Math 111, Math & Society. Probability

Transcription

1 Math 111, Math & Society Probability 1 Counting Probability consists in the assignment of likelihoods to the possible outcomes of an experiment, activity, or phenomenon. Correctly calculating probabilities requires that we be able to employ a number of important skills and principles. The most fundamental of these required skills is counting. 1.1 Multiplication and Addition Rules Multiplication Rule If you have n options followed by m options, then the total number of options is n m. For example, if you have three pairs of trousers and five shirts, then you have fifteen options for how to dress. This generalizes: if you have three pairs of trousers, five shirts, and four pairs of socks, then you have = 60 ways to dress. Note that in this example we are assuming that you are wearing one shirt and one pair of trousers and one pair of socks. If you were going to wear a pair of trousers or a shirt, then you would use the addition rule (below) Addition Rule If you must pick one of two sets of options, and the first set has n options and the second set has m options, then the total number of options is n + m 1

2 Probability 2 For example, if you are buying a shirt and the cotton ones come in four colors and the polyester ones come in six colors, then you have 10 choices of color. Note that this assumes you are buying either a cotton shirt or a polyester shirt. If you were buying a cotton shirt and a polyester shirt, then the multiplication rule would apply. 1.2 Power and Subset Rules Power Rule If there are n objects, and for each one of them there are m possible dispositions, then there are a total of m n options. This is actually a special case of the Multiplication Rule. For example, if you have 11 kitchen utensils, and you have three drawers to keep them in, then the total number of ways you could organize them into the drawers is 3 11 = 177, 147 ways. This is because you have three options for the first utensil, and three options for the second utensil, and three options for the third, and so on, so by the Multiplication Rule you must multiply three times itself, eleven times Subset Rule If there are n objects, and you may choose some, none, or all of them, then there are 2 n options. This is a special case of the power rule, where for each of the n objects there are just two dispositions, to be chosen or not to be chosen. In effect, we are just counting the number of subsets of a set of n objects, which we know to be 2 n. For example, if I have a bank of 7 light switches, then I have 2 7 = 128 ways that I can set the lighting, including all of them off, all of them on, and everything in-between.

3 Probability Permutations and Combinations Factorials If n is a non-negative integer, then the factorial of n, denoted by n!, is defined as follows: { 1 if n = 0 n! = n(n 1)! if n > 0 This is an example of definition by recursion. To know the factorial of a number, it is only necessary to multiply it by the factorial of the number that comes before it. Thus, 4! = 4 3!. To find 3!, multiply 3 times 2!, which is 2 times 1, which is 1 times 0!, which is 1. In other words, 4! = 4 3! = 4 3 2! = ! = ! = = 24 So 5! is 120, since it is 5 4!. For any positive integer n, this is the same as to say that, n! = n (n 1) 2. Numbers get very large very quickly when you take their factorial. For example, if you put 100! in your calculator, it will almost certainly return an overflow error Permutations The number of ways of permuting (arranging in order) a set of n objects is n!. More generally, the number of ways of permuting (arranging in order) a

4 Probability 4 selection of r objects from a collection of n objects, called n permute r and denoted by npr, is given by npr = n! (n r)! This is, once again, just the Multiplication Rule in disguise. For suppose I wish to arrange four objects in order from first to fourth. I have four choices for the object that will be first. Once it is chosen, I then have three choices for the object that will be second, then two remaining choices for the object that will be third, and finally just one choice (the remaining object) to put in last place. Consequently, there are = 4! ways of doing this. Now suppose I wish only to select a first and a second object from my collection of four objects. Then I have four choices for the first object, and once it is chosen I have three choices for the second object. Consequently I have 4 3 = 12 ways of ordering a selection of two objects from a set of four objects. This result matches what I would obtain from the formula above: 4P 2 = 4! (4 2)! = 4! 2! = 4 3 2! 2! = 4 3 = 12 As another example, suppose I have six garden beds, and I wish to use one of them for corn, one for tomatoes, and one for beans. This is the same as to order three

5 Probability 5 of the beds from my collection of six. Using the formula above, we have 6P 3 = 6! (6 3)! = 6! 3! = ! 3! = = Combinations The number of ways of choosing (without respect to order) a selection of r objects from( a collection ) of n objects, called n choose r and denoted by n ncr or by, is given by r ncr = ( n r ) = n! r!(n r)! ( ) n The notation ncr is what is generally found on calculators, but is what is r most commonly use in written mathematics. This rule differs from the Permutations rule only by the addition of r! in the denominator. This is easily understood if we think of beginning with permuting r objects from the collection of n objects, but then noticing that for each choice of r objects we counted all of the ways those objects could be arranged. So, if we don t wish to pay attention to the order in which they are arranged but only to which objects were chosen, we must divide off the ways they can be arranged, which of course means dividing off r!.

6 Probability 6 For example, suppose I have six garden beds, and I wish to use three of them for corn. Here I am choosing three of the beds from my collection of six without regard to the order in which they are chosen, since they will all be planted with the same crop. Using the formula above, we have ( ) 6 6! = 3 3!(6 3)! = 6! 3! 3! = ! 3! 3! = ! = = 20 2 Odds In probability we seek to assign likelihoods to outcomes. To avoid confusion, we define the following terms so that we may use them without ambiguity in what follows:

7 Probability 7 Definition: Sample Spaces, Events Suppose an experiment (or activity, or phenomenon) has a fixed number of possible outcomes, one of which must occur, and no two of which can occur simultaneously. We call the set of these possible outcomes the sample space of the experiment. A nonempty subset of the sample space is called an event For example, suppose I flip two coins, a quarter and a dime. There are four possible outcomes, corresponding to (1) heads on both coins, (2) tails on both coins, (3) heads on the quarter and tails on the dime, and (4) tails on the quarter and heads on the dime. These four outcomes comprise the sample space for this experiment. The event one heads corresponds to the set containing outcomes 2 and 3; the event not all tails corresponds to the set containing outcomes 1, 2, and 3; the event no heads corresponds to the set containing just outcome 4. One way of expressing the likelihood of an event is as a ratio of two numbers, called the odds: Definition: Odds Suppose an experiment has a fixed number of equally likely outcomes, and that an event E consists of a of those outcomes, while the remaining number b of outcomes are not in the set E. Then the odds in favor of E are a to b, denoted a:b, and the odds against E are b to a, denoted b:a. Example: Consider the experiment of flipping one quarter and one dime, let E be the event one heads. Since there are four possible outcomes and two of these are in E, and two are not in E, the odds in favor of E are 1:1. (Note: when stating odds, the ratio is always expressed in lowest terms. Consequently we reduce the ratio of 2:2 to 1:1 by canceling the common factor 2.) On the other hand, if E is the event all heads then there is only one outcome in E (both coins coming up

8 Probability 8 heads), and three outcomes not in E. Therefore the odds in favor of E are 1:3, and the odds against E are 3:1. Note that when we calculated the odds in the above example, we implicitly assumed that each outcome was equally likely, i.e., that they were fair coins. Example: Suppose we roll a fair six-sided die, and note the number of spots facing up. The possible outcomes are 1, 2, 3, 4, 5, and 6. Then the odds in favor of rolling a 6 are 1:5, the odds against rolling a number less than 5 are 2:4, and the odds in favor (or against) rolling an even number are 1:1. Example: Suppose we roll two fair six-sided dice, one red and one white, and note the sum of the spots facing up. Then the sample space consists of 36 distinct, equally likely outcomes. These are represented in the diagram below. The odds in favor of rolling doubles are 1:5, because there are 6 outcomes that correspond to doubles (the main diagonal) and 30 outcomes that don t correspond to doubles. (And 6 to 30 reduces to 1 to 5. ) What are the odds in favor of rolling higher than a 7? In favor of rolling a 10? Against rolling an odd sum? You will benefit from working out the answers to these questions, and practicing by creating more of your own. Example: Suppose I have five marbles in a bag; three red ones and two green ones. I reach in without looking and select one marble at random. What are the odds of selecting a red marble? Answer: Because there are three ways of selecting a red marble and two ways of selecting a not-red marble, the odds in favor of selecting a red marble are 3:2.

9 Probability 9 Example: Suppose I have five marbles in a bag; three red ones and two green ones. I reach in without looking and select two marbles at random. What are the odds of selecting one red and one green? Of selecting two red? Of selecting two green? Answer: To answer the remaining questions it will be useful to( apply ) our counting 5 principles. First we note that the size of the sample space is, which is Also, there are 3 2 ways of selecting one red and one green marble, which is 6. Consequently the odds in favor of one red and one green are 3:2 (reduced from 6:4). ( ) 3 Since there are ways of picking two red marbles, which is 3, the odds in 2 favor of picking two red marbles is 3:7. Finally, since there are only two green marbles there is only one way to pick two green marbles (i.e., to pick them both), and consequently the odds in favor of selecting two green marbles is 1:9. 3 Probability In the previous section we looked at how the likelihood of an event can be expressed as odds. Another and now more common way of expressing likelihood is as a probability. Definition: Probability A probability is a number between 0 and 1, inclusive. How we may reasonably assign such a number, and what it means, are the topics of this section.

10 Probability Discrete Probabilities Simple Discrete Probabilities Recall that the sample space for an experiment is the set of all possible outcomes, and an event is any subset of the sample space. If an experiment (activity, phenomenon) has a finite number of equally likely outcomes, then the probability of any event is simple to calculate: Definition: Simple Discrete Probability Suppose that an experiment has n equally likely outcomes, and that an event E includes a of those outcomes. Then the probability of E, denoted by P(E), is defined by P(E) = a n It is easy to see that for simple discrete probabilities, the probability of an event is directly related to the odds in favor of the event: Theorem If the odds in favor of an event are a:b, then the probability of a the event is a + b. Example: Suppose we flip three fair ( coins. ) Then there are 2 3 = 8 equally likely 3 possible outcomes. Since there are = 3 ways of getting exactly one head, 1 the odds in favor of exactly one head are 3:5. The probability P(1H) of exactly

11 Probability 11 one head is 3 8. Example: Consider the experiment of tossing two fair six-sided dice. What is the probability of rolling a sum of 7? What are the odds in favor? What are the probability and odds for rolling a sum of 5? Of a red three? of rolling 7 and a red three? Answer: We observe that there are 36 equally likely outcomes. Of these, six correspond to a roll of 7. Therefore the probability of rolling a 7 is P(7) = 6 36 = 1 6, and the odds in favor are 1:5. Similarly, P(5) = 4 36 = 1, and the odds in favor 9 of a 5 are 1:8. Since six outcomes correspond to showing three spots on the red die, the probability and odds are the same as for rolling a 7. To roll a 7 and a red three, it is necessary to have the (unique) outcome with three spots on the red and four on the white. Therefore we have P(7 and Red three) = 1, and the odds in 36 favor are 1:35. Suppose, in the previous example, we had asked for the probability of rolling a 5 and doubles simultaneously. Since no outcome meets both conditions, we would have to say that the probability was 0, i.e., 0. On the other hand, if we had asked 36 for the probability of rolling a 2 or greater, every outcome satisfies that condition. In that case, the probability is 36, i.e., 1. So, events that cannot occur (because 36 they don t contain any outcomes) always have probability 0, and events that must occur (because they include every outcome) always have probability 1.

12 Probability Discrete Probability Models In the examples used so far, all of the outcomes are equally likely. However, there are many situations where the outcomes of an event are not equally likely. Consider for example a weighted die, in which the probability of a 6 might be 3 8, and the probability of every other number (1 through 5) is just 1. Then the probability of rolling an even number, for instance, would be 5, while the probability of 8 8 rolling an odd number would be 3 8. Note that the probabilities assigned in the weighted die example sum to 1. This would have to be the case, because we would want the event that includes all possible outcomes to be a certainty, i.e., to have probability 1. This brings us to a very important definition. Definition: Discrete Probability Model. Suppose that an experiment has a finite number of possible outcomes. A probability model for the experiment is an assignment of a number between 0 and 1 to each possible outcome of the experiment, such that the sum of the assigned probabilities is 1. Example: Suppose that a bridge across a stream has three arches over the water. The middle arch (B) is twice as wide as the other two (A and C). Consider the experiment of dropping a stick in the water upstream, and noting which arch it passes under. Assume the likelihood of passing under an arch is determined solely by its width. Find a probability model for this experiment. Answer: If the arches were of equal width the probabilities could each just be one-third, but in this case we need arch B to be twice as likely as arch A, and also twice as likely as arch C, while the likelihoods for arches A and C must be the same. If we divide up the total into four units, one each for A and C and two for

13 Probability 13 B, we see that the probability model should be given by P(A) = P(C) = 1 4, and P(B) = Probability Rules There are three rules (and two definititions) which together provide a nice set of tools for determining the correct probability for any event from a corresonding probability model Complement Rule The fact that all the probabilities in a model must sum to 1 means that if we know the probability P(E) of an event E, then we immediately know the probability of the complement of E. (Recall that the complement of E, denoted E, is the set of objects that are not in E.) Specifically, because P(E) + P(E) = 1, we must have that P(E) = 1 P(E). Probability Rule #1: Complement Rule. Let E be an event and let P(E) be the probability of E. Then, P(E) = 1 P(E). Example: Let E be the event of rolling a sum of 7 on two fair six-sided dice. It was seen in the previous section that P(E) = 1. Consequently, the probability 6 P(E) of rolling anything but a 7 would be given by 1 1 6, which is 5. This is 6 easily verified by observing that there are 30 outcomes (out of 36 total) that do not correspond to a sum of 7 on the dice.

14 Probability Addition Rule Because events in probability are actually sets (of outcomes), it is reasonable to use set notation and operations. If A and B are events, then the union of the two events, A B, would correspond to the compound event A occurs or B occurs or both occur. Similarly, the intersection of the two events, A B, would be the compound event both A and B occur. For example, let A be the event doubles and let B be the event sum of 8 on a roll two fair six-sided dice. We could then consider the probability P(A B) of the compound event doubles or a sum of 8, where A B represents the union of the two events. (That is, an outcome is in the compound event if it is a doubles or an 8 or both.) From the diagram we can count the outcomes in this compound event: there are 6 outcomes corresonding to doubles, and 5 outcomes corresonding to a sum of 8. However, one of these outcomes, a red four with a white four, is both a doubles and an 8, and we ll get an incorrect answer if we count it twice. Consequently, the correct probability for this compound event is 10 out of 36, or As this example shows, we can always find the probability of the union of two events by adding their individual probabilities, but then subtracting the probability that they both occur, since this will have gotten counted twice.

15 Probability 15 Probability Rule #2: Addition Rule. Let A and B be events. Then the probability P(A B) that either A occurs or B occurs (or both) is given by, P(A B) = P(A) + P(B) P(A B) It sometimes happens that two events cannot both occur. For instance, if A is the event roll doubles and B is the event roll a 7, then the probability P(A B) that they both occur is 0. In this case we say that the two events are disjoint. Definition: Disjoint Events. Let A and B be events. If P(A B) = 0 then A and B are said to be disjoint events. It is easy to see that in the case of disjoint events, the addition rule simplifies to being simply the sum of the individual probabilities. For instance, the probability of roll doubles or a 7 is just the sum = Conditional Probability and the Multiplication Rule Suppose two fair dice are rolled, and we know that there are three spots showing on the red die. What is the probability that the sum is 8?

16 Probability 16 We can see from the third row of the diagram that there are six outcomes with three spots on the red die. We can also see that just one of these, red three and white five, corresponds to a sum of 8. Consequently, we would say that the conditional probability of a sum of 8 given a red three is 1 out of 6. In general, if A and B are events, we can find the conditional probability that B occurs given that A occurs by restricting the sample space to the outcomes in A, and then counting how many of those are also in B. This conditional probability is denoted by P(B A). Definition: Conditional Probability. Let A and B be events. The conditional probability of B given A, denoted P(B A), is given by P(B A) = P(A B). P(A) Note that P(B A) is not the same as P(A B), and usually these two well have different values. For example, returning to the example with dice, if A = red three and B = sum of 8 then P(A B) is given by, P(A B) = P(A B) P(B) = = 1 5,

17 Probability 17 whereas, as we saw above, P(B A) = P(A B) P(A) = = 1 6. Although sometimes they may be equal, usually, as in this case, P(A B) P(B A). The definition of conditional probability leads immediately to our third rule, which is really just a restatement of the definition except that we state the probability of the intersection of two events as a product: Probability Rule #3: Multiplication Rule. Let A and B be events. Then the probability P(A B) that both A and B occur is given by, P(A B) = P(A) P(B A) = P(B) P(A B) Example: Suppose a basketball player is 50% likely to make his first shot, and if he makes it then he is 60% likely to make his second shot. What is the probability he makes both shots? Answer: Let A = makes first shot and B = makes second shot. Then since we are given both P(A) and P(B A) we can calculate that P(A B) = P(A) P(B A) = = 0.3 = 30%.

18 Probability Independent Events Consider the events E = doubles and F = three on red on the toss of two fair six-sided dice. It is easy to veryify that P(E) = 1. However, if we assume that 6 we have rolled a three on red, and restrict our sample space to these outcomes, then the probability of doubles is still one in six. That is, P(E F ) = 1 6 = P(E). When this happens we say that E is independent of F, and it will always happen then that F is independent of E too, i.e., P(F E) = P(F ). Definition: Independent Events. Events A and B are said to be independent if P(B A) = P(B). (Equivalently, if P(A B) = P(A) Note that if two events E and F are independent, then the Multiplication Rule is simpler: P(E F ) = P(E) P(F ). Example: Are the events E = roll doubles and F = roll an odd on red independent? Answer: We recall that P(E) = 1 6. Now suppose we know that we have rolled an odd number on the red die. From the diagram it is clear that there are 18 outcomes corresonding to this event, and that of those there are 3 that are also doubles. So

19 Probability 19 P(E F ) = 3 18 = 1. Since P(E) = P(E F ), these events are independent. 6 Example: Suppose that in a certain drawer there are twelve pairs of socks. Three are red, and four are made of wool. One of the wool socks is red. Consider the experiment of selecting one pair of socks at random. Are the events select a red pair and select a wool pair independent? Answer: If we carefully fill in a Venn diagram, we see that the probability of selecting a red pair is 3 in 12, or 1/4, and the probability of selecting a red pair given that it is wool is 1 in 4, also 1/4. Therefore selecting a red pair is independent of selecting a wool pair. Checking it the other way around: the probability of selecting a wool pair is 4 in 12, or 1/3, and the probability of selecting a wool pair given that it is red is 1 in 3, also 1/ Continuous Probability Models In the previous sections we considered probabilities for discrete outcomes, that is, situations in which the number of possible outcomes was fixed and finite. However, some quantities do not vary by jumping from one possibility to the next, but by continuously moving among an indeterminable number of outcomes. The following definition makes this distinction precise.

20 Probability 20 Definition: Continuous variable. If a quantity varies in such a way that between any two possible values one may find other possible values, then the quantity is said to be a continuous variable. A variable that is not continuous is said to be discrete. The three Probability Rules of the previous section will still apply, but we cannot use the methods of discrete probability models to calculate the probabilities. In particular, if a variable can take on an effectively infinite number of possible values, then the probability of any particular value must, at least in most cases, be vanishingly small. Consequently, we will need a new kind of probability model for continuous variables, one that allows us to assign a probability to a range of values rather than to particular values Density Curves We need a way of assigning probabilities to a continuous range of values among the possible outcomes of an experiment, while ensuring that whenever the range of values is divided into one, two, or more segments, then the probabilities for all the individual segments sum to 1. We do this geometrically, by designating an area over the whole range of values to represent the sum of the probabilities. The probability of any part of the range then corresponds to the area over that part. Definition: Density Curve. A density curve is a curve in the Cartesian plane that lies entirely above the horizontal-axis across the range of a continous variable, in such a way that the area between the curve and the horizontal axis is 1.

21 Probability 21 For example, suppose that the outcomes of an experiment are the possible values of a continuous variable whose least value is 0.5 and highest value is 2, and that the outcomes (values) are equally likely to fall anywhere along its range. A probability model for this experiment is displayed below. Note that the curve (in this case a straight line) extends across the range. Also, because the area between it and the horizontal axis (the gray rectangle) must be 1, and the length of the line is 3 2, the height of the line has to be 2 3 (because = 1.) In this model, the probability of an outcome falling in the range 1 to 1.5, which we would denote by P(1 x 1.5), is 1 3. Similarly, P(1 x 2) = = 2 3. In general, for any a and b in the range, we can calculate that P(a x b) = (b a) 2 3. Of course, not all density curves are straight lines. Any shape is possible, even discontinuous curves, provided the area trapped against the horizontal axis across the whole range of possible values is The Normal Distribution The most important of the continuous probability models is the Normal Distribution. This is the distribution of a continuous variable whose values cluster symmetrically around a central value, called the mean. You should expect to see a normal distribution in any situation where there is a typical value, and any variance from that typical value is owing solely to natural variability and not to any other influence. The typical distance from the typical value that other values tend

22 Probability 22 to fall is known as the standard deviation. These two values, the mean and standard deviation, together completely determine a normal distribution. Usually, the mean is denoted by µ (the lower-case Greek letter mu), and the standard deviation is denoted by σ (the lower-case Greek letter sigma). Although there is a precise, mathematical definition of the normal distribution, we will content ourselves in this course with the following: Definition: Normal Distribution. A normal distribution is a bell-shaped density curve that is determined by its mean µ and standard deviation σ. As you see in the representation above, the normal distribution has a bell shape, with the mean µ in the center. The distance σ is the average distance that scores tend to fall from the mean, and it happens that this will always occur where the curvature of the bell shape changes from concave down to concave up, as it begins to flatten out toward the tail. When the standard deviation is small, the bell curve will be tall and narrow. When the standard deviation is large, the bell curve will be squat and wide. In every case, however, the mean is in the center, and the standard deviation is the average distance from the mean, at the point where the curvature changes.

23 Probability 23 For a variable to be normally distributed, there can be nothing influencing the values apart from the attraction of the mean and random variability. Any other factor will skew or otherwise distort the distribution so that it is no longer symmetric and/or no longer bell-shaped. Consequently, although many variables are normally distributed, many are not. Example: Which of the following are likely to be normally distributed? 1. Length that a kangaroo can jump. 2. Salaries of professional football players. 3. Student scores on an exam. 4. Heights of college women. 5. Heights of spectators at a sporting event. Answer: The distance a kangaroo can jump is likely to be normally distributed, because the only influence on the distance is natural variability in the size and vigor of the kangaroos. The salaries of athletes, however, is strongly influenced by the celebrity status of a small number of athletes, who make salaries often many times larger than the average salary, so this would not be a normally distributed variable. Student scores on an exam are influenced by study habits. Those with good study habits will have scores that make a high hump in the distribution, and those with poor study habits make a low hump, resulting in what is called a bimodal distribution. The heights of college women is normally distributed. However, the heights of spectators at a sporting event will not be normal, but, like the student scores, bimodal. (Why?) The Rule No matter what the mean and standard deviation are, the values of a normal distribution form a predictable pattern, always with the same proportion of scores clustered around the middle and the same proportion farther from the middle, in each tail. Specifically, we have:

24 Probability 24 The Rule. For any normally distributed variable, about 68% of the population will have values that fall within 1 standard deviation of the mean, about 95% will have values that fall within 2 standard deviations, and about 99.7% will have values that fall within 3 standard deviations. Thus, the majority of scores just over two-thirds fall withing the average distance from the mean, and only a very, very small number, about three-tenths of a percent, will fall farther than three times the average distance from the mean. Example: Most people blink several times per minute, and the time between blinks is normally distributed with a mean of 6 seconds and a standard deviation of 2 seconds. What proportion of all blinks occur between 4 and 8 seconds apart? What proportion occur more than 8 seconds apart? What proportion occur more than 10 seconds apart?

25 Probability 25 Answer: Because the mean is 6 seconds and the standard deviation (stdv) is 2 seconds, then 4 seconds is one stdv below the mean and 8 seconds is one stdv above the mean, so the Rule tells us that about 68% of all intervals between blinks fall within this range. To answer the second question, note that when there is 68% between 4 and 8 seconds, there must be 32% that are either less than 4 or greater than 8. Since the normal distribution is symmetric, exactly half these, or 16%, must be greater than 8. Similar reasoning shows that because 10 seconds is two stdv above the mean, and 95% of all intervals between blinks fall within two stdev, then half of 5% or just 2.5% of intervals between blinks are greater than 10 seconds The Standard Score (z-score) Because the area under the normal curve within a given distance of the mean as measured by standard deviations is the same regardless of the actual value of the mean or the standard deviation, so too the frequency of values that may be expected in a normal distribution to fall within a given number of standard deviations is not dependent on the actual value of the mean or standard deviation. Consequently, it is useful to consider a standard normal distribution, one with mean equal to 0 and standard deviation equal to 1, in which the value of a variable is exactly its distance from the mean as measured in standard deviations. The variable for the standard normal distribution is denoted by z, and any normal distribution may be standardized by replacing each value of the variable with its distance from the mean in standard deviations what is called its z-score.

26 Probability 26 Definition: The Standard Score (z-score). Suppose x is a normally distributed variable with mean µ and standard deviation σ. Then given any value x 0 of the variable, the z-score of x 0 is given by. z = x 0 µ σ Example: Suppose heights of college men is a normally distributed variable with mean µ = 69.5 inches and standard deviation σ = 2.5 inches. If Mike (a college male) is 72 inches tall, what is his z-score for height? Answer: We observe that 72 inches is 2.5 inches taller than the mean 69.5, and 2.5 inches is 1 standard deviation above the mean, so Mike s z-score is z = 1. Example:If in the previous example Mike were 66 inches tall, what would be his z-score for height? Answer: We observe that 66 inches is 2.5 inches shorter than the mean 69.5, so 1 standard deviation below the mean. Since it is in the negative direction from the mean, Mike s z-score is z = 1. Example: If in the previous example Mike were 71 inches tall, what would be his z-score for height? Answer: Since 71 isn t a whole number of standard deviations from the mean, we use the formula: z = 71 µ σ = = = 3 5 = The Cumulative Distribution Function Recall that the normal distribution is an example of a continuous probability model, in which the probability of a variable falling within a range of values is

27 Probability 27 the area under the bell curve in that range. It happens that the probabilities are completely determined by the z-scores of the values. These probabilities can always be calculated using the cumulative distribution function. Definition: The Cumulative Distribution Function. Let z 0 be a score (value) on the standard normal distribution. Then the value of the Cumulative Distribution Function of z 0, denoted cdf(z 0 ), is the area under the curve from to z 0, corresponding to the probability that a randomly selected individual will have a z-score in that range. In other words, the cumulative distribution of z 0 is the probability of a score being less than or equal to z 0, and corresponds to the area under the curve to the left of z 0. Given a z-score z 0, we may find the value of cdf(z 0 ) either by using a statistical table or by using a calculator or computer app. A suitable app can be found at the bottom of the Platonic Realms article on the Normal Distribution, at the following link: Just click on the line between the dark and light areas under the normal curve to change the z. The proportion to the left that is shaded, the dark area corresponding to cdf(z), is displayed at the lower right of the graph. Note that when z = 0 it is in the exact middle, corresponding to when the value is equal to the mean, and the proportion to the left is exactly 0.5. Drag it to the right, and the z-value becomes a positive number; to the left it is negative. For positive z s the proportion is greater than 0.5, for negative z s it is smaller than 0.5. Example: What is the value of cdf(1)? Answer: A z of 1 corresponds to being 1 standard deviation above the mean. From our previous work with the Rule we know that the area to the left of this should be about 84%. Using the app we linked to above and carefullly sliding the line to z = 1, we see that the proportion is actually 0.841, or about 84.1%.

28 Probability 28 Example: Going back to our example with height of college men, what is the probability that a randomly selected college male will be no more than 71 inches tall? Answer: We noted above that the corresponding z-score (see Section 3.3.4) is 0.6. From the Cumulative Distribution Function app we can observe that the area to the left of z = 0.6 is approximately 72.6% Since this area covers precisely those values for heights of no more than 71 inches, the probability that a randomly selected college male will not be taller than 71 inches is 72.6%. Example: What is the probability that a randomly selected college male will be between 69 and 71 inches tall?

29 Probability 29 Answer: For this question we need to calculate both z-scores. We already know the z-score for 71 inches is 0.6, and the z-score for 69 inches is z = = = 0.2 Using the app, we find that cdf(0.6)=72.6%, and cdf(-0.2)=42.1%. Since what we want is the area between these scores, we subtract 42.1 from 72.6 to find that the probability P(69 < x < 71) of a randomly selected college male being between 69 and 71 inches tall is 30.5% Sampling Distributions and the Central Limit Theorem We can only apply the methods we learned in the previous two sections to normal distributions, and unfortunately most variables that interest us are not normally distributed. However, it turns out that distributions that are far from normal, even heavily skewed, can be analyzed using the standard normal, provided the population is large and we can take random samples from it.

30 Probability 30 Definition: The Sampling Distribution. If x is a continuous random variable, then the sampling distribution of sample means, for samples of a fixed size n, is the distribution of the means (arithmetic averages) of all the possible samples of size n. Suppose for example that we wanted to study the distribution of the variable x = height of a college man. We could take random samples of, say, 20 men at a time, and get the average height for each sample. We would denote such averages by x. For one sample, the mean might be x = Another might give us x = 70.8, and so on. If we consider the distribution of all the averages we would get from all the possible samples of 20 college men at a time, that would be the sampling distribution of sample means. It turns out that if the population is large and the samples are of a well-chosen size, not too small but still much smaller than the population size, then the sampling distribution will be very close to normal even if the variable itself isn t normally distributed in the population. This very important fact is known as The Central Limit Theorem. The Central Limit Theorem If x is a continuous random variable, with mean µ and standard deviation σ, then the sampling distribution of sample means x for samples of size n is approximately normal, with the same mean µ and a standard deviation equal to σ n. Example: Suppose the variable height of a college man has a mean µ = 69.5 inches and a standard deviation of σ = 2.7 inches. Let x be the variable average height of a sample of 20 college men. What is the mean and standard deviation of x?

31 Probability 31 Answer: In this case we are considering the sampling distribution of sample means of size 20. The mean for the sampling distribution is the same as the mean for the underlying variable, in this case 69.5 inches. The standard deviation of the sampling distribution is the standard deviation of the underlying variable divided by the square root of the sample size. Thus, the standard deviation of x is Example: Again suppose the variable height of a college man has a mean µ = 69.5 inches and a standard deviation of σ = 2.7 inches, but suppose this time x is the variable average height of a sample of n college men. How big would the sample size n need to be to ensure that the standard deviation of the sampling distribution was no more than one quarter inch? Answer: We know that the standard deviation of the sampling distribution is given by the formula σ n, so we can set up the following equation: 2.7 n = 0.25 Solving this for n gives n = ( ) Therefore, the sample size would need to be at least 117, that is, we would need to sample 117 men at a time to ensure that our sample mean x had a standard deviation of no more than 0.25 inches Interval Estimation It often happens that we don t know the average value of a variable, but that we can make a reasonable estimate of its standard deviation, i.e., of its variability. In this case we can use the fact that the sampling distribution of sample means is normally distributed to choose an interval in which the average is likely to lie. Here s how it works. Suppose we don t know the average height of college men, or even if the height is a normally distributed variable, but that we have good reason

32 Probability 32 to believe that the standard deviation is 2.7 inches. We know that the sampling distribution is normally distributed, with the same average as the variable (in this case the average height of college men). Consequently, if we take a sample of size 20, we know from the example in the previous section that the standard deviation of the sampling distribution is approximately 0.6. From all this we can deduce that the mean x that we get from our sample is 95% likely to fall within two standard deviations, or 1.2 inches, of the real average height. Now, we take a sample of size 20 and get an average height for our sample of, say, 69 inches. We can deduce that the real average is 95% likely to lie between 67.8 inches and 70.2 inches. This interval, (67.8, 70.2), is the 95% confidence interval for the average height of college men. Definition: 95% Confidence Interval Estimate. If x is a continuous variable with an unknown mean µ and a known standard deviation σ, then the 95% Confidence Interval Estimate for µ from a sample of size n with sample mean x is ( x 2σ, x + 2σ ) n n The center of the interval estimate is always the sample mean, and the distance 2σ from the middle to either end of the interval,, is called the margin of error. n Example: Suppose we want to know the average length of time x that students spend watching televised sports each week. (Note, this is not a normally distributed variable.) Assume that the standard deviation is one half hour. We take a sample of 45 students and find that the average time they spend watching televised sports is 2.75 hours. Calculate a 95% confidence interval estimate of the average time students spend watching televised sports each week. Answer: Here our sample mean x is 2.75, the standard deviation σ is 0.5, and the

33 Probability 33 sample size n is 45. Consequently the 95% confidence interval estimate is given by ( x 2σ, x + 2σ ) ( = , ) = (2.6, 2.9) n n So we are 95% confident that the real average for all students is between 2 hours 36 minutes and 2 hours 54 minutes. Example: What was the margin of error in the example worked above? How large a sample would be required to reduce this margin of error to 5 minutes? Answer: The margin of error for a 95% confidence interval estimate is given by 2σ n, which in this problem worked out to approximately 0.15 hours, i.e., about 9 minutes: To get it down to 5 minutes, we would need to set the margin of error equal to 1 12 (because 5 minutes is one-twelfth of an hour), and solve for the sample size: Solving for n: n = 1 12 ( ) n = = Therefore a sample size of 144 students would be needed to reduce the margin of error of our 95% confidence interval estimate to 5 minutes.