2. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Chapter 6 Probability Distributions Section 6.1 Summarizing Possible Outcomes and Their Probabilities

3. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 3 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 Randomness

4. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 4 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 Random Variable

5. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 5 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. Probability Distribution

6. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 6 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. Probability Distribution of a Discrete Random Variable

7. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 7 What is the estimated probability of at least three home runs? 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

8. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 8 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 Table 6.1 Probability Distribution of Number of Home Runs in a Game for San Francisco Giants Example: Number of Home Runs in a Game

9. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 9 The mean of a probability distribution for a discrete random variable is: 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). Mean of a Discrete Probability Distribution

10. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 10 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. Expected Value of X

11. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 11 Find the mean of this probability distribution. 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

12. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 12 The mean: = 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 Example: Number of Home Runs in a Game

13. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 13 The standard deviation of a probability distribution, denoted by the parameter,, measures variability from the mean.  Larger values of correspond to greater spread.  Roughly, describes how far the random variable falls, on the average, from the mean of its distribution. The Standard Deviation of a Probability Distribution

14. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 14  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. Continuous Random Variable

15. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 15 A continuous random variable has possible values that form an interval. Its probability distribution is specified by a curve.  Each interval has probability between 0 and 1.  The interval containing all possible values has probability equal to 1. Probability Distribution of a Continuous Random Variable

16. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 16 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. Smooth curve approximation Probability Distribution of a Continuous Random Variable

17. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Chapter 6 Probability Distributions Section 6.2 Probabilities for Bell-Shaped Distributions

18. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 18 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. Normal Distribution

19. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 19 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. Normal Distribution

20. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 20 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? Normal Distribution

21. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 21 ≈ 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. Normal Distribution: 68-95-99.7 Rule for Any Normal Curve 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?

22. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 22 Heights of adult women can be approximated by a normal distribution, inches; inches 68-95-99.7 Rule for women’s heights: 68% are between 61.5 and 68.5 inches  95% are between 58 and 72 inches  99.7% are between 54.5 and 75.5 inches  Example : 68-95-99.7% Rule

23. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 23 The z-score for a value x of a random variable is the number of standard deviations that x falls from the mean. 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. Z-Scores and the Standard Normal Distribution

24. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 24 A standard normal distribution has mean and standard deviation. 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. Z-Scores and the Standard Normal Distribution

25. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 25 Table A enables us to find normal probabilities.  It tabulates the normal cumulative probabilities falling below the point. 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. Table A: Standard Normal Probabilities

26. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 26 Find the probability that a normal random variable takes a value less than 1.43 standard deviations above ; Example: Using Table A

27. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 27 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, so 0.9236 is the probability less than 1.43 standard deviations above the mean of any normal distribution (that is, below ). The complement probability of 0.0764 is the probability above in the right tail. Example: Using Table A

28. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 28 Find the probability that a normal random variable assumes a value within 1.43 standard deviations of.  Probability below  Example: Using Table A

29. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 29 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. How Can We Find the Value of z for a Certain Cumulative Probability?

30. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 30 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 to. Thus, the probability that a normal random variable falls at least 1.96 standard deviations below the mean is 0.025. How Can We Find the Value of z for a Certain Cumulative Probability?

31. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 31  If we’re given a value x and need to find a probability, convert x to a z-score using, 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. SUMMARY: Using Z-Scores to Find Normal Probabilities or Random Variable x Values

32. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 32 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 and and 30 on the ACT which has and. How can we compare these scores to tell which score is relatively higher? Example: Comparing Test Scores That Use Different Scales

33. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 33  Compare z-scores: SAT: ACT: Since your z-score is greater for the ACT, you performed relatively better on this exam. Using Z-scores to Compare Distributions

34. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. Chapter 6 Probability Distributions Section 6.3 Probabilities When Each Observation Has Two Possible Outcomes

35. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 35 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. The Binomial Distribution: Probabilities for Counts with Binary Data

36. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 36  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.  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. Conditions for the Binomial Distribution

37. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 37 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:

38. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 38 Rules for factorials:  For example,  Factorials

39. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 39 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. Example: An ESP Experiment

40. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 40 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.  Each of these has probability:.  The total probability of two correct guesses is. Example: An ESP Experiment

41. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 41 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. Example: An ESP Experiment

42. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 42 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: 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. Example: Testing for Gender Bias in Promotions

43. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 43 Before using the binomial distribution, check that its three conditions apply:  Binary data (success or failure)  The same probability of success for each trial (denoted by p)  Independent trials Example: Testing for Gender Bias in Promotions

44. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 44 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. Example: Testing for Gender Bias in Promotions

45. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 45 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: Mean and Standard Deviation of the Binomial Distribution

46. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 46 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? Example: Checking for Racial Profiling

47. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 47 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: Example: Checking for Racial Profiling

48. Copyright © 2013, 2009, and 2007, Pearson Education, Inc. 48 Recall: Empirical Rule  When a distribution is bell-shaped, close to 100% of the observations fall within 3 standard deviations of the mean: 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 non- white. 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. Example: Checking for Racial Profiling