know the three parts of a bernoulli trial

Know the three parts of a Bernoulli trial

Use Geometric model when interested in the number of Bernoulli trials until the next success

AP Statistics Objectives Ch17

Use Binomial model when interested in the number of successes in a certain number of Bernoulli trials

Use Normal model to approximate a Binomial model when expecting at least 10 successes and 10 failures

AP Statistics Objectives Ch17

Bernoulli trialGeometric probability modelBinomial probability modelSuccess/Failure conditionExpected Value

Vocabulary

Chapter 17 Notes

Chapter 17 Assignment

Chapter 16Part I Chapter 17

AnswersChapter 16Part II

Additional Examples

Chapter 17 Assignment

Part I: Page 398 #2, 6

Part II: Pages 398 – 400 #14&22, 24

Part III: Page 400 #30

Chapter 17

1. Bernoulli Trial – Events that meet the following 3 conditions:

i. There are only two outcomes.ii. The probability of success is constant.iii. The trials are independent.

(Must meet all 3 conditions above.)

EXAMPLES: flipping heads on a coin, EXAMPLES: flipping heads on a coin, finding a defect, EXAMPLES: flipping heads on a coin, finding a defect, finding a prize, EXAMPLES: flipping heads on a coin, finding a defect, finding a prize, making a basket, EXAMPLES: flipping heads on a coin, finding a defect, finding a prize, making a basket, getting a hit.

Chapter 172. 10% Condition –

Recall that for the events A & B to be independent, P(A) = P(A|B). This isn’t true when we sample without replacement. However, the change is insignificant if the sample is smaller than 10% of the population.

Chapter 17

Geometric Probability Model – Counts the number Bernoulli Trials until the first Success.

Binomial Probability Model – Counts the number of successes within a fixed number of Bernoulli Trials.

Chapter 17

Counts the number Bernoulli Trials until the first Success.

P(X=x) =

E(X) = =

Geom(p)

Memorize this.

p = probability of success <------ ONLY PARAMETERq = probability of failure (q=1-p)X = number of trials until the first success occurs

3. Geometric Probability Model – Geom(p)

SD(X) = =

EXAMPLE 1A new sales gimmick has 30% of the M&M’s covered with speckles. These “groovy” candies are mixed randomly with the normal candies as they are put into the bags for distribution and sale. You buy a bag and remove candies one at a time looking for the speckles.

I sample without replacement, but the sample is less than 10% of all M&M’s.

This is a Bernoulli Trial becausei. There are only two outcomes which are…

ii. The probability of success is constant.

iii. The trials are not independent, because…

1) Find speckled 2) Don’t find speckled

p = .30 Success: Speckled

A new sales gimmick has 30% of the M&M’s covered with speckles. These “groovy” candies are mixed randomly with the normal candies as they are put into the bags for distribution and sale. You buy a bag and remove candies one at a time looking for the speckles.

a) How many M&M’s do you expect to check before finding a speckled M&M?• Note that you keep checking until you find a speckled

M&M (success). This is how we identify that we need the Geometric Probability Model.

• Also note that we are asked for an expected #. This is how we identify that we need the Expected Value for Geom(0.3).

E(X) =

b) What is the probability that the first speckled M&M will be the third one that you check?• Note that if the third is the first success, then you started

with two failures. • We are still using the Geometric Probability Model

so P(X=x) =

P(X=3) = c) What is the probability that the first speckled M&M will be the fourth one that you check?

P(X=4) =

Geom(0.3)

d) What is the probability that the first speckled M&M will be one of the first three that you check?• Note that includes the third is the first success, second is

the first success, OR a success the first time you check.

P(X3) = P(X=3) + P(X=2) + P(X=1)

= (.7)(.7)(.3) + (.7)(.3) + .3 =

Geom(0.3)

f) What is the probability that the first speckled M&M will be one of the first four that you check?

P(X<4) = = .7599

Geom(0.3)

4. TI-84 steps to calculate the P(X=x) or P(X<x) for Geom(p):

2nd VARS

For P(X = x) --- a specific value use…

E: geometpdf(

For P(X < x) --- a specific value OR FEWER use…

F: geometcdf(

Don’t use CALCULATOR SPEAK in your work.

TI-84 and Geom(p)

P(X=x)

Must add xP(X=3) for Geom(.1)

2nd VARS

E: geometpdf(

F: geometcdf(

Practice using Geom(.3) for the values of X that we just calculated.P(X = 3) = P(X = 4) =.𝟏𝟒𝟕 .𝟏𝟎𝟐𝟗

TI-84 and Geom(p)

P(X<x)

Must add xP(X<4) for Geom(.1)

2nd VARS

E: geometpdf(

F: geometcdf(

Practice using Geom(.3) for the values of X that we just calculated.P(X = 3) = P(X = 4) =

P(X < 3) = P(X < 4) =

.𝟏𝟒𝟕 .𝟏𝟎𝟐𝟗.657 .7599

Factorial5. 3! is read “3 factorial” is 3*2*1 = 6

6. What is 5! ?5*4*3*2*1 = 120

Combinations7. What is the formula for calculating a combination of n things taken k at a time? (Don’t worry. This is background information only.)

“ A combination of n things taken k at a time”

Combinations8. How many combinations of 3 can you get from 4 items?

“ A combination of 4 things taken 3 at a time” = = = = 4

Combinations8. How many combinations of 3 can you get from 4 items?

= = = = 4

“ A combination of n things taken k at a time”

Combinations9. How many combinations of 5 can you get from 8 items? = = =

= = 8*7 = 56

10. TI-84 and Combinationsi. Type n first for

iv. Then type k

iii. Under PRB choose

Geometric Probability Model – Counts the number Bernoulli Trials until the first Success.11. Binomial Probability Model Binom(n,p)

P(X=x) = Binom(n,p)

E(X) = = np

n = number of trials <------ 1 of 2 parametersp = probability of success <------ 2 of 2 parameters

Remember:

Counts the number of successes within a fixed number of Bernoulli Trials.

q = probability of failure (q=1-p)X = number of successes in n trials

SD(X) = =

12. TI-84 steps to calculate the P(X=x) or P(X<x) for Binom(n,p):

2nd VARS

For P(X = x) --- a specific value use… A: binompdf(

B: Binomcdf(

a) n = 12, p = 0.2, find P(2 successes)

P(X=2) =

Binom(12, 0.2)

b) n = 10, p = 0.4, find P(1 successes)

P(X=1) =

Binom(10, 0.4)

USE TI-84’s binompdf , but don’t write that

c) n = 20, p = 0.5, find P(10 successes)

P(X=10) =

Binom(20, 0.5)

d) n = 15, p = 0.9, find P(11 successes)

P(X=11) =

Binom(15, 0.9)

.𝟎𝟒𝟑

On Your Own

e) n = 7, p = , find P(4 successes)

.𝟏𝟐𝟖P(X=4) =

Binom(7, 1/3 )

f) n = 11, p = 0.05, find P(3 failures)

P(X=8) =

Binom(11, .05 )

.𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟓3 failures is the same as 8 successes

OR q = 0.95 so

Binom(11, .95 )P(X=3) =.𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟓

g) n = 15, p = 0.99, find P(1 failure)

P(X=14) =

Binom(15, .99 )

.𝟏𝟑1 failures is the same as 14 successes

OR q = 0.01

Binom(15, .01 )P(X=1) =.𝟏𝟑

On Your Own

a) n = 6, p = 0.35, find P(at least 3 successes)

P(X>3) =

= 1 – P(X<2)

P(X=3) OR P(X=4) OR P(X=5) OR P(X=6)

Binom(6, .35)

USE TI-84’s binomcdf , but don’t write that

down.Careful! This is a discrete model…

limited values for X.X can only be equal to …

0,1, 2, 3, 4, 5, or 6P(X>3)P(X<2)

b) n = 100, p = 0.01, find P(no more than 3 successes)

P(X<3) = P(X=3) OR P(X=2) OR P(X=1) OR P(X=0)

Binom(100, .01)

Chapter 17 – Binomial

This is a Bernoulli Trial becausei. There are only two outcomes which are

iii. The trials are independent, if

1) Answer Right 2) Answer Wrong

p = .20 Success: Answering correctly

he chooses randomly.

The model is Binom(10,0.2)

P(X > 7) = 1 - P(X < 6) =

In history class, Colin takes a multiple choice quiz.15.

0.00086

1) Defective 2) Not Defective

p = .01 Success: Defective

the components are randomly selected

The model is Binom(50,.01)P(X = 0) =

Binom(50,.99)P(X = 50) =

NOTEa) None defectiveb) At least 1 defective

0.605 0.605

Binom(50,.01)P(X > 1) = 1 – P(X = 0)

= 0.395

= 1 – 0.605

P(X = 0) = 0.481

1) Cracked 2) Not Cracked

p = .03 Success: Cracked egg

the eggs are randomly selected

The model is Binom(24,.03)

Chapter 17 – Binomial WS II

Binom(24,.03)P(X > 1) =

= 0.519

1 – P(X = 0)

= 1 – 0.481

Binom(24,.03)P(X = 2) = 0.127

Binom(20,)

P(X = 6) = 0.014

iii. The trials are independent, because

1) Sum is 5 2) Sum is not 5

p = Success: Sum of 5

the dice rolls are independent of each other

The model is

Chapter 17 – Binomial WS

Binom(20,)

P(X > 4) = 0.1751 – P(X < 3) =

Careful! This is a discrete model… limited values for X.

X can only be equal to …0,1, 2, 3, 4, 5, 6, 7, 8, 9, …, 19, 20

P(X>4)P(X<3)

Binom(20,)

P(X < 5) = 0.982

The model is

P(X = 21) = 0.157If the shots are independent of each other.

Binom(30, .7)

iii. The trials are independent?

1) Sinks shot 2) Doesn’t sink shot

p = Success: sinks shot

May not be true. Since Tim may get tired or “warmed up” the more he shots.

Binom(30, .7)P(X > 21) = 1 – P(X < 20) = 0.589

If the shots are independent of each other.

Careful! This is a discrete model… limited values for X.

X can only be equal to …0,1, 2, 3,…20, 21, 22, 23, …, 29, 30

P(X>21)P(X<20)

Binom(30, .7)

P(X < 21) = 0.568

Binom(30, .7)

P(18 < X < 20)

= P(X < 20) – P(X < 17)0.327

= 0.411 – 0.084 =

0,1, 2, 3,…17, 18, 19, 20, 21, 22, … 29, 30

P(X<17)

P(X<20)

0,1, 2, 3,…17, 18, 19, 20, 21, 22, … 29, 300,1, 2, 3,…17, 18, 19, 20, 21, 22, … 29, 30

P(X = 15) = 0.042

The model is Binom(50, .4)

iii.The trials are independent, because

1) Red drawn 2) Red not drawn

p = Success: Red drawn

the marbles are replaced after each draw.

Binom(50, .4)

P(X > 15) = 0.9461 – P(X < 14) =

Binom(50, .4)

P(X < 20) = 0.561

Binom(50, .4)P(17 < X < 25)

= P(X < 25) – P(X < 16) =0.787

= 0.943 – 0.156 =

1, 2, 3,…16, 17, 18, 19, 20, 21, 22, 23, 24, 25, … 49, 50

P(X<16)

P(X<25)

1, 2, 3,…16, 17, 18, 19, 20, 21, 22, 23, 24, 25, … 49, 501, 2, 3,…16, 17, 18, 19, 20, 21, 22, 23, 24, 25, … 49, 50

Chapter 1721. Binomial Model can be approximated

using the Normal Model if the number of trials is large enough. HOW LARGE?

• The SUCCESS/FAILURE CONDITION: A binomial model is approximately Normal if we expect at least 10 successes and 10 failures:

np > 10 and nq > 10

Chapter 17

a) n = 600, p = 0.35, find P(at least 220 successes)

SUCCESS/FAILURE CONDITION: np = and nq =

22.Given the number of trials and the probability of success, determine the probability indicated using the Normal Model as an estimate.

(600)(0.35) = 210(600)(0.65) = 390

(600)(0.35) = 210

= 11.68

z = = 0.856 P(z > ) = 0.196

> 10> 10

Chapter 17

23. Given the number of trials and the probability of success, determine the probability indicated using the Normal Model as an estimate.

(1000)(0.48) = 480(1000)(0.52) = 520

(1000)(0.48) = 480

= 15.80

z = = -1.90 P(z < ) = 0.029

> 10> 10

Chapter 17

23. Given the number of trials and the probability of success, determine the probability indicated using the Normal Model as an estimate.

(1000)(0.48) = 480(1000)(0.52) = 520

(1000)(0.48) = 480

= 15.80

z = = -1.90 P(z < ) = 0.029

> 10> 10

Bell Work Suppose a used car dealer runs autos through a two-stage process to get them ready to sell. The mechanical checkup costs $50 per hour and takes an average of 90 minutes, with a standard deviation of 15 minutes. The appearance prep (wash, polish, etc.) costs $6 per hour and takes an average of 60 minutes, with a standard deviation of 5 minutes.1. What are the mean and standard deviation of the total time spent preparing a car? (Note that we cannot find the standard deviation if we do not believe that the two phases of the process are independent, an important assumption to check.)

Chapter 16 Part I

x 0 1 2P(X=x) 0.2 0.4 0.4

2&10. Find the expected value and standard deviation of each random variable. (Show work)a)

Chapter 16 Part I2&10. Find the expected value and standard deviation of each random variable. (Show work)b) x 100 200 300 400

P(X=x) 0.1 0.2 0.5 0.2

Chapter 16 Part I18. An insurance policy costs $100 and will pay policyholders $10,000 if they suffer a major injury (resulting in hospitalization) or $3000 if they suffer a minor injury (resulting in lost time from work). The company estimates that each year 1 in every 2000 policyholders may have a major injury, and 1 in 500 a minor injury.a) Create a probability model for the profit on a policy.

Chapter 16 Part I18. An insurance policy costs $100 and will pay policyholders $10,000 if they suffer a major injury (resulting in hospitalization) or $3000 if they suffer a minor injury (resulting in lost time from work). The company estimates that each year 1 in every 2000 policyholders may have a major injury, and 1 in 500 a minor injury.b) What’s the company’s expected profit on this policy?

Chapter 16 Part I18. An insurance policy costs $100 and will pay policyholders $10,000 if they suffer a major injury (resulting in hospitalization) or $3000 if they suffer a minor injury (resulting in lost time from work). The company estimates that each year 1 in every 2000 policyholders may have a major injury, and 1 in 500 a minor injury.c) What’s the standard deviation for profit?

Chapter 16 Part I20. Your company bids for two contracts. You believe the probability you get contract #1 is 0.8. If you get contract #1, the probability you also get contract #2 will be 0.2, and if you do not get #1, the probability you get #2 will be 0.3.a) Are the two contracts independent? Explain.

Chapter 16 Part I20. Your company bids for two contracts. You believe the probability you get contract #1 is 0.8. If you get contract #1, the probability you also get contract #2 will be 0.2, and if you do not get #1, the probability you get #2 will be 0.3.b) Find the probability you get both contracts.

Chapter 16 Part I20. Your company bids for two contracts. You believe the probability you get contract #1 is 0.8. If you get contract #1, the probability you also get contract #2 will be 0.2, and if you do not get #1, the probability you get #2 will be 0.3.c) Find the probability you get no contract.

Chapter 16 Part I20. Your company bids for two contracts. You believe the probability you get contract #1 is 0.8. If you get contract #1, the probability you also get contract #2 will be 0.2, and if you do not get #1, the probability you get #2 will be 0.3.d) Let X be the number of contacts you get. Find the probability model for X.

Chapter 16 Part I20. Your company bids for two contracts. You believe the probability you get contract #1 is 0.8. If you get contract #1, the probability you also get contract #2 will be 0.2, and if you do not get #1, the probability you get #2 will be 0.3.e) Find the expected value and standard deviation of X.

Chapter 16 Part I24. Given independent random variables with means and standard deviations as shown, find the mean and standard deviation of each of these variables. (SHOW WORK)

a) X – 20

Mean SDX 80 12Y 12 3

b) 0.5Y

c) X + Y

d) X – Y

Chapter 16 Part II27. A grocery supplier believes that in a dozen eggs, the mean number of broken ones is 0.6 with a standard deviation of 0.5 eggs. You buy 3 dozen eggs without checking them.a) How many broken eggs do you expect to get?

Chapter 16 Part II27. A grocery supplier believes that in a dozen eggs, the mean number of broken ones is 0.6 with a standard deviation of 0.5 eggs. You buy 3 dozen eggs without checking them.b) What’s the standard deviation?

Chapter 16 Part II27. A grocery supplier believes that in a dozen eggs, the mean number of broken ones is 0.6 with a standard deviation of 0.5 eggs. You buy 3 dozen eggs without checking them.c) What assumptions did you have to make about the eggs in order to answer this question?

The cartons of eggs must be independent of each other.

Chapter 16 Part II32. A casino knows that people play the slot machines in hopes of hitting the jackpot, but that most of them lose their dollar. Suppose a certain machine pays out an average of $0.92, with a standard deviation of $120.a) Why is the standard deviation so large?

Chapter 16 Part II32. A casino knows that people play the slot machines in hopes of hitting the jackpot, but that most of them lose their dollar. Suppose a certain machine pays out an average of $0.92, with a standard deviation of $120.b) If you play 5 times, what are the mean and standard deviation of the casino’s profit?

Chapter 16 Part II32. A casino knows that people play the slot machines in hopes of hitting the jackpot, but that most of them lose their dollar. Suppose a certain machine pays out an average of $0.92, with a standard deviation of $120.c) If gamblers play this machine 1000 times in a day, what are the mean and standard deviation of the casino’s profit?

Chapter 16 Part II32. A casino knows that people play the slot machines in hopes of hitting the jackpot, but that most of them lose their dollar. Suppose a certain machine pays out an average of $0.92, with a standard deviation of $120.d) Do you think the casino is likely to be profitable? Explain.

Chapter 16 Part II34&36. The American Veterinary Association claims that the annual cost of medical care for dogs averages $100, with a standard deviation of $30, and for cats averages $120, with a standard deviation of $35.a) Answer the following questions:i. What’s the expected difference in the cost of medical care for dogs and cats?

Chapter 16 Part II34&36. The American Veterinary Association claims that the annual cost of medical care for dogs averages $100, with a standard deviation of $30, and for cats averages $120, with a standard deviation of $35.a) Answer the following questions:ii. What’s the standard deviation of that difference?

Chapter 16 Part II34&36. The American Veterinary Association claims that the annual cost of medical care for dogs averages $100, with a standard deviation of $30, and for cats averages $120, with a standard deviation of $35.a) Answer the following questions:iii. If the difference in costs can be described by a Normal model, what’s the probability that medical expenses are higher for someone’s dog than for her cat?

P(z>0.4338) = 0.332

Chapter 16 Part II34&36. The American Veterinary Association claims that the annual cost of medical care for dogs averages $100, with a standard deviation of $30, and for cats averages $120, with a standard deviation of $35.b) You’re thinking about getting two dogs and a cat. Assume the annual veterinary expenses are independent. Answer the following questions:

i. Define appropriate variables and express the total annual veterinary costs you may have.

ii. Describe the model for this total cost. Be sure to specify its name, expected value, and standard deviation.

iii. What’s the probability that your total expenses will exceed $400?

P(z>1.45) = 0.073

Chapter 16 Part II38. Bicycles arrive at a bike shop in boxes. Before they can be sold, they must be unpacked, assembled, and turned (lubricated, adjusted, etc.). Based on past experience, the shop manager makes the following assumptions about how long this may take:*The times for each setup phase are independent.*The times for each phase follow a Normal model.*The means and standard deviations of the times (in minutes) are as shown:

a) What are the mean and standard deviation for the total bicycle setup time?

Phase Mean SDUnpacking 3.5 0.7Assembly 21.8 2.4

Tuning 12.3 2.7

a) What are the mean and standard deviation for the total bicycle setup time?

Tuning 12.3 2.7

b) A customer decides to buy a bike like one of the display models but wants a different color. The shop has one, still in the box. The manager says they can have it read in half an hour. Do you think the bike will be set up and ready to go as promised? Explain.

Tuning 12.3 2.7

Chapter 16 Part II40. The bicycle shop will be offering 2 specially priced children’s models at a sidewalk sale. The basic model will sell for $120 and the deluxe model for $150. Past experience indicates that sales of the basic model will have a mean of 5.4 bikes with a standard deviation of 1.2, and sales of the deluxe model will have a mean of 3.2 bikes with a standard deviation of 0.8 bikes. The cost of setting up for the sidewalk sale is $200.a) Define random variables and use them to express the bicycle shop’s net income.

Chapter 16 Part II40. The bicycle shop will be offering 2 specially priced children’s models at a sidewalk sale. The basic model will sell for $120 and the deluxe model for $150. Past experience indicates that sales of the basic model will have a mean of 5.4 bikes with a standard deviation of 1.2, and sales of the deluxe model will have a mean of 3.2 bikes with a standard deviation of 0.8 bikes. The cost of setting up for the sidewalk sale is $200.b) What’s the mean of the net income?

Chapter 16 Part II40. The bicycle shop will be offering 2 specially priced children’s models at a sidewalk sale. The basic model will sell for $120 and the deluxe model for $150. Past experience indicates that sales of the basic model will have a mean of 5.4 bikes with a standard deviation of 1.2, and sales of the deluxe model will have a mean of 3.2 bikes with a standard deviation of 0.8 bikes. The cost of setting up for the sidewalk sale is $200.c) What’s the standard deviation of the net income?

Chapter 16 Part II40. The bicycle shop will be offering 2 specially priced children’s models at a sidewalk sale. The basic model will sell for $120 and the deluxe model for $150. Past experience indicates that sales of the basic model will have a mean of 5.4 bikes with a standard deviation of 1.2, and sales of the deluxe model will have a mean of 3.2 bikes with a standard deviation of 0.8 bikes. The cost of setting up for the sidewalk sale is $200.d) Do you need to make any assumptions in calculating the mean? How about the standard deviation?

Chapter 17

2. Can we use probability models based on Bernoulli trials to investigate the following situations? Explain by listing all conditions met and those that are not met.a) You are rolling 5 dice and need to get at least two 6’s to win the game.This can be a Bernoulli trial, because…

1) There are only 2 outcomes- Roll a 6 - Don’t roll a 6

2) The probability of success is constantp = 1/6 Success: Roll a 6

3) The rolls are independent of each other.

2. Can we use probability models based on Bernoulli trials to investigate the following situations? Explain by listing all conditions met and those that are not met.b) We record the eye colors found in a group of 500 people.

Chapter 17

This cannot be a Bernoulli trial, because…1) There are more than two possible outcomes for eye color.

2. Can we use probability models based on Bernoulli trials to investigate the following situations? Explain by listing all conditions met and those that are not met.c) A manufacturer recalls a doll because about 3% have buttons that are not properly attached. Customers return 37 of these dolls to the local toy store. Is the manufacturer likely to find any dangerous buttons?

Chapter 17

This can be a Bernoulli trial, because…1) There are only 2 outcomes - Dangerous button - Not dangerous button2) The probability of success is constant

p = .03 Success: Find dangerous button3) The trials are not independent, since the total number of dolls is finite, but 37 is probably less than 10% of all dolls.

2. Can we use probability models based on Bernoulli trials to investigate the following situations? Explain by listing all conditions met and those that are not met.d) A city council of 11 Republicans and 8 Democrats picks a committee of 4 at random. What’s the probability they choose all Democrats?

Chapter 17

This cannot be a Bernoulli trial, because…the probability of choosing a Democrat changes depending on who has already been chosen. The 10% condition is not met, because 4 of 19 people is more than 10%.

2. Can we use probability models based on Bernoulli trials to investigate the following situations? Explain by listing all conditions met and those that are not met.e) A 2002 Rutgers University study found that 74% of high-school students have cheated on a test at least once. Your local high-school principal conducts a survey in homerooms and gets responses that admit to cheating from 322 of the 481 students.

Chapter 17

This can be a Bernoulli trial, because…1) There are only 2 outcomes - Admit cheating - Don’t admit cheating2) The probability of success is constant

p = .74 Success: Admit cheating3) The trials are not independent, since the total number of students is finite, but 481 is less than 10% of all students.

6. Suppose 75% of all drivers always wear their seatbelts. Let’s investigate how many of the drivers might be belted among five cars waiting at a traffic light.a) Describe how you would simulate the number of seatbelt-wearing drivers among the five cars.

Chapter 17

Component: Check one driverOutcomes: 00 – 74 Driver wearing seatbelt

75 – 99 Driver is not wearing seatbeltTrail: Check five driversResponse variable: Number of drivers wearing seatbelts out of 5 drivers

Chapter 176. Suppose 75% of all drivers always wear their seatbelts. Let’s investigate how many of the drivers might be belted among five cars waiting at a traffic light.b) Run 30 trials

Wearing Seatbelt

# Wearing Seatbelt3

5 5 3 3 5

Chapter 176. Suppose 75% of all drivers always wear their seatbelts. Let’s investigate how many of the drivers might be belted among five cars waiting at a traffic light.c) Based on you simulation, estimate the probabilities there are no belted drivers, exactly 1, exactly 2, exactly 3, exactly 4, and exactly 5.

x 0 1 2 3 4 5P(X=x) =0 =0

Chapter 176. Suppose 75% of all drivers always wear their seatbelts. Let’s investigate how many of the drivers might be belted among five cars waiting at a traffic light.d) Calculate the actual probability model.

Binom(5, .75)

P(X=0) P(X=1) P(X=2) P(X=3) P(X=4) P(X=5)

Chapter 176. Suppose 75% of all drivers always wear their seatbelts. Let’s investigate how many of the drivers might be belted among five cars waiting at a traffic light.e) Compare the distribution of outcomes in your simulation to the probability model.

x 0 1 2 3 4 5P(X=x) 0 0

x 0 1 2 3 4 5P(X=x) 0 .001

The differences were small (the largest being 6.7% for 4 cars) and by the Law of Large Numbers they would get even smaller as the number of trials increased.

14 & 22. An Olympic archer is able to hit the bull’s-eye 80% of the time. Assume each shot is independent of the others.

Chapter 17

This can be a Bernoulli trial, because…1) There are only 2 outcomes - Hit bull’s-eye - Don’t hit bull’s-eye2) The probability of success is constant

p = .80 Success: Hit bull’s-eye3) We are told to assume each shot is independent.

Chapter 1714. An Olympic archer is able to hit the bull’s-eye 80% of the time. Assume each shot is independent of the others. i. If she shoots 6 arrows, what’s the probability of each result described below. Be sure to name the model and use proper notation each time. (First two are set-up for you.)

a) Her first bull’s-eye comes on the third arrow.

Geom(______) P(X = 3) =0.80 0.032

b) She misses the bull’s-eye at least once.

Binom(_____,_____) P(Y > 1) = 1 – P(Y = 0) =6 0.20 0.738

c) Her first bull’s-eye comes on the fourth or fifth arrow.

= 0.006 + 0.001

Geom(______) P(X = 4 OR X =5) = P(X = 4) + P(X = 5)

= 0.007

d) She gets exactly 4 bull’s-eyes.

Binom(_____,_____) P(Y = 4) = 6 0.80 0.246

e) She gets at least 4 bull’s-eyes.

Binom(_____,_____) P(Y > 4) = 1 – P(X < 3) =6 0.80 0.901

f) She gets at most 4 bull’s-eyes.

Binom(_____,_____) P(Y < 4) = 6 0.80 0.345

Chapter 1722. An Olympic archer is able to hit the bull’s-eye 80% of the time. Assume each shot is independent of the others. ii. If she shoots 200 arrows, what’s the probability of each result described below.

a) What are the mean and standard deviation of the number of bull’s-eyes she might get? (Make sure that you name the model first.)

Binom(_____,_____)200 0.80

E(Y) = np = 200(0.8) = 160 bull’s eyes

SD(Y) = = 5.66 bull’s eyes

b) Verify that you can use a Normal model to approximate the distribution of the number of good first serves.

Since np = 160 and nq = 40 are both greater than 10,

Binom(200, 0.8) may be approximated by N(160, 5.66).

c) Use the 68-95-99.7 Rule to describe this distribution.

d) Would you be surprised if she made only 140 bull’s-eyes? Explain.

I would be surprised if she only made 140 bull’s-eyes out of 200, because that should happen only about 0.02% of the time.

Chapter 1724. An orchard owner knows that he’ll have to use about 6% of the apples he harvests for cider because they will have bruises or blemishes. He expects a tree to produce about 300 apples.

a) Describe an appropriate model for the number of cider apples that may come from that tree. Justify your model. (You justify the model by showing its conditions are met.)

The model is Binom(300, .06)

iii.The trials are not independent, because

1) Apple for cider 2) Apple not for cider

p = Success: Apple for cider

there are a finite number of apples, but 300 apples is probably less than 10% of all the farmer’s apples.

a) Describe an appropriate model for the number of cider apples that may come from that tree. Justify your model. (You justify the model by showing its conditions are met.)

For Binom(300, .06) … E(X) = np = 300(0. cider apples

Since np = 18 and nq = 282 are both greater than 10,

SD(X) = = 4.11 cider apples

Binom(300, .06) can be approximated by N(18, 4.11).

b) Find the probability there will be no more than a dozen cider apples.

Binom(300, .06)

P(X<12) 0.085

b) Find the probability there will be no more than a dozen cider apples.

c) Is it likely there will be more than 50 cider apples? Explain.

Binom(300, 0.06) P(X > 50) = 1 - P(X < 50) = 2.2 x

N(18, 4.11) z = = 7.79 P(z > 50) = 3.4 x

It is nearly impossible for there to be more than 50 cider apples, because for both models the probability is nearly zero.

Chapter 1730. Vitamin D is essential for strong, healthy bones. Our bodies produce vitamin D naturally when sunlight falls upon the skin, or it can be taken as a dietary supplement. Although the bone disease rickets was largely eliminated in England during the 1950s, some people there are concern that this generation of children is at increased risk because they are more likely to watch TV or play computer games than spend time outdoors. Recent research indicated that about 20% of British children are deficient in vitamin D. Suppose doctors test a group of elementary school children.

iii.The trials are not independent, because

1) Vitamin D deficient 2) Not vitamin D deficient

p = Success: Being Vitamin D deficient

there are a finite number of British children, but children at this school represent fewer than 10% of all British children.

a) What’s the probability that the first vitamin D-deficient child is the 8th one tested?

Geom(0.2) P(X = 8) 0.042

b) What’s the probability that the first 10 children tested are all okay?

Binom(10, 0.8) P(X = 10) 0.107

c) How many kids do they expect to test before finding one who has this vitamin deficiency?

Geom(0.2)

E(X) = = = 5 children

d) They will test 50 students at the third grade level. Find the mean and standard deviation of the number who may be deficient in vitamin D.

Binom(50, 0.2)

E(X) = np = 50(0.2) = 10 children

SD(X) = = 2.83 children

e) If they test 320 children at this school, what’s the probability that no more than 50 of them have the vitamin deficiency?

Binom(320, 0.2) P(X < 50) = 0.027

e) If they test 320 children at this school, what’s the probability that no more than 50 of them have the vitamin deficiency?

Binom(320, 0.2) P(X < 50) = 0.027

Additional ExamplesWhich of the following are true statements?

I. The histogram of a binomial distribution with p = .5 is always symmetric no matter what the value of n, the number of trials.

II. The histogram of a binomial distribution with p = .2 is skewed to the left.

III. The histogram of a binomial distribution with p = .9 looks more and more symmetric, the larger the value of n.

I and III

Additional ExamplesFive managers and five employees are on a grievance committee. A three-person subcommittee is formed by a random selection from the tem committee members. What is the probability that all three members of the committee are managers?

Additional ExamplesAmong the 125 at a small college, 75 are registered Democrats, 35 are registered Republicans, and the rest are independents. If a 10-person committee is randomly picked, what is the probability that at least two independents are chosen.

know the three parts of a bernoulli trial

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