statistics and modelling course 2011. probability distributions (cont.) achievement standard 90646...

Statistics and Modelling Course

2011

Probability Distributions (cont.)

Achievement Standard 90646

Solve Probability Distribution Models to solve straightforward

problems

4 CreditsExternally Assessed

NuLake Pages 278 322

Lesson 8: The Binomial Distribution (an introduction)

Learning outcomes:• Learn what the Binomial Distribution is & its

parameters, n and .• Learn the 4 requirements for using it (you will

be memorising these).• Calculate Binomial Distribution probabilities

using the formula.

Work:Notes (sheet to fill in), cards activity & examples as class.Sigma (old – 2nd ed): Pg. 67 – Ex. 5.1. Finish for HW.

Experiment: Drawing cards from a pack of 52 with replacement.

Winner: The person who draws the most hearts in 4 attempts.

Record the number of hearts that you get (0, 1, 2, 3 or 4)

Later we will calculate the theoretical probability of getting each number of hearts (0, 1, 2, 3 or 4).

Experiment: Drawing cards from a pack of 52 with replacement.

Winner: The person who draws the most hearts in 4 attempts.

Record the number of hearts that you get (0, 1, 2, 3 or 4)

Later we will calculate the theoretical probability of getting each number of hearts (0, 1, 2, 3 or 4).

The BINOMIAL DISTRIBUTION:

Used for situations where we’re running a series of trials, each of which has TWO POSSIBLE OUTCOMES – success and failure (e.g. tossing a coin – if you call ‘heads’ each time).

We use the Binomial Distribution to work out the probability of getting a particular number of successful outcomes (e.g. 6 heads if we toss a coin 10 times).

The BINOMIAL DISTRIBUTION:

Used for situations where we’re running a series of trials, each of which has TWO POSSIBLE OUTCOMES – success and failure (e.g. tossing a coin – if you call ‘heads’ each time).

We use the Binomial Distribution to work out the probability of getting a particular number of successful outcomes (e.g. 6 heads if we toss a coin 10 times).

The 4 requirements for using the Binomial Distribution are:

(1.) Fixed number, n, of identical trials.

(2.) 2 Possible Outcomes for each trial

(success/failure)

(3.) Trials are INDEPENDENT

(4.) Probability of success, π, is the same for each

trial.

HINT: Use International Fight Club 2

We use the Binomial Distribution to work out the probability of getting a particular number of successful trials (e.g. 6 heads if we toss a coin 10 times).

The 4 requirements for using the Binomial Distribution are:

(1.) Fixed number, n, of identical trials.

(2.) 2 Possible Outcomes for each trial (success/failure)

(3.) Trials are INDEPENDENT

(4.) Probability of success, π, is the same for each trial.

HINT: Use International Fight Club 2

E.g. Experiment: Drawing cards from a pack of 52 with replacement.

4 trials by each class-member (with replacement).1 Trial: Drawing a card once.2 Outcomes for each trial: heart or not a heart

The 4 requirements for using the Binomial Distribution are:(1.) Fixed number, n, of identical trials.(2.) 2 Possible Outcomes for each trial (success/failure)(3.) Trials are INDEPENDENT(4.) Probability of success, π, is the same for each trial.HINT: Use International Fight Club 2

E.g. Experiment: Drawing cards from a pack of 52 with replacement.

4 trials by each class-member (with replacement).1 Trial: Drawing a card once.2 Outcomes for each trial: heart or not a heart.

Discrete random variable X: The number of hearts, x, drawn in the 4 independent trials.

Winner: The person who draws the most hearts.

E.g. Experiment: Drawing cards from a pack of 52 with replacement.

4 trials by each class-member (with replacement).1 Trial: Drawing a card once.2 Outcomes for each trial: heart or not a heart.

Discrete random variable X: The number of hearts, x, drawn in the 4 independent trials.

Winner: The person who draws the most hearts.

Success: Drawing a heart.

P(success) = 0.25.

P(failure) = 0.75.

E.g. Experiment: Drawing cards from a pack of 52 with replacement.

4 trials by each class-member (with replacement).1 Trial: Drawing a card once.2 Outcomes for each trial: heart or not a heart.

Discrete random variable X: The number of hearts, x, drawn in the 4 independent trials.

Winner: The person who draws the most hearts.

Success: Drawing a heart.

P(success) = 0.25. We use “” to represent

P(Success)

P(failure) = 0.75.

E.g. Experiment: Drawing cards from a pack of 52 with replacement.

4 trials by each class-member (with replacement).1 Trial: Drawing a card once.2 Outcomes for each trial: heart or not a heart.

Discrete random variable X: The number of hearts, x, drawn in the 4 independent trials.

Winner: The person who draws the most hearts.

Success: Drawing a heart.

P(success) = 0.25. We use “” to represent

P(Success)

P(failure) = 0.75. We use “1- π” to represent

P(Failure)

Discrete random variable X: The number of hearts, x, drawn in the 4 independent trials.

Winner: The person who draws the most hearts.

Success: Drawing a heart.P(success) = 0.25. We use “” to represent

P(Success)P(failure) = 0.75. We use “1- π” to represent

P(Failure)

Q. What is the probability, in theory, that exactly three of the 4 cards that you draw will be hearts?

A.Well, the probability that the first 3 are hearts and the other 1 is not is:

P(H, H, H, N) = 0.25 × 0.25 × 0.25 × 0.75

BUT. . .

Winner: The person who draws the most hearts.

Success: Drawing a heart.P(success) = 0.25. We use “” to represent

P(Success)P(failure) = 0.75. We use “1- π” to represent

P(Failure)

Q. What is the probability, in theory, that exactly three of the 4 cards that you draw will be hearts?

i.e. P(X=3)

A.Well, the probability that the first 3 are hearts and the other 1 is not is:

P(H, H, H, N) = 0.25 × 0.25 × 0.25 × 0.75

BUT. . .the 3 hearts can be selected in 4 different orders

(4C3):H,H,H,N or H,H,N,H or H,N,H,H or N,H,H,H

Q. What is the probability, in theory, that exactly three of the 4 cards that you draw will be hearts?

A.Well, the probability that the first 3 are hearts and the other 1 is not is:

P(H, H, H, N) = 0.25 × 0.25 × 0.25 × 0.75

BUT. . .the 3 hearts can be selected in 4 different orders

(4C3):H,H,H,N or H,H,N,H or H,N,H,H or N,H,H,H

So P(X=3) = 4 × 0.25 × 0.25 × 0.25 × 0.75

= 4C3 × 0.253 × 0.75

= 0.046875

Q. What is the probability, in theory, that exactly three of the 4 cards that you draw will be hearts?

A.Well, the probability that the first 3 are hearts and the other 1 is not is:

P(H, H, H, N) = 0.25 × 0.25 × 0.25 × 0.75

BUT. . .the 3 hearts can be selected in 4 different orders

(4C3):H,H,H,N or H,H,N,H or H,N,H,H or N,H,H,H

So P(X=3) = 4 × 0.25 × 0.25 × 0.25 × 0.75

= 4C3 × 0.253 × 0.75

= 0.046875

Q: What is this saying about how often we would expect to get 3 hearts when running this experiment?

BUT. . .the 3 hearts can be selected in 4 different orders

(4C3):H,H,H,N or H,H,N,H or H,N,H,H or N,H,H,H

So P(X=3) = 4 × 0.25 × 0.25 × 0.25 × 0.75

= 4C3 × 0.253 × 0.75

= 0.046875

Q: What is this saying about how often we would expect to get 3 hearts when running this experiment?

BUT. . .the 3 hearts can be selected in 4 different orders

(4C3):H,H,H,N or H,H,N,H or H,N,H,H or N,H,H,H

So P(X=3) = 4 × 0.25 × 0.25 × 0.25 × 0.75

= 4C3 × 0.253 × 0.75

= 0.046875

We define the Binomial variable, X, by the formula:

BUT. . .the 3 hearts can be selected in 4 different orders

(4C3):H,H,H,N or H,H,N,H or H,N,H,H or N,H,H,H

So P(X=3) = 4 × 0.25 × 0.25 × 0.25 × 0.75

= 4C3 × 0.253 × 0.75

= 0.046875

We define the Binomial variable, X, by the formula:

P(X = x) =

BUT. . .the 3 hearts can be selected in 4 different orders

(4C3):H,H,H,N or H,H,N,H or H,N,H,H or N,H,H,H

So P(X=3) = 4 × 0.25 × 0.25 × 0.25 × 0.75

= 4C3 × 0.253 × 0.75

= 0.046875

We define the Binomial variable, X, by the formula:

P(X = x) = πx (1–π)n–x xnC

BUT. . .the 3 hearts can be selected in 4 different orders

(4C3):H,H,H,N or H,H,N,H or H,N,H,H or N,H,H,H

So P(X=3) = 4 × 0.25 × 0.25 × 0.25 × 0.75

= 4C3 × 0.253 × 0.75

= 0.046875

We define the Binomial variable, X, by the formula:

P(X = x) = πx (1–π)n–x

where x is the number of successes

xnC

BUT. . .the 3 hearts can be selected in 4 different orders

(4C3):H,H,H,N or H,H,N,H or H,N,H,H or N,H,H,H

So P(X=3) = 4 × 0.25 × 0.25 × 0.25 × 0.75

= 4C3 × 0.253 × 0.75

= 0.046875

We define the Binomial variable, X, by the formula:

P(X = x) = πx (1–π)n–x

where x is the number of successes out of n independent trials.

xnC

BUT. . .the 3 hearts can be selected in 4 different orders

(4C3):H,H,H,N or H,H,N,H or H,N,H,H or N,H,H,H

So P(X=3) = 4 × 0.25 × 0.25 × 0.25 × 0.75

= 4C3 × 0.253 × 0.75

= 0.046875

We define the Binomial variable, X, by the formula:

P(X = x) = πx (1–π)n–x

where x is the number of successes out of n independent trials.

is the probability of “success” in each individual trial.

xnC

BUT. . .the 3 hearts can be selected in 4 different orders

(4C3):H,H,H,N or H,H,N,H or H,N,H,H or N,H,H,H

So P(X=3) = 4 × 0.25 × 0.25 × 0.25 × 0.75

= 4C3 × 0.253 × 0.75

= 0.046875

We define the Binomial variable, X, by the formula:

P(X = x) = πx (1–π)n–x

where x is the number of successes out of n independent trials.

is the probability of “success” in each individual trial.

x

n

BUT. . .the 3 hearts can be selected in 4 different orders

(4C3):H,H,H,N or H,H,N,H or H,N,H,H or N,H,H,H

So P(X=3) = 4 × 0.25 × 0.25 × 0.25 × 0.75

= 4C3 × 0.253 × 0.75

= 0.046875

We define the Binomial variable, X, by the formula:

P(X = x) = πx (1–π)n–x

where x is the number of successes out of n independent trials.

is the probability of “success” in each individual trial.

xnC

So P(X=3) = 4 × 0.25 × 0.25 × 0.25 × 0.75

= 4C3 × 0.253 × 0.75

= 0.046875

Activity:1. Use the Binomial Distribution formula to calculate the

theoretical probability of getting each of 0, 1, 2, 3 or 4 hearts when 4 cards are selected at random with replacement.

2.Show these probabilities on a probability distribution table.

We define the Binomial variable, X, by the formula:

P(X = x) = πx (1–π)n–x

where x is the number of successes out of n independent trials.

is the probability of “success” in each individual trial.

xnC

the 3 hearts can be selected in 4 different orders (4C3):

H,H,H,N or H,H,N,H or H,N,H,H or N,H,H,H

So P(X=3) = 4 × 0.25 × 0.25 × 0.25 × 0.75 = 4C3 × 0.253 × 0.75

Activity:1. Use the Binomial Distribution formula to calculate the

theoretical probability of getting each of 0, 1, 2, 3 or 4 hearts when 4 cards are selected at random with replacement.

2.Show these probabilities on a probability distribution table.

HW: Do Sigma (old – 2nd ed): p67 – Ex. 5.1. SKIP Q10

We define the Binomial variable, X, by the formula:

P(X = x) = πx (1–π)n–x

where x is the number of successes out of n independent trials.

is the probability of “success” in each individual trial.

xnC

x 0 1 2 3 4

P(X=x)

Probability Distribution Table of Number of Hearts obtained (X) when 4 cards are selected with replacement from a standard deck

We define the Binomial variable, X, by the formula:

P(X = x) = πx (1–π)n–x

where x is the number of successes out of n independent trials.

is the probability of “success” in each individual trial.

xnC

HW: Do Sigma (old – 2nd ed): p67 – Ex. 5.1. SKIP Q10

Lesson 9: Use the Binomial Distribution to calculate probabilities.

Learning outcome: Calculate binomial probabilities for individual and combined events using the formula & tables.

STARTER: Warm-up quiz (handouts to write answers on).

Combined events e.g.Do NuLake p285 and 286 (finish for HW).

Year 13 today:

Getting binomial distribution sorted!

Do warm-up quiz (answers on projector).

Copy notes on how to calculate Binomial probabilities for more than 1 value of X.

E.g. P(X<4)

Do NuLake pg. 285 and 286: Q411.

*Extension people go right through to Q19.

Warm-up Quiz:A police officer checks five cars in succession. She knows from experience that the probability of a car not having a warrant of fitness (WOF) is 1/6.(1.) Write out the 4 requirements for the use of the Binomial Distn.

(2.) If we use the Binomial Distn to model the number of cars not having a WOF, which of the 4 requirements are definitely met? Do any require us to make questionable assumptions?

(3.) Use the binomial distribution to calculate the probability that three of the cars have not got warrants of fitness.

A police officer checks five cars in succession. The probability of a car not havingWOF is 1/6.(1.) Write out the 4 requirements for the use of the Binomial Distn.

There are four conditions to check

• only two possible outcomes (success or failure) at each trial

• a fixed number of trials

• probability of success at each trial is constant

• each trial is independent

A police officer checks five cars in succession. The probability of a car not havingWOF is 1/6.(2.)If we use the Bin. Distn to model the number of cars not having a WOF, whichof the 4 requirements are definitely met? Which require us to make assumptions?

There are four conditions to check

• only two possible outcomes (success or failure) at each trial

• a fixed number of trials

• probability of success at each trial is constant

• each trial is independent

Yes. A car either has a WOF or does not

Yes. 5 cars are checked

Making an assumption that whether one car has a WOF is independent of whether another one does. Should be a valid assumption.

A police officer checks five cars in succession. The probability of a car not havingWOF is 1/6.(2.)If we use the Bin. Distn to model the number of cars not having a WOF, whichof the 4 requirements are definitely met? Which require us to make assumptions?

There are four conditions to check

• only two possible outcomes (success or failure) at each trial

• a fixed number of trials

• probability of success at each trial is constant

• each trial is independent

Yes. A car either has a WOF or does not

Yes. 5 cars are checked

Yes, again so long as we don’t get cars travelling in convoy.

Making an assumption that whether one car has a WOF is independent of whether another one does. Should be a valid assumption. There is the possibility that successive cars could be travelling in convoy! But this would be rare.

A police officer checks five cars in succession. The probability of a car not having a WOF is 1/6.3. Find the probability that three of the cars have not got WOFs.

Give the values of the two parameters.n = 5 16

Calculate the probability of failure(i.e. a given car fails its WOF),1-π.

1-π 56

3. π =

15.01 A police officer checks five cars in succession. The probability of a car not having a WOF is 1/6.3. Find the probability that three of the cars have not got WOFs.

n = 5 16 1–π =

563.

What are we trying to findand what formula will we use?

π =

Give the values of the two parameters.

Calculate the probability of failure(i.e. a given car fails its WOF),1-π.

15.01 A police officer checks five cars in succession. The probability of a car not having a WOF is 1/6.3. Find the probability that three of the cars have not got WOFs.

n = 5 16 1–π =

563.

Find P(X = 3) using P(X = x) = nCx π x (1- π )n–x

( 3)P X 3 2

53

1 56 6C

2507776

Calculate P(X=3)

π =

= 0.03215 (4sf)

How to calculate Binomial probabilities for multiple X values:

E.g. Calculate P (X<2) for n=10, =0.3

How to calculate Binomial probabilities for multiple X values:

( 3)P X 3 2

53

1 56 6C

2507776

Calculate P(X=3)

= 0.03215 (4sf)

1–π = 563.

Find P(X = 3) using P(X = x) = nCx π x (1- π )n–x

π = n = 5 16

E.g. Calculate P (X<2) for n=10, =0.3

P(X<2) = P(X=0) + P(X=1) + P(X=2) ~ look up on tables

= _______ + ______ + _______

How to calculate Binomial probabilities for multiple X values:

( 3)P X 3 2

53

1 56 6C

2507776

Calculate P(X=3)

= 0.03215 (4sf)

E.g. Calculate P (X<2) for n=10, =0.3

P(X<2) = P(X=0) + P(X=1) + P(X=2) ~ look up on tables

= _______ + ______ + _______

How to calculate Binomial probabilities for multiple X values:

E.g. Calculate P (X<2) for n=10, =0.3

P(X<2) = P(X=0) + P(X=1) + P(X=2) ~ look up on tables

= _______ + ______ + _______

How to calculate Binomial probabilities for multiple X values:

Parameters: n=10, =0.3To find P(X<2):

P(X= 0) = + P(X= 1) = + P(X= 2) =

Parameters: n=10, =0.3To find P(X<2):

P(X= 0) = + P(X= 1) = + P(X= 2) =

Parameters: n=10, =0.3To find P(X<2):

P(X= 0) = 0.0282+ P(X= 1) = + P(X= 2) =

Parameters: n=10, =0.3To find P(X<2):

P(X= 0) = 0.0282+ P(X= 1) = + P(X= 2) =

Parameters: n=10, =0.3To find P(X<2):

P(X= 0) = 0.0282+ P(X= 1) = 0.1211+ P(X= 2) =

Parameters: n=10, =0.3To find P(X<2):

P(X= 0) = 0.0282+ P(X= 1) = 0.1211+ P(X= 2) =

Parameters: n=10, =0.3To find P(X<2):

P(X= 0) = 0.0282+ P(X= 1) = 0.1211+ P(X= 2) = 0.2335

Parameters: n=10, =0.3To find P(X<2):

P(X= 0) = 0.0282+ P(X= 1) = 0.1211+ P(X= 2) = 0.2335 P(X<2) =

Parameters: n=10, =0.3To find P(X<2):

P(X= 0) = 0.0282+ P(X= 1) = 0.1211+ P(X= 2) = 0.2335 P(X<2) = 0.3828

E.g. Calculate P (X<2) for n=10, =0.3

P(X<2) = P(X=0) + P(X=1) + P(X=2) ~ look up on tables

= _______ + ______ + _______

How to calculate Binomial probabilities for multiple X values:

E.g. Calculate P (X<2) for n=10, =0.3

P(X<2) = P(X=0) + P(X=1) + P(X=2) ~ look up on tables

= 0.0282 + ______ + _______

How to calculate Binomial probabilities for multiple X values:

E.g. Calculate P (X<2) for n=10, =0.3

P(X<2) = P(X=0) + P(X=1) + P(X=2) ~ look up on tables

= 0.0282 + ______ + _______

How to calculate Binomial probabilities for multiple X values:

E.g. Calculate P (X<2) for n=10, =0.3

P(X<2) = P(X=0) + P(X=1) + P(X=2) ~ look up on tables

= 0.0282 + ______ + _______

How to calculate Binomial probabilities for multiple X values:

E.g. Calculate P (X<2) for n=10, =0.3

P(X<2) = P(X=0) + P(X=1) + P(X=2) ~ look up on tables

= 0.0282 + 0.1211 + _______

How to calculate Binomial probabilities for multiple X values:

How to calculate Binomial probabilities for multiple X values:

E.g. Calculate P (X<2) for n=10, =0.3

P(X<2) = P(X=0) + P(X=1) + P(X=2) ~ look up on tables

= 0.0282 + 0.1211 + _______

How to calculate Binomial probabilities for multiple X values:

E.g. Calculate P (X<2) for n=10, =0.3

P(X<2) = P(X=0) + P(X=1) + P(X=2) ~ look up on tables

= 0.0282 + 0.1211 + _______

How to calculate Binomial probabilities for multiple X values:

E.g. Calculate P (X<2) for n=10, =0.3

P(X<2) = P(X=0) + P(X=1) + P(X=2) ~ look up on tables

= 0.0282 + 0.1211 + 0.2335

How to calculate Binomial probabilities for multiple X values:

E.g. Calculate P (X<2) for n=10, =0.3

P(X<2) = P(X=0) + P(X=1) + P(X=2) ~ look up on tables

= 0.0282 + 0.1211 + 0.2335

= 0.3828 answer

• Do NuLake pg. 285 and 286: Q411.

Extension people: Q419. Q1219 involve intersections and unions with 2 or more independent events.

Lesson 10: Practice calculating Binomial probabilities.

Learning Outcomes: Practice calculating Binomial probabilities for combined events and become confident at this.

• How to use your G.C. to calculate binomial probabilities.

• Finish NuLake pg. 285288.

• Do Sigma (must be in old edition) - Ex. 5.3

How to calculate Binomial probabilities for multiple X values:

Yesterday’s example: Calculate P (X<2) for n=10, =0.3

P(X<2) = P(X=0) + P(X=1) + P(X=2) ~ look up on tables

= 0.0282 + 0.1211 + 0.2335

= 0.3828 answer

BUT… there is a much faster way…

How to calculate Binomial probabilities for multiple X values on your G. Calculator

Yesterday’s example: Calculate P (X<2) for n=10, =0.3

P(X<2) = P(X=0) + P(X=1) + P(X=2)

= 0.0282 + 0.1211 + 0.2335

= 0.3828 answer

BUT… there is a much faster way…

How to calculate Binomial probabilities for multiple X values on your G. Calculator

Yesterday’s example: Calculate P (X<2) for n=10, =0.3

P(X<2) = P(X=0) + P(X=1) + P(X=2)

= 0.0282 + 0.1211 + 0.2335

= 0.3828 answer

BUT… there is a much faster way…

On your Graphics Calculator:• STAT, DIST (like you did for the Normal Dist)• BINM (F5) (Binomial distribution)• Bcd Cumulative binomial probabilities, eg P(X ≤ 2)• Data: Variable • “x” means X < Enter what X is LESS THAN OR

EQUAL TO.

Note: Bpd gives individual binomial probabilities, eg P(X=2), Bcd gives cumulative binomial probabilities, eg P(X ≤

2)

How to calculate Binomial probabilities for multiple X values on your G. Calculator

On your Graphics Calculator:• STAT, DIST (like you did for the Normal Dist)• BINM (F5) (Binomial distribution)• Bcd Cumulative binomial probabilities, eg P(X ≤ 2)

Data: Variable • “x” means X < Enter what X is LESS THAN OR

EQUAL TO.

Note: Bpd gives individual binomial probabilities, eg P(X=2), Bcd gives cumulative binomial probabilities, eg P(X ≤

2)E.g. for previous example: Calculate P (X<2) for n=10, =0.3

Under Bcd, Enter x: 2. This means X < 2.

Type execute. Check that you get 0.38278

Finish NuLake pg. 286288 (right up to Q19).Then Sigma (must be in old edition): Ex. 5.3.

Lesson 11: Solve inverse Binomial Distribution problems.

• Calculate the number of trials required for the probability of getting at least x successes to be at a certain level.

• Calculate , the probability of success in a single trial if told the probability of x successes in n trials.

2 Examples as class, then:

Old Sigma (2nd ed): Pg. 73 – Ex. 5.4

E.g.1: A teacher tosses a pair of dice at the start of every lesson and tells his class that they get a free period if he gets double sixes.(a)Name the 2 parameters of the Binomial Distribution.

(b)What would count as one trial? Calculate the probability of success in a single trial.

(c)How many times must the teacher do this for there to be at least a 50% chance of getting at least one free period?

Inverse Binomial problems

Let X: Number of trials that result in double sixes.

Calculate n if P(X >1) is 50% i.e. P(X >1) = 0.5.

Cannot do it this way, so take the complement – i.e. P(X = 0) =0.5.

Answer: n (number of trials) and (prob. of success in any single trial)

Answer: One trial: tossing the pair of dice once. = P(double sixes) = 1/36 assuming the dice are both fair.

5.036

35

36

1)0(0

0

nnC

5.036

35 1 1

n

(b) What would count as one trial? Calculate the probability of success in a single trial.

(c) How many times must the teacher do this for there to be at least a 50% chance of getting at least one free period?

Let X: Number of trials that result in double sixes.

Calculate n if P(X >1) is 50% i.e. P(X >1) = 0.5.

Cannot do it this way, so take the complement – i.e. P(X = 0) =0.5.

Answer: One trial: tossing the pair of dice once. = P(double sixes) = 1/36 assuming the dice are both fair.

5.036

35

36

1)0(0

0

nnC

5.036

35

n

5.036

35 1 1

n

5.0log36

35log

n

(c) How many times must the teacher do this for there to be at least a 50% chance of getting at least one free period?

Let X: Number of trials that result in double sixes.

Calculate n if P(X >1) is 50% i.e. P(X >1) = 0.5.

Cannot do it this way, so take the complement – i.e. P(X = 0) =0.5.

Answer: Must be a whole number. So n = 25 is the minimum number of trials before P(X=0) drops below 0.5 – i.e. P(X>1) exceeds 0.5.

5.036

35

36

1)0(0

0

nnC

5.036

35

n

5.036

35 1 1

n

5.0log36

35log

n

3635

log

5.0logn

...6.24n

Let X: Number of trials that result in double sixes.

Calculate n if P(X >1) is 50% i.e. P(X >1) = 0.5.

Cannot do it this way, so take the complement – i.e. P(X = 0) =0.5.

Answer: Must be a whole number. So n = 25 is the minimum number of trials before P(X=0) drops below 0.5 – i.e. P(X>1) exceeds 0.5.(d) Another teacher flips a pair of coins every lesson, giving out chocolates to

everyone whenever both come up tails. He only has enough chocolates to give them out to the class 3 times. He wants to keep the prob. of it happening more than 3 times to below 0.1. What’s the greatest number of trials he can afford to do?

5.036

35

36

1)0(0

0

nnC

5.036

35

n

5.036

35 1 1

n

5.0log36

35log

n

3635

log

5.0logn

...6.24n

(d) Another teacher flips a pair of coins every lesson, giving out chocolates to everyone whenever both come up tails. He only has enough chocolates to give them out to the class 3 times. He wants to keep the prob. of it happening more than 3 times to below 0.1. What’s the greatest number of trials he can afford to do?

Calculate the largest possible number of trials, n, such that:P(X >3) remains below 0.1 , where = P(Both tails) = 0.25.

i.e. P(X<3) remains above 0.9. P(X=0) + P(X=1) + P(X=2) + P(X=3) > 0.9.Problem: We can’t solve this algebraically using the formula.

Solution: Refer to your tables or GC and use guess-and-check. Starting with n =3, calculate P(X< 3). Do the same for n =4, then n =5 etc.Stop when you get an n value where P(X< 3) drops below 0.9.

Answer: Must be a whole number. So n = 25 is the minimum number of trials before P(X=0) drops below 0.5 – i.e. P(X>1) exceeds 0.5.

5.0log36

35log

n

3635

log

5.0logn

...6.24n

(d) Another teacher flips a pair of coins every lesson, giving out chocolates to everyone whenever both come up tails. He only has enough chocolates to give them out to the class 3 times. He wants to keep the prob. of it happening more than 3 times to below 0.1. What’s the greatest number of trials he can afford to do?

Calculate the largest possible number of trials, n, such that:P(X >3) remains below 0.1 , where = P(Both tails) = 0.25.

i.e. P(X<3) remains above 0.9. P(X=0) + P(X=1) + P(X=2) + P(X=3) > 0.9.Problem: We can’t solve this algebraically using the formula.

Solution: Refer to your tables or GC and use guess-and-check. Starting with n =3, calculate P(X< 3). Do the same for n =4, then n =5 etc.Stop when you get an n value where P(X< 3) drops below 0.9.

For n=3,

(d) Another teacher flips a pair of coins every lesson, giving out chocolates to everyone whenever both come up tails. He only has enough chocolates to give them out to the class 3 times. He wants to keep the prob. of it happening more than 3 times to below 0.1. What’s the greatest number of trials he can afford to do?

Calculate the largest possible number of trials, n, such that:P(X >3) remains below 0.1 , where = P(Both tails) = 0.25.

i.e. P(X<3) remains above 0.9. P(X=0) + P(X=1) + P(X=2) + P(X=3) > 0.9.Problem: We can’t solve this algebraically using the formula.

Solution: Refer to your tables or GC and use guess-and-check. Starting with n =3, calculate P(X< 3). Do the same for n =4, then n =5 etc.Stop when you get an n value where P(X< 3) drops below 0.9.

For n=3, P(X<3) = 1 n=4, P(X<3) = 0.99609 n=5, P(X<3) = 0.98437. Try something a bit higher… n=7, P(X<3) = 0.92944. Getting close now… n=8, P(X<3) = 0.88618. STOP. Below 0.9.

(d) Another teacher flips a pair of coins every lesson, giving out chocolates to everyone whenever both come up tails. He only has enough chocolates to give them out to the class 3 times. He wants to keep the prob. of it happening more than 3 times to below 0.1. What’s the greatest number of trials he can afford to do?

Calculate the largest possible number of trials, n, such that:P(X >3) remains below 0.1 , where = P(Both tails) = 0.25.

i.e. P(X<3) remains above 0.9. P(X=0) + P(X=1) + P(X=2) + P(X=3) > 0.9.Problem: We can’t solve this algebraically using the formula.

Solution: Refer to your tables or GC and use guess-and-check. Starting with n =3, calculate P(X< 3). Do the same for n =4, then n =5 etc.Stop when you get an n value where P(X< 3) drops below 0.9.

For n=3, P(X<3) = 1 n=4, P(X<3) = 0.99609 n=5, P(X<3) = 0.98437. Try something a bit higher… n=7, P(X<3) = 0.92944. Getting close now… n=8, P(X<3) = 0.88618. STOP. Below 0.9. i.e. P(X>3) = 1 – 0.88618

(d) Another teacher flips a pair of coins every lesson, giving out chocolates to everyone whenever both come up tails. He only has enough chocolates to give them out to the class 3 times. He wants to keep the prob. of it happening more than 3 times to below 0.1. What’s the greatest number of trials he can afford to do?

Calculate the largest possible number of trials, n, such that:P(X >3) remains below 0.1 , where = P(Both tails) = 0.25.

i.e. P(X<3) remains above 0.9. P(X=0) + P(X=1) + P(X=2) + P(X=3) > 0.9.Problem: We can’t solve this algebraically using the formula.

Solution: Refer to your tables or GC and use guess-and-check. Starting with n =3, calculate P(X< 3). Do the same for n =4, then n =5 etc.Stop when you get an n value where P(X< 3) drops below 0.9.

For n=3, P(X<3) = 1 n=4, P(X<3) = 0.99609 n=5, P(X<3) = 0.98437. Try something a bit higher… n=7, P(X<3) = 0.92944. Getting close now… n=8, P(X<3) = 0.88618. STOP. Below 0.9. i.e. P(X>3) = 0.11382Answer: The max. number of times the teacher can flip the pair of coins is n = 7.After this, the probability of getting double tails more than 3 times exceeds 0.1.

• Copy down e.g. & working then do:

Old Sigma (2nd ed): Pg. 73 – Ex. 5.4

Or new Sigma: Pg. 326 – Ex. 15.03.

Complete for HW.

Lesson 12: Approximations 1: Binomial approximated by Normal

Learning outcome:Solve problems where the binomial distribution

must be approximated by the normal distribution.

STARTER: Re-cap of 4 requirements for use of the Binomial Distribution.

Notes and demo using comparison of the 2 distributions on Excel.

Do Sigma (old) – pg. 118 - Ex. 8.1.

Go into “play mode” to view slideshow as intended.

Hold down SHIFT and click F5.

Warm-up quiz:Q1: List the 4 requirements for a discrete random variable to be

modelled by the Binomial Distribution.HINT: International Fight Club 2

The 4 requirements for a Binomial Distribution are:

(1.) Fixed number, n, of identical trials.

(2.) 2 Possible Outcomes for each trial (success/failure)

(3.) Trials are INDEPENDENT

(4.) Probability of success, π, is the same for each trial.

Q2: If you toss a coin 100 times, use your G.C. to calculate the probability of getting a tail less than 49 times.Why can’t you use your tables for this?

Q3: Now try to calculate the probability of getting less than 490 tails from 1000 coin tosses. What happens?

Warm-up quiz:Q2: If you toss a coin 100 times, use your G.C. to calculate the probability of getting a tail less than 49 times.Why can’t you use your tables for this?

Q3: Now try to calculate the probability of getting less than 490 tails from 1000 coin tosses. What happens?

We can’t calculate this using the Binomial Distribution with either our tables or G.C. as neither stores probabilities for n values this large.

But…

Using the Normal Distribution to approximate the Binomial Distribution

E.g. You flip a fair coin 1000 times. What’s the probability of getting a tail less than 490 times?

Using the Normal Distribution to approximate the Binomial Distribution

E.g. You flip a fair coin 1000 times. What’s the probability of getting a tail less than 490 times?

We can’t calculate this using the Binomial Distribution with either our tables or G.C. as neither stores probabilities for n values this large.

But… we can approximate it using the Normal Distribution.

Using the Normal Distribution to approximate the Binomial Distribution

E.g. You flip a fair coin 1000 times. What’s the probability of getting a tail less than 490 times?

We can’t calculate this using the Binomial Distribution with either our tables or G.C. as neither stores probabilities for n values this large.

But… we can approximate it using the Normal Distribution.

We can use the Normal Distribution to approximate Binomial probabilities when:

1. n is very large.2. is NOT extreme (not too close to 0 or 1) – so

distribution is symmetrical, not skewed.CHECK: The conditions that ensure this are that both n and n(1-) are > 5

E.g. You flip a fair coin 1000 times. What’s the probability of getting a tail less than 490 times?

We can’t calculate this using the Binomial Distribution with either our tables or G.C. as neither stores probabilities for n values this large.

But… we can approximate it using the Normal Distribution.

We can use the Normal Distribution to approximate Binomial probabilities when:

1. n is very large.2. is NOT extreme (not too close to 0 or 1) – so

distribution is symmetrical, not skewed.CHECK: The conditions that ensure this are that both n and n(1-) are > 5

** NOTE: The Normal approximation requires a Continuity

Correction.

We can’t calculate this using the Binomial Distribution with either our tables or G.C. as neither stores probabilities for n values this large.

But… we can approximate it using the Normal Distribution.

We can use the Normal Distribution to approximate Binomial probabilities when:

1. n is very large.2. is NOT extreme (not too close to 0 or 1) – so

distribution is symmetrical, not skewed.CHECK: The conditions that ensure this are that both n and n(1-) are > 5

** NOTE: The Normal approximation requires a Continuity

Correction.

Normal Distn Parameters: n = )1( n

We can use the Normal Distribution to approximate Binomial probabilities when:

1. n is very large.2. is NOT extreme (not too close to 0 or 1) – so

distribution is symmetrical, not skewed.CHECK: The conditions that ensure this are that both n and n(1-) are > 5

** NOTE: The Normal approximation requires a Continuity

Correction.Normal Distn Parameters: n = )1( n

We can use the Normal Distribution to approximate Binomial probabilities when:

1. n is very large.2. is NOT extreme (no too close to 0 or 1) – so

distribution is symmetrical, not skewed.CHECK: The conditions that ensure this are that both n and n(1-) are > 5** NOTE: The Normal approximation requires a

Continuity Correction.Normal Distn Parameters: n = )1( n

So to calculate the probability of getting a tail less than 490 times out of 1000 coin tosses:

Use a normal approximation with parameters and : = n = 1000 × 0.5 = 500

We can use the Normal Distribution to approximate Binomial probabilities when:

1. n is very large.2. is NOT extreme (not too close to 0 or 1) – so

distribution is symmetrical, not skewed.CHECK: The conditions that ensure this are that both n and n(1-) are > 5** NOTE: The Normal approximation requires a

Continuity Correction.Normal Distn Parameters: n = )1( n

So to calculate the probability of getting a tail less than 490 times out of 1000 coin tosses:

Use a normal approximation with parameters and : = n = = 1000 × 0.5 = 500

)1( n

We can use the Normal Distribution to approximate Binomial probabilities when:

1. n is very large.2. is NOT extreme (not too close to 0 or 1) – so

distribution is symmetrical, not skewed.CHECK: The conditions that ensure this are that both n and n(1-) are > 5** NOTE: The Normal approximation requires a

Continuity Correction.Normal Distn Parameters: n = )1( n

So to calculate the probability of getting a tail less than 490 times out of 1000 coin tosses:

Use a normal approximation with parameters and : = n = = 1000 × 0.5 = 500 =

)1( n

5.05.01000

CHECK: The conditions that ensure this are that both n and n(1-) are > 5** NOTE: The Normal approximation requires a

Continuity Correction.Normal Distn Parameters: n = )1( n

So to calculate the probability of getting a tail less than 490 times out of 1000 coin tosses:

Use a normal approximation with parameters and : = n = = 1000 × 0.5 = 500 =

= 15.8114

So the expected number of tails in 1000 throws is 500 with a standard deviation of 15.8114.

)1( n

5.05.01000

Normal Distn Parameters: n = )1( n

So to calculate the probability of getting a tail less than 490 times out of 1000 coin tosses:

Use a normal approximation with parameters and : = n = = 1000 × 0.5 = 500 =

= 15.8114

So the expected number of tails in 1000 throws is 500 with a standard deviation of 15.8114.

P(X < 490) ≈ P( ? ) (using continuity correction)

)1( n

5.05.01000

Normal Distn Parameters: n = )1( n

So to calculate the probability of getting a tail less than 490 times out of 1000 coin tosses:

Use a normal approximation with parameters and : = n = = 1000 × 0.5 = 500 =

= 15.8114

So the expected number of tails in 1000 throws is 500 with a standard deviation of 15.8114.

P(X < 490) ≈ P(X < 489.5) (using continuity correction)

)1( n

5.05.01000

So to calculate the probability of getting a tail less than 490 times out of 1000 coin tosses:

Use a normal approximation with parameters and : = n = = 1000 × 0.5 = 500 =

= 15.8114

So the expected number of tails in 1000 throws is 500 with a standard deviation of 15.8114.

P(X < 490) ≈ P(X < 489.5) (using continuity correction)

≈ P(Z < 489.5 – 500)

15.8114

≈ 0.2533 (4sf) answer

)1( n

5.05.01000

Do old Sigma (2nd ed): Pg. 118 – Ex. 8.1.

Complete for HW.

So 490 tails would be just above the lower quartile.