Probability PPT for Statistics - Unit 2 & 3

Statistics - PPT

1. Course Name: Statistics for Business Course Code: MBA 134 Total number of hours: 30 Hours Credits: 3 Unit - II

2. Need for Probability in Business (Analysis of uncertainties) 1 . What are the chances that sales will decrease if we increase prices? 2. What is the likelihood a new assembly method will increase productivity? 3. How likely is it that the project will be finished on time? 4. What is the chance that a new investment will be profitable ? 5. Why Probability in Business ? Probability - Numerical measure of the likelihood that an event will occur Measures of the degree of uncertainty - Strength of belief / Degree of chance or likelihood of occurrence in the occurrence of an uncertain event Probability values - Scale from 0 to 1 Near 0 - E vent is unlikely to occur N ear 1 - Event is almost certain to occur

3. For example, we consider the event “rain tomorrow,” we understand that when the weather report indicates “a near-zero probability of rain,” it means almost no chance of rain. If a .90 probability of rain is reported- rain is likely to occur . A .50 probability indicates that rain is just as likely to occur as not View of probability as a numerical measure of the likelihood of an event occurring

4. Experiment Is a process that generates well-defined outcomes. Ex. Experiment Experimental outcomes Toss a coin Head, Tail Conduct a sales call Purchase, No purchase Roll a die 1,2,3,4,5,6 Play a cricket game Win, lose, tie

5. Sample space The sample space for an experiment is the set of all experimental outcomes. Ex. Tossing a coin s= (head, tail) Rolling a die s=(1,2,3,4,5,6) Workout- Game An experimental outcome is also called as a sample point.

6. Characteristics: Well - defined outcomes 2. On one trial, only one of the possible outcome 3. Outcome is based on chance Ex: Tossing a coin S - Sample space 6 - Sample Points Rolling a Dice

7. Counting Rules, Combinations and Permutations Multiple-step experiment The experiment of tossing two coins can be thought of as two-step experiment. Sample space S= (H,H),(H,T),(T,H),(T,T) Tossing two coin n1=2 and n2=2 then using counting rule (2)(2)=4. Four experimental outcomes are possible The counting rule for multiple-step experiment makes it possible to determine the number of experimental outcomes witho ut listing them. Counting rule for multiple-step experiment If an experiment can be described as a sequence of k steps with n1 possible outcomes on the first step , n2 possible outcomes on the second step an so on , then the total number of experimental outcome is given by (n1) (n2)….( nk ) The number of experimental outcomes involving tossing six coins is (2)(2)(2)(2)(2)(2) = 64.

8. Tree diagram - Graphical representation - Visualizing a multiple-step experiment

9. Combinations A second useful counting rule enables us to count the number of experimental outcomes when n objects are to be selected from a set of N objects . Simple random sampling uses a sample of size n from a population of size N to obtain data that can be used to make inferences about the characteristics of a population. where: N ! = N ( N - 1)( N - 2) . . . (2)(1) n ! = n ( n - 1)( n - 2) . . . (2)(1) 0! = 1 Ex: In a group of five parts how many combinations of two parts can be selected 10 outcomes are possible. If five parts are A,B,C,D,E then 10 combinations of outcomes are AB,AC,AD,AE,BC,BD,BE,CD,CE,DE

10. Ex:1 A quality control procedure in which an inspector randomly selects two of five parts to test for defects. In a group of five parts, how many combinations of two parts can be selected? N = 5 and n = 2, we have Thus , 10 outcomes are possible for the experiment of randomly selecting two parts from a group of five. If we label the five parts as A, b, C, D, and E, 10 combinations or experimental outcomes can be identified as Ab , AC, AD, AE, bC , bD , bE , CD, CE, and DE Ex.2 Suppose that, from a population of 50 bank accounts, we want to take a random sample of four accounts in order to learn about the population. How many different random samples of four accounts are possible?.

11. Permutations - Third counting rule - to compute the number of experimental outcomes when n objects are to be selected from a set of n objects where the order of selection is important. The same n objects selected in a different order are considered as a different experimental outcome. The number of permutations of N objects taken n at a time is given by where: N! = N(N - 1)(N - 2) . . . (2)(1) n! = n(n - 1)(n - 2) . . . (2)(1) 0! = 1 Ex. Same problem as combinations Ordering is important ex: AB,BA,AC,CA,AD,DA,AE,EA,BC.CB.BD.DB.BE.EB.CD.DC.CE.EC.DE.ED .

12. Assigning Probabilities Three approaches - classical, relative frequency, and subjective methods C lassical method - All the experimental outcomes are equally likely Ex: Tossing a coin, Rolling a Die

13. Classical Method Appropriate when all the experimental outcomes are equally likely. If n experimental outcomes are possible, a probability of 1/n is assigned to each experimental outcome. Ex. Tossing a coin and rolling a die Relative frequency When data are available to estimate the proportion of the time the experimental outcome will occur if the experiment is repeated a large number of times. Ex Number waiting Number of days outcome occurred Relative frequency 0 2 . 10 1 5 .25 2 6 . 30 3 4 . 20 4 3 .15 total 20

14. Subjective method Subjective probability result from intuition, educated guesses, and estimates. For instance, given a patient’s health and extent of injuries, a doctor may feel a patient has 90% chance of a full recovery. A business analyst may predict that the chance of the employees of a certain company going on strike is .25 Ex. The probability of your phone ringing during the class is is 0.1 This probability is most likely based on an educated guess. It is an example of subjective probability. Event Collection of outcomes having a common characteristic E.g.: Even number A = {2,4,6} Event A occurs if an outcome in the set A occurs Probability of an event Sum of the probabilities of the outcomes of which it consists P(A) = P(2) + P(4) + P(6)

15. Permutation : Permutation means  arrangement  of things. The word  arrangement  is used, if the order of things  is considered . Combination: Combination means  selection  of things. The word  selection  is used, when the order of things has  no importance . Example:      Suppose we have to form a number of consisting of three digits using the digits  1,2,3,4 , To form this number the digits have to be  arranged . Different numbers will get formed depending upon the order in which we arrange the digits. This is an example of  Permutation . Now suppose that we have to make a team of 11 players out of 20 players, This is an example of  combination , because the order of players in the team will not result in a change in the team. No matter in which order we list out the players the team will remain the same!

16. Exercise Tossing a coin three times a. develop a tree diagram b. list the experimental outcomes c. What is the probability for each experimental outcome. What method you use. 2. An experiment with three outcomes has been repeated 50 times and it was learned that E1 occurred 20 times E2 occurred 13 times and E3 occurred 17 times. Assign probabilities to the outcomes. What method did you use?. 3. A decision maker subjectively assigned the following probabilities to the four outcomes of an experiment P(E1)=.10, P(E2)=.15, P(E3)=.40 and P(E4)=.20. Are the probability assignment valid?

17. Complement of a Set A S Venn Diagram illustrating the Complement of an event

18. Intersection (And) a set containing all elements in both A and B Union (Or) a set containing all elements in A or B or both Complements - Probability of not A Basic Definitions (Continued)

19. Sets: A Intersecting with B A B S

20. Sets: A Union B A B S 2- 22

21. Example: Page_No : 191, 192

22. Ex.1 Study by personal manager of a software company. The study shows 30% of the employees who left the firm within two years did so primarily because they were dissatisfied with their salary, 20% left for work assignment and 12% indicated dissatisfaction with both their salary and work assignment. What is the prob. that an employee who leaves within two years does so because of dissatisfaction with salary, work assignment or both?

23. Mutually Exclusive or Disjoint Sets A B S Sets have nothing in common 2- 25

24. Ex. We have a sample space with five equally likely experimental outcomes. E1, E2, E3,E4,E5 and let A=(E1, E2) B =(E3, E4) C =(E2, E3,E5) Find P(A), P(B), P(C) b. Find P(A or B). Are A and B mutually exclusive Find Ac, Cc, P(Ac) and P(Cc) d.P (A or Bc ) Ex.2 A survey of magazine subscribers showed that 45.8% rented a car during the past 12 months for business reasons, 54% rented a car during the past 12 months for personal reasons, and 30% rented a car during the past 12 months for both business and personal reasons. What is the probability that a subscriber rented a car during the past 12 months for business or personal reasons? What is the probability that a subscriber did not rent a car during the past 12 months for either business or personal reasons?

25. Conditional probability: 2.A conditional probability is the probability of one event, given that another event has occurred The condition al proba bility of A given that B has occurred. A given B The conditional probability of B given that A has occurred. B given A Where P(A and B) = joint probability of A and B The police force consists of 1200 officers, 960 men and 240 women. over the past two years, 324 officers on the police force received promotions. 288 male officers had received promotions, but only 36 female officers had received promotions .

26. Rules of conditional probability: If events A and B are statistically independent : so so Rules of Probability: General Addition Rule P(A or B) = P(A) + P(B) - P(A and B) If A and B are mutually exclusive , then P(A and B) = 0, so the rule can be simplified P(A or B) = P(A) + P(B)

27. From the marginal probabilities, we see that 80% of the force is male, 20% of the force is female, 27% of all officers received promotions, and 73% were not promoted

28. Multiplication Rule– for two events A and B Two events A and B are statistically independent if the probability of one event is unchanged by the knowledge that other even occurred. That is: Then the multiplication rule for two statistically independent events is: Consider the situation of a service station manager who knows from past experience that 80% of the customers use a credit card when they purchase gasoline. What is the probability that the next two customers purchasing gasoline will each use a credit card?

29. Prior probability P osterior probabilities Bayes’ theorem Ex: 202 P(A ∩ B) and P(A|B) are very closely related. Their only difference is that the conditional probability assumes that we  already know something  -- that B is true. The intersection doesn't assume that we know anything. So for P(A ∩ B), we will receive a probability between 0, impossible, and 1, certain. For P(A|B), however, we will receive a probability between 0, if A cannot happen when B is true, and P(B), if A is always true when B is true . So the only difference between P(A ∩ B) and P(A|B) is the number P(B). Four candidates A, B, C, and D are running for a  political office . Each has an equal chance of winning: 25%. However, if candidate A drops out of the race due to ill health, the probability will change: P(Win | One candidate drops out) = 33.33%. If there is a pregnant women, the  probability of having a boy or girl  is the same: 50%. However, if she already have one child (say, a boy), it will change. If first child is a boy, having another boy drop to one third (33.33%). The reason for this is that the  sample space  for the event you have one boy out of two is S = {BB, BG, GB}. If you have one boy, the only possible event within this space is BB, which is one third of the sample space.

30. Consider a newspaper circulation department 84 % of the households in a particular neighborhood subscribe to the daily edition of the paper. If we let D denote the event that a household subscribes to the daily edition, P ( D ) = .84. In addition, it is known that the probability that a household that already holds a daily subscription also subscribes to the Sunday edition (event s ) is .75; that is, P ( s ∣ D ) = .75. What is the probability that a household subscribes to both the Sunday and daily editions of the newspaper ? using the multiplication law, we compute the desired P ( s ∩ D ) as We now know that 63% of the households subscribe to both the Sunday and daily editions.

31. Some more definitions Set - a collection of elements or objects of interest Empty set (denoted by ) a set containing no elements Universal set (denoted by S) a set containing all possible elements Complement (Not). The complement of A is a set containing all elements of S not in A

32. Multiplication rule Examples When asked to find the probability of A and B, we want to find out the probability of events A and B happening. Suppose we roll one die followed by another and want to find the probability of rolling a 4 on the first die and rolling an even number on the second die. Notice in this problem we are not dealing with the sum of both dice. We are only dealing with the probability of 4 on one die only and then, as a separate event, the probability of an even number on one die only.    

33. P(4) = 1/6  P(even) = 3/6   So P(4 even) = (1/6)(3/6) = 3/36 = 1/12 While the rule can be applied regardless of dependence or independence of events, we should note here that rolling a 4 on one die followed by rolling an even number on the second die are  independent  events. Each die is treated as a separate thing and what happens on the first die does not influence or effect what happens on the second die. This is our basic definition of independent events: the outcome of one event does not influence or effect the outcome of another event.

34. Suppose you have a box with 3 blue marbles, 2 red marbles, and 4 yellow marbles. You are going to pull out one marble, record its color, put it back in the box and draw another marble. What is the probability of pulling out a red marble followed by a blue marble?   The multiplication rule says we need to find P(red)   P(blue).   P(red) = 2/9 P(blue) = 3/9 P(red blue) = (2/9)(3/9) = 6/81 = 2/27 Independent or not ? 

35. Workout 1.There are 11 marbles in a bag. Two are yellow, five are pink and four are green. Suppose you pull out one marble, record its color, put it back in the bag and then pull out another marble. What is the probability of P(yellow and pink)

36. Consider the same box of marbles as in the previous example. However in this case, we are going to pull out the first marble, leave it out, and then pull out another marble. What is the probability of pulling out a red marble followed by a blue marble?   We can still use the multiplication rule which says we need to find P(red)   P(blue). But be aware that in this case when we go to pull out the second marble, there will only be 8 marbles left in the bag.   P(red) = 2/9 P(blue) = 3/8 P(red blue) = (2/9)(3/8) = 6/72 = 1/12   The events in this example were dependent. When the first marble was pulled out and kept out, it effected the probability of the second event. This is what is meant by dependent events.

37. Rule of Subtraction The probability of an event ranges from 0 to 1. The sum of probabilities of all possible events equals 1. The rule of subtraction follows directly from these properties. Rule of Subtraction  The probability that event A will occur is equal to 1 minus the probability that event A will  not  occur. P(A) = 1 - P(A') Suppose, for example, the probability that Bill will graduate from college is 0.80. What is the probability that Bill will not graduate from college? Based on the rule of subtraction, the probability that Bill will not graduate is 1.00 - 0.80 or 0.20.

38. Summary Two  events  are  mutually exclusive  or  disjoint  if they cannot occur at the same time. The probability that Event A occurs, given that Event B has occurred, is called a  conditional probability . The conditional probability of Event A, given Event B, is denoted by the symbol P(A|B). The  complement  of an event is the event not occuring . The probability that Event A will  not occur is denoted by P(A'). The probability that Events A and B  both  occur is the probability of the  intersection  of A and B. The probability of the intersection of Events A and B is denoted by P(A ∩ B). If Events A and B are mutually exclusive, P(A ∩ B) = 0.

39. Random Variables A random variable is a numerical description of the outcome of an experiment. It is a set of possible values from a random experiment. Ex. Height, Weight or age etc A discrete random variable may assume either a finite number of values or an infinite sequence of values . consider the experiment of cars arriving at a tollbooth. The random variable of interest is x = the number of cars arriving during a one-day period. The possible values for x come from the sequence of integers 0,1 , 2, and so on .

40. A continuous random variable may assume any numerical value in an interval or collection of intervals . Which of the following is a discrete random variable? I. The average height of a randomly selected group of boys. II. The annual number of volleyball winners from New York City. III. The number of presidential elections in the 20th century.

41. Discrete Probability Distributions The probability distribution for a random variable describes how probabilities are distributed over the values of the random variable. We can describe a discrete probability distribution with a table, graph, or formula. The probability distribution is defined by a probability function , denoted by f ( x ), which provides the probability for each value of the random variable.

42. The required conditions for a discrete probability function are: f ( x ) >  f ( x ) = 1 Ex. Probability distribution of TV sales Number Units Sold of Days x f(x) 0 80 0 .40 1 50 1 .25 2 40 2 .20 3 10 3 .05 4 20 4 .10 200 1 The use of the relative frequency method to develop discrete probability distributions leads to what is called an empirical discrete distribution. We define the random variable x to be the number of dots on the upward face. for this experiment, n = 6 values are possible for the random variable; x = 1, 2, 3, 4, 5, 6 .

43. The expected value , or mean, of a random variable is a measure of its central location. E ( x ) =  =  xf ( x ) The expected value is a weighted average of the values the random variable may assume

44. The variance summarizes the variability in the values of a random variable. Var ( x ) =  2 =  ( x -  ) 2 f ( x ) The variance is a weighted average of the squared deviations of a random variable from its mean. The weights are the probabilities. The standard deviation ,  , is defined as the positive square root of the variance.

45. Expected Value x f ( x ) xf ( x ) 0 .40 .00 1 .25 .25 2 .20 .40 3 .05 .15 4 .10 .40 E ( x ) = 1.20expected number of TVs sold in a day

46. 1 2 3 4 -1.2 -0.2 0.8 1.8 2.8 1.44 0.04 0.64 3.24 7.84 .40 .25 .20 .05 .10 .576 .010 .128 .162 .784 x -  ( x -  ) 2 f ( x ) ( x -  ) 2 f ( x ) Variance of daily sales = s 2 = 1.660 x TVs squared Standard deviation of daily sales = 1.2884 TVs Variance

47. Workout Children Couples p(X) 0 1 1 4 2 3 3 2 4 2 Total 12

48. H.W Probability distribution of a random variable y Y f(y) 2 .20 .30 .40 .10 Compute E(y) Compute Var (y) and s.d

49. Discrete probability distributions - The binomial - The Poisson

50. Binomial Probability Distributions A coin-tossing experiment is a simple example of an important discrete random variable called the binomial random variable . Other situations that are similar to the coin-tossing experiment: - A sociologist is interested in the proportion of elementary school teachers who are men. - A soft-drink marketer is interested in the proportion of cola drinkers who prefer her brand.

51. Definition: A binomial experiment is one that has these four characteristics: 1. The experiment consists of n identical trials. 2. Each trial results in one of two outcomes: one outcome is called a success, S , and the other a failure, F. 3. The probability of success on a single trial is equal to p and remains the same from trial to trial. The probability of failure is equal to (1 - p ) = q. 4. The trials are independent.

52. Ex. An insurance sales person who visits 10 randomly selected families. The outcome associated with each visit is classified as a success if the family purchases an insurance policy and a failure if the family does not. From past experience, the sales person knows the probability of a randomly selected family will purchase policy is .10. Check the properties of Binomial experiment.

53. 1. The experiment consists of 10 identical trials, each trial involves contacting one family. 2. Two outcomes are possible on each trial : the family purchases a policy (success) or the family does not purchase a policy (failure) 3. The probability of a purchase and a non purchase are assumed to be the same for each sales call, with p=.10 and 1-p=.90. 4. The trials are independent because the families are randomly selected.

54. The Binomial Probability function x = the number of successes p = the probability of a success on one trial n = the number of trials f ( x ) = the probability of x successes in n trials

55. Mean and Standard Deviation for the Binomial Random Variable: Mean: m = np Variance: s 2 = npq Standard deviation:

56. Binomial Formula.  Suppose a binomial experiment consists of  n  trials and results in  x  successes. If the probability of success on an individual trial is  P , then the binomial probability is: b( x ;  n, P ) =  n C x  * P x  * (1 - P) n – x EXAMPLE 2 Suppose a die is tossed 5 times. What is the probability of getting exactly 2 fours? Solution:  This is a binomial experiment in which the number of trials is equal to 5, the number of successes is equal to 2, and the probability of success on a single trial is 1/6 or about 0.167. Therefore, the binomial probability is: b(2; 5, 0.167) =  5 C 2  * (0.167) 2  * (0.833) 3   b(2; 5, 0.167) = 0.161

57. Ex3. In city x, 30% of workers take public transportation daily. A. In a sample of 10 workers, what is the probability that exactly three workers take public transportation daily. b. In sample of 10 workers, what is the probability that at least three workers take public transportation daily.

58. Ex.4. A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 students registered for the course. Compute the probability that two or fewer will withdraw. Compute the probability that exactly four will withdraw. Compute the probability that more than three will withdraw. Compute the expected number of withdrawals.

59. Ex.5 Nine percent of undergraduate students carry credit card balances greater that 7000. Suppose 10 undergraduate students are selected randomly to be interviewed about credit card usage. Is the selection of 10 students a binomial experiment? Explain? What is the probability that two of the students will have a credit balance greater than 7000. What is the probability that none will have a credit card balance greater than 7000. What is the probability that at least three will have a credit card balance greater than 7000.

60. Example The probability that a student is accepted to a prestigious college is 0.3. If 5 students from the same school apply, what is the probability that at most 2 are accepted? Solution:  To solve this problem, we compute 3 individual probabilities, using the binomial formula. The sum of all these probabilities is the answer we seek. Thus, b(x  <  2; 5, 0.3) = b(x = 0; 5, 0.3) + b(x = 1; 5, 0.3) + b(x = 2; 5, 0.3) b(x  <  2; 5, 0.3) = 0.1681 + 0.3601 + 0.3087  b(x  <  2; 5, 0.3) = 0.8369

61. Less than 3 f(0)+f(1)+f(2) Greater than or more than 3 1- f(0)+f(1)+f(2)+f(3) At least 3 (p(x>=3)) 1- f(0)+f(1)+f(2) At most 3 (p(x<=3) f(0)+f(1)+f(2)+f(3) 3 or fewer f(0)+f(1)+f(2)+f(3) Exactly 3 f(3)

62. Poisson Probability Distributions The Poisson probability distribution is a good model for data that represent the number of occurrences of a specified event in a given unit of time or space. Some examples of Poisson random variables: - The number of calls received by a switchboard during a given period of time. - The number of customer arrivals at a checkout counter during a given minute - The number of machine breakdowns during a given day - The number of traffic accidents at a given intersection during a given time period

63. A  Poisson experiment  is a  statistical experiment  that has the following properties: The experiment results in outcomes that can be classified as successes or failures. The average number of successes (μ) that occurs in a specified region is known. The probability that a success will occur is proportional to the size of the region. The probability that a success will occur in an extremely small region is virtually zero. Note that the specified region could take many forms. For instance, it could be a length, an area, a volume, a period of time, etc.

64. Notation The following notation is helpful, when we talk about the Poisson distribution. e : A constant equal to approximately 2.71828. (Actually,  e  is the base of the natural logarithm system.) μ: The mean number of successes that occur in a specified region. x : The actual number of successes that occur in a specified region. P( x ; μ): The  Poisson probability  that  exactly   x  successes occur in a Poisson experiment, when the mean number of successes is μ.

65. Poisson Formula.  Suppose we conduct a Poisson experiment, in which the average number of successes within a given region is μ. Then, the Poisson probability is: P( x ; μ) = (e -μ ) ( μ x ) / x! where  x  is the actual number of successes that result from the experiment, and  e is approximately equal to 2.71828.

66. Example 1 The average number of homes sold by the Acme Realty company is 2 homes per day. What is the probability that exactly 3 homes will be sold tomorrow? Solution:  This is a Poisson experiment in which we know the following: μ = 2; since 2 homes are sold per day, on average. x = 3; since we want to find the likelihood that 3 homes will be sold tomorrow. e = 2.71828; since  e  is a constant equal to approximately 2.71828. We plug these values into the Poisson formula as follows: P( x ; μ) = (e -μ ) ( μ x ) / x!  P(3; 2) = (2.71828 -2 ) (2 3 ) / 3!  P(3; 2) = (0.13534) (8) / 6  P(3; 2) = 0.180  Thus, the probability of selling 3 homes tomorrow is 0.180 .

67. Ex.2. An average 15 aircraft accidents occur each year. Compute the mean number of accidents per month Compute the probabilities of no accidents during a month Compute the probability of exactly one accident per month Compute the probability of more than one accident during a month.

68. Ex.3 Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 10 passengers per minute. Compute the probability of no arrivals in a one-minute period. Compute the probability that three or fewer passengers arrive in a one-minute period. Compute the probability of no arrivals in a 15 second period. Compute the probability of at least one arrival in a 15 second period.

69. Continuous Probability Distributions A continuous variable is a variable that can assume any value on a continuum (can assume an uncountable number of values) thickness of an item time required to complete a task temperature of a solution height, in inches

70. a is the smallest value b is the largest value

71. What is the probability that the flight time is between 120 and 130 minutes? P(120 ≤ x ≤ 130)? Consider the area under the graph of f (x) in the interval from 120 to 130. The area is rectangular, and the area of a rectangle= l * b Width of the interval 130 − 120 = 10 Height = value of the probability density function f (x) = 1/20 area = width × height = 10(1/20) = 10/20 = .50. The area under the graph of f(x ) and probability are identical!

72. What is the probability of a flight time between 128 and 136 minutes? The width of the interval is 136 − 128 = 8. with the uniform height of f ( x ) = 1/20, P (128 ≤ x ≤ 136) = 8(1/20) = . 40

73. Bell Shaped Symmetrical Mean, Median and Mode are Equal Location is determined by the mean, μ Spread is determined by the standard deviation, σ Heights and weights of people , test scores, scientific measurements , amounts of rainfall, The random variable has an infinite theoretical range: +  to   Mean = Median = Mode X f(X) μ σ The Normal Distribution

74. The Normal Distribution Density Function The formula for the normal probability density function is Where e = the mathematical constant approximated by 2.71828 π = the mathematical constant approximated by 3.14159 μ = the population mean σ = the population standard deviation X = any value of the continuous variable

75. A B C A and B have the same mean but different standard deviations. B and C have different means and different standard deviations. By varying the parameters μ and σ , we obtain different normal distributions

76. The Normal Distribution Shape X f(X) μ σ Changing μ shifts the distribution left or right. Changing σ increases or decreases the spread.

77. The Standardized Normal Any normal distribution (with any mean and standard deviation combination) can be transformed into the standardized normal distribution (Z) To compute normal probabilities need to transform X units into Z units The standardized normal distribution ( Z ) has a mean of 0 and a standard deviation of 1

78. Translation to the Standardized Normal Distribution Translate from X to the standardized normal (the “Z” distribution) by subtracting the mean of X and dividing by its standard deviation : The Z distribution always has mean = 0 and standard deviation = 1

79. The Standardized Normal Probability Density Function The formula for the standardized normal probability density function is Where e = the mathematical constant approximated by 2.71828 π = the mathematical constant approximated by 3.14159 Z = any value of the standardized normal distribution

80. The Standardized Normal Distribution Also known as the “Z” distribution Mean is 0 Standard Deviation is 1 Z f(Z) 1 Values above the mean have positive Z-values. Values below the mean have negative Z-values.

81. Example If X is distributed normally with mean of $100 and standard deviation of $50 , the Z value for X = $200 is This says that X = $200 is two standard deviations (2 increments of $50 units) above the mean of $100.

82. Comparing X and Z units Note that the shape of the distribution is the same, only the scale has changed. We can express the problem in the original units (X in dollars) or in standardized units (Z) Z $100 2.0 $200 $X ( μ = $100, σ = $50) ( μ = 0, σ = 1)

83. Probability is measured by the area under the curve a b X f(X) P a X b ( ) ≤ ≤ P a X b ( ) < < = (Note that the probability of any individual value is zero) Finding Normal Probabilities

84. Probability as Area Under the Curve The total area under the curve is 1.0 , and the curve is symmetric, so half is above the mean, half is below f(X) X μ 0.5 0.5

85. The Standardized Normal Table The Cumulative Standardized Normal table in the textbook (Appendix table E.2) gives the probability less than a desired value of Z (i.e., from negative infinity to Z) Z 2.00 0.9772 Example: P(Z < 2.00) = 0.9772

86. The Standardized Normal Table The value within the table gives the probability from Z =   up to the desired Z value .9772 2.0 P(Z < 2.00) = 0.9772 The row shows the value of Z to the first decimal point The column gives the value of Z to the second decimal point 2.0 . . . (continued) Z 0.00 0.01 0.02 … 0.0 0.1

87. General Procedure for Finding Normal Probabilities Draw the normal curve for the problem in terms of X Translate X-values to Z-values Use the Standardized Normal Table To find P(a < X < b) when X is distributed normally:

88. Finding Normal Probabilities Let X represent the time it takes (in seconds) to download an image file from the internet. Suppose X is normal with a mean of18.0 seconds and a standard deviation of 5.0 seconds. Find P(X < 18.6) 18.6 X 18.0

89. Finding Normal Probabilities Let X represent the time it takes, in seconds to download an image file from the internet. Suppose X is normal with a mean of 18.0 seconds and a standard deviation of 5.0 seconds. Find P(X < 18.6) Z 0.12 X 18.6 18 μ = 18 σ = 5 μ = 0 σ = 1 P(X < 18.6) P(Z < 0.12)

90. Z 0.12 0.5478 Standardized Normal Probability Table (Portion) 0.00 = P( Z < 0.12) P( X < 18.6) Z .00 .01 0.0 .5000 .5040 .5080 .5398 .5438 0.2 .5793 .5832 .5871 0.3 .6179 .6217 .6255 .02 0.1 . 5478 Solution: Finding P(Z < 0.12)

91. Finding Normal Upper Tail Probabilities Suppose X is normal with mean 18.0 and standard deviation 5.0. Now Find P(X > 18.6) X 18.6 18.0

92. Finding Normal Upper Tail Probabilities Now Find P(X > 18.6)… (continued) Z 0.12 Z 0.12 0.5478 1.000 1.0 - 0.5478 = 0.4522 P(X > 18.6) = P(Z > 0.12) = 1.0 - P(Z ≤ 0.12) = 1.0 - 0.5478 = 0.4522

93. Finding a Normal Probability Between Two Values Suppose X is normal with mean 18.0 and standard deviation 5.0. Find P(18 < X < 18.6) P(1 8 < X < 18.6) = P( 0 < Z < 0.12) Z 0.12 X 18.6 18 Calculate Z-values:

94. Z 0.12 0.0478 0.00 = P( 0 < Z < 0.12) P(1 8 < X < 18.6) = P( Z < 0.12) – P(Z ≤ 0) = 0.5478 - 0.5000 = 0.0478 0.5000 Z .00 .01 0.0 . 5000 .5040 .5080 .5398 .5438 0.2 .5793 .5832 .5871 0.3 .6179 .6217 .6255 .02 0.1 . 5478 Standardized Normal Probability Table (Portion) Solution: Finding P(0 < Z < 0.12)

95. Probabilities in the Lower Tail Suppose X is normal with mean 18.0 and standard deviation 5.0. Now Find P(17.4 < X < 18) X 17.4 18.0

96. Probabilities in the Lower Tail Now Find P(17.4 < X < 18)… X 17.4 18.0 P(17.4 < X < 18) = P(-0.12 < Z < 0) = P(Z < 0) – P(Z ≤ -0.12) = 0.5000 - 0.4522 = 0.0478 (continued) 0.0478 0.4522 Z -0.12 The Normal distribution is symmetric, so this probability is the same as P(0 < Z < 0.12)

97. Empirical Rule μ ± 1 σ encloses about 68.26% of X’s  f(X) X μ μ +1 σ μ -1 σ What can we say about the distribution of values around the mean? For any normal distribution: σ σ 68.26%

98. The Empirical Rule μ ± 2 σ covers about 95.44% of X’s μ ± 3 σ covers about 99.73% of X’s x μ 2 σ 2 σ x μ 3 σ 3 σ 95.44% 99.73% (continued)

99. Given a Normal Probability Find the X Value Steps to find the X value for a known probability: 1. Find the Z value for the known probability 2. Convert to X units using the formula:

100. Finding the X value for a Known Probability Example: Let X represent the time it takes (in seconds) to download an image file from the internet. Suppose X is normal with mean 18.0 and standard deviation 5.0 Find X such that 20% of download times are less than X. X ? 18.0 0.2000 Z ? (continued)

101. Find the Z value for 20% in the Lower Tail 20% area in the lower tail is consistent with a Z value of -0.84 Z .03 -0.9 .1762 .1736 .2033 -0.7 .2327 .2296 .04 -0.8 . 2005 Standardized Normal Probability Table (Portion) .05 .1711 .1977 .2266 … … … … X ? 18.0 0.2000 Z -0.84 1. Find the Z value for the known probability

102. Finding the X value 2. Convert to X units using the formula: So 20% of the values from a distribution with mean 18.0 and standard deviation 5.0 are less than 13.80