1. Introduction: Random Variables What are random variables? Intro to probability distributions Discrete Continuous

2. What are random variables? A Random variable, X, represents a quantitative variable that is measured or observed in an experiment. The value that X takes on in a given experiment is a random outcome. Counting the number of defective light bulbs in a case of bulbs Measuring daily rainfall in inches Measuring the average depression score of computer science majors

3.  Sample means, standard deviations, proportions, and frequencies are all Random variables.  Ex:Distribution of a sample mean  Say you have 100 people and you select 20 people randomly from the 100. If you selected 20 people randomly over and over again and recorded the mean age of each group, you could compare the means of each sample.  Since the mean age is different for each sample, Sample Means are Random Variables.

4. Two types of random variables The observations can take only a finite, countable number of values. The observations can take on any of the countless number of values in an interval The number of heads in four coin tosses The number of anorexics in a random sample of 500 people The average response time of a random sample of 200 depressed patients The average IQ of a random sample of 22 statistics students

5. In general, averages are continuous and counts are discrete. The average anger response The number of juvenile delinquents

6. What is a probability distribution?  Tells you the probability that a random variable (frequency, sample mean, etc.) will either BE a certain value or fall in a certain interval or range.  Can either be: The probabilities associated with each specific value of the RV. The probabilities associated with a range of values of the RV.

7. 5.1 and 5.2

8. Two balls are randomly chosen from an urn of blue and red balls. We win $1 for every blue and lose $1 for every red. Let X = our total winnings. X is a random variable taking on one of the values -2, 0, 2 Discrete Probability Distributions- The Sample Space Suppose that we toss three coins. Let X = the number of heads appearing. X is a random variable taking on one of the values 0,1,2,3

9. The probability distribution of X lists the values in the sample space and their associated probabilities.  In Chapter 4, we learned that the relative frequencies obtained from and experiment or a sample space can be used to approximate probabilities. However, when the relative frequencies represent the population, they give the actual (theoretical) probabilities of the outcomes. Suppose that we toss a fair die. Let X = the outcome of the toss. X is a random variable taking on one of the values 1, 2, 3, 4, 5, 6 x i p i 1 1/6 2 1/6 3 1/6 4 1/6 5 1/6 6 1/6 Probability Distributions

10. probability outcome x i p i 1 1/6 2 1/6 3 1/6 4 1/6 5 1/6 6 1/6 Probability Distributions  The probability distribution of X lists the values in the sample space and their associated probabilities. Again, the relative frequencies we did in Chapter 4 are the probabilities of an outcome, if they represent the POPULATION.

11.  Probability Distribution  Values, x Probabilities, p ( x )  01/4 =.25  12/4 =.50  21/4 =.25 Experiment: Toss 2 Coins. Count # Tails. © 1984-1994 T/Maker Co.

12. Suppose that we toss two coins. Let X = the number of heads. Make the probability distribution and a bar graph. We can have outcomes: HH, TT, HT, or TH. x i p i 0 1/4 1 2/4 2 1/4 probability outcome Probability Distributions

13. Cryptography: Frequencies of letters in a 1000 letter sample x i p i A73/1000 B9/1000 C30/1000 D44/1000 E130/1000 F28/1000 probability outcome Sometimes you can estimate discrete probability distributions using a really large sample, but again, relative frequencies approximate probabilities unless they represent the WHOLE population. Probability Distributions

14.  0  p(x)  1   p(x) = 1 xP(x) 7.70 8.50 9-.20 Although the sum of all probabilities is equal to 1, we can not have a negative probability. Therefore, this is NOT a valid probability distribution.

15.  Quality Control for DVDs When manufacturing DVDs for Sony, batches of DVDs are randomly selected and the number of defects x is found for each batch. SOLUTION:Total of P(x) = 0.977  1.00 and therefore this is not a “valid” probability distribution. SOLUTION:Total of P(x) = 1.00 and therefore this is a “valid” probability distribution. Prior Sentences When randomly selecting a jail inmate convicted of DWI (driving while intoxicated), the probability distribution of the number x of prior DWI sentences is as described in the accompanying table (based on data from the U.S. Department of Justice).

16.  A) Present this probability distribution graphically.  B) Find the probability that the number of breakdowns for this machine during a given week is › Exactly 2 › P (x=2) =.35 › 0 to 2 › P( ) = P (x = 0) + P (x = 1) + P (x = 2) =.15 +.20 +.35 =.70 › More than 1 › P ( x > 1) = P (x = 2) + P ( x = 3) =.35 +.30 =.65 › At most 1 › P ( ) = P ( x = 0) + P ( x = 1) = 15 +.20 =.35 Breakdowns per week 0123 Probability.15.20.35.30

18.  According to a survey, 60% of all students suffer from math anxiety. 2 students are selected at random. Let x denote the # of students in this sample who suffer from math anxiety. Develop the probability distribution of x.  Let N= the student does NOT suffer  M = the student DOES suffer

19.  P ( x = 0) = P (NN) =.16  P ( x = 1) = P (NM or MN) = P(NM) + P (MN) =.24 +.24 =.48  P ( x = 2) = P (MM) =.36 Thus we have : xP (x) 0.16 1.48 2.32

20.  Page 196 (2, 3)  Page 202 (8-14)

21. The mean of a discrete probability distribution (called the “expected value”) can be found using this formula It is a weighted average of the possible values of X, each value being weighted by its probability of occurrence. 5.3 Expected Values

22. x i p i 1 1/6 2 1/6 3 1/6 4 1/6 5 1/6 6 1/6 Suppose that we toss a fair die. Let X = the outcome of the toss. X is a random variable taking on one of the values 1, 2, 3, 4, 5, 6 Expected Values What is the expected value?

23. Suppose we draw one marble out of a bowl containing 3 green and 7 black marbles. We win $10 if we draw a green marble but we lose $2 if we draw a black marble. Let X = our winnings. What is the expected value of X? Should you play this game? x i p i 10 -2 3/10 7/10 Expected Values

24. The standard deviation of a discrete probability distribution can be found using this formula Standard deviation measures the spread of a discrete variable x’s probability distribution. 5.4 Standard Deviation of a Discrete Random Variable

25. Remember μ = 3.5 What is σ = ? Suppose that we toss a fair die. Let X = the outcome of the toss. X is a random variable taking on one of the values 1, 2, 3, 4, 5, 6 xP (x)x*P (x)x2x2 X 2 *P(x) 11/61*1/6=1/61 21/62 * 1/6 =2/644 * 1/6 = 4/6 31/63 *1/6 = 3/699 * 1/6 = 9/6 41/6 5 6 Total

26. μ x = 3.5 σ = ? Suppose that we toss a fair die. Let X = the outcome of the toss. X is a random variable taking on one of the values 1, 2, 3, 4, 5, 6

27. Example 2: Suppose that we toss a fair coin. Let X = the random variable of tossing a HEADS. Make a probability distribution for X. What is the expected value of X? What is the standard deviation? μ = ? σ = ? xP (x)x*P (x)x2x2 X 2 *P(x) 0 1

28. H OMEWORK P. 210 #s 23 and 24

29. 5.5 Factorials and Combinations Factorials n! = n * (n – 1) * (n – 2) * (n – 3)… 3 * 2 * 1 and0 ! = 1 Examples: 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5,040 (12 – 4) ! = 8 ! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 40, 320 (5 – 5) ! = 0 ! = 1 Appendix C Table II p. C6 has list of all factorials. You can also use calculator to do factorials.

30. Combinations Combinations- tells you how many different ways you can select k elements from n total elements. Notation and Formula- “The number of combinations of n elements selected k at a time.”

31. Example You want to buy 2 flavors of ice cream, but you have 6 flavors to choose from. How many possible combinations are there for you to order? n = 6 possible flavors k = 2 since you are choosing 2 at a time. Thus, there are 15 ways to choose 2 different flavors out of 6.

32. Some more examples… 2.) Use Appendix C Table III to find the number of ways a company can select three applicants out of a possible 10. 3.)Use formula to find 5 C 5 = ? 4.)Use calculator to find 8 C 0 = ?

33. Homework Page 217 – 218 ( 40 – 45)

34. Looking ahead to 5.6 and 5.8 We will learn about two Known Discrete Distributions The Binomial (# heads in n tosses)- success or failure. The Poisson (# customers entering a post office in a day)- number of occurrences in a certain interval.

35.  -One of the most widely used discrete probability distributions.  It is applied to find the probability that an outcome will occur x times in n performances of an experiment.  Counting success and failure.

36.  We may be interested in finding the probability that a baseball player with a batting average of.250 will have NO hits in 10 trips to the plate.  Each repetition (trips to the plate) of the experiment is called a trial or Bernoulli trial.

37.  1. There are n identical trials.  2. Each trial has 2 outcomes (success and failure).  3. Probability of success is p and that of failure is q, where p + q = 1. Also, p and q stay constant for each trial.  4. The trials are independent. Meaning the outcomes of 1 trial do not affect the outcomes of another.

38.  Success and failure does NOT mean outcomes that are favorable or not. Rather, the outcome to which the question refers is called success and the outcome it does not is failure.  Example: Binomial or Not?  Tossing a Coin 10 times.  1. Identical trials?  2. Only 2 outcomes?  3. Does p + q = 1 and remain constant for each trial?  4. Are the trials independent?

39.  For a binomial experiment, the probability of exactly x successes in n trials is given by the binomial formula:  Where  n = total # of trials,  p = probability of success,  q = 1 – p = the probability of failure,  x = # of successes in n trials, and  n – x = the number of failures in n trials.  n C x = the # of ways to obtain x successes in n trials.

40.  n and p are called Binomial parameters.  They are all you need to find the probability of x successes in n trials for a Binomial experiment.  Table IV of Appendix C has all the Probabilities for x successes in n trials of a Binomial Experiment.  You could use this table instead of formula to find answers.

41.  5% of all VCRS manufactured by a large electronics company are defective. A quality control inspector randomly selects 3 VCRS from the production line. What is the probability that exactly one of these 3 VCRS is defective?  First before using formula, make sure you realize why this is BINOMIAL experiment!!!  Next, declare all your information you know.  Finally, plug in formula.

42.  n = total # of trials = 3  p = probability of success =.05  q = 1 – p = the probability of failure = 1 -.05 =.95  x = # of successes in n trials = 1  n – x = the number of failures in n trials = 3 – 1 = 2  n C x = 3 C 1 = 3! / 1! ( 3 – 1)! = 3 * 2 *1 / 1 * 2 * 1 = 3  Use Formula:  Thus there is a 13.54% chance that 1 out of 3 VCRS will be defective!

43.  What is the probability that at most 1 of these 3 VCRS are defective?  P( 0 or 1) = P(x = 0) + P(x = 1) =?

44.  Page 229 (53 by hand, 55(a) use table, 59 by hand, 61 use table, 62 by hand)

45.  Ex: According to an Allstate Survey, 56% of Baby Boomers have car loan payments. Let x = the # in a random sample of 3 Baby Boomers who are making payments on their loans. Write a Probability Distribution of x and draw a bar graph.  First, check to make sure it meets all 4 conditions to be a Binomial Experiment.

46.  Next, define all info you know:  n = total trials = 3  p = Probability of successes, P ( a Baby boomer is making a payment) =.56  q = P( Baby boomer is NOT making a payment)= 1 -.56 =.44  How may x = the # in random sample of 3 baby boomers be distributed?  x may be equal to 0, 1, 2, 3. This will be our first column in our Probability Distribution table.

47.  Use Formula or Appendix to find all Binomial Probabilities.  P(x = 0) =  P(x = 1) =  P(x = 2) =  P(x = 3) =

48.  Therefore, the probability distribution for x is:  The Bar graph is also easy to make: xP(x) 0.0852 1.3252 2.4140 3.1756

49.  The binomial probability distribution is:  Symmetric if p =.50  Skewed Right ifp <.50  Skewed Left ifp >.50

50.  We could use formulas for probability distributions learned in sections 5.3 to 5.4, but if distribution is Binomial, we can use the following “easier” formulas instead:

51.  Find the mean and standard deviation for previous example about choosing 3 Baby Boomers and recording whether or not they are making payments on loans.  Information:  n = 3, p =.56, and q =.44  Therefore,

52.  Page 230 – 231 (56, 67, 69)  We will start 5.8 notes first, and there will be a few more homework problems to add to above.

53.  Another important discrete random variable distribution is the Poisson Distribution.  The Poisson Probability Distribution tells you the probability that you will have x occurrences with respect to a certain interval (time, space, or volume).

54.  Conditions for a distribution to be Poisson:  Occurrences are random and independent.  Occurrence are always with respect to an interval.  Occurrences are random. No pattern and thus are unpredictable.  Note: Each occurrence is a trial.

55.  The Number of telemarketing calls received by a household during a given day.  The # of defects in a 5- foot long iron rod.  The # of TVs sold at a store in a given week.

56.  The Probability of x occurrences in an interval is given from the formula:  Where:  λ= lambda = the mean # of occurrences in that interval. Lambda is called the Poisson parameter.  e = a constant = 2.71828…  Note: e -λ can be found in Table V (Appendix C) or by using calculator e button.

57.  An average household receives 9.5 telemarketing calls per week. Find the probability that a randomly selected household receives exactly 6 telemarketing phone calls during a week.  Information from problem:  λ=9.5 = mean # of calls  x = the # of calls received during the week.  Use formula:  They want us to find P(x = 6) = ?  Thus, there is a 7.64 % chance that a household will receive 6 phone calls in a week.

58.  Rest of homework for today:  Page 242 (82, 86, 88)

59.  Ex: A washing machine breaks down on an average of 3 times per month. Find the Probability that in the next month the machine will have:  i.) at most one breakdown (using formula) P(x = 0) + P( x = 1) = ?  ii.) at least 7 breakdowns (using table VI appendix C) P(x = 7) + P(x = 6) + P( x = 5) + P(x = 4) + P( x = 3) + P( x = 2) + P( x = 1) + P( x = 0) = ?

60.  We could use formulas for probability distributions learned in sections 5.3 to 5.4, but if distribution is Poisson, we can use the following “easier” formulas instead:  Mean:  Standard Deviation:  Variance:  Note : Mean and Variance are equal for Poisson!

61.  Find mean and standard deviation per month for previous example with a washing machine that breaks down on average 3 times per month.

62.  Page 242 (89, 91, 93, 96)  Tomorrow we will review Chapter 5 and test the following day.

63.  Page 251 (3 – 8, 13 – 15 all, 16 a, 18 a)