Download presentation

Presentation is loading. Please wait.

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
**RandInt (1,2,6) where 1 = Left and 2 = Right**

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
**LLLLLL RLLLLL LRLLLL RRLLLL LLRLLL RLRLLL LRRLLL RRRLLL**

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 1 2 3 4 5 6 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 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 X 1 2 3 4 5 6 P(X) .0156 .0938 .2344 .3125 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 X 1 2 3 4 5 6 P(X) .0156 .0938 .2344 .3125 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… μx = 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. 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
**Community College Problem**

(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) = 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**

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**

31
**Enter binomial probabilities into list L2**

Enter binomial probabilities into list L2. Define Plot 1 to be a histogram with XList: L1 and Freq: L2

32
**Set windown to be X[0,11]1 and Y[-.15,.5]1. Press TRACE**

33
**To calculate the cumulative prob. Highlight list L3**

To calculate the cumulative prob. Highlight list L3. Press 2nd DISTR and select A: binomcdf(10,.1, L1) Press ENTER

35
**Turn off Plot 1 and turn on Plot 2**

Turn off Plot 1 and turn on Plot 2. Define Plot 2 to be a histogram with XList: L1 and YList: L3. In the window set Ymin = -.3 and Y max= 1.2.

36
**This command simulates 12 free throw attempts with an accuracy of 75%.**

37
**Calculates the number of throws that hit the basket.**

38
Totals of ten games

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-1p -- 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 2nd 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)=

44
**Hawaiian Villager Problem**

When this data is graphed what do you notice?

45
**Characteristics of a Geometric Distribution**

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.

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 2. the probability of a success is p**

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) μx = 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) = 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) = 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 μx =

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-1p 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. 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 4th 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**

67
Enter the numbers 1 to 10 into list L1. Enter the probabilities into L2. Highlight L2 then press 2ndVARS (DISTR). Scroll down and select D:geometricpdf(1/6, L1) and ENTER.

68
**Window dimensions x[0,11] 1 and Y[-. 05,. 2]**

Window dimensions x[0,11] 1 and Y[-.05, .2].1 Define histogram PLOT1, Xlist: L1 and Freq: L2

69
**Install cdf as L3. In the STAT/Edit window, place the cursor on L3**

Install cdf as L3. In the STAT/Edit window, place the cursor on L3. Enter the formula geometcdf(1/6, L1) and ENTER

70
To plot the cumulative distribution histogram, first specify the window: X[0.11]1 and Y[-3,1]1 Then deselect Plot1 and define Plot2 to be a histogram with Xlist: L1 and Freq: L3 Press GRAPH

Similar presentations

Presentation is loading. Please wait....

OK

Chapter 6: Random Variables

Chapter 6: Random Variables

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Complete ppt on hepatitis b Convert free pdf to ppt online convertor Ppt on weather and climate for grade 5 Download ppt on world population day Ppt on tata trucks india Ppt on different types of birds Ppt on power quality and control Ppt on cross-sectional study Ppt on national education day images Ppt on pollution in india free download