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DOES THIS QUESTION RESULT IN A BERNOULLI TRIAL? 1. 2. 3. Yes No Who was Bernoulli again? 33% 33% 33% Slide 1- 1 1 2 3 BERNOULLI, GEOMETRIC, BINOMIAL Bernoulli – single trial Geometric – number of trials until success Binomial – number of success in n trials Poisson – number of success with large n and very small p Slide 1- 2 THE GEOMETRIC MODEL (CONT.) Geometric probability model for Bernoulli trials: Geom(p) p = probability of success q = 1 – p = probability of failure X = number of trials until the first success occurs x-1 P(X = x) = q p 1 E(X) p q p2 Slide 1- 3 THE BINOMIAL MODEL (CONT.) Binomial probability model for Bernoulli trials: Binom(n,p) n = number of trials p = probability of success q = 1 – p = probability of failure X = number of successes in n trials n! n x n x n P( X x) p q where x x x !(n x)! np npq Slide 1- 4 THE NORMAL MODEL TO THE RESCUE! When dealing with a large number of trials in a Binomial situation, making direct calculations of the probabilities becomes tedious (or outright impossible). Fortunately, the Normal model comes to the rescue… Slide 1- 5 THE NORMAL MODEL TO THE RESCUE (CONT.) As long as the Success/Failure Condition holds, we can use the Normal model to approximate Binomial probabilities. Success/failure condition: A Binomial model is approximately Normal if we expect at least 10 successes and 10 failures: np ≥ 10 and nq ≥ 10. Slide 1- 6 CONTINUOUS RANDOM VARIABLES When we use the Normal model to approximate the Binomial model, we are using a continuous random variable to approximate a discrete random variable. So, when we use the Normal model, we no longer calculate the probability that the random variable equals a particular value, but only that it lies between two values. Slide 1- 7 APPLYING THE NORMAL MODEL A lecture hall has 190 seats with folding arm tablets, 25 of which are designed for left-handers. The average size of classes that meet there is 177. We can assume that about 12% of students are left-handed. What’s the probability that a right-handed student in one of these classes is forced to use a lefty arm tablet? CAN WE USE THE NORMAL MODEL? 1. 2. 3. 4. Yes, n*p ≥ 10 and n*(1-p) ≥ 10 and we are examining a range of possible outcomes. Yes, n*p ≥ 10 and n*(1-p) ≥ 10 and we are examining one possible outcome. No, n*p ≥ 10 and n*(1-p) ≥ 10 and we are examining a range of possible outcomes. No, n*p ≥ 10 and n*(1-p) ≥ 10 and we are examining one possible outcome. Slide 1- 9 WHAT IS THE EXPECTED NUMBER OF RIGHT HANDED STUDENTS IN THE AVERAGE CLASS? 1. 2. 3. 4. 177*(.12) 177*(.88) 190*(.12) 190*(.88) 0% 0% 0% 0%Slide 1- 10 1 2 3 4 WHAT IS THE STANDARD DEVIATION OF RIGHT HANDED STUDENTS IN THE AVERAGE CLASS? 25% 1. 2. 3. 4. 25% 25% 25% 177*(.12)*(.88) sqrt(177*(.12)*(.88)) 190*(.12)*(.88) sqrt(190*(.12)*(.88)) Slide 1- 11 1 2 3 4 WHAT HAS TO HAPPEN FOR A RIGHT-HANDED STUDENT TO HAVE TO SIT IN A LEFTY CHAIR? 1. 2. 3. 4. X ≥ 190 X<190 X ≥ 166 X<166 0% 1 0% 2 0% 3 Slide 0% 1- 12 4 WHAT’S THE PROBABILITY THAT A RIGHTHANDED STUDENT HAS TO SIT IN A CHAIR WITH A LEFT-HANDED TABLET? 1. 2. 3. 4. .0089 .9911 .0166 .9834 0% 0% 0% 0%Slide 1- 13 1. 2. 3. 4. PROCESS Determine X Check Normality Assumptions Find E(X) and SD(X) based on the Binomial model Determine z-score based on a fixed level or fixed standard given by the problem. Apply to z-tables to find probability Slide 1- 14 USING STATISTICS TO TEST THE VALIDITY OF CLAIMS. A newly hired telemarketer is told he will probably make a sale on about 12% of his phone calls. The first week he called 300 people, but only made 12 sales. Should he suspect he was misled about the true success rate? Explain. Slide 1- 15 SHOULD HE SUSPECT HE WAS MISLED? 0% 1. 2. 0% 3. 0% 0% 4. Twelve sales is less than 1 SD below the mean. He was probably misled. Twelve sales is less than 1 SD below the mean. He was probably NOT misled. Twelve sales is less than 5 SD below the mean. He was probably NOT misled. Twelve sales is more than 3 SD below the mean. He was probably misled. THE POISSON MODEL The Poisson probability model was originally derived to approximate the Binomial model when the probability of success, p, is very small and the number of trials, n, is very large. The parameter for the Poisson model is λ. To approximate a Binomial model with a Poisson model, just make their means match: λ = np. Slide 1- 17 THE POISSON MODEL (CONT.) Poisson probability model for successes: Poisson(λ) λ = mean number of successes or n*p X = number of successes e is an important mathematical constant (approximately 2.71828) x e P X x x! EX SD X Slide 1- 18 APPLYING THE POISSON MODEL The probability of contracting a disease is small, with p about 0.0005 for a new case in a given year. In a town of 6,000 people, what is the expected number of new cases? Slide 1- 19 WHAT IS THE EXPECTED NUMBER OF NEW CASES IN A TOWN OF 6,000? 1. 2. 3. 4. 6,000*0.0005 6,000*0.9995 6,000*0.0005*0.9995 We need more information. 0% 0% 0% 0%Slide 1- 20 1 2 3 4 USE THE POISSON MODEL TO APPROXIMATE THE PROBABILITY THAT THERE WILL BE AT LEAST ONE NEW CASE OF THE DISEASE NEXT YEAR 1. 2. 3. 4. Exp(-3)*31/1! Exp(-3)*30/0! 1-Exp(-3)*31/1! 1-Exp(-3)*30/0! 0% 1 0% 0% 2 3 0% 4 Slide 0% 1- 21 5 WHAT’S THE PROBABILITY THAT THERE WILL BE MORE THAN TEN NEW CASES NEXT YEAR? 1. 2. 3. 4. Exp(-3)*310/10! 0.9901 1-Exp(-3)*310/10! 1-0.9901 0% 1 0% 0% 2 3 0% 4 Slide 0% 1- 22 5 APPLYING THE POISSON MODEL Hurricanes in a particular place arrive with a mean of 2.35 per year. Suppose the number of hurricanes can be modeled by a Poisson distribution with this mean. Slide 1- 23 WHAT’S THE PROBABILITY OF NO HURRICANES NEXT YEAR? 1. 2. 3. 4. Exp(2.35)*2.35 Exp(-2.35)*2.35 Exp(2.35)*2.350 Exp(-2.35)*2.350 0% 0% 0% 0%Slide 1- 24 1 2 3 4 WHAT’S THE PROBABILITY THAT DURING THE NEXT TWO YEARS, THERE IS EXACTLY ONE HURRICANE? 1. 2. 3. 4. Exp(-2.35)*2.35 Exp(-2.35)*2.350 Exp(-2.35)*2.35*Exp(-2.35)*2.350 Exp(-2.35)*2.35*Exp(-2.35)*2.350 +Exp(-2.35)*2.35*Exp(-2.35)*2.350 0% 0% 0% 0%Slide 1- 25 1 2 3 4 UPCOMING Homework 7 due Sunday Review for Exam on Tuesday Exam 1 on Thursday Slide 1- 26