Statistics and Modelling Course 2011. Topic 5: Probability Distributions Achievement Standard 90646 Solve Probability Distribution Models to solve straightforward.

Statistics and Modelling Course 2011

Topic 5: Probability Distributions Achievement Standard 90646 Solve Probability Distribution Models to solve straightforward problems 4 Credits Externally Assessed NuLake Pages 278  322

Binomial Distribution revision question: NCEA 2009: Tom and Tane are surfers. Tom successfully rides a wave on average 7 out of every 10 attempts, and Tane 6 out of every 10. Calculate the probability that Tom & Tane each ride 2 of the next 5 waves. What assumptions must be made to use the Binomial Distn here?

Lesson 1: Intro to the Poisson Distribution. Learning outcomes:  Learn about what the Poisson Distribution is and its parameter,.  Learn the 4 requirements for the use of the Poisson Distribution.  Calculate probabilities using the Poisson Distribution. 1.Warm-up: 2009 NCEA q. on the Binomial Distn. 2.The Poisson Distribution: Notes & examples using the formula & tables (most of lesson). HW: Sigma - Old edition: Ex. 6.1 or New: Ex. 16.01

The Poisson Distribution Deals with rate of occurrence in an interval. We use the Poisson Distribution when counting the number of occurrences of a discrete event over a given continuous interval (time/space).  Number of earthquakes per year (excluding after-shocks).  Population density - Number of people per square km.  House burglaries reported per week in Christchurch.

The Poisson Distribution Deals with rate of occurrence in an interval. We use the Poisson Distribution when counting the number of occurrences of a discrete event over a given continuous interval (time/space).  Number of earthquakes per year.  Population density - Number of people per square km.  House burglaries reported per week in Christchurch. The Poisson Distribution is also known as the Distribution of Rare Events.

We use the Poisson Distribution when counting the number of occurrences of a discrete event over a given continuous interval (time/space).  Number of earthquakes per year.  Population density - Number of people per square km.  House burglaries reported per week in Christchurch. The Poisson Distribution is also known as the Distribution of Rare Events. You’ll be using the Poisson Distribution to calculate the probability of an event occurring x times in a given interval. You will always be given the average number of occurrences,, per interval. is the Greek symbol “Lambda”

 Population density - Number of people per square km.  House burglaries reported per week in Christchurch. The Poisson Distribution is also known as the Distribution of Rare Events. is the Greek symbol “Lambda” = Mean rate of occurrence. You’ll be using the Poisson Distribution to calculate the probability of an event occurring x times in a given interval. You will always be given the average number of occurrences,, per interval.

The Poisson Distribution is also known as the Distribution of Rare Events. The 4 requirements for a Poisson Distribution are (RIPS): (1.)Random: Events occur at random (unpredictable). (2.)Independent: Each event occurs completely independently of any others. is the Greek symbol “Lambda” = Mean rate of occurrence. You’ll be using Poisson Distribution to calculate the probability of an event occurring x times in a given interval. You will always be given the average number of occurrences,, per interval.

The 4 requirements for a Poisson Distribution are (RIPS): (1.)Random: Events occur at random (unpredictable). (2.)Independent: Each event occurs completely independently of any others. (3.)Proportional: The probability of an event occurring is proportional to the size of the interval. (4.)Simultaneous: Events CANNOT occur simultaneously (at the same time or in exactly the same spot). is the Greek symbol “Lambda” = Mean rate of occurrence. You’ll be using the Poisson Distribution to calculate the probability of an event occurring x times in a given interval. You will always be given the average number of occurrences,, per interval.

The 4 requirements for a Poisson Distribution are (RIPS): (1.)Random: Events occur at random (unpredictable). (2.)Independent: Each event occurs completely independently of any others. (3.)Proportional: The probability of an event occurring is proportional to the size of the interval. (4.)Simultaneous: Events CANNOT occur simultaneously (at the same time or in exactly the same spot). is the Greek symbol “Lambda” = Mean rate of occurrence. We define the Poisson variable, X, by the formula: P(X = x ) = e   x x!

(1.)Random: Events occur at random (unpredictable). (2.)Independent: Each event occurs completely independently of any others. (3.)Proportional: The probability of an event occurring is proportional to the size of the interval. (4.)Simultaneous: Events cannot occur simultaneously (at the same time or in exactly the same spot). We define the Poisson variable, X, by the formula: P(X = x ) = e   x x! Eg: The average number of accidents at a particular intersection every year is 18. (a) Calculate the probability that there are exactly 2 accidents there this month.

Eg: The average number of accidents at a particular intersection every year is 18. (a) Calculate the probability that there are exactly 2 accidents there this month. P(X = 2) = ? (b) Calculate the probability that there is at least one accident this month. = 1.5 accidents per month = 0.2510 (4sf) answer There are 12 months in a year, so  =

P(X = 3) = (b) Calculate the probability that there is at least one accident this month. = 1.5 accidents per month = 0.2510 (4sf) answer There are 12 months in a year, so  = P(X ≥ 1 ) = P(X=1) + P(X=2) + P(X=3) + …. Infinite. So…

P(X = 3) = (b) Calculate the probability that there is at least one accident this month. = 1.5 accidents per month = 0.2510 (4sf) answer There are 12 months in a year, so  = P(X ≥ 1 ) = P(X=1) + P(X=2) + P(X=3) + …. Infinite. So… Take the complement: P(X=0)

P(X = 3) = (b) Calculate the probability that there is at least one accident this month. = 0.2510 (4sf) answer P(X ≥ 1 ) = P(X=1) + P(X=2) + P(X=3) + …. Infinite. So… Take the complement: P(X=0)

(b) Calculate the probability that there is at least one accident this month. = 0.2510 (4sf) answer P(X ≥ 1 ) = P(X=1) + P(X=2) + P(X=3) + …. Infinite. So… Take the complement: P(X=0) = 0.223130…

(b) Calculate the probability that there is at least one accident this month. P(X ≥ 1 ) = P(X=1) + P(X=2) + P(X=3) + …. Infinite. So… Take the complement: P(X=0) = 0.223130… So P(X>1) = 1 – P(X=0) = 1 – 0.223130… = 0.7769 (4sf) answer

P(X ≥ 1 ) = P(X=1) + P(X=2) + P(X=3) + …. Infinite. So… Take the complement: P(X=0) = 0.223130… So P(X>1) = 1 – P(X=0) = 1 – 0.223130… = 0.7769 (4sf) answer (c) What is the probability that there are more than 2 accidents in a particular month

= 0.223130… So P(X>1) = 1 – P(X=0) = 1 – 0.223130… = 0.7769 (4sf) answer (c) What is the probability that there are more than 2 accidents in a particular month P(X > 2 ) = There are an infinite number of cases so instead consider X < 2 1 – P(X < 2) = 1.5

So P(X>1) = 1 – P(X=0) = 1 – 0.223130… = 0.7769 (4sf) answer (c) What is the probability that there are more than 2 accidents in a particular month P(X > 2 ) = There are an infinite number of cases so instead consider X < 2 1 – P(X < 2) = 1.5 = 1 – [ P(X =0) + P(X=1) + P(X=2)] Look up Poisson Tables = 1 – [ _____ + ______ + ______]

= 1 – [ P(X =0) + P(X=1) + P(X=2)] b P(X > 2 ) = There are an infinite number of cases so instead consider X < 2 1 – P(X < 2) = 1.5 Look up Poisson Tables = 1 – [ _____ + ______ + ______] c What is the probability that there are more than 2 accidents in a particular month = 1 – 0.223130… = 0.7769 (4sf) answer

If  = 1.5, P(X = 0) = ?

If  = 1.5, P(X = 0) = 0.2231

= 1 – [ P(X =0) + P(X=1) + P(X=2)] b P(X > 2 ) = There are an infinite number of cases so instead consider X < 2 1 – P(X < 2) = 1.5 Look up Poisson Tables = 1 – [ _____ + ______ + ______] c What is the probability that there are more than 2 accidents in a particular month = 1 – 0.223130… = 0.7769 (4sf) answer

b P(X > 2 ) = There are an infinite number of cases so instead consider X < 2 1 – P(X < 2) = 1.5 Look up Poisson Tables = 1 – [ 0.2231 + ______ + ______] c What is the probability that there are more than 2 accidents in a particular month = 1 – [ P(X =0) + P(X=1) + P(X=2)] = 1 – 0.223130… = 0.7769 (4sf) answer

If  = 1.5, P(X = 0) = 0.2231

If  = 1.5, P(X = 0) = 0.2231 + P(X = 1) = ?

If  = 1.5, P(X = 0) = 0.2231 + P(X = 1) = 0.3347

= 1 – [ P(X =0) + P(X=1) + P(X=2)] b P(X > 2 ) = There are an infinite number of cases so instead consider X < 2 1 – P(X < 2) = 1.5 Look up Poisson Tables = 1 – [ 0.2231 + ______ + ______] c What is the probability that there are more than 2 accidents in a particular month = 1 – 0.223130… = 0.7769 (4sf) answer

b P(X > 2 ) = There are an infinite number of cases so instead consider X < 2 1 – P(X < 2) = 1.5 Look up Poisson Tables = 1 – [ 0.2231 + 0.3347 + ______] c What is the probability that there are more than 2 accidents in a particular month = 1 – [ P(X =0) + P(X=1) + P(X=2)] = 1 – 0.223130… = 0.7769 (4sf) answer

If  = 1.5, P(X = 0) = 0.2231 + P(X = 1) = 0.3347

If  = 1.5, P(X = 0) = 0.2231 + P(X = 1) = 0.3347 + P(X = 2) = ?

If  = 1.5, P(X = 0) = 0.2231 + P(X = 1) = 0.3347 + P(X = 2) = 0.2510

b P(X > 2 ) = There are an infinite number of cases so instead consider X < 2 1 – P(X < 2) = 1.5 Look up Poisson Tables = 1 – [ 0.2231 + 0.3347 + ______] c What is the probability that there are more than 2 accidents in a particular month = 1 – [ P(X =0) + P(X=1) + P(X=2)] = 1 – 0.223130… = 0.7769 (4sf) answer

b P(X > 2 ) = There are an infinite number of cases so instead consider X < 2 1 – P(X < 2) = 1.5 Look up Poisson Tables = 1 – [ 0.2231 + 0.3347 + 0.2510] c What is the probability that there are more than 2 accidents in a particular month = 1 – [ P(X =0) + P(X=1) + P(X=2)] = 1 – 0.223130… = 0.7769 (4sf) answer

If  = 1.5, P(X = 0) = 0.2231 + P(X = 1) = 0.3347 + P(X = 2) = 0.2510 P(X < 2) =

If  = 1.5, P(X = 0) = 0.2231 + P(X = 1) = 0.3347 + P(X = 2) = 0.2510 P(X < 2) = 0.8088

b P(X > 2 ) = There are an infinite number of cases so instead consider X < 2 1 – P(X < 2) = 0.1912 HW: Sigma: Old edition: Ex. 6.1 or New edition: Ex. 16.01 = 1.5 Look up Poisson Tables = 1 – [ 0.2231 + 0.3347 + 0.2510] c What is the probability that there are more than 2 accidents in a particular month = 1 – [ P(X =0) + P(X=1) + P(X=2)] = 1 – 0.8088 = 1 – 0.223130… = 0.7769 (4sf) answer

Lesson 2: Poisson probabilities using a Graphics calculator Learning outcome: Practice calculating probabilities of combined events using the Poisson Distribution, using both tables and GC. Work: 1.HW qs. 2.Problem on board – how to use your GC. 3.Do Sigma (old – 2 nd ed): p84 – Ex. 6.2 or in new version: p340 – Ex. 16.02

Poisson probabilities: Tables vs Graphics Calculator E.g. Calculate P (X>3) for =4 USING TABLES. Do not copy this. P(X>3) = 1 – P(X<3) = 1 – [P(X=0) + P(X=1) + P(X=2) + P(X=3)]

Poisson probabilities: Tables vs Graphics Calculator E.g. Calculate P (X>3) for =4 USING TABLES. Do not copy this. P(X>3) = 1 – P(X<3) = 1 – [P(X=0) + P(X=1) + P(X=2) + P(X=3)] = 1 – [ 0.018315 + 0.073262 + 0.14652 + 0.19536]

Poisson probabilities: Tables vs Graphics Calculator E.g. Calculate P (X>3) for =4 USING TABLES. Do not copy this. P(X>3) = 1 – P(X<3) = 1 – [P(X=0) + P(X=1) + P(X=2) + P(X=3)] = 1 – [ 0.018315 + 0.073262 + 0.14652 + 0.19536 ] = 1 – 0.433457 = 0.5665(4sf)

Poisson probabilities: Tables vs Graphics Calculator E.g. Calculate P (X>3) for =4. Using tables (do not copy). P(X>3) = 1 – P(X<3) = 1 – [P(X=0) + P(X=1) + P(X=2) + P(X=3)] = 1 – [ 0.018315 + 0.073262 + 0.14652 + 0.19536 ] = 1 – 0.433457 = 0.5665(4sf) BUT… there is a much faster way… On your Graphics Calculator: (copy)

Poisson probabilities: Tables vs Graphics Calculator E.g. Calculate P (X>3) for =4. Using tables (do not copy). P(X>3) = 1 – P(X<3) = 1 – [P(X=0) + P(X=1) + P(X=2) + P(X=3)] = 1 – [ 0.018315 + 0.073262 + 0.14652 + 0.19536 ] = 1 – 0.433457 = 0.5665(4sf) BUT… there is a much faster way… On your Graphics Calculator: (copy) STAT, DIST (same as before)

Poisson probabilities: Tables vs Graphics Calculator E.g. Calculate P (X>3) for =4. Using tables (do not copy). P(X>3) = 1 – P(X<3) = 1 – [P(X=0) + P(X=1) + P(X=2) + P(X=3)] = 1 – [ 0.018315 + 0.073262 + 0.14652 + 0.19536 ] = 1 – 0.433457 = 0.5665(4sf) BUT… there is a much faster way… On your Graphics Calculator: (copy) STAT, DIST (same as before) POISN (F6, F1) (same as before)

Poisson probabilities: Tables vs Graphics Calculator E.g. Calculate P (X>3) for =4. Using tables (do not copy). P(X>3) = 1 – P(X<3) = 1 – [P(X=0) + P(X=1) + P(X=2) + P(X=3)] = 1 – [ 0.018315 + 0.073262 + 0.14652 + 0.19536 ] = 1 – 0.433457 = 0.5665(4sf) BUT… there is a much faster way… On your Graphics Calculator: (copy) STAT, DIST (same as before) POISN (F6, F1) (same as before) Pcd (c for Cumulative probabilities – more than 1 value)

Poisson probabilities: Tables vs Graphics Calculator E.g. Calculate P (X>3) for =4. Using tables (do not copy). P(X>3) = 1 – P(X<3) = 1 – [P(X=0) + P(X=1) + P(X=2) + P(X=3)] = 1 – [ 0.018315 + 0.073262 + 0.14652 + 0.19536 ] = 1 – 0.433457 = 0.5665(4sf) BUT… there is a much faster way… On your Graphics Calculator: (copy) STAT, DIST (same as before) POISN (F6, F1) (same as before) Pcd (c for Cumulative probabilities – more than 1 value) Data: Variable (as before)

Poisson probabilities: Tables vs Graphics Calculator E.g. Calculate P (X>3) for =4. Using tables (do not copy). P(X>3) = 1 – P(X<3) = 1 – [P(X=0) + P(X=1) + P(X=2) + P(X=3)] = 1 – [ 0.018315 + 0.073262 + 0.14652 + 0.19536 ] = 1 – 0.433457 = 0.5665(4sf) BUT… there is a much faster way… On your Graphics Calculator: (copy) STAT, DIST (same as before) POISN (F6, F1) (same as before) Pcd (c for Cumulative probabilities – more than 1 value) Data: Variable (as before) “ x ” means X < Enter what X is LESS THAN OR EQUAL TO.

Poisson probabilities: Tables vs Graphics Calculator E.g. Calculate P (X>3) for =4. Using tables (do not copy). P(X>3) = 1 – P(X<3) = 1 – [P(X=0) + P(X=1) + P(X=2) + P(X=3)] = 1 – [ 0.018315 + 0.073262 + 0.14652 + 0.19536 ] = 1 – 0.433457 = 0.5665(4sf) BUT… there is a much faster way… On your Graphics Calculator: (copy) STAT, DIST (same as before) POISN (F6, F1) (same as before) Pcd (c for Cumulative probabilities – more than 1 value) Data: Variable (as before) “ x ” means X < Enter what X is LESS THAN OR EQUAL TO. Note: Ppd gives individual Poisson probabilities, eg P(X=2), Pcd gives cumulative Poisson probabilities, eg P(X ≤ 2)

Poisson probabilities: Tables vs Graphics Calculator Setting out of working when using your Graphics Calculator: E.g. Calculate P ( X > 3) for =4 P( X > 3)= 1 – P( X < 3) = 1 – 0.433457 (G.C.) = 0.5665(4sf) On your Graphics Calculator: (copy) STAT, DIST (same as before) POISN (F6, F1) (same as before) Pcd (c for Cumulative probabilities – more than 1 value) Data: Variable (as before) “ x ” means X < Enter what X is LESS THAN OR EQUAL TO. Note: Ppd gives individual Poisson probabilities, eg P(X=2), Pcd gives cumulative Poisson probabilities, eg P(X ≤ 2)

16.02B The mean number of volcanic eruptions on White Island is 1.2 per year. Calculate the probability that there are more than 2 in one year. P(X > 2) = 1– P(X ≤ 2) As this is infinite, rewrite > in terms of ≤ = 1– 0.879 48 (G.C.) = 0.1205 (4sf) ==  Do old Sigma p84: Ex. 6.2 In new version: p340 – Ex. 16.02

Lesson 3: Inverse Poisson problems. Learning outcomes:  Calculate the variance & standard deviation of a Poisson probability distribution.  Solve inverse Poisson problems. STARTER: Look at a graph of the Poisson Distribution (see Sigma workbook CD) 1.Variance & SD of Poisson – notes & e.gs Sigma 2 nd ed: p89 – Ex. 6.3; New Sigma: p344 – Ex. 16.03 2. Inverse Poisson problems – notes & e.gs Sigma 2 nd ed: p90 – Ex. 6.4; New Sigma: p347 – Ex. 16.04

E.g.: A Poisson distribution has mean = 4.6. What is the standard deviation? Standard deviation = Do old Sigma p89: Ex. 6.3 – Q3 & 4 In new version: p345 – Ex. 16.03 – Q4 & 5 Variance & Standard Deviation of a Poisson Distn = mean and also  =variance. So standard deviation,  =

How to solve Inverse Poisson Questions E.g.1: For a Poisson distribution, P(X = 0) = 0.0123. Calculate. Write out the Poisson formula for the information given.

E.g.1: For a Poisson distribution, P(X = 0) = 0.0123. Calculate. Write out the Poisson formula for the information given. To calculate, take logs of both sides. Simplify, using 0 and 0! both equal 1. Eg2: In samples of material taken from a textile factory, 40% had at least one fault. Estimate the mean number of faults per sample. Simplify, using log e e = 1 (cancels out). - = ln(0.0123) = 4.398 (4sf)

E.g.1: For a Poisson distribution, P(X = 0) = 0.0123. Calculate. Write out the Poisson formula for the information given. To calculate, take logs of both sides. Simplify, using 0 and 0! both equal 1. Eg2: In samples of material taken from a textile factory, 40% had at least one fault. Estimate the mean number of faults per sample. First choose the random variable, X. Simplify, using log e e = 1 (cancels out). - = ln(0.0123) = 4.398 (4sf)

Eg2: In samples of material taken from a textile factory, 40% had at least one fault. Estimate the mean number of faults per sample. Let X be ‘the number of faults per sample’ X has a Poisson distribution. Given P(X ≥ 1) = 0.4 P(X = 0) = 0.6 Simplify, using 0 and 0! both equal 1 First choose the random variable, X. Write out the Poisson formula for the information given. To calculate, take logs of both sides. As ≥ is infinite (no upper limit), we consider the complement, knowing P(X>1) = 1- P(X=0) - = ln(0.0123) = 4.398 (4sf) Simplify, using log e e = 1 (cancels out).

Eg2: In samples of material taken from a textile factory, 40% had at least one fault. Estimate the mean number of faults per sample. Let X be ‘the number of faults per sample’ X has a Poisson distribution. Given P(X ≥ 1) = 0.4 P(X = 0) = 0.6 To calculate, take logs of both sides. Simplify, using log e e = 1 (cancels out). As ≥ is infinite (no upper limit), we consider the complement, knowing P(X>1) = 1- P(X=0) = 0.5108 (4sf) - = ln(0.6) Do old Sigma p90: Ex. 6.4 In new version: p347 – Ex. 16.04 Complete for HW

Lesson 4: Problems that involve the use of more than 1 type of distribution Learning outcome: Solve problems involving a mixture of all 3 types of distribution. Key skill: identifying the appropriate distribution to use. 1. STARTER: Go over HW qs from Inverse Poisson chapter in Sigma. 2. Revision worksheet for all 3 Probability Distribution. 3. Do Sigma (NEW): p384 – Ex. 18.03. Finish Q1  3 for HW.

Problems that require the use of more than one type of probability distribution

Problems that require the use of more than one type of probability distribution Sometimes probabilities calculated from one distribution can be used as parameters in another distribution.

Problems that require the use of more than one type of probability distribution Sometimes probabilities calculated from one distribution can be used as parameters in another distribution. A typical e.g. is when a Normal Distribution probability is used as the probability of an individual trial  in a Binomial Distribution. Do Sigma (NEW): pg. 384 – Ex. 18.03

Problems that require the use of more than one type of probability distribution Sometimes probabilities calculated from one distribution can be used as parameters in another distribution. A typical e.g. is when a Normal Distribution probability is used as the probability of an individual trial  in a Binomial Distribution. Do Sigma (NEW): pg. 384 – Ex. 18.03 1. At an athletics meeting, the highest successful attempt at the high- jump by different athletes is known to be normally distributed with a mean of 1.4m and a standard deviation of 10cm. (a)What is the probability that an athlete chosen at random can jump 1.5m?. (b)Calculate the prob. that 2 consecutive athletes chosen at random both fail to jump 1.3m. (c)If 3 athletes are chosen at random, calculate the prob. that all three of them have a highest successful jump between 1.2 & 1.6m

Sometimes probabilities calculated from one distribution can be used as parameters in another distribution. A typical e.g. is when a Normal Distribution probability is used as the probability of an individual trial  in a Binomial Distribution. Do Sigma (NEW): pg. 384 – Ex. 18.03 1. At an athletics meeting, the highest successful attempt at the high- jump by different athletes is known to be normally distributed with a mean of 1.4m and a standard deviation of 10cm. (a)What is the probability that an athlete chosen at random can jump 1.5m?. (b)Calculate the prob. that 2 consecutive athletes chosen at random both fail to jump 1.3m. (c)If 3 athletes are chosen at random, calculate the prob. that all three of them have a highest successful jump between 1.2 & 1.6m (d)What is the probability that only one out of three athletes chosen at random can jump over 1.6m?

EXTENSION SESSION: Approximations – approximating between all 3 distributions. Learning outcome: Solve problems involving a mixture of all 3 types of distribution. Key skill: identifying the appropriate distribution to use. 1.Handout on approximations (flow-chart). View applet (graphical demonstration using bar graphs) 2.Problems to get the hang of approximating – all in Sigma old edition: Approximating between: - the Binomial with the Poisson: p121 – Ex. 8.2 Q1 & 2 only. - the Poisson with the Normal: p123– Ex. 8.3 Q1 & 2 only. - mixed approximation problems: p125 – Ex. 8.4. All. HW: Challenge problem set.

APPROXIMATIONS Only in scholarship exam now – see first version of NuLake (2004) for concise notes. Do chapter in Sigma 2 nd edition.

Approximating from the Binomial Distribution to the Poisson Can only use when:  n is very large.   is extreme (probabilities are very close to 0 or 1) Question: Why must n be large and  be close to 0 or 1? Hint: What do the Binomial and Poisson Dists have in common? Answer: Both model how often an event occurs (counting – discrete). Another hint:  n  Do “Comparing Models” handout, JUST THE FIRST PAGE. (from Achieving in Statistics workbook).

Approximating from the Binomial Distribution to the Poisson Can only use when:  n is very large.   is extreme (probabilities are very close to 0 or 1) Question: Why must n be large and  be close to 0 or 1? Hint: What do the Binomial and Poisson Dists have in common? Answer: Both model how often an event occurs (counting – discrete). Another hint: What is different about the Binomial & Poisson distributions? Answer: Binomial used when event occurs within a fixed number of trials (discrete). Poisson used when event occurs within a continuous interval.  n 

Approximations challenge problems Problem 1: In a certain suburb, 30% of new homes are built with an outdoor bbq area, and 2% of new homes require alterations after construction. (a)A random sample of 50 homes is to be taken. Find the probability that no more than 20 homes in this sample will be built with an outdoor bbq area. (b)Another sample of n new homes is to be taken. Find the value of n so that the probability that the sample contains no homes requiring alterations is approximately 0.6.

Approximations challenge problems Problem 2:

TEST Formative assessment of AS90646 Lesson 5:

Probability Distribution Scholarship slides

The Normal Distribution Scholarship slides

P( Z  ) = 0.01 STARTER: A product has a normally distributed weight with μ = 1kg. If the risk of marketing a product more than 50g underweight must be less than 1%, what is the maximum allowable standard deviation? Solution: Let X be a random variable representing the weight of the product in grams. We want  such that P( X  950 ) = 0.01 P( Z  ) = 0.01 For the maximum value of  take P( Z  ) = 0.01 And, we can use tables or graphics calc to find, based on prob of 0.01, z=-2.33… Re-arranging to solve for  …

P( Z  ) = 0.01 STARTER: A product has a normally distributed weight with μ = 1kg. If the risk of marketing a product more than 50g underweight must be less than 1%, what is the maximum allowable standard deviation? Solution: Let X be a random variable representing the weight of the product in grams. We want  such that P( X  950 ) = 0.01 P( Z  ) = 0.01 For the maximum value of  take P( Z  ) = 0.01   =  = 21.5 so 21.5 is the maximum allowable standard deviation And, we can use tables or graphics calc to find, based on prob of 0.01, z=-2.33… Re-arranging to solve for  …

17.05A Eg1: The weights of suitcases received by an airline at check-in can be modelled by a normal distribution with mean 17 kg and standard deviation 4 kg. The airline charges for excess baggage if a suitcase is ‘overweight’—defined as weighing more than 20 kg. a Find the probability that two consecutive passengers both have to pay for excess baggage. b What proportion of overweight suitcases weigh less than 22 kg?

Use your graphics calculator to work out the probability that a single suitcase is overweight to 4 sf. Eg1: The weights of suitcases received by an airline at check-in can be modelled by a normal distribution with mean 17 kg and standard deviation 4 kg. The airline charges for excess baggage if a suitcase is ‘overweight’—defined as weighing more than 20 kg. a Find the probability that two consecutive passengers both have to pay for excess baggage.

P(excess) = 0.2266 = P(X > 20) Eg1: The weights of suitcases received by an airline at check-in can be modelled by a normal distribution with mean 17 kg and standard deviation 4 kg. The airline charges for excess baggage if a suitcase is ‘overweight’—defined as weighing more than 20 kg. a Find the probability that two consecutive passengers both have to pay for excess baggage.

Draw a tree diagram to show the possibilities for two suitcases. P(X > 20) = 0.2266 Eg1: The weights of suitcases received by an airline at check-in can be modelled by a normal distribution with mean 17 kg and standard deviation 4 kg. The airline charges for excess baggage if a suitcase is ‘overweight’—defined as weighing more than 20 kg. a Find the probability that two consecutive passengers both have to pay for excess baggage.

Eg1: The weights of suitcases received by an airline at check-in can be modelled by a normal distribution with mean 17 kg and standard deviation 4 kg. The airline charges for excess baggage if a suitcase is ‘overweight’—defined as weighing more than 20 kg. a Find the probability that two consecutive passengers both have to pay for excess baggage. P(two cases overweight) = 0.2266  0.2266 = 0.0513 P(X > 20) = 0.2266 Over weight Under weight Over weight Under weight Over weight Under weight 0.2266 0.7734 0.2266 0.7734 *

Eg1: The weights of suitcases received by an airline at check-in can be modelled by a normal distribution with mean 17 kg and standard deviation 4 kg. The airline charges for excess baggage if a suitcase is ‘overweight’—defined as weighing more than 20 kg. b What proportion of overweight suitcases weigh less than 22 kg? This is a conditional probability: P(X 20) = = 0.5338 (4 sf) that X 20 Do NuLake qs on “Combined Events for Poisson, Normal and Binomial” Q31  34 (p303  305 in 2007 edition)

a Draw a probability tree. Eg2: Over a long period of time, the number of accidents per day on a certain road has been found to have a mean of 2 on weekdays and 3 at the weekend. a What is the probability that there are 3 accidents on a day chosen at random? b If there are no accidents on a particular day, calculate the probability that it is a weekday.

a weekday weekend 5/7 2/7 3 accidents other than 3 accidents 3 accidents other than 3 accidents ? ? Calculate using Poisson = 2, x = 3 Eg2: Over a long period of time, the number of accidents per day on a certain road has been found to have a mean of 2 on weekdays and 3 at the weekend. a What is the probability that there are 3 accidents on a day chosen at random? b If there are no accidents on a particular day, calculate the probability that it is a weekday.

a weekday weekend 5/7 2/7 3 accidents other than 3 accidents 3 accidents other than 3 accidents 0.1804 ? Calculate using Poisson = 3, x = 3 Eg2: Over a long period of time, the number of accidents per day on a certain road has been found to have a mean of 2 on weekdays and 3 at the weekend. a What is the probability that there are 3 accidents on a day chosen at random? b If there are no accidents on a particular day, calculate the probability that it is a weekday.

a weekday weekend 5/7 2/7 3 accidents other than 3 accidents 3 accidents other than 3 accidents 0.1804 0.2240 P(3 accidents) = 5/7  0.1804+ 2/7  0.2240 = 0.19 (2 dp) Eg2: Over a long period of time, the number of accidents per day on a certain road has been found to have a mean of 2 on weekdays and 3 at the weekend. a What is the probability that there are 3 accidents on a day chosen at random? b If there are no accidents on a particular day, calculate the probability that it is a weekday.

b weekday weekend 5/7 2/7 0 accidents other than 0 accidents 0 accidents other than 0 accidents ? ? Calculate using Poisson = 2, x = 0 Eg2: Over a long period of time, the number of accidents per day on a certain road has been found to have a mean of 2 on weekdays and 3 at the weekend. a What is the probability that there are 3 accidents on a day chosen at random? b If there are no accidents on a particular day, calculate the probability that it is a weekday.

b weekday weekend 5/7 2/7 0 accidents other than 0 accidents 0 accidents other than 0 accidents 0.1353 ? Calculate using Poisson = 3, x = 0 Eg2: Over a long period of time, the number of accidents per day on a certain road has been found to have a mean of 2 on weekdays and 3 at the weekend. a What is the probability that there are 3 accidents on a day chosen at random? b If there are no accidents on a particular day, calculate the probability that it is a weekday.

b weekday weekend 5/7 2/7 0 accidents other than 0 accidents 0 accidents other than 0 accidents 0.1353 0.0498 P(weekday/0 accidents) = P(weekday AND 0 accidents) P( 0 accidents) Eg2: Over a long period of time, the number of accidents per day on a certain road has been found to have a mean of 2 on weekdays and 3 at the weekend. a What is the probability that there are 3 accidents on a day chosen at random? b If there are no accidents on a particular day, calculate the probability that it is a weekday.

b weekday weekend 5/7 2/7 0 accidents other than 0 accidents 0 accidents other than 0 accidents 0.1353 0.0498 P(weekday/0 accidents) = 5/7  0.1353 P(0 accidents) Eg2: Over a long period of time, the number of accidents per day on a certain road has been found to have a mean of 2 on weekdays and 3 at the weekend. a What is the probability that there are 3 accidents on a day chosen at random? b If there are no accidents on a particular day, calculate the probability that it is a weekday.

b weekday weekend 5/7 2/7 0 accidents other than 0 accidents 0 accidents other than 0 accidents 0.1353 0.0498 P(weekday/0 accidents) = 5/7  0.1353 (5/7  0.1353 + 2/7  0.0498) Eg2: Over a long period of time, the number of accidents per day on a certain road has been found to have a mean of 2 on weekdays and 3 at the weekend. a What is the probability that there are 3 accidents on a day chosen at random? b If there are no accidents on a particular day, calculate the probability that it is a weekday.

b weekday weekend 5/7 2/7 0 accidents other than 0 accidents 0 accidents other than 0 accidents 0.1353 0.0498 P(weekday/0 accidents) = 0.0966 0.1109 Eg2: Over a long period of time, the number of accidents per day on a certain road has been found to have a mean of 2 on weekdays and 3 at the weekend. a What is the probability that there are 3 accidents on a day chosen at random? b If there are no accidents on a particular day, calculate the probability that it is a weekday.

b weekday weekend 5/7 2/7 0 accidents other than 0 accidents 0 accidents other than 0 accidents 0.1353 0.0498 P(weekday/0 accidents) =0.8717 (4sf) Eg2: Over a long period of time, the number of accidents per day on a certain road has been found to have a mean of 2 on weekdays and 3 at the weekend. a What is the probability that there are 3 accidents on a day chosen at random? b If there are no accidents on a particular day, calculate the probability that it is a weekday.

Eg3: A travel agent receives faxes from either within New Zealand or overseas. The probability that any given fax arrives from a New Zealand number is 0.75. One day the travel agent receives 10 faxes. If at least 6 are from New Zealand numbers, calculate the probability that only 1 of the 10 faxes is from overseas. n = 10 and  = 0.25 Suppose X = number of faxes from overseas State the random variable involved. Give the values of the two parameters. What are we trying to find? We want the probability that 1 fax is from overseas given that 4 or fewer are from overseas = 0.18771 0.92187 = 0.2036 (4sf) Using Graphics Calculator

E.g.4: A traffic research unit has established that the speeds of both cars and motorbikes on a particular point on a road can be modelled by normal distributions. The speeds of cars (C) have mean 98 km/h and standard deviation 7 km/h and the speeds of motorbikes (M) have mean 102 km/h and standard deviation 4 km/h. The Police ask the unit to determine a speed at which each type of vehicle is equally likely to be ticketed. 80859095100105110115120 C M The problem is to determine k such that P(C > k) = P(M > k). The graph shows the relationship between the two distributions.

80859095100105110115120 C M The problem is to determine k such that P(C > k) = P(M > k). The required value has the same area to the right of it for each curve. k i.e. the two shaded areas must be the same. E.g.4: A traffic research unit has established that the speeds of both cars and motorbikes on a particular point on a road can be modelled by normal distributions. The speeds of cars (C) have mean 98 km/h and standard deviation 7 km/h and the speeds of motorbikes (M) have mean 102 km/h and standard deviation 4 km/h. The Police ask the unit to determine a speed at which each type of vehicle is equally likely to be ticketed.

80859095100105110115120 C M The problem is to determine k such that P(C > k) = P(M > k). Note – this is not given by the point where the curves intersect! The required value has the same area to the right of it for each curve. k E.g.4: A traffic research unit has established that the speeds of both cars and motorbikes on a particular point on a road can be modelled by normal distributions. The speeds of cars (C) have mean 98 km/h and standard deviation 7 km/h and the speeds of motorbikes (M) have mean 102 km/h and standard deviation 4 km/h. The Police ask the unit to determine a speed at which each type of vehicle is equally likely to be ticketed.

80859095100105110115120 C M The problem is to determine k such that P(C > k) = P(M > k). For the two probabilities to be equal, each curve must convert to the same z-value as the other k E.g.4: A traffic research unit has established that the speeds of both cars and motorbikes on a particular point on a road can be modelled by normal distributions. The speeds of cars (C) have mean 98 km/h and standard deviation 7 km/h and the speeds of motorbikes (M) have mean 102 km/h and standard deviation 4 km/h. The Police ask the unit to determine a speed at which each type of vehicle is equally likely to be ticketed.

80859095100105110115120 C M k Each type of vehicle is equally likely to be ticketed at 107.3 km/h. Solve for k.. E.g.4: A traffic research unit has established that the speeds of both cars and motorbikes on a particular point on a road can be modelled by normal distributions. The speeds of cars (C) have mean 98 km/h and standard deviation 7 km/h and the speeds of motorbikes (M) have mean 102 km/h and standard deviation 4 km/h. The Police ask the unit to determine a speed at which each type of vehicle is equally likely to be ticketed.

The Binomial Distribution Scholarship slides

Eg5 – University Stage 2 Probability exam q: In a particular game where I can either win or lose (assuming: a) the independence among the outcomes of the games and; b) the same probability to win in each game); if the probability to win 3 games out of 3 is, what is the probability to win 2 out of 4? Solution: Probability of winning a single game,  is: So prob. of winning 2 out of 4 games is found by the binomial distribution formula where n=4, x=2, and  =1/3.

The geometric distribution: probability that the 1 st success occurs in the nth trial. The negative binomial distribution: probability that the kth success occurs on the nth trial.

Statistics and Modelling Course 2011. Topic 5: Probability Distributions Achievement Standard 90646 Solve Probability Distribution Models to solve straightforward.

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Statistics and Modelling Course 2011. Topic 5: Probability Distributions Achievement Standard 90646 Solve Probability Distribution Models to solve straightforward.

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Presentation on theme: "Statistics and Modelling Course 2011. Topic 5: Probability Distributions Achievement Standard 90646 Solve Probability Distribution Models to solve straightforward."— Presentation transcript:

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