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Chapter 6 Probability Distributions Section 6.1 Summarizing Possible Outcomes and Their Probabilities Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Randomness A random variable is a numerical measurement of the outcome of a random phenomenon. Often, the randomness results from selecting a random sample for a population or performing a randomized experiment 3 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Random Variable Use letters near the end of the alphabet, such as x, to symbolize: variables a particular value of the random variable Use a capital letter, such as X, to refer to the random variable itself. Example: Flip a coin three times X= number of heads in the 3 flips; defines the random variable x=2; represents a possible value of the random variable 4 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Probability Distribution The probability distribution of a random variable specifies its possible values and their probabilities. Note: It is the randomness of the variable that allows us to specify probabilities for the outcomes. 5 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Probability Distribution of a Discrete Random Variable A discrete random variable X takes a set of separate values (such as 0,1,2,…) as its possible outcomes. Its probability distribution assigns a probability P(x) to each possible value x: For each x, the probability P(x) falls between 0 and 1. The sum of the probabilities for all the possible x values equals 1. 6 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Example: Number of Home Runs in a Game What is the estimated probability of at least three home runs? Table 6.1 Probability Distribution of Number of Home Runs in a Game for San Francisco Giants 7 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Example: Number of Home Runs in a Game Table 6.1 Probability Distribution of Number of Home Runs in a Game for San Francisco Giants The probability of at least three home runs in a game is P(3)+P(4)+P(5 or more)= 0.0556 + 0.0185 + 0 = 0.0741 8 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Mean of a Discrete Probability Distribution The mean of a probability distribution for a discrete random variable is: x p(x) where the sum is taken over all possible values of x. The mean of a probability distribution is denoted by the parameter, . The mean is a weighted average; values of x that are more likely receive greater weight P(x). 9 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Expected Value of X The mean of a probability distribution of a random variable X is also called the expected value of X. The expected value reflects not what we’ll observe in a single observation, but rather what we expect for the average in a long run of observations. It is not unusual for the expected value of a random variable to equal a number that is NOT a possible outcome. 10 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Example: Number of Home Runs in a Game Find the mean of this probability distribution. Table 6.1 Probability Distribution of Number of Home Runs in a Game for San Francisco Giants 11 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Example: Number of Home Runs in a Game The mean: x p(x) = 0(0.3889) + 1(0.3148) + 2(0.2222) + 3(0.0556) + 4(0.0185) = 0 * P(0) + 1 * P(1) + 2 * P(2) + 3 * P(3) + 4 * P(4) =1 12 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. The Standard Deviation of a Probability Distribution The standard deviation of a probability distribution, denoted by the parameter, , measures variability from the mean. 13 Larger values of correspond to greater spread. Roughly, describes how far the random variable falls, on the average, from the mean of its distribution. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Continuous Random Variable A continuous random variable has an infinite continuum of possible values in an interval. Examples are: time, age and size measures such as height and weight. Continuous variables are usually measured in a discrete manner because of rounding. 14 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Probability Distribution of a Continuous Random Variable A continuous random variable has possible values that form an interval. Its probability distribution is specified by a curve. 15 Each interval has probability between 0 and 1. The interval containing all possible values has probability equal to 1. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Probability Distribution of a Continuous Random Variable Smooth curve approximation Figure 6.2 Probability Distribution of Commuting Time. The area under the curve for values higher than 45 is 0.15. Question: Identify the area under the curve represented by the probability that commuting time is less than 15 minutes, which equals 0.29. 16 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Chapter 6 Probability Distributions Section 6.2 Probabilities for Bell-Shaped Distributions Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Normal Distribution The normal distribution is symmetric, bell-shaped and characterized by its mean and standard deviation . The normal distribution is the most important distribution in statistics. Many distributions have an approximately normal distribution. The normal distribution also can approximate many discrete distributions well when there are a large number of possible outcomes. Many statistical methods use it even when the data are not bell shaped. 18 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Normal Distribution Normal distributions are Bell shaped Symmetric around the mean The mean ( ) and the standard deviation ( ) completely describe the density curve. Increasing/decreasing moves the curve along the horizontal axis. Increasing/decreasing controls the spread of the curve. 19 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Normal Distribution Within what interval do almost all of the men’s heights fall? Women’s height? Figure 6.4 Normal Distributions for Women’s Height and Men’s Height. For each different combination of and values, there is a normal distribution with mean and standard deviation . Question: Given that = 70 and = 4, within what interval do almost all of the men’s heights fall? 20 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Normal Distribution: 68-95-99.7 Rule for Any Normal Curve ≈ 68% of the observations fall within one standard deviation of the mean. ≈ 95% of the observations fall within two standard deviations of the mean. ≈ 99.7% of the observations fall within three standard deviations of the mean. Figure 6.5 The Normal Distribution. The probability equals approximately 0.68 within 1 standard deviation of the mean, approximately 0.95 within 2 standard deviations, and approximately 0.997 within 3 standard deviations. Question: How do these probabilities relate to the empirical rule? 21 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Example : 68-95-99.7% Rule Heights of adult women can be approximated by a normal distribution, 65 inches; 3.5 inches 68-95-99.7 Rule for women’s heights: 68% are between 61.5 and 68.5 inches [ 65 3.5] 95% are between 58 and 72 inches [ 2 65 2(3.5) 65 7] 99.7% are between 54.5 and 75.5 inches [ 3 65 3(3.5) 65 10.5] 22 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Z-Scores and the Standard Normal Distribution The z-score for a value x of a random variable is the number of standard deviations that x falls from the mean. z x A negative (positive) z-score indicates that the value is below (above) the mean. Z-scores can be used to calculate the probabilities of a normal random variable using the normal tables in Table A in the back of the book. 23 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Z-Scores and the Standard Normal Distribution A standard normal distribution has mean 0 standard deviation 1 . and When a random variable has a normal distribution and its values are converted to z-scores by subtracting the mean and dividing by the standard deviation, the z-scores follow the standard normal distribution. 24 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Table A: Standard Normal Probabilities Table A enables us to find normal probabilities. It tabulates the normal cumulative probabilities falling below the point z . To use the table: Find the corresponding z-score. Look up the closest standardized score (z) in the table. First column gives z to the first decimal place. First row gives the second decimal place of z. The corresponding probability found in the body of the table gives the probability of falling below the z-score. 25 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Example: Using Table A Find the probability that a normal random variable takes a value less than 1.43 standard deviations above ; P( z 1.43) 0.9236 26 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Example: Using Table A Figure 6.7 The Normal Cumulative Probability, Less than z Standard Deviations above the Mean. Table A lists a cumulative probability of 0.9236 for z 1.43, so 0.9236 is the probability less than 1.43 standard deviations above the mean of any normal distribution (that is, below 1.43 ). The complement probability of 0.0764 is the probability above 1.43 in the right tail. 27 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Example: Using Table A Find the probability that a normal random variable assumes a value within 1.43 standard deviations of . 28 Probability below 1.43 0.9236 Probability below 1.43 0.0764 P(1.43 z 1.43) 0.9236 0.0764 0.8472 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. How Can We Find the Value of z for a Certain Cumulative Probability? To solve some of our problems, we will need to find the value of z that corresponds to a certain normal cumulative probability. To do so, we use Table A in reverse. Rather than finding z using the first column (value of z up to one decimal) and the first row (second decimal of z). Find the probability in the body of the table. The z-score is given by the corresponding values in the first column and row. 29 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. How Can We Find the Value of z for a Certain Cumulative Probability? Example: Find the value of z for a cumulative probability of 0.025. Look up the cumulative probability of 0.025 in the body of Table A. A cumulative probability of 0.025 corresponds zto 1.96 . Thus, the probability that a normal random variable falls at least 1.96 standard deviations below the mean is 0.025. 30 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. SUMMARY: Using Z-Scores to Find Normal Probabilities or Random Variable x Values If we’re given a value x and need to find a probability, convert x to a z-score using z ( x ) / , use a table of normal probabilities (or software, or a calculator) to get a cumulative probability and then convert it to the probability of interest If we’re given a probability and need to find the value of x , convert the probability to the related cumulative probability, find the z-score using a normal table (or software, or a calculator), and then evaluate x z . 31 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Example: Comparing Test Scores That Use Different Scales Z-scores can be used to compare observations from different normal distributions. Picture the Scenario: There are two primary standardized tests used by college admissions, the SAT and the ACT. You score 650 on the SAT which has 500 and 100 and 30 on the ACT which has 21.0 and 4.7. How can we compare these scores to tell which score is relatively higher? 32 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Using Z-scores to Compare Distributions Compare z-scores: SAT: z 650 500 1.5 100 30 21 1.91 ACT: z 4.7 Since your z-score is greater for the ACT, you performed relatively better on this exam. 33 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Chapter 6 Probability Distributions Section 6.3 Probabilities When Each Observation Has Two Possible Outcomes Copyright © 2013, 2009, and 2007, Pearson Education, Inc. The Binomial Distribution: Probabilities for Counts with Binary Data We use the binomial distribution when each observation is binary: it has one of two possible outcomes. Examples: Accept or decline an offer from a bank for a credit card. Have or do not have health insurance. Vote yes or no on a referendum. 35 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Conditions for the Binomial Distribution Each of n trials has two possible outcomes: “success” or “failure”. Each trial has the same probability of success, denoted by p, so the probability of a failure is denoted by 1 p . The n trials are independent, That is, the result for one trial does not depend on the results of other trials. The binomial random variable X is the number of successes in the n trials. 36 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Probabilities for a Binomial Distribution When the number of trials n is large, it’s tedious to write out all the possible outcomes in the sample space. There is a formula you can use to find binomial probabilities for any n. Denote the probability of success on a trial by p. For n independent trials, the probability of x successes equals: n! P(x) p x (1 p) nx , x 0,1,2,...,n x!(n - x)! 37 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Factorials Rules for factorials: n! n *(n 1) *(n 2)...2*1 1! 1 0! 1 For example, 4! 4*3* 2*1 24 38 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Example: An ESP Experiment John Doe claims to possess extrasensory perception (ESP). An experiment is conducted: A person in one room picks one of the integers 1, 2, 3, 4, 5 at random. In another room, John Doe identifies the number he believes was picked. Three trials are performed for the experiment. John Doe got the correct answer twice. 39 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Example: An ESP Experiment If John Doe does not actually have ESP and is actually guessing the number, what is the probability that he’d make a correct guess on two of the three trials? The three ways John Doe could make two correct guesses in three trials are: SSF, SFS, and FSS. 2 Each of these has probability: (0.2) (0.8) 0.032 . The total probability of two correct guesses is 3(0.032) 0.096. 40 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Example: An ESP Experiment The probability of exactly 2 correct guesses is the binomial probability with n = 3 trials, x = 2 correct guesses and p = 0.2 probability of a correct guess. 3! 2 1 P(2) (0.2) (0.8) 3(0.04)(0.8) 0.096 2!1! 41 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Example: Testing for Gender Bias in Promotions A group of female employees has claimed that female employees are less likely than male employees of similar qualifications to be promoted. Of 1000 employees, 50% are female. None of the 10 employees chosen for management training were female. The probability that no females are chosen is: 10! P(0) (0.50)0 (0.50)10 0.001 0!10! It is very unlikely (one chance in a thousand) that none of the 10 selected for management training would be female if the employees were chosen randomly. 42 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Example: Testing for Gender Bias in Promotions Before using the binomial distribution, check that its three conditions apply: 43 Binary data (success or failure) The same probability of success for each trial (denoted by p) Independent trials Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Example: Testing for Gender Bias in Promotions Do the Binomial Conditions Apply? The data are binary (male, female). If employees are selected randomly, the probability of selecting a female on any given trial is 0.50. With random sampling of 10 employees from a large population, the outcome for one trial doesn’t depend on the outcome of another trial. 44 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Mean and Standard Deviation of the Binomial Distribution Binomial Mean and Standard Deviation: The binomial probability distribution for n trials with probability p of success on each trial has mean and standard deviation given by: np, 45 np(1-p) Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Example: Checking for Racial Profiling In 2006, the New York City Police Department (NYPD) confronted approximately 500,000 pedestrians for suspected criminal violations. 88.9% were non-white. Meanwhile, according to the 2006 American Community Survey conducted by the U.S. Census Bureau, of the more than 8 million individuals living in New York City, 44.6% were white. Are the data presented above evidence of racial profiling in police officers’ decisions to confront particular individuals? 46 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Example: Checking for Racial Profiling Assume: 500,000 confrontations as n = 500,000 trials P(driver is non-white) is p = 0.554 Calculate the mean and standard deviation of this binomial distribution: 500,000(0.554) 277,000 500,000(0.554)(0.446) 351 47 Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Example: Checking for Racial Profiling Recall: Empirical Rule When a distribution is bell-shaped, close to 100% of the observations fall within 3 standard deviations of the mean: u - 3 277,000 - 3(351) 275,947 3 277,000 3(351) 278,053 If no racial profiling is taking place, we would not be surprised if between about 275,947 and 278,053 of the 500,000 people stopped were nonwhite. However, 88.9% of all stops, or 500,000(0.889) = 444,500 involved non-whites. This suggests that the number of non-whites stopped is much higher than we would expect if the probability of confronting a pedestrian were the same for each resident, regardless of their race. 48 Copyright © 2013, 2009, and 2007, Pearson Education, Inc.