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# Chapter 8 The Geometric Distributions

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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 to the Right 3 times and to the Left 3 times in any order.

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

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 _____________________________

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

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

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)

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

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

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

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)

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”.

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

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?

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

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

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

(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

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

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

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

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

Community College Problem

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)

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

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)

Credit Card Example Is np ≥ 10 ? 2500(.60) = 1500 ≥ 10
Yes, it approximates a normal distribution. μ = np = 1500

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

Do problems 8.2, 8.8 and 8.16

Technology Toolbox Exploring binomial distributions

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

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

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

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.

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

Calculates the number of throws that hit the basket.

Totals of ten games

Chapter 8 Section 8.2 The Geometric Distribution

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

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

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…

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)=

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

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.

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

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

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

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

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

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.

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.)

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?

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?

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?

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?

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

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

Sophie (2c) calculate and interpret the mean and standard deviation of X μx =

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)

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

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

#37 in the textbook Which are binomial or geometric?

# 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) =

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

Technology Toolbox Exploring geometric distributions

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.

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

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

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

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