Statistics and Modelling Course 2011

Probability Distributions (cont.) Achievement Standard Solve Probability Distribution Models to solve straightforward problems 4 Credits Externally 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 – 2 nd ed): Pg. 67 – Ex 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) = 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) = 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) = We use “  ” to represent P(Success) P(failure) = 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) = We use “  ” to represent P(Success) P(failure) = 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) = We use “  ” to represent P(Success) P(failure) = 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 ( 4 C 3 ): 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 ( 4 C 3 ): 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 = 4 C 3 × × 0.75 =

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 ( 4 C 3 ): 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 = 4 C 3 × × 0.75 = 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 ( 4 C 3 ): 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 = 4 C 3 × × 0.75 = 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 ( 4 C 3 ): 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 = 4 C 3 × × 0.75 = We define the Binomial variable, X, by the formula:

BUT... the 3 hearts can be selected in 4 different orders ( 4 C 3 ): 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 = 4 C 3 × × 0.75 = We define the Binomial variable, X, by the formula: P(X = x ) =

BUT... the 3 hearts can be selected in 4 different orders ( 4 C 3 ): 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 = 4 C 3 × × 0.75 = We define the Binomial variable, X, by the formula: P(X = x ) = π x (1– π ) n– x

BUT... the 3 hearts can be selected in 4 different orders ( 4 C 3 ): 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 = 4 C 3 × × 0.75 = We define the Binomial variable, X, by the formula: P(X = x ) = π x (1– π ) n– x where x is the number of successes

BUT... the 3 hearts can be selected in 4 different orders ( 4 C 3 ): 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 = 4 C 3 × × 0.75 = We define the Binomial variable, X, by the formula: P(X = x ) = π x (1– π ) n– x where x is the number of successes out of n independent trials.

BUT... the 3 hearts can be selected in 4 different orders ( 4 C 3 ): 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 = 4 C 3 × × 0.75 = We define the Binomial variable, X, by the formula: P(X = x ) = π x (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.

BUT... the 3 hearts can be selected in 4 different orders ( 4 C 3 ): 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 = 4 C 3 × × 0.75 = We define the Binomial variable, X, by the formula: P(X = x ) = π x (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.

BUT... the 3 hearts can be selected in 4 different orders ( 4 C 3 ): 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 = 4 C 3 × × 0.75 = We define the Binomial variable, X, by the formula: P(X = x ) = π x (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.

So P(X=3) = 4 × 0.25 × 0.25 × 0.25 × 0.75 = 4 C 3 × × 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. We define the Binomial variable, X, by the formula: P(X = x ) = π x (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.

the 3 hearts can be selected in 4 different orders ( 4 C 3 ): 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 = 4 C 3 × × 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 – 2 nd ed): p67 – Ex SKIP Q10 We define the Binomial variable, X, by the formula: P(X = x ) = π x (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 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 ) = π x (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. HW: Do Sigma (old – 2 nd ed): p67 – Ex 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: Q4  11. *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 having WOF 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 having WOF is 1/6. (2.)If we use the Bin. Distn to model the number of cars not having a WOF, which of 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 having WOF is 1/6. (2.)If we use the Bin. Distn to model the number of cars not having a WOF, which of 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 Calculate the probability of failure (i.e. a given car fails its WOF), 1-π. 1-π 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 1–π = 3. What are we trying to find and 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 1–π = 3. Find P(X = 3) using P(X = x) = n C x π x (1- π ) n–x Calculate P(X=3) π = = (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: Calculate P(X=3) = (4sf) 1–π = 3. Find P(X = 3) using P(X = x) = n C x π x (1- π ) n–x π = n =5

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: Calculate P(X=3) = (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,  =0.3 To find P(X < 2): P(X= 0) = + P(X= 1) = + P(X= 2) =

Parameters: n=10,  =0.3 To find P(X < 2): P(X= 0) = + P(X= 1) = + P(X= 2) =

Parameters: n=10,  =0.3 To find P(X < 2): P(X= 0) = P(X= 1) = + P(X= 2) =

Parameters: n=10,  =0.3 To find P(X < 2): P(X= 0) = P(X= 1) = + P(X= 2) =

Parameters: n=10,  =0.3 To find P(X < 2): P(X= 0) = P(X= 1) = P(X= 2) =

Parameters: n=10,  =0.3 To find P(X < 2): P(X= 0) = P(X= 1) = P(X= 2) =

Parameters: n=10,  =0.3 To find P(X < 2): P(X= 0) = P(X= 1) = P(X= 2) =

Parameters: n=10,  =0.3 To find P(X < 2): P(X= 0) = P(X= 1) = P(X= 2) = P(X < 2) =

Parameters: n=10,  =0.3 To find P(X < 2): P(X= 0) = P(X= 1) = P(X= 2) = P(X < 2) =

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:

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:

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: 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: 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 = = answer Do NuLake pg. 285 and 286: Q4  11. Extension people: Q4  19. Q12  19 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. 285  288. 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 = = 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) = = 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) = = 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 Finish NuLake pg. 286  288 (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 (2 nd 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.

(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.

(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.

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?

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

(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) = 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) = 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) = n=5, P(X<3) = Try something a bit higher… n=7, P(X<3) = Getting close now… n=8, P(X<3) = 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) = 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) = n=5, P(X<3) = Try something a bit higher… n=7, P(X<3) = Getting close now… n=8, P(X 3) = 1 –

(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) = 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) = n=5, P(X<3) = Try something a bit higher… n=7, P(X<3) = Getting close now… n=8, P(X 3) = Answer: 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 (2 nd ed): Pg. 73 – Ex. 5.4 Or new Sigma: Pg. 326 – Ex 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 Ex 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 Dist n Parameters:   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 Dist n Parameters:   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 Dist n Parameters:   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 Dist n Parameters:   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 Dist n Parameters:   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 =

CHECK: The conditions that ensure this are that both n  and n (1-  ) are > 5 ** NOTE: The Normal approximation requires a Continuity Correction. Normal Dist n Parameters:   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 = = So the expected number of tails in 1000 throws is 500 with a standard deviation of

Normal Dist n Parameters:   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 = = So the expected number of tails in 1000 throws is 500 with a standard deviation of P(X < 490) ≈P(?) (using continuity correction)

Normal Dist n Parameters:   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 = = So the expected number of tails in 1000 throws is 500 with a standard deviation of P(X < 490) ≈ P(X < 489.5) (using continuity correction)

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 = = So the expected number of tails in 1000 throws is 500 with a standard deviation of P(X < 490)≈ P(X < 489.5) (using continuity correction) ≈ P(Z < – 500) ≈ (4sf) answer Do old Sigma (2 nd ed): Pg. 118 – Ex Complete for HW. So 490 tails would be just above the lower quartile.