March 21, Fractal Friday: Sign up & pay now!

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1 Welcome to the Fourth Quarter. We did Discrete Random Variables at the start of HL1 (1516) Q4 We did Continuous RVs at the end of HL2 (1617) Q3 and start of Q4 7½ weeks plus finals Three Units: Discrete Random Variables Trigonometry Complex Numbers It will go fast...very fast Keys to success I want to I can I will I do Fractal Friday: Sign up & pay now!

2 1. Understand the meaning of a discrete random variable 25A: #2,3,4 (Discrete RVs) 25B: #2,4,6-8,11,12 (Distributions) A random variable is a letter (variable) that represents all the possible outcomes of an event in number form. Examples: X represents the result of the roll of a die. G represents the position of a pig after rolling it (no, not a real pig...) M represents the number of machine malfunctions over a 24 hour period T represents the times at which runners finish a race. E represents the distance by which a dart player misses the bulls eye. A discrete random variable is a random variable that can take on a finite set of distinct values x 1, x 2, x 3,... Generally speaking, discrete random variables represent things that are counted. Examples X and M above. Notice that while M can take on very many different values, there is a finite set of possibilities. Therefore M is discrete. By contrast a continuous random variable can take on any value in an interval. Generally speaking, continuous random variables represent things that are measured. Examples T and E above. (Notice that the "continuity" of something depends on the degree of precision of the measuring device. But even though we may not be able to measure a distinction, we assume that the underlying phenomenon is continuous and thus so is the random variable assigned to it.)

3 1. Understand and use properties of discrete probability distributions A random variable has an associated probability distribution that is a description of the probability that the variable will take on any particular value. For example: X is the result of the roll of one die. The variable X can take on one of six values. The probability of each is the same so the probability distribution function would look like: x P(X = x) 1 Notice that the capital X stands for the variable. Lower case x or xi represents the values that X can take on. We can also graph probability distribution functions: A more interesting example: X is the number of times you get a sum of 5 if you roll a pair of dice 6 times. We will look at this case in more detail soon... The variable X can take on any value from 0 to 6 with differing probabilities. The graph of the probability distribution function would be: Can you graph the probability distribution for the random variable that represents the number of heads in two flips of a fair coin? Properties of discrete probability distributions The probability of any individual outcome must obey: 0 pi 1 The total probability must sum to one: Some practice: We've seen probability distributions presented as tables and as graphs. They can also be given as a function where P(x) means P(X = x) in which case, the function P is called the probability distribution function or the probability mass function the random variable. 25A: #2,3,4 (Discrete RVs) 25B: #2,4,6-8,11,12 (Distributions)

4 Fractal Friday: Sign up (and pay!) now! 25A: #2,3,4 (Discrete RVs) 25B: #2,4,6-8, (Distributions) Present 6-8, Understand and use the expected value of a discrete probability distribution 25C: #1-11 odd (Expectation) You roll a dice 120 times. How many times do you expect to roll a six? The probability of rolling a 6 on any individual roll is The rolls are independent You are rolling 120 times, therefore you would expect to get of 120 or 20 sixes. Expectation In n trials of an experiment with an outcome that has a probability of p of occurring, the expected number of those outcomes is np. Note that this is not the same as the question: What is the probability that you will roll 20 sixes? Another example: There are 20 different color poker chips in a box and you pay $5 to pull one out. The white chips are worth $1, the red worth $5, the blue worth $10 and the black worth $20. There are 10 white chips, 5 red chips, and 3 blue chips, and 2 black chips. Would you play this game? What would you expect the outcome of the game to be? Let X be the value of the chip you draw Outcome Value of Probability i of Outcome Likely payout of 1 White $1 2 Red $5 3 Blue $10 4 Black $20 Expected Value $5.25 Expected Value A random variable X that takes on n values xi each of which has probability pi has an expected value of: Super important: Notice that E(X) 2 E(X 2 ). Why not? Try one: The table gives the probabilities that a customer buys x magazines from a shop. What is the expected number of magazines that a customer would buy? How do we interpret the fact that the result is not an integer? A random variable represents a population of outcomes with various probabilities. Considered as a population, it will have measures of central tendency, measures of dispersion and other associated statistics. The Mean of a Discrete Random Variable: Recall the definition of the population mean in general for values xi, each of which occurs with frequency fi This can be written as: where N is the total number of samples. We can break apart the sum, and rewrite the population mean as = E(X)! Mean of a Discrete Random Variable The population mean of a discrete random variable is its expected value. That is: E(X) = µ 25C: #1-11 odd (Expectation)

5 25C: #1-11 odd (Expectation) Present D: #2-7 all (Variance & SD) 1. Understand and use the variance and standard deviation as measures of spread of a discrete probability distribution We've talked about the mean of a random variable. What about measures of spread? What are the variance and standard deviation of a random variable? The Variance of a Discrete Random Variable: Recall the definition of the variance in general for values xi, each of which occurs with frequency fi This can be written as: where N is the total number of samples. Again, we can break apart the sum, and rewrite as Variance of a Random Variable The population variance of a random variable is given by: The variance of X can be thought of as the expected value of the related random variable (X µ) 2 But be careful: We are talking about E[(X µ) 2 ], which is not the same as [E(X µ)] 2 Let's look more closely at how we might evaluate variance: The original formula Expand the square Distribute the probabilities Rewrite as a sum of sums and factor out constants The second sum is µ and the sum of the probabilities is 1 Sometimes it's easier to evaluate As you might imagine, we also define the standard deviation as the root of the variance: Standard Deviation of a Random Variable The population standard deviation of a random variable is given by: or its alternative form: The median of a random variable is the middle value. It occurs when the data are ordered and the cumulative probability is 0.5 This can be found by examining the expected frequency in 100 trials The mode of a random variable is the value that occurs most often. It is the value that has the highest probability of occurring Let's try these out: The table gives the probabilities for a random variable X representing the number of magazines a random customer buys from a shop. Find the mean, median, mode, variance, and standard deviation of the random variable X. 25D: #2-7 all (Variance & SD)

6 25D: #2-7 all (Variance & SD) Questions? 1. Understand and use properties of expected value and variance Recall our results from last time: 25E: #3-13 odd,14 (Properties of E(x) & Var(x)) QB: #6,10,12 (Mean & SD of RVs) Next Monday! Unit Exam on Discrete Random Variables Let's develop some algebraic rules for manipulating random variables Prove the following for a discrete random variable X: Properties of Expectation E(k) = k for any constant k E(kX) = ke(x) for any constant k E(f(X) + g(x)) = E(f(X)) + E(g(X)) for any functions f and g Prove the following for a discrete random variable X: A Property of Variance Var(X) = E(X 2 ) {E(X)} 2 or Var(X) = E(X 2 ) µ 2 Prove the following properties of a linear function of a random variable X. Properties of Linear Functions of a Random Variable Let X be a random variable and a and b be constants. E(aX + b) = ae(x) + b Var(aX + b) = a 2 Var(X) Next Monday! Unit Exam on Discrete Random Variables 25E: #3-13 odd,14 (Properties of E(x) & Var(x)) QB: #6,10,12 (Mean & SD of RVs)

7 6C1 March 21, E: #3-13 odd,14 (Properties of E(x) & Var(x)) Present 7,9,13 QB: #6,10,12 (Mean & SD of RVs) Present all 1. Understand and use features of the binomial distribution 25F.1: #2,4,5 (Binomial distribution) 25F.2: #2-14 even (Binomial dist function) 25F.3: #3-5 all (Mean & SD of binomial) QB: #2,8,15 (Binomial distributions) Monday! Unit Exam on Discrete Random Variables Joe throws darts a at a dartboard 6 times. On each throw, he has a 20% change of hitting a bullseye. What is the likelihood that he hits the bullseye: 6 times? 6C0(0.2) 6 (0.8) 0 5 times? 6 (0.2) 5 (0.8) 1 (the 6 is because there are 6 ways to do this) 4 times? 6C2(0.2) 4 (0.8) 2 3 times? 6C3(0.2) 3 (0.8) 3 2 times? 6C4(0.2) 2 (0.8) 4 1 time? 6C5(0.2) 1 (0.8) 5 6C6(0.2) 0 (0.8) 6 0 times? This is our favorite type of experiment! A binomial random variable obeys the conditions of a binomial experiment: Binomial Experiments There are only two possible outcomes, success and failure. Thus, the probability of success is 1 minus the probability of failure. Multiple trials of the experiment must be independent. Each trial must be one with replacement. Let X be the number of successful outcomes in n trials. We have discussed the probability of getting r successes in n trials. The binomial distribution function is a description of the probabilities of all the possible number of successes, r, that can occur in n trials. Recall that for a binomial experiment with a probability of success given by p, the probability of getting r successes in n trials is given by: P(r successes in n trials) = Using the notation of random variables, we write: P(X = r) is called the binomial probability distribution function. If X is a random variable of a binomial experiment with parameters n and p we say that X is distributed as B(n, p) and write X ~ B(n, p) You will often be using a calculator in these types of problems. You should know how to use ncr, binompdf(n, p, r) and binomcdf(n, p, r) You should also be able to calculate ncr values using the following technique which is very fast without a calculator: Mean of a Binomial Random Variable Recall that the mean of a random variable is given by Since, in the case of a binomial random variable the value of X is equal to the number of successes, r, we have: Hmmm. This looks rather complex. Can we simplify it? What do you think it simplifies to? Intuitively, since the probability of a success is constant, p, and all trails are independent, then in then in n trials we expect to get np successes. So the above should simplify to np. Does it? (...Yes...an exercise for the reader...) Mean of a Binomial Random Variable The mean or expected value of a binomial random variable is np where n is the number of trials and p is the probability of success on each trial. Standard Deviation of a Binomial Random Variable Recall that the variance of any discrete random variable is given by: The probabilities are given by the binomial expansion and µ = np. Try deriving the variance algebraically...(another exercise for the reader...) It turns out that: Variance and SD of a Binomial Random Variable Given a random variable X ~ B(n, p): Var(X) = σ 2 = np(1 p) = npq where q = 1 p Monday! Unit Exam on Discrete Random Variables 25F.1: #2,4,5 (Binomial distribution) 25F.2: #2-14 even (Binomial dist function) 25F.3: #3-5 all (Mean & SD of binomial) QB: #2,8,15 (Binomial distributions)

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10 25F.1: #2,4,5 (Binomial distribution) 25F.2: #2-14 even (Binomial dist function) 25F.3: #3-5 all (Mean & SD of binomial) QB: #2,8,15 (Binomial distributions) Present as needed Present all Quick Quiz E(Y) = 13 Var(Y) = 16 M1A1 M1A1 E(Y) = -7 Var(Y) = 16 M1A1 M1A1 E(Y) = 0 Var(Y) = 1 M1A1 M1A1 1. Understand and use features of the Poisson distribution 25G: #2-10 even (Poisson distribution) QB: #14,16,18,20 (Poisson distribution) Monday! Unit Exam on Discrete Random Variables The French mathematician Simeon-Denis Poisson was sitting outside a saloon in Paris one day, watching the horses go by. His friend bet him that exactly 10 horses would pass before the saloon closed, an hour later. Brilliant math guy that he was, Poisson quickly came up with a formula to find the probability that this would happen. He recognized that the situation could be thought of as a binomial distribution as the number of trials gets infinitely large. How? Well, he began by breaking the hour into minutes and considering each minute as a trial. Recall that Binomial probabilities Let's use x for the number of horses and 60 for the number of trials (minutes). We need p. But recall that for a binomial distribution Thus: Considering each minute as a trial, the probability of x horses passing is But what if two horses pass in one minute - we miss one. So we can make the interval smaller, say a second. Well, the same argument could be made again. So what we really want to do is break the hour into infinitesimally small moments to ensure that only one success can happen on a given trial. Mathematically we can do this with limits: Now for the fun! Write out the binomial coefficient: But notice that : Substitute: Rewrite and reorganize: Use the product of limits rule: Evaluate limits: Done! The Poisson Distribution where x = 0, 1, 2, 3,... m is called the parameter of the distribution and is the expected number of occurrences in the interval. Properties and Conditions for a Poisson Distribution Properties m is the mean and the variance! µ = m and σ 2 = m We write X ~ Po(m) Conditions 1. The average number of occurrences in each interval, µ, is the same for all intervals. 2. The numbers of occurrences in any interval is "small". 3. The numbers of occurrences in disjoint intervals are independent. In practice, we usually don't know the true mean so we estimate. For example, suppose we count the horses for the same hour of the day in front of the same saloon for a year and the average is horses. Then the probability that 8 horses would come through the intersection is given by: 25G: #2-10 even (Poisson distribution) QB: #14,16,18,20 (Poisson distribution) Monday! Unit Exam on Discrete Random Variables

11 A random variable is a letter (variable) that represents all the possible outcomes of a non-deterministic event in number form. A discrete random variable is a random variable that can take on a finite set of distinct values x1, x2, x3,... Generally speaking, discrete random variables represent things that are counted. By contrast a continuous random variable can take on any value in an interval. Generally speaking, continuous random variables represent things that are measured. A random variable has an associated probability distribution that is a description of the probability that the variable will take on any particular value. For example: X is the number of times you get a sum of 5 if you roll a pair of dice 6 times. The variable X can take on any value from 0 to 6 with differing probabilities. The graph of the probability distribution function would be: Properties of discrete probability distributions The probability of any individual outcome must obey: 0 pi 1 The total probability must sum to one: Expectation In n trials of an experiment with an outcome that has a fixed probability, p, of occurring, the expected number of those outcomes is np. More generally, if X is a discrete random variable where the probability that xi occurs is pi, the expected value of X is given by: Statistics of a Discrete Random Variable Population mean is its expected value: µ = E(X) Population variance Population Standard Deviation or The median of a random variable is the middle value. It occurs when the data are ordered and the cumulative probability is 0.5 This can be found by examining the expected frequency in 100 trials The mode of a random variable is the value that occurs most often. It is the value that has the highest probability of occurring Properties of Statistics of Random Variables Expected Value E(k) = k for any constant k E(kX) = ke(x) for any constant k E(f(X) + g(x)) = E(f(X)) + E(g(X)) for any functions f and g Variance Var(X) = E(X 2 ) {E(X)} 2 or Var(X) = E(X 2 ) µ 2 Linear functions of discrete RVs Let X be a random variable and a and b be constants. E(aX + b) = ae(x) + b Var(aX + b) = a 2 Var(X) Binomial Random Variables There are only two possible outcomes, success and failure. Thus, the probability of success is 1 minus the probability of failure. Multiple trials of the experiment must be independent. Each trial must be one with replacement. X = the number of successful outcomes in n trials. n = # of trials p = probability of success r = # of successes X ~ B(n, p) Mean Variance: σ 2 = np(1 p) = npq = µq where q = 1 p Standard Deviation The Poisson Distribution Describes the distribution of events happening over an interval (of time) where The average number of occurrences in each interval, µ, is the same for all intervals. The numbers of occurrences in any interval is "small". The numbers of occurrences in disjoint intervals are independent. m, (or λ) parameter, is the expected number of occurrences in the interval. X ~ Po(m) where x = 0, 1, 2, 3,... Mean µ = m Variance: σ 2 = m Standard Deviation Your calculator supports these two distributions very well: ncr find just the binomial coefficient P(X = r) binompdf (n,p,r) function or poissonpdf (m,r) P(X r) binomcdf (n,p,r) function or poissoncdf (m,r) P(X r) 1 binomcdf (n,p,r 1) function or 1 poissoncdf (m,r 1)

12 We did Discrete Random Variables at the start of 1516Q4 (HL1) We now begin Continuous RVs 1. Understand and use features of continuous random variables 26A: #2-10 even (Continuous RVs) QB: #4,7,9,14,15,16 (Cont RVs) Recall that a continuous random variable can take on any value in an interval. Generally speaking, continuous random variables represent things that are measured. Consider a random variable X that represents the length of an adult arm span (as opposed to, say, the number of M&M's in a bag which would be discrete) In statistics we looked at cumulative frequency diagrams of these variables. The gradient of this function at x describes how many data values are close to x which is, in essence the gradient of a cumulative frequency curve is P(X = x) If we differentiate the cumulative frequency curve, we get a probability density function (PDF) of the random variable X. In most cases that we look at we can describe the PDF with a mathematical function and look at its associated graph. Here is a common example, Recall the properties of discrete PDFs. What are the analogous conditions for continuous PDFs? Properties of a Continuous Probability Density Function Given a probability density function f(x) that describes P(X = x) 1) Any individual value of x must obey 2) The total probability on some interval, [a,b], must equal one: Technically, the probability of X taking on an exact value is zero! Why? Suppose the probability of X being exactly m were > 0. Since there are an infinite number of possible values m in the continuous interval from a to b, there is no way the total probability between a and b could add to one. Thus, the probability of X being exactly m must be zero. The value (height) of the PDF at some value x is not the same as the probability that the exact value of x will occur. In practice, we are always interested in the probability that a variable takes on a value within some range. Thus: Properties of a Continuous Probability Density Function The probability that the variable lies between the values c and d is given by the area under the PDF curve between c and d. As with discrete random variables, we have measures of central tendency: Measures of a Continuous Random Variable Notice the parallels to the discrete case and to the same ideas in descriptive statistics in general. Here's a little summary that's also on the HL Formula Sheet. 26A: #2-10 even (Continuous RVs) QB: #4,7,9,14,15,16 (Cont RVs)

13 26A: #2-10 even (Continuous RVs) Present one each, discuss as needed QB: #4,7,9,14,15,16 (Cont RVs) 1. Understand and use the general features of normal distributions. 26B: #1-13 odd (Normal Distribution) The most common continuous PDF is the bell-curve or normal distribution. It describes phenomena that are symmetrically centered around a mean, with a bell shaped spread. It occurs in many natural situations. Mathematically, the PDF for a normal distribution is given by: The Normal Probability Density Function A normally distributed random variable X has a PDF given by: Where µ is the mean of the random variable and σ is the standard deviation of the random variable This looks complex, but it's really just a transformation of Find the value of x where the function is a maximum. Find the value(s) of x where the function has inflection point(s) Properties of Normal PDFs µ and σ are called the parameters of the curve. They shift and stretch the curve respectively For a given pair of parameters we say that X ~ N(µ, σ 2 ) The total area underneath the curve is always one The curve is centered at µ The value of f(x) at µ is The inflection points occur at µ ± σ Because it is so common, we need to memorize some facts about its proportions: Knowing these allows us to easily answer such questions as: The arm spans of 1000 individuals are normally distributed with a mean of 180 cm and standard deviation of 20 cm. How many people do you expect to have arm spans above 200 cm? How many people do you expect to have arm spans between 140 and 160 cm?...and a million variations on these types of questions... One can also find the mean and/or standard deviation given other information. 26B: #1-13 odd (Normal Distribution)

14 26B: #1-13 odd (Normal Distribution) Present any questions 1. Use a calculator to find probabilities involving a normal distribution. 26C: #2-8 even (Calculator (Normal Dist)) QB: #2,5,8,13 (Normal RVs) 26D: #2,4,5,6ghi,7 (Standard Normal) QB: #6,11 (Standard Normal RVs) Fortunately, one can use a calculator to find probabilities for normal distributions: TI-84 Help with Normal Distributions For a normally distributed random variable with mean µ and standard deviation σ the probability that the variable lies between c and d is given by normalcdf(c, d, µ, σ) One can also find the value, a, such that P(X < a) = p by using the inverse normal PDF function: invnorm(p, µ, σ) For both functions, if µ and σ are omitted, they are assumed to be 1 and 0 respectively (the Standard Normal Distribution) More on the second function later. In this section you will work with normalcdf.

15 1. Understand and use the standard normal distribution. The normal distribution describes the variability of many natural phenomena. Suppose you go shopping and pick a random watermelon and a random apricot from their respective bins and ask the question which is the "better value". It makes no sense to compare the weights of the watermelon and the apricot. But we still might want to know which is the more "unique". When two variables are normally distributed we can compare the relative "uniqueness" of two samples by finding the number of standard deviations that each is from the mean. Let's take an example: For the watermelons, the mean weight is 6 kg and the standard deviation is 500 grams. The mean weight of the apricots is 80 grams and the standard deviation is 10 grams. Suppose you choose a 6.2 kg watermelon and a 85 gram apricot. How many standard deviations is your watermelon from the mean? How many standard deviations is your apricot from the mean? Z -Score The number of standard deviations that a sample is from the mean is called the sample's z-score. It is calculated as Recall the formula for the normal distribution? If we make the substitution we get a new distribution. Try it. So z has a normal distribution with a mean of zero and standard deviation of one. Z ~ N(0,1) is called the Standard Normal Distribution whose formula is: Standard Normal Distribution The probability density function for the Z-distribution is given by: The standard normal distribution has µ = 0 and σ = 1 Important interpretation: The z score is the number of σ's away from µ It's useful because we can find probabilities with z-scores instead of with the regular variable: We can also ask the question, what fraction of the apricots would be less than or equal to the one I chose? This would be calculated as the fraction of the area under the Z-distribution at or below a z-score of 0.5 This area is the cumulative standard normal distribution function and is given by Notice that these areas depend only on the value of a. Prior to calculators, people used tables of values to look up cumulative z-values. For your historical interest, IB just dropped using tables in 2014! Have a wonderful break. I suggest you do some of this HW in the next few days, then go back and do the rest a few days before we return. We will finish this unit and have the unit test on the Friday after you return. 26C: #2-8 even (Calculator (Normal Dist)) QB: #2,5,8,13 (Normal RVs) 26D: #2,4,5,6ghi,7 (Standard Normal) QB: #6,11 (Standard Normal RVs) HL Folks - If you have time, I suggest you begin your review by going through the handouts. The goal is mostly to brush off cobwebs and identify areas that we want to look at in more detail.

16 Q4: Today's lesson, review Wed, test Friday, cumulative review starts next week. 26C: #2-8 even (Calculator (Normal Dist)) Discuss as needed QB: #2,5,8,13 (Normal RVs) 26D: #2,4,5,6ghi,7 (Standard Normal) QB: #6,11 (Standard Normal RVs).1 1. Use a calculator to find quantiles from a normal distribution. 2. Use probabilities under the standard normal distribution to find µ & σ 26E.1: #2-8 even (Quantiles) 26E.2: #2-8 even (Finding µ & σ) QB: #1,3,10,12,17 (Find µ & σ) Problems that involve ranking of things o3en use the idea of a percen8le or quan8le. You have seen these in test scores. The 95th percen8le represents the test score below which 95% of the sample fell. In other words, there is some value k such that P(X k) is The value k is the 95th quan%le. The quan8le is also known as the percen8le or the k-value. Suppose that a set of test scores are normally distributed with a mean score of 48 and standard devia8on of 10. What would be the score that corresponds to the 75th percen8le? P(X k) = 0.75? What score lies at the 75th percen8le of these data? To answer this ques8on, we can use the inverse normal distribu%on func8on on our calculators or use the inverse standard normal distribu%on given in the table below. Let's try the calculator first: In general, the syntax is invnorm(k, μ, σ) where μ and σ are op8onal (default to 0, 1) For our problem we use invnorm(.75, 48, 10) = So a score of 54.7 represents the 75th percen8le.

17 .2 1. Use probabilities under the standard normal distribution to find µ & σ As we have seen, we can find quantiles in the standard normal distribution too. Using the mapping from the x world to the z world can then allow you to solve problems such as the following. A more indirect example (of which there are many variants) Since we don't know the mean and standard deviation, we must convert the random variable to z-scores! If you knew either the mean or the standard deviation, you would not need to solve simultaneous equations to find the other. This is a good time to remind you about the power of PLYSMLT2 when you have a calculator is the first year with no tables of z-scores. This means the questions are likely to change - to ones more like those above. 26E.1: #2-8 even (Quantiles) 26E.2: #2-8 even (Finding µ & σ) QB: #1,3,10,12,17 (Find µ & σ) Coming up: Review Wed, test on Friday

18 26E.1: #2-8 even (Quantiles) 26E.2: #2-8 even (Finding µ & σ) QB: #1,3,10,12,17 (Find µ & σ) Discuss as needed Continuous Random Variables Also, don't forget discrete random variables, particularly binomial. Review Set 26C #1-11 all QB Review all