8 Probability. An event occurs if, when the experiment is performed, the outcome is in that event.

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1 8 Probability When an experiment is performed, an outcome is observed. A set of possible outcomes is an event. The sample space for the experiment is the set of all possible outcomes. An experiment must have an outcome. Nevertheless we regard the empty set /0 as an event the impossible event. E.g. Roll a die once, observe the number of pips on the top face. Possible outcomes: 1,2,3,4,5,6 Some possible events: {}, {1}, {5}, {2,4,6}, {1,2,3,4,5,6} An event occurs if, when the experiment is performed, the outcome is in that event. E occurred E did not occur E.g. Roll one die. Let E = {get an even number}, F = {get a number less than 4}. If you get a 1, then F has occurred but E has not. If you get a 4, then E has occurred but F has not. If you get a 2, both E and F have occurred. If you get a 5, neither E nor F has occurred. Venn diagrams When we use Venn diagrams to understand probability, we think of shapes (rectangles, circles, etc.) as representing events. Points inside a shape represent outcomes in the corresponding event, and points outside the shape represent outcomes not in the event. The rectangular frame represents the sample space and so contains all the possible outcomepoints. E F We may indicate individual outcomes in events by drawing points in the circles. Whole events are shown by shading the corresponding regions. For example, the diagram below is meant to indicate the event consisting of all the outcomes that are in both event E and event F: E F 23

2 Probability The probability of an event E is a number between 0 and 1 that expresses how likely E is to occur. If event F is more likely to occur than event E, then P(F) > P(E). If P(E) is near zero, then E is unlikely to occur; if P(E) is near 1, then E is likely to occur. In order to model what happens in the physical world, we accept the following rules about probability: P(/0) = 0 P(sample space) = 1 If x and y are different outcomes, then P({x,y}) = P(x) + P(y) Note that {x,y} is the event containing the outcomes x and y The last rule can be thought of as saying P(x or y) = P(x) + P(y). Note that the rules imply that: The probability of an event is equal to the sum of the probabilities of the outcomes in that event. If x 1,x 2,...,x n are all possible outcomes (i.e., all the outcomes in the sample space), then P(x 1 ) + P(x 2 ) + + P(x n ) = 1. These statements in turn imply that if a sample space consists of n equally likely outcomes, then the probability of any particular outcome is 1 n. E.g. Roll one fair, six-sided die. The sample space for this experiment is {1,2,3,4,5,6}. Note. In this course, dice are always fair and six-sided unless something says otherwise. E.g. Roll two dice, one red and one green. Then the sample space consists of equally likely outcomes: (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) Thus each outcome has probability 1. E.g. What s the probability of rolling a 9 with two dice? The event roll a sum of 9 contains four outcomes: it is the set {(3,6),(4,5),(5,4),(6,3)}. So P(sum = 9) = P({3,6}) + P({4,5}) + P({5,4}) + P({6,3}) = = 4 Theoretical probability When there are only finitely many outcomes and all are equally likely, we define the theoretical probability of an event E by: # outcomes in the event P(E) = total # possible outcomes Can only be used in ideal/mathematical situations 24

3 E.g. A bag holds ten identical marbles numbered 1,2,3,...,10. One is drawn out. What s P(get marble #5)? all outcomes equally likely one outcome in event: get #5 = {5} ten outcomes total P(get #5) = 1 10 E.g. A fair die is rolled once. What s P(get an odd number)? all outcomes equally likely three outcomes in event: {1, 3, 5} six outcomes total P(get an odd number) = 3 6 = 0.5 E.g. The probability of drawing an ace from a standard deck is E.g. If a data point x is chosen at random from any data set, then Probability(x < Q 1 ) = Probability(Q 1 < x < Q 3 ) = In the real world, the probability of rolling a 1 with a fair die is 1 6 does not mean that one in every six rolls will be a 1. It means that if we roll more and more times, the fraction or rolls that yield a 1 will come closer and closer to 1 6. Empirical probability Suppose that an experiment is performed repeatedly under very similar conditions. Then we define the empirical probability of an event E by: number of times E occurs P(E) = total # of observations I.e., an empirical probability is a relative frequency. E.g. The table below shows the times at which Jill has gotten up on ten different school mornings. Based on this data, what s the probability that she ll get up before 7:05 on any given school morning? event: Jill gets up before 7:05 # outcomes in which event occurred: 6 total # observations = 10 prob = 6 10 = 0.6 Day Time 7:01 7:03 7:01 7:00 7:10 7:07 7:03 7:01 7:15 7:06 The Law of Large Numbers says that as an experiment is repeated independently more and more times under the same conditions, the empirical probability of events approaches their theoretical probability. Subjective probability Subjective probability is a probability judgment. E.g. The doctor says there s a 60% chance John will survive the surgery. Subjective probability is used when an experiment cannot be repeated or in a situation too complicated to be compared directly to other cases. 25

4 Properties of probability All probabilities are between 0 and 1. E impossible: P(E) = 0 E just as likely to happen as not: P(E) = 1 2 E certain: P(E) = 1 (or 100%) 26

5 9 More Probability Complement of an event Suppose that, for some experiment, you have the sample space S = {x 1,x 2,x 3,x 4,x 5 } with five equally likely outcomes, and let E be the event {x 1,x 2,x 4 }. Then E occurred means that one of the outcomes x 1, x 2, or x 4 has happened, and E didn t occur means that either x 3 or x 5 has happened. We call the set of outcomes outside E the complement of E, written E c. E is shaded E c is shaded Note that E and E c don t overlap, and that together they contain all outcomes. Thus, P(E) + P(E c ) = 1. That is, if the probability that an event E will occur is p, then the probability of E c is 1 p, and vice versa: P(E c ) = 1 P(E) P(E) = 1 P(E c ) Probability of the complement Obviously if an outcome is in E, then it can t be in E c, so if E occurs, then E c doesn t occur, and vice versa. On the other hand, one or the other of them must occur, because the outcome has to be somewhere. E.g. A coin is tossed. If H is the event get heads, what is H c, and what is P(H c )? E.g. A die is thrown. If E is the event get an odd number, what is E c and what is P(E c )? Intersection of events The intersection of two events E and F is the event containing all outcomes that are in both E and F. The intersection of E and F is denoted E F. Notice that P(E F) = P(E and F), because an outcome is in E F if and only if it is in both E and F. 27

6 Union of events The union of two events E and F is the event containing all outcomes that are either in E or in F (or both). The union of E and F is denoted E F. Notice that P(E F) = P(E or F), because an outcome is in E F if and only if it is in E or F or both. Diagram A Consider the sample space represented by the diagram above. When trying to compute P(E F), it s natural to start with P(E) + P(F), but the diagram makes it obvious that this sum counts outcomes 4 and 5 twice. That is, it counts the outcomes in E F twice. Thus, the general formula for the probability of a union, called the Addition Rule for Probability, is P(E F) = P(E) + P(F) P(E F) The Addition Rule for probability E.g. Roll a single die, and let E = {get a number less than 4} and F = {get an odd number}. What is P(E F)? 28

7 Mutually exclusive events Events E and F are mutually exclusive if there is no outcome that is in both. Two mutually exclusive events cannot both happen at the same time. E.g. Roll a single die. Let E = get an even number, F = get an odd number. Then E and F are mutually exclusive. E.g. Roll a single die. Let E = get an even number, F = get a number greater than 3. Then E and F are not mutually exclusive, because there is at least one outcome that makes both occur (e.g., {4}). If E and F are mutually exclusive events, then P(E occurs or F occurs) = P(E) + P(F) Or here is the inclusive or : it means one or the other or both. Mutual exclusivity is not defined in terms of probability. It is a set property. When events E and F are mutually exclusive, E F is empty, so P(E F) = 0. So, when E and F are mutually exclusive, we have a modified addition rule: P(E F) = P(E) + P(F) The Addition Rule for mutually exclusive events E.g. Suppose the spinner pictured, on which each area is 1 /3 of the whole, is spun once. Let E = {get a 3}, F = {get a 5}. What is P(E F)? 29

8 Contingency tables The following table shows the results found when researchers examined a possible link between a specific gene and high IQ. Gene found Gene not found Total IQ high IQ not high Total First, because each of the outcomes is described by two characteristics, we must write the sample space as {IQ high/gene found, IQ high/gene not found, IQ not high/gene found, IQ not high/gene not found} Now if we choose a child at random from among those studied: (a) What is the probability that the child had the gene? (b) What is the probability that the child does not have a high IQ? (c) Find the probability that the child does not have the gene or does not have a high IQ. (d) Find the probability that the child has both the gene and a high IQ. 30

9 10 Conditional Probability and Independent Events The conditional probability that an event F occurs if an event E occurs is denoted P(F E). Note that if the event E has occurred, then we already know that the only outcomes that could have occurred are those in E. So P(F E) is the probability that the outcome was in F if we already know that it was in E. We sometimes say that we have reduced the sample space to E. Here s how to compute conditional probabilities: P(F E) = P(F E) P(E) The Conditional Probability Rule E.g. An urn contains six red marbles, labeled 1 6, and six green marbles, also labeled 1 6. Let E be the event get a green marble and F be the event get a number higher than 4. One marble is chosen. Find (a) P(E F) and (b) P(F E). We compute a few quantities first: All possible outcomes: {R1,...,R6,G1,...,G6} (12 outcomes) P(E and F) = P({G5,G6}) = 2 12 = 1 6 P(E) = P({G1,...,G6}) = 6 12 = 1 2 P(F) = P({R5,R6,G5,G6}) = 4 12 = 1 3 So (a) P(E F) = (b) P(F E) = P(E and F) P(F) = 1 /6 1/3 = 1 2 and P(F and E) P(E) = 1 /6 1/2 = 1 3 E.g. The following table concerns the NY Mets 2017 season: Wins Losses Total Home Away Total Suppose that a particular game is randomly chosen from that season. (a) What is the probability that the Mets won the game? (b) What is the probability that they won, given that they played at home? 31

10 (c) If they lost the game, what is the probability that they were playing away? Independent events Two events are statistically independent (or just independent) if P(F E) = P(F) and P(E F) = P(E). Note that P(F E) = P(F) says that the probability of F is the same whether E happens or not i.e., the occurrence or non-occurrence of E doesn t change the probability that F will occur. So E and F are independent if the occurrence or non-occurrence of either one of them doesn t affect the probability that the other will occur. Intuitively, events E and F are independent if they have nothing to do with each other. True fact: events E,F are independent if and only if P(both E and F occur) = P(E)P(F). That is, E and F are independent if and only if P(E F) = P(E)P(F) Condition for independence E.g. Roll one fair die and flip one fair coin. What s the probability of getting heads and a 4? Solution 1. All possible outcomes: {1H,2H,3H,4H,5H,6H,1T,2T,3T,4T,5T,6T } There are 12 possible outcomes, all equally likely. The event get heads and a 4 contains the single outcome 4H, so its probability is Solution 2. P(H) = 1 2, P(4) = 1 6, and the two events are independent, so P(E and F) = = Note that independence is defined in terms of probability. Unless we are given probabilities (or can compute them), our method for determining independence is educated guess. Dependent events Two events are dependent if they aren t independent. For any two events E and F, whether dependent or independent, it is true that P(E F) = P(E) P(F E) = P(F) P(E F) Probability of E F We often get dependent events when we draw an object from a group of objects and then draw a second object without replacing the first. E.g. A bag contains five red marbles and four blue marbles. John draws out a marble and then draws out another marble without replacing the first. What is the probability that both marbles drawn are red? Let E = {the first marble is red} and F = {the second marble is red}; then we re looking for P(E F). Clearly, P(E) = 5 9. Also, once we know that E has occurred i.e., once we know that the first marble is red then we are left with four red marbles and four blue marbles, so P(F E) = 4 8. By the formula above, then, P(E F) = P(E)P(F E) = =

11 11 Counting The Fundamental Counting Principle The Fundamental Counting Principle, sometimes called the Multiplication Rule, says that if there are m ways to do Task 1 and n ways to do Task 2, then the number of ways to do first Task 1, then Task 2, is m n. This can be extended to any number of tasks. For example, if there are v ways to do Task 3, then there are m n v ways to do first Task 1, then Task 2, then Task 3. E.g. The Wednesday special at Meatsa Pizza is a one-person pizza with one topping and a soft drink for $8.95. If there are twelve different toppings to choose from and six different soft drinks, how many different Wednesday specials can be ordered? Think of ordering a special as completing two tasks: Task 1 is to choose a topping, and Task 2 is to choose a soft drink. There are twelve ways to do Task 1 and six ways to do Task 2, so there are 12 6 = 72 ways to choose altogether. In other words, there are 72 different Wednesday specials. E.g. The Thursday special at Meatsa Pizza is a large pizza with any combination of toppings (but no more than one of each). The available toppings are pepperoni, mushrooms, onions, sausage, peppers, olives, and pineapple. How many different Thursday specials can be ordered? pepperoni mushrooms onions sausage peppers olives pineapple Permutations Suppose you have n different objects in a line. A permutation of those objects is an arrangement of them in a definite order. Two different permutations of four billiard balls A question that often arises is, how many permutations of n different objects are there? To answer this in the case of the pool balls shown above, think of having a box with four compartments in a row and choosing a ball to go in each compartment. We think of putting the balls in the box as four separate tasks. There are four different balls, so there are four ways to choose the leftmost ball. Once that one is in place, there are only three balls to choose from, so there are three ways to choose a ball for the second slot. Similarly, there are only two ways to choose for the third slot, and only one way for the last. By the Fundamental Counting Principle, then, there are = 24 ways to arrange the balls. That is, there are 24 permutations of these four different objects. We write the product as 4! and call it four factorial.. We can use the factorial notation with any integer; e.g., 6! = It s pretty obvious from the pool-ball example that there are exactly n! permutations of n different objects. We ll take this one step further. Suppose we have five pool balls and a box with only three slots. If we choose three of the five balls to put in the box, how many different arrangements could we make? 33

12 We can think about this in the same way as the earlier problem: there are five choices for the first slot, four for the second, and three for the third, so there are = 60 choices altogether. Notice that = = 5! 2! = 5! (5 3)!. The sort of analysis we ve done here works in general. If we have n different objects, and we want to select r of them in n! order, then the number of ways to make the selection is (n r)!. This is called the number of permutations of r things chosen from among n and is denoted n P r or just P(n,r). We have P(n,r) = n! (n r)! Number of permutations E.g. Twelve horses are running in a race. How many different first, second, and third place winners are possible? Combinations In the pool-ball examples, the order in which the balls ended up was important. Sometimes, however, the order is not important. For example, suppose we were given the four pool balls pictured above, and we wanted to choose three of them to throw in a bag. How many different choices of three balls could we make? In this course, we re just going to tell you the answer 1 4! : there are (4 3)!3! = 4 ways to choose. In general, the number of ways to choose r objects from among n different objects when ( ) order doesn t matter is called the number of n combinations of r things chosen from among n and is denoted n C r,, or just C(n,r). We have r C(n,r) = n! (n r)!r! Number of combinations E.g. A certain restaurant allows a customer to choose three different side dishes when he buys an entrée. If there are twelve side dishes to choose from, how many different choices of side dishes could he make? Permutations and combinations on the TI calculator To compute n P r : 1. Enter n 2. Click MATH 3. Arrow over to PRB 4. Arrow down to 2:nPr and press ENTER 5. Enter r 6. Press ENTER To compute n C r, follow the same instructions except: at Step 4, arrow down to 3:nCr. To compute n!, enter n and then go to MATH > PRB > 4:! and press ENTER twice. 1 For the curious: Define C(n,r) to be the number of ways to choose r items from among n when order doesn t matter, and choose r balls from among n in order in two steps: first, choose the r balls without regard to order, and then order them. There are, by definition, C(n,r) ways to do the first step, and we know that there are r! ways to do the second step, so by the Fundamental Counting Principle, there are C(n,r)r! ways to do the whole task. But we already know that there are P(n,r) ways to do it, so it must be that C(n,r)r! = P(n,r). Thus, C(n,r) = P(n,r) r! = n! (n r)!r!. 34

13 When to use what When faced with a counting problem, ask yourself two questions. First, is it possible to choose an item twice? If so, you must use the Fundamental Counting Principle. If items cannot be chosen more than once, does the order of the chosen items matter when all is said and done? If order does matter, you ll need either the Fundamental Counting Principle or np r. If order doesn t matter, you ll need n C r. E.g. A club with ten members must elect a president and a vice-president. How many different choices could be made? One way to do this problem is to think of choosing a president and then a vice-president. There are 10 ways to choose a person to be president, and after that, there are 9 ways to choose a vice-president, so by the Fundamental Counting Principle, there are 10 9 = 90 possible choices altogether. Another approach is to think of choosing an ordered pair of people and making the first person in the pair the president and the second person the vice-president. Clearly, order matters here, so we are looking for the number of permutations of two objects chosen from among ten, which is 10 P 2 = 90. E.g. A combination lock has four rotators, each of which contains the digits 0 9. How many different combinations are possible? Note that the problem does not say that the digits have to be different; e.g., is a valid combination. Because digits can be chosen more than once, we must use the Fundamental Counting Principle. We have ten choices for the first number, ten for the second, ten for the third, and ten for the fourth, giving a total of = possibilities altogether. E.g. How many different five-card hands can be dealt from a standard deck? 35

14 12 Random Variables; Discrete Probability Distributions; Expected Value A random variable is a variable that takes on numerical values associated with the outcomes of an experiment. E.g. Choose an apple at random and let X = weight of the apple. Choose a box of blueberries at the supermarket, let X = the number of (whole) blueberries in the box. Choose a sample of ten apples at random, and let X = the mean weight for that sample. Note that each time the experiment is performed, the value of the RV may change. Continuous vs. Discrete Just like data distributions. A RV is discrete if all of its (possibly infinitely many) possible values can be listed, at least in principle. A RV is continuous if it can take on any value in some interval of real numbers. Probability distributions If X is a discrete RV, its probability distribution is a table that lists each possible value of X and the probability that X takes on that value. A discrete probability distribution is usually a table of relative frequencies. If X is a continuous RV, its probability distribution is a function from which it is possible to compute the probability that X takes on a value in any given interval. (We ll come back to these.) E.g. Two (fair, six-sided) dice are rolled, and the value of X is the sum of the dice. Construct a probability distribution for X. E.g. A psychological test was administered to 150 people. Possible final scores were 1, 2, 3, 4, and 5. Results: X (score) Frequency Note that the RV X takes on a numerical value for each person who took the test. (a) Construct a discrete probability distribution. (b) What is the probability that a randomly chosen participant s score was (i) 3? (ii) greater than 3? E.g. Assuming that the sample space for an experiment is {a, b, c, d, e}, what value for P(d) makes the following table into a probability distribution? Outcome a b c d e Probability ? 0.25

15 Expected Value If the possible values of a RV X are x 1,x 2,..., and if the probabilities of these are p 1, p 2,... respectively, then the expected value or mean of X is E[X] = p 1 x 1 + p 2 x 2 + E.g. Find the expected value of the following probability distribution. Value Probability Extend the table with a column for (Value Prob) and sum these: Value Prob Value Prob Σ(Value Prob) = 80 E[x] = 80 Variance and standard deviation of a random variable We will not compute these, but either one can be regarded as a measure of risk in certain circumstances. E.g. The prices of two stocks, A and B, vary every few seconds as shares are bought, sold, and traded. If the variance of A is smaller than the variance of B, which would you recommend to your nervous old granny in Topeka? Independent random variables Two RVs are independent if neither variable s probability distribution depends on the value of the other s. The RV equivalent of E occurs is X takes on a value. Our method of determining independence is once again educated guess. E.g. Choose a person at random. X = number of atoms in the person s left hand Y = the person s SAT score E.g. Choose a location on Earth. X = average temperature in that place on Jan 1 Y = average snowfall in that place on Jan 1 37

16 13 The Binomial Distribution A binomial experiment is made up of smaller experiments called trials. The trials are all supposed to be the same. E.g., rolling the same die many times in the same way and on the same surface. An experiment is binomial if It consists of a fixed number of independent trials (called Bernoulli trials) The only possible outcomes for each trial are success (S) and failure (F) The probability p of success is the same in each trial Some variables have standard names in the context of binomial experiments: n = the number of trials p = the probability of success = P(S) q = 1 p = P(F) X = the number of successes in n trials E.g. An experiment consists of flipping one fair coin once. Success is defined as getting heads. What are all the possible outcomes? What are n, p, and q? What are the possible values of X? Here there s only one trial, so n = 1, and the possible outcomes of the experiment are just H and T. The coin is fair, so p = 1 2, whence q = 1 p = 1 2. The possible values of X i.e., the possible numbers of successes are 0 and 1. E.g. An experiment consists of flipping one fair coin twice. Success is defined as getting heads. Find each of the following: (a) n = (d) All possible outcomes of one trial: (b) p = (e) All possible outcomes of the experiment: (c) q = (f) All possible values of X: Remember that success and failure are defined only for each trial, not the whole experiment. E.g. Each trial of an experiment consists of flipping a coin and throwing a die. The experiment itself consists of two trials. List all possible outcomes of one trial, and all possible outcomes of the whole experiment. Possible outcomes of one trial: Possible outcomes of the experiment: H1,H2,H3,H4,H5,H6,T 1,T 2,T 3,T 4,T 5,T 6 (H1, H1) (H1, H2) (H1, H3) (H1, H4) (H1, H5) (H1, H6) (H1,T 1) (H1,T 2) (H1,T 3) (H1,T 4) (H1,T 5) (H1,T 6) (H2, H1) (H2, H2) (H2, H3) (H2, H4) (H2, H5) (H2, H6). (T 6,T 1) (T 6,T 2) (T 6,T 3) (T 6,T 4) (T 6,T 5) (T 6,T 6) Note that the possible outcomes of the experiment are just all the possible ordered lists of outcomes of the n trials. 38

17 Examples of binomial distributions The interesting question is, for any given probability of success on each trial, what s the probability distribution of X? That is, what s the probability of no successes, one success, two successes, and so on? E.g. A fair coin is flipped five times. S = get heads. Here n = 5, p = 1 2, q = 1 2. The possible values of X are 0,1,2,3,4, and 5. Probability distribution: x P(X = x) Experiment: flip coin 5 times Trial: one flip p = 1 2 q = 1 2 n = 5 Value of X = number of successes E.g. Same, for 10, 20, 50 flips: Ain t that purty. 39

18 E.g. A fair, six-sided die is thrown ten times. Success = get a 1 or 3 Here n = 10, p = 2 6, q = 4 6 Distribution (to four decimal places) and histogram: x P(x) Same, for 50 throws: 40

19 Finding binomial probabilities with the TI calculator For a binomial experiment with n trials in which the probability of success on any one trial is p: binompdf(n, p,x) = probability of exactly x successes binomcdf(n, p,x) = probability of at most x successes 1 binomcdf(n, p, x 1) = probability of at least x successes Think of the last item as saying that if you want the probability of at least successes, you have to calculate the probability of at most successes and then subtract that from 1. E.g. A certain surgery is successful in 80% of cases. (a) The surgery is performed four times. What s the probability of exactly two successes? (b) The surgery is performed sixty times. What s the probability of at most 50 successes? (c) The surgery is performed forty times. What s the probability of at least 32 successes? Remember: no more than x means at most x no fewer than x means at least x Expected value of the binomial distribution What s E[X] for the X of the experiment in the example just above? It s E[X] = 0 (1 p) p(1 p) p 2 (1 p) + 3 p 3 = 3p It s possible to show that for a binomial experiment with n trials, E[X] = np. It s also possible to show that the standard deviation is σ = npq. 41

20 42

21 Index Symbols x, 14 /0, 24 1-Var Stats, 20 common shapes, 9 frequency, 7 probability, 37 relative frequency, 8 B bar graph, 11 bimodal, 15 binomcdf, 42 Binomial distribution, 39 expected value, 42 experiment, 39 binompdf, 42 bivariate, 3 box plot, 23 C categorical data, 5 Chart Pareto, 11 pie, 10 time series, 12 classes, in frequency distributions, 7 Combinations, 35 complement, 28 continuous data, 5 Cumulative frequency, 10 D Data, 2 bivariate, 3 categorical, 5 continuous, 5 discrete, 5 grouped, 7 numerical, 5 numerical data, 5 qualtitative, 5 quantitative, 5 ungrouped, 7 univariate, 3 deviation, 18 discrete data, 5 dispersion, 18 Distribution binomial, 39 E Error sampling, 6 Estimator unbiased, 14 Event, 24 complement of, 28 impossible, 24 independent, 33 intersection of events, 28 mutually exclusive, 30 union of events, 29 Expected value, 38 of binomial, 42 Experiment, 3 binomial, 39 F factorial, 34 Five-number summary, 22 Frequency, 7 cumulative, 10 distribution, 7 histogram, 8 relative, 8 Fundamental Counting Principle, 34 G Graph bar, 11 line, 12 scatter plot, 13 grouped data, 7 H Histogram frequency, 8 relative frequency, 9 I Independent events, 33 random variables, 38 Interquartile range, 23 intersection of events, 28 1

22 IQR, 23 L line graph, 12 M Mean, 14 arithmetic, 20 expected value, 38 weighted, 20 Measure of central tendency, 14 Median, 14 Mode, 15 multimodal, 15 mutually exclusive events, 30 O Outcome, 24 Outlier, 23 intuitive definition, 16 P Parameter, 2 Pareto chart, 11 Percentiles, 21 Permutations, 34 pie chart, 10 Plot box, 23 Population, 2 Probability, 24 classical, 25 empirical, 26 properties of, 27 subjective, 26 theoretical, 25 Q qualtitative, 5 quantitative, 5 Quartiles, 21 R Random variable, 37 continuous, 37 discrete, 37 independent, 38 Range, 18 relative frequency, 8 relative frequency distribution, 8 relative frequency histogram, 9 Risk, 38 S Sample, 2 biased, 6 census, 6 simple random, 6 Sample space, 24 Sampling error, 6 Scatter plot, 13 simple random sample, 6 spread, 18 Standard deviation of a data set, 19 Statistic, 2 Statistics definition, 2 descriptive, 2 inferential, 2 T TI 1-Var Stats, 20 time series chart, 12 Trial, 39 Bernoulli, 39 U unbiased estimator, 14 ungrouped data, 7 union of events, 29 univariate, 3 V variability, 18 Variable, 2 Variance, 18 as risk, 38 population, 19 sample, 19 W weighted mean, 20 Z z-score, 23 2