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Chapter 8 The Geometric Distributions. PLINKO Our goal is to determine the probability that a ball will land in slot “D”. A win occurs when the ball falls.

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Presentation on theme: "Chapter 8 The Geometric Distributions. PLINKO Our goal is to determine the probability that a ball will land in slot “D”. A win occurs when the ball falls."— Presentation transcript:

1 Chapter 8 The Geometric Distributions

2 PLINKO Our goal is to determine the probability that a ball will land in slot “D”. A win occurs when the ball falls to the Right 3 times and to the Left 3 times in any order.

3 Let the random variable X = number of times the ball falls to the right. Our goal is to find P (X=3)

4 Use the random number generator on your calculator to generate six numbers representing the positions A, B, C, E, F, and G. _ A__ ___B___ ___C___ ___D__ ___E___ ___F___ ___G___ RandInt (1,2,6) where 1 = Left and 2 = Right Lands in D __________________________ Not in D _____________________________

5 Lands in D _____________________________ Not in D ______________________________ CLASS TOTALS in D _________ Not in D _________

6 Determine the total number of possible outcomes for Plinko…list them systematically HINT: How many positions to left and right? Use powers of 2 to find to combinations of R’s and L’s I’ll get you started…see the board LLLLLL RLLLLL LRLLLL RRLLLL LLRLLL RLRLLL LRRLLL RRRLLL

7 Complete the probability distribution for the random variable X = # times the ball falls to the right (i.e.) 0 R’s, 1 R, 2 R’s, # R’s etc. X P(X)

8 The features of this experiment are as follows: There are two outcomes (L, R) or heads/tails, evens/odds (success/failure) 6 digits are drawn for a single trial The flips or draws are independent (one outcome has no influence on the next) The probability of success (falling to right) is the same each trial

9 A situation in which these four conditions are satisfied is called a binomial setting To reiterate these characteristics…

10 The Binomial Setting 1.each observation falls into one of two categories (success and failure) 2.there is a fixed number n of observations 3.The n observations are all independent. 4. the probability of success, p, is the same for each observation

11 The distribution of the count X of successes in the binomial setting is the binomial distribution with parameters n and p. The parameters n is the number of observations, and p is the probability of a success on any one observation. The possible values of X are the whole numbers from 0 to n. B ( n, p)

12 Continuing with PLINKO Suppose 5 PLINKO balls are dropped down the board in succession. Find the probability that all of them will land in slot “D”. Find the probability that exactly 2 of them land in slot “D”.

13 PLINKO P(lands in slot D) =.3125 So all 5 in slot D would be P(all in slot D) = (.3125) 5 = P(2 in slot D) = 2 in slot D & 3 not in D = (.3125) 2 ( ) 3 =.0317

14 PLINKO If a ball landing in slot “A” or “G” pays $50, a ball landing in “B” or “F” pays $25, a ball in “C” and “E” pay $10, and a ball landing in “D” pays $5, find the expected winnings(mean)when 5 balls are dropped. What is the standard deviation of the total amount won?

15 PLINKO Pay out $50$25$10$5$10$25$50 X P(X) Expected value = 50(.0156)+25(.0938)+10(.2344)+5(.3125)+10(.2344)+25(.0938)+50 (.0156) = $12.50

16 PLINKO Pay out $50$25$10$5$10$25$50 X P(X) VAR (X) = ( ) 2 (.0156) + ( ) 2 (.0938) + ( )2(. 2344) + ( )2(. 3125) + ( )2(. 2344) + ( ) 2 (.0938) + ( ) 2 (.0156) VAR (X) = STD DEV(X) = $9.68

17 Example 1 Suppose Dolores is a 65% free throw shooter. If we assume that the repeated shots are independent “what is the probability that Dolores makes exactly 7 of her next 10 free throws?” If X is the binomial random variable that gives us the count of successes for the experiment, we say X has B(10,.65) The question is to find P(X=7) use n C r under MATH PRB where r = x 10 C 7 (.65) 7 (.35) 3 =.252

18 (b) What is Dolores’ probability that she makes no more than 5 free throws? This time the question is asking what is P (x ≤ 5)? P (x ≤ 5) = P(x=0) + P(x=1) + P(x=2) + P(x=3) + P(x=4) + P(x =5) = 10 C 0 (.65) 0 (.35) C 1 (.65) 1 (.35) C 2 (.65) 2 (.35) C 3 (.65) 3 (.35) C 4 (.65) 4 (.35) C 5 (.65) 5 (.35) 5 =.249 about 25% chance

19 What is the probability that Dolores will make at least 6 free throws? Do we have to redo all of the calculations or is there another way? If she makes at least 6 how many will she miss? Hmmm…????? How is this related to the previous problem? P(x ≥ 6) is the same as P( 1- P(x ≤ 5)) = = =.751

20 Mean and Standard Deviation Clearly we can calculate the mean and standard deviation of a binomial random variable using the methods from Chapter 7 but there is another way… μ x = np

21 Normal Approximation for Binomial Distribution When np and n(1-p) are SUFFICIENTLY LARGE i.e. both are ≥ 10, the binomial random variable X has an approximately normal distribution. The mean μ = np and

22 Community College Problem Nationally, 15% of community college students live more than 6 miles from campus. Data from a simple random sample of 400 students at one community college is analyzed. (a)What are the mean and standard deviation for the number of students in the sample trial? X has B(400,.15) μ = np = 400(.15) = 60

23 Community College Problem

24 (b) Use a normal approximation to calculate probability that at least 65 of the students in the sample live more than 6 miles from campus. Because 400(.15) = 60 ≥10 and 400(.85) = 340 ≥ 10, we can use the normal approximation to the binomial with N(60,7.14)

25 Community College Problem Find P (x ≥ 65) P (z ≥ (65-60)/7.14 =.70 From Table A the P (z <.70) =.7580 so P (z ≥.70 ) = ( )=.242

26 Credit Card Example Suppose 60% of adults have credit card debt. If we survey 2500 adults, what is the probability more than 1520 would have credit card debt? X = # adults who have credit card debt out of 2500 X is B(2500,.60) We want to find P(X > 1520)

27 Credit Card Example Is np ≥ 10 ? 2500(.60) = 1500 ≥ 10 Is n(1-p) ≥ 10? 2500(.40) = 1000 ≥ 10 Yes, it approximates a normal distribution. μ = np = 1500

28 Credit Card Example We want to find P (x >1520) P (z >

29 Do problems 8.2, 8.8 and 8.16

30 Technology Toolbox Exploring binomial distributions

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39 Chapter 8 Section 8.2 The Geometric Distribution

40 Hawaiian Villager Problem On the island of Oahu in the village of Nankuli, 80% of the residents are of Hawaiian ancestry. If you visit Nanakuli, what is the probability the first village you meet is Hawaiian? X = # villagers you must meet P(X = 1) P(X = n) = (1 – p) n-1 p -- probability it is (p) * probability it is not (1-p) P (X = 1) = (1-.8) 1-1 (.8) =.8

41 RULE FOR CALCULATING GEOMETRIC PROBABILITIES If X has a geometric distribution with probability p of success and (1-p) of failure on each observation, the possible values of X are 1,2,3,…If n is any one of these, the probability that the first success occurs on the nth trial is

42 Hawaiian Villager Problem What is the P( you don’t meet a Hawaiian until the 2 nd villager?) P (X =2) (1-.8) 2-1 (.8)=.16 Let’s extend this concept for third, fourth, fifth villagers…

43 Hawaiian Villager Problem P (X =1) (1-.8) 1-1 (.8)=.8 P (X =2) (1-.8) 2-1 (.8)=.16 P (X =3) (1-.8) 3-1 (.8)=.032 P (X =4) (1-.8) 4-1 (.8)=.0064 P (X =5) (1-.8) 5-1 (.8)=.00128

44 Hawaiian Villager Problem When this data is graphed what do you notice?

45 Graphs of Geometric Distributions have a ‘step ladder” appearance since you are multiplying the height of each bar by a number less than 1. Each bar will be shorter than the previous bar. The histogram is ALWAYS right skewed. Characteristics of a Geometric Distribution

46 Hawaiian Villager Problem Find the probability it will take more than 4 villagers to meet a native Hawaiian. P(x > 4) = (1-p) n = (1-.8) 4 = (.2) 4 =.0016

47 Hawaiian Villager Problem Find the average number of villagers it will take to meet a native Hawaiian.

48 Hawaiian Villager Problem How much variability is there in the number of villagers required to meet a Hawaiian?

49 The Geometric Setting 1. each observation falls into one of two categories, success or failure 2. the probability of a success is p 3. the observations are all independent 4. the variable of interest is the number of trials required to obtain the first success

50 HANDY DANDY FORMULAS If X is a Geometric random variable with P(success) = p these formulas apply: P(X=n) = (1- p) n-1 (p) μ x = P(X > n) = (1- p) n

51 How Can You Tell Geometric from Binomial? Both of these models must meet the 3 conditions often called the Bernoulli trials. (1) there are two possible outcomes (2) the probability of a success is constant (3) the trials are independent The distinguishing characteristic is: A binomial probability model is appropriate for a random variable that counts the # of successes in a fixed number of trials.

52 How Can You Tell Geometric from Binomial? While a geometric probability model is appropriate for a random variable that counts the # of trials until the first success. (there could be an unlimited number of trials.)

53 Which are these? (1) The Los Angeles Times reported that 80% of airline passengers prefer to sleep on long flights rather than watch movies, read, etc. Consider randomly selecting 25 passengers from a particular long flight. Defind a random variable X, calculate P( X=12). Is this binomial or geometric?

54 Which are these? (2) Sophie is a dog who loves to play catch. Unfortunately, she isn’t very good, and the P(catches a ball) = 0.1. Define X=# tosses required until Sophie to catches the ball. Is this binomial or geometric?

55 Which are these? (3) You are to take a multiple choice exam of 100 questions with five possible responses (A,B,C,D,E). Suppose you have not studied and decide to guess randomly on each question. Let X = # correct responses. Is this binomial or geometric?

56 Which are these? (4) Suppose 5% of cereal boxes contain a prize. You are determined to buy cereal boxes until you win a prize. Is this binomial or geometric?

57 Let’s Explore the Sophie problem and the cereal problem in more depth.

58 The Sophie problem (2) Sophie is a dog who loves to play catch. Unfortunately, she isn’t very good, and the P(catches a ball) = 0.1. Define X=# tosses required until Sophie to catches the ball. (a) calculate and interpret P(X=2) P (X = n) = (1-p) n-1 (p) P (X =2) (1-.1) 2-1 (.1)=.09 (b) calculate and interpret P(X ≥ 3) P(X > n) = (1- p) n P(X ≥ 3) = (1-.1) 3 =.729

59 Sophie (2c) calculate and interpret the mean and standard deviation of X μ x =

60 Cereal Problem Suppose 5% of cereal boxes contain a prize. You are determined to buy cereal boxes until you win a prize. (a) What is the probability you will have to buy at most 2 boxes? (X ≤ 2) (b) What is the probability you will have to buy exactly 4 boxes? ( X = 4) (c) What is the probability you will have to buy more than 4 boxes? (X ≥ 4)

61 Cereal Problem (a) What is the probability you will have to buy at most 2 boxes? X = # boxes you will buy until you win a prize. Find P (X ≤ 2) is the same as P (1- complement) P( X > 2) P(X > n) = (1- p)n P (X ≤ 2) = (1-.5) 2 =.25

62 Cereal Problem (b) What is the probability you will have to buy exactly 4 boxes? ( X = 4) P(X = n) = (1- p) n-1 p P(X = 4) = (1-.5) 4-1 (.5) =.0625 (c) What is the probability you will have to buy more than 4 boxes? (X ≥ 4) P(X > n) = (1- p) n P(X ≥ 4) = (1-.5) 4 =.0625

63 #37 in the textbook Which are binomial or geometric?

64 # 37 in the book. (a)yes, geometric X = success (tail) failure (head) a trial is one flip of the coin P(tail) =.5 (b) Not independent (c) X = success of getting a Jack a trial is drawing a card with replacement P(J) =

65 Dolores the Basketball Player Remember Dolores the basketball player whose free throw shooting percentage was.65? What is the probability that the first free throw she hits is on her 4 th attempt? P(X = 4) (1-.65) 4-1 (.65)= (.35) 3 (.65)=.028 Using the TI 83/84 geometpdf (p,n) geometpdf (.65,4) =.028

66 Technology Toolbox Exploring geometric distributions

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