 10 2/29/2012."— Presentation transcript:

1 Statistics Class 10 2/29/2012

2 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. Find the expected value for the $5 bet that the outcome is 0 or 00 or 1 or 2 or 3. 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?

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

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

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

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

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

8 Binomial Probability Distributions
Note on Independence Often 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.

9 Binomial Probability Distributions
Notation for a Binomial Probability Distribution S and F (success and failure) denote the two possible categories of outcomes.

10 Binomial Probability Distributions
Notation for a Binomial Probability Distribution S and F (success and failure) denote the two possible categories of outcomes. P(S)=p p=probability of success

11 Binomial Probability Distributions
Notation for a Binomial Probability Distribution S 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

12 Binomial Probability Distributions
Notation for a Binomial Probability Distribution S 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

13 Binomial Probability Distributions
Notation for a Binomial Probability Distribution S 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

14 Binomial Probability Distributions
Notation for a Binomial Probability Distribution S 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

15 Binomial Probability Distributions
Notation for a Binomial Probability Distribution S 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

16 Binomial Probability Distributions
Notation for a Binomial Probability Distribution S 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

17 Binomial Probability Distributions
Consider 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 ¾. That is P(green pod) = Suppose we want to find the probability that exactly 3 of the 5 offspring peas have a green pod. Does this procedure result in a binomial distribution? If this procedure does result in a binomial distribution, identify the values of 𝑛, 𝑥, 𝑝, 𝑎𝑛𝑑 𝑞. the word success is used arbitrary two describe one of the categories

18 Binomial Probability Distributions
Consider 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 ¾. That is P(green pod) = Suppose we want to find the probability that exactly 3 of the 5 offspring peas have a green pod. Does this procedure result in a binomial distribution? Yes If this procedure does result in a binomial distribution, identify the values of 𝑛, 𝑥, 𝑝, 𝑎𝑛𝑑 𝑞. n=5, x=3, p=0.75, and q=0.25 the word success is used arbitrary two describe one of the categories

19 Binomial Probability Distributions
Determine 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

20 Binomial Probability Distributions
Determine 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

21 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.

22 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

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

24 Binomial Probability Distributions
Determine whether the given procedure results in a binomial distribution. Recording the number of children in 250 families Not binomial, there are more than two outcomes.

25 Binomial Probability Distributions
Determine 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.

26 Binomial Probability Distributions
Determine 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!

27 Binomial Probability Distributions
Determine 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.

28 Binomial Probability Distributions
Determine 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.

29 Binomial Probability Distributions
Binomial Probability Formula In a binomial Probability distribution, probabilities can be calculated by using the binomial probability formula. the word success is used arbitrary two describe one of the categories

30 Binomial Probability Distributions
Binomial Probability Formula In 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. the word success is used arbitrary two describe one of the categories

31 Binomial Probability Distributions
Binomial Probability Formula In 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 4!=4∙3∙2∙1 and 0!=1. the word success is used arbitrary two describe one of the categories

32 Binomial Probability Distributions
Binomial Probability Formula In 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 4!=4∙3∙2∙1 and 0!=1. 𝑃 𝑥 = 𝑛! 𝑛−𝑥 !𝑥! ∙ 𝑝 𝑥 ∙ 𝑞 𝑛−𝑥 for 𝑥=0,1,2,…,𝑛 where n=number of trials x=number of success among n trials (p=probability of success/q=probability of failure) in any one trial the word success is used arbitrary two describe one of the categories

33 Binomial Probability Distributions
The probability of an offspring pea will have a green pod is ¾. That is P(green pod) = Let’s use the binomial probability formula to find probability that exactly 3 of the 5 offspring peas have a green pod. the word success is used arbitrary two describe one of the categories

34 Binomial Probability Distributions
The probability of an offspring pea will have a green pod is ¾. That is P(green pod) = Let’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 𝑃 3 = 5! 5−3 !3! ⋅ ⋅ −3 = the word success is used arbitrary two describe one of the categories

35 Binomial Probability Distributions
Assume 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. the word success is used arbitrary two describe one of the categories

36 Binomial Probability Distributions
Assume 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. the word success is used arbitrary two describe one of the categories

37 Binomial Probability Distributions
The 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. the word success is used arbitrary two describe one of the categories

38 Homework!!! 5-3: 1-8,13, 33, 35, and 39.

39 𝜇, 𝜎 2 , 𝑎𝑛𝑑 𝜎 for Binomial Distributions
For a Binomial Distribution 𝜇, 𝜎 2 , 𝑎𝑛𝑑 𝜎 are given by the following formulas: the word success is used arbitrary two describe one of the categories

40 𝜇, 𝜎 2 , 𝑎𝑛𝑑 𝜎 for Binomial Distributions
For a Binomial Distribution 𝜇, 𝜎 2 , 𝑎𝑛𝑑 𝜎 are given by the following formulas: 𝜇=𝑛∙𝑝 the word success is used arbitrary two describe one of the categories

41 𝜇, 𝜎 2 , 𝑎𝑛𝑑 𝜎 for Binomial Distributions
For a Binomial Distribution 𝜇, 𝜎 2 , 𝑎𝑛𝑑 𝜎 are given by the following formulas: 𝜇=𝑛∙𝑝 𝜎 2 =𝑛∙𝑝∙𝑞 the word success is used arbitrary two describe one of the categories

42 𝜇, 𝜎 2 , 𝑎𝑛𝑑 𝜎 for Binomial Distributions
For a Binomial Distribution 𝜇, 𝜎 2 , 𝑎𝑛𝑑 𝜎 are given by the following formulas: 𝜇=𝑛∙𝑝 𝜎 2 =𝑛∙𝑝∙𝑞 𝜎= 𝑛∙𝑝∙𝑞 the word success is used arbitrary two describe one of the categories

43 𝜇, 𝜎 2 , 𝑎𝑛𝑑 𝜎 for Binomial Distributions
For a Binomial Distribution 𝜇, 𝜎 2 , 𝑎𝑛𝑑 𝜎 are given by the following formulas: 𝜇=𝑛∙𝑝 𝜎 2 =𝑛∙𝑝∙𝑞 𝜎= 𝑛∙𝑝∙𝑞 Now lets do example 1 and 2, then do problem 6 on the worksheet the word success is used arbitrary two describe one of the categories

44 𝜇, 𝜎 2 , 𝑎𝑛𝑑 𝜎 for Binomial Distributions
Use 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.  the word success is used arbitrary two describe one of the categories

45 𝜇, 𝜎 2 , 𝑎𝑛𝑑 𝜎 for Binomial Distributions
Use 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) = Min usual value: μ - 2σ = 15 – 2(3.354) = Max usual value: μ + 2σ = (3.354) = the word success is used arbitrary two describe one of the categories

46 𝜇, 𝜎 2 , 𝑎𝑛𝑑 𝜎 for Binomial Distributions
Several 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) = 5 Standard deviation: σ = √(25 *.2 *.8) = 2 the word success is used arbitrary two describe one of the categories

47 𝜇, 𝜎 2 , 𝑎𝑛𝑑 𝜎 for Binomial Distributions
232 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 the word success is used arbitrary two describe one of the categories

48 𝜇, 𝜎 2 , 𝑎𝑛𝑑 𝜎 for Binomial Distributions
Data 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 = 4.3 The max usual values: (4.3) = 32.6 M&Ms The min usual values: 24 – 2(4.3) = 15.4 M&Ms the word success is used arbitrary two describe one of the categories

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


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