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1. Probability and Probability Distributions

2. 1. Introduction What is probability? Probability is a mathematical construction that determines the occurrence of events that are subject to chance. Chance of observing a particular outcome Likelihood of an event O utcome of an event is the result of single trial experiment Probability and Probability Distributions

3. The theory of probability developed from studying games of chance like throwing dice and playing cards And also it is fundamental for quantifying the uncertainty that is helpful in the decision-making process. Probability theory is a foundation for statistical inference & allow us to draw conclusions. Introduction…

4. Basic terms Experiment is any process with an uncertain outcome. An experiment is a trial and all possible outcomes are events l ike counts, measurements or responses. Sample space : The set of all possible outcomes for an experiment Example: The sample space for the sex of newborns when two mothers are in the gynecology ward to give birth is:{MM, MF, FM, FF} Tossing coin (Head, Tail )

5. Outcome: It is the result of a single trial in a probability or a result of an experiment An event: Consists of one or more outcomes and is a subset of the sample space and represented by uppercase letters such as A , B , C , etc. From the previous example, an event consisting of at least one female is E = {MF, FM, FF} Basic terms…

6. Probability Can be defined as the number of times in which that event occurs in a very large number of trials Probability = Probability of an e vent E A number between 0 and 1 representing the proportion of event E expected to happen when the experiment is done over and over under the same conditions   Basic terms…

7. Objective and Subjective Probabilities . Objective probability is the probability an event will occur based on an analysis in which each measure is based on a recorded observation In contrast,  subjective probability  allows the observer to gain insight by referencing things they've learned and their own experience Objective probability Classical probability & Relative frequency probability. 7 Two Categories of Probability

8. Classical Probability It is used when each outcome in a sample space is equally likely to occur. Is based on gambling ideas Rolling a die - There are 6 possible outcomes: – Total ways = {1, 2, 3, 4, 5, 6}. Each is equally likely to occur – P(i) = 1/6, i=1,2,...,6.  P(1) = 1/6  P(2) = 1/6  …….  P(6) = 1/6 SUM = 1 8

9. – Another “equally likely” setting is the tossing of a coin . – There are 2 possible outcomes in the set of all possible outcomes {H, T}.  P(H) = 0.5  P(T) = 0.5 SUM = 1.0 9

10. Relative Frequency Probability Definition : If a process is repeated a large number of times ( n ), and if an event with the characteristic E occurs m times, the relative frequency of E , Probability of E = P(E) = m/n 10

11. If yo u toss a coin 10 0 times and head comes up 40 times, P(H) = 40/100 = 0.4. If we toss a coin 10,000 times and the head comes up 5562, P(H) = 0.5562. Therefore, the longer the series and the longer sample size, the closer the estimate to the true value ( 0.5 ). 11

12. Since trials or experimen ts cannot be repeated for an infinite number of times, theoretical probabilities are often estimated by empirical (statistical) probabilities based on a finite amount of data Example: Of 158 people who attended a dinner party, 99 were ill. P (Illness) = 99/158 = 0.63 = 63%. 12

13. Probability of a male livebirth during the period 1965 ‐ 1974 Time Period No. of male livebirths Total no. of livebirths Empirical prob of a male livebirth 1965 1,927,054 3,760,358 0.51447 1965-1969 9,219,202 17,989,361 . 51348 1965-1974 17,857,857 34,832,051 0.51268 These are empirical probabilities or, relative frequency probability based on a finite amount of data. If the sample size increased indefinitely, more precise estimate (0.5) can be obtained 13

14. Subjective Probability P erson a l i s tic ( a n o p inio n o r judgme n t b y a d ec i s i on maker about the likelihood of an event ). Persona l assessment of whic h is m ore effectiv e to provide p ure – traditional probability. Also uses classical and relative frequency methods to assess the likelihood of an event, but does not rely on repeatability of any process. 14

15. Examples A physician might say that, on the basis of her diagnosis, there is a 30% chance the patient will need an operation. If someone says that he is 95% certain that a cure for AIDS will be discovered within 5 years. P(discovery of cure for AIDS within 5 years) = 95% = 0.95

16. Mutually Exclusive Events Two events A and B are mutually exclusive if they cannot both happen at the same time P (A ∩ B) = 16

17. A coin toss cannot produce head and tail simultaneously. Weight of an individual can’t be classified simultaneously as “underweight”, “normal”, and “overweight” 17 Examples

18. Independent Events Two events A and B are independent if the probability of the first one happening has no matter how the second one turns out. OR: The outcome of one event has no effect on the occurrence or non‐occurrence of the other. That is, A and B are independent if : P (B |A) = P (B) or if P (A |B) = P (A) P(A∩B) = P(A) x P(B) (Independent events) Example: – The outcomes on the first and second coin tosses are independent 18

19. Dependent Events Occurrence of one event affects the probability of the other P(A∩B) ≠ P(A) x P(B) Example: Consider the DBP measurements from a mother and her first‐born child. Let: A = {mother’s DBP≥95} B = {first‐born child’s DBP≥80} • Suppose P{A∩B} = 0.05 P{A} = 0.1 P{B} = 0.2 Then P{A∩B} = 0.05 > P{A} x P{B} = 0.02 And Events A and B would be dependent. 19

20. Intersection and union The intersection of two events A and B , is the event that A and B happen simultaneously P ( A and B ) = P (A ∩ B ) Example : Let A represent the event that a randomly selected newborn is LBW, and B the event that he or she is from a multiple birth What is (A ∩ B) & (A U B)? The intersection of A ∩ B is the event that the infant is both LBW and from a multiple birth 20

21. The union of A and B , A U B , is the event that either A happens or B happens or they both happen simultaneously P ( A or B ) = P ( A U B ) Here, the union of A and B is the event that the newborn is either LBW or from a multiple birth, or both 21

22. Properties of Probability The numerical valu e of a probability always lies between 0 and 1, inclusive.  P(E)  1 A value means the event can not occur A value 1 means the event definitely will occur A value of 0.5 means that the probability that the event will occur is the same as the probability that it will not occur. 22

23. … 11/28/2022 By habtamu.A. 2. The sum of the probabilities of all mutually exclusive outcomes is equal to 1 P(E 1 ) + P(E 2 ) + .... + P( E n ) = 1 3. For two mutually exclusive events A and B, P(A or B ) = P(AUB)= P(A) + P(B). If not mutually exclusive: P(AUB) = P(A) + P(B) ‐ P( AnB )

24. 4. The complement of an event A , denoted by Ā or A c , is the event that A does not occur This c onsists of all the outcomes in which event A does NOT occur P(Ā) = P(not A) = 1 – P(A) Ā occurs only when A does not occur. 24

25. Basic Probability Rules Addition rule If events A and B are mutually exclusive : P(A U B) = P(A) + P(B) P( A n B ) = If not : P(A U B) = P(A) + P(B) - P(A n B) P (event A or event B occurs or they both occur) 25

26. If E1 and E2 are mutually exclusive , then E1 E2 P(E1 and E2) = So P(E 1 U E 2 ) = P(E 1 ) + P(E 2 ) - P(E 1 and E 2 ) = P(E 1 ) + P(E 2 ) 26

27. Example : the probability below represent years of schooling completed by mothers of newborn infants Mother’s education Probability  8 years 0.056 9 to 11 years 0.159 12 years 0.321 13 to 15 years 0.218  16 years 0.230 Not reported 0.016 27

28. What is the probability that a mother has completed < 12 years of schooling? P(  8 years) = 0 . 056 and P(9‐11 years) = . 159 Since these two events are mutually exclusive, P(  8 or 9‐11) = P(  8 U 9‐11) = P(  8) + P(9‐11) = . 056+0 . 159 = . 215 28

29. Wha t is the p robabilit y that a mothe r has completed 12 or more years of schooling? P(  12) = P(12 or 13‐15 or  16) = P(12 U 13‐15 U  16) = P(12)+P(13‐15)+P(  16) = . 321+0 . 218+0 . 230 = . 769 29

30. 2. Multiplication rule If A and B are independent events, then P( A ∩ B ) = P( A ) × P( B ) ‐ P (A and B) denotes the probability that A and B both occur at the same time (simultaneous occurrence). More generally independent eve nts: P(A ∩ B) = P(A.B) = P(A) P(B|A) = P(B) P(A|B) 30

31. Conditional Probability Refers to the probability of an event (E 2 ), given that another event is known to have occurred (E 1 ). The conditional probability is given as: Hint ‐ When thinking about conditional probabilities, think in stages . E.g. T hink of the two events A and B occurring chronologically, one after the other, either in time or space. 31

32. Example 1 11/28/2022 A study investigating the effect of exposure to bright light on retina prolonged premature infants damaged in

33. 11/28/2022 The probability of developing retinopathy is: P (Retinopathy) = No. of infants with retinopathy Total No. of infants = (18+21)/(21+39) = 0.65 What is the conditional probability of retinopathy, given exposure to bright light ?

34. 11/28/2022 P(Retinopathy/exposure to bright light ) = No. of infants with retinopathy exposed to bright light No . of infants exposed to bright light = 18/21 = 0.86 What is the conditional probability of retinopathy, given exposure to reduced light ? P(Retinopathy/exposure to reduced light ) = # of infants with retinopathy exposed to reduced light No . of infants exposed to reduced light = 21/39 = 0.54

35. Example 2 11/28/2022 Suppose in country X the chance that an infant lives to age 25 is .95, whereas the chance that he lives to age 65 is .65. For the latter, it is understood that to survive to age 65 means to survive both from birth to age 25 and from age 25 to 65. What is the chance that a person 25 years of age survives to age 65?

36. 11/28/2022 Notation Event Probability A Survive birth to age 25 0.95 A and B Survive both birth to age 25 0.65 and age 25 to 65 is B/A Survive age 25 to 65 given ? survival to age 25 ? ► Then, Pr (B/A) = Pr (A n B ) / Pr (A) = .65/.95 = .684 ► That is, a person aged 25 has a 68.4% chance of living to age 65.

37. Test for Independence 11/28/2022 Two events A and B are independent if: P(B|A )= P(B) or P(A and B) = P(A) • P(B ) Two events A and B are dependent if P(B|A ) ≠ P(B) or P(A and B) ≠ P(A) • P(B)

38. Example 11/28/2022 In a study of optic-nerve degeneration in Alzheimer’s disease, postmortem examinations were conducted on 10 Alzheimer’s patients . The following table shows the distribution of these patients according to sex and evidence of optic-nerve degeneration. Are the events “a patient has optic-nerve degeneration” and “patient is female” independent for the sample of 10 patients ?

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40. Solution P(Optic-nerve degeneration/Female) = No. of females with optic-nerve degeneration No. of females = 4/5 = 0.80 P(Optic-nerve degeneration) = No of patients with optic-nerve degeneration Total No. of patients = 8/10 = 0.80 The events are independent for this sample . 40

41. Counting Rules 11/28/2022 Counting rule is a way to figure out the number of outcomes in a probability problem. We have three different counting rules. Basic multiplication rule Permutations Combinations

42. Multiplication Rule 11/28/2022 If we have an experiment with k parts (such as k tosses), and Each part has n possible outcomes (such as heads & tails), then The total number of possible outcomes for the experiment is n k  This is the simplest multiplication rule .

43. Permutations 11/28/2022 The number of possible permutations is the number of orders in which particular events occur. The number of possible permutations are:- where r is the number of events in the series, n is the number of possible events, and n! denotes the factorial of n = the product of all the positive integers from 1 to n

44. Combinations 11/28/2022 When the order in which the events occurred is of no interest, we are dealing with combinations. The number of possible combinations is Where r is the number of events in the series, n is the number of possible events , and n! denotes the factorial of n = the product of all the positive integers from 1 to n .

45. 11/28/2022 Probability Distributions

46. Probability Distributions A probability distribution is a device used to describe the behavior of random variable by applying the theory of probability . I t is a function that assigns probability for each element of random variable . It is the way that data are distributed, in order to draw conclusions about a set of data Random Variable is a ny quantity that is able to assume a number of different values such that any particular outcome determined by chance 46

47. Random variables can be either discrete or continuous A discrete random variable is able to assume only a finite or countable number of outcomes A continuous random variable can take on any value in a specified interval 47

48. 11/28/2022 Probability distribution classified as: Discrete probability distribution C ontinues probability distribution. This classification depends on the possible value of the random variables, whether it is discrete or continuous. Discrete probability distributions are; B inomial distn ., Poisson distn ., Bernoulli distn . and etc. C ontinues probability distributions are; Normal distn ., chi-square distn ., t-distribution, F-distribution and etc.

49. A. Discrete Probability Distributions In discrete probability distribution random variable can take only limited number of values Example number of heads in tossing of two coins.

50. We represent a potential outcome of the random variable X by x 0 ≤ P(X = x) ≤ 1 ∑ P(X = x) = 1 P(X = x) means Probability that X has value x (or x is the value of random variable X) 50

51. The following data shows the number of diagnostic services a patient receives The sum of all the individual probabilities is 1. x P(X=x) 0.671 1 0.229 2 0.053 3 0.031 4 0.010 5 0.006 1.00 51 Example

52. 11/28/2022 Find the following values of probabilities. a) What is the probability that a patient receives exactly 3 diagnostic services? b) What is the probability that a patient receives at most one diagnostic service ? c) What is the probability that a patient receives at least four diagnostic service?

53. What is the probability that a patient receives exactly 3 diagnostic services? P(X=3) = 0.031 What is the probability that a patient receives at most one diagnostic service? P(X≤1) = P(X = 0) + P(X = 1) = 0.671 + 0.229 = 0.900 53

54. What is the p robabilit y that a patient d iagnostic receive s at least four services? P(X≥4) = P(X = 4) + P(X = 5) = 0.010 + 0.006 = 0.016 54

55. The Expected Value of a Discrete Random variable If a random variable is able to take on a large number of values, then a probability mass function might not be the most useful way to summarize behavior of the distribution of data. Instead, measures of location and dispersion can be calculated (as long as the data are not categorical) 55

56. The average value assumed by a random variable is called its expected value , or the population mean and i t is represented by E(X) or µ To obtain the expected value of a discrete random variable X, we multiply each possible outcome by its associated probability and sum all values with a probability greater than 56

57. Mean (X) = E(X) = μ = ∑ xi P (X=xi) For example the mean for diagnostic service data: Mean (X) = 0(0.671) +1(0.229) +2(0.053) +3(0.031) +4(0.010) +5(0.006) = 0.498 ≈ 0.5 57

58. The Variance of a Discrete random variable The variance of a random variable X is called the population variance and is represented by Var(X) or σ 2 It quantifies the dispersion of the possible outcomes of X around the expected value μ 58

59. σ 2 = ∑(xi- μ) 2 P(X=xi) = (0− 0.5) 2 (0.671) +(1 − 0.5) 2 (0.229) +(2 − 0.5) 2 (0.053) +(3 − 0.5) 2 (0.031) +(4 − 0.5) 2 (0.010) +(5 − 0.5) 2 (0.006) = 0.782 Standard deviation = σ = √0.782 = 0.884 59

60. Examples of discrete probability distributions are: Binomial distribution Poisson distribution 60

61. 1. Binomial Distribution It is one of the most widely encountered discrete probability distributions. It summarizes the likelihood that a value will take one of two independent values under a given set of parameters Consider dichotomous (binary) random variable – When a single trial of an experiment can result in only one of two mutually exclusive outcomes (success or failure; dead or alive; sick or well, male/female, etc) 61

62. Characteristics of a Binomial Distribution The experiment consist of n identical trials. Only two possible outcomes on each trial. The probability of A ( success ), denoted by p, remains the same from trial to trial. The probability of B ( failure ), denoted by q, q = 1‐ p . 62

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64. •Represents the number of ways of selecting x objects out of n where the order of selection does not matter (Combination) 64

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66. Example: Suppose we know that 40% of a certain population are cigarette smokers. If we take a random sample of 10 people from this population, what is the probability that we will have exactly 4 smokers in our sample? 66

67. If the probability that any individual in the population is a smoker to be P=.40 , then the probability that x=4 smokers out of n=10 subjects selected is: P(X=4) = 10 C 4 (0.4) 4 (1‐0.4) 10‐4 = 10 C 4 (0.4) 4 (0.6) 6 = 210(.0256)(.04666) = 0.25 The probability of obtaining exactly 4 smokers in the sample is 0.25. 67

68. We can compute the probability of observing zero smokers out of 10 subjects selected at random, exactly 1 smoker, and so on, and display the results in a table, as given, below. The third column, P(X ≤ x), gives the cumulative probability. E.g. the probability of selecting 3 or fewer smokers into the sample of 10 subjects is P(X ≤ 3) =.3823, or about 38%. 68

69. Binomial Probability Distribution for n=10, p=0.4. x P(X=x) P(X  x) 0.0060 0.0060 1 0.0404 0.0464 2 0.1209 0.1673 3 0.2150 0.3823 4 0.2508 0.6331 5 0.2007 0.8338 6 0.1114 0.9452 7 0.0452 0.9877 8 0.0106 0.9983 9 0.0016 0.9999 10 0.0001 1.0000 1.0000 69

70. The Mean and Variance of a Binomial Distribution Once n and P are specified, we can compute the proportion of success. P = x/n 70

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72. Example: 70% of a certain population has been immunized for polio. If a sample of size 50 is taken, what is the “expected total number”, in the sample who have been immunized? µ = np = 50(0.70) = 35 This tells us that “on the average” we expect to see 35 immunized subjects in a sample of 50 from this population. 72

73. If repeated samples of size 10 are selected from the population of infants born, the mean number of children per sample who survive to age 7 2 would be µ = np = (10)(0.72) = 7.2 The variance would be npq = (10)(0.72)(0.28) = 2.02 and the SD would be √ 2.02 = 1.42 73

74. 11/28/2022 Exercise Suppose that in a certain malarious area past experience indicates that the probability of a person with a high fever will be positive for malaria is 0.7. Consider 3 randomly selected patients (with high fever) in that same area . a) What is the probability that no patient will be positive for malaria? b) What is the probability that exactly one patient will be positive for malaria? c) What is the probability that exactly two of the patients will be positive for malaria? d) What is the probability that all patients will be positive for malaria?

75. Poisson distribution Poisson distribution is a  discrete probability distribution used to show how many times an event is likely to occur within a specified period of time When the probability of “success” is very small e.g., the probability of a mutation, then P x and (1-P) 1-x become too small to calculate exactly with binomial distribution Poisson distribution becomes useful

76. Suppose events happen randomly and independently in time at a constant rate . If events happen with rate  events per unit time, the probability of x events happening in unit time is: Poisson distribution…

77. where x = 0, 1, 2, . . . ∞ x is a potential outcome of X The constant λ (lambda) represents the rate at which the event occurs, or the expected number of events per unit time e = 2.71828 It depends up on just one parameter, which is the µ number of occurrences ( λ ). Poisson distribution….

78. Where x is the number of success, λ is the expected number of successes in a process consisting of n trials, i.e., λ = np. The mean and variance of a Poisson distributed variable are given by: mean = variance = λ Poisson distribution…

79. Example The daily number of new registrations of cancer case is 2.2 on average. What is the probability of Getting no new cases Getting 1 case Getting 2 cases Getting 3 cases Getting 4 cases

80. Solutions a) b) P(X=1) = 0.244 c) P(X=2) = 0.268 d) P(X=3) = 0.197 e) P(X=4) = 0.108

81. 0 1 2 3 4 5 6 7 0.3 0.2 0.1 0.0 Probability Poisson distribution with mean 2.2

82. B. Continuous Probability Distributions A continuous random variable X can take on any value in a specified interval or range The probability distribution of X is represented by a smooth curve called a probability density function . 82

83. 11/28/2022 Distribution of serum triglyceride The area under the smooth curve is equal to 1 The area under the curve between any two points x1 and x2 is the probability that X takes a value between x1 and x2

84. 11/28/2022 The probability distribution of a continuous random variable is called a continuous probability distribution. Continuous Probability Distributions

85. Continuous Probability Distributions 11/28/2022 There are infinite number of continuous random variables In case of continuous probability distribution instead of assigning probabilities to specific outcomes of the random variable X, probabilities are assigned to ranges of values The probability associated with any one particular value is equal to 0, P(X=x) = Therefore, P(X ≥ x) = P(X > x)

86. We can calculate the probability as : Pr [ a < X < b] , the probability of an interval of values of X. For this reason the probability associated summation notation at specific value of X have no values = 0 86

87. The Normal distribution The ND is the most important probability distribution in statistics Frequently called the “Gaussian distribution” or bell‐shape curve. Variables such as blood pressure , weight, height, serum cholesterol level, and IQ score are approximately normally distributed 87

88. A random variable X is said to have a normal distribution if it has a probability distribution that is symmetric and bell-shaped 88

89. 1. The mean µ tells you about location - Increase µ - Location shifts right Decrease µ – Location shifts left Shape is unchanged The variance σ 2 tells you about narrowness or flatness of the bell - Increase σ 2 - Bell flattens. Extreme values are more likely Decrease σ 2 - Bell narrows. Extreme values are less likely Location is unchanged 89

90. Properties of normal distribution 11/28/2022 It is symmetrical about its mean, m. The mean, the median and mode are almost equal. It is unimodal. The total area under the curve about the x‐ axis is 1 The curve never touches the x‐axis. As the value of σ 2 increases, the curve becomes more and more flat and vice versa.

91. Properties of normal distribution 11/28/2022 ND is more important in statistical work, most estimation procedures and hypothesis tests undergo with ND The concept of “probability of X=x” in the discrete probability distribution is replaced by the “probability density function f(x) The ND is also an approximating distribution to other distributions (e.g., binomial)

92. Properties of normal distribution 11/28/2022 The formula that generates the normal probability distribution is:- where -  < x <  . This is a bell shaped curve with different centers and spreads depending on  and  This is a bell shaped curve with different centers and spreads depending on  and 

93. Properties of normal distribution 11/28/2022 π (pi) = 3.14159 e = 2.71828, x = Value of X Range of possible values of X: -∞ to +∞ µ = Expected value of X (“the long run average”) σ 2 = Variance of X. µ and σ are the parameters of the normal distribution — they completely define its shape & location

94. 8. 94 Properties of normal distribution The normal distribution is described by two parameters: its mean and its standard deviation . Increasing the mean shifts the curve to the right…

95. 8. 95 Properties of normal distribution Increasing the standard deviation “ flattens ” the curve…

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

97. The area under the Normal curve The area under a curve can be obtained: a. By taking the integral of an interval, (a, b)

98. The area under the curve… b. By preparing a tables containing areas for each curve However, both of these are not good solutions because: i . Either it requires us to have some knowledge of calculus or ii. Preparing tables for the infinite family of normal curves is impossible

99. Standard Normal Distribution It is a normal distribution that has a mean equal to 0 and a SD equal to 1, and is denoted by N(0, 1). Standardization solves the above two problems The main idea is to standardize all the data that is given by using Z‐scores. These Z‐scores can then be used to find the area (and thus the probability) under the normal curve. 99

100. Z - Transformation If a r andom v ariable X ~N (  ,  ) t h e n w e c a n transform it to a SN D with the help of Z ‐ T ransformation Z represents the Z-score for a given x value 100

101. The Standard Normal Distribution... The resulting distribution will be the standard normal with a mean of 0 and a standard deviation of 1

102. The Standard Normal Distribution… The area that falls in the interval under the unstandardized normal curve (the x-values) is the same as the area under the standard normal curve(within the corresponding z-boundaries) That means standardization preserves area. After the formula is used to transform an x-value into a z-score, a Standard Normal Table can be used to find the cumulative area under the curve .

103. Properties of the Standard Normal Distribution 1. The cumulative area is close to 0 for z-scores close to z = −3.49. 2. The cumulative area increases as the z-scores increase. 3. The cumulative area for z = 0 is 0.5000. 4. The cumulative area is close to 1 for z-scores close to z = 3.49

104. Properties of the Standard Normal Distribution

105. Approximately 68% of the area under the standard normal curve lies between ± 1, about 95% between ± 2, and about 99% between ± 2.5

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108. This process is known as standardization and gives the position on a normal curve with μ=0 and σ=1, i.e., the SND, Z. Z‐score is the number of standard deviations that a given x value is above or below the mean (μ). 108

109. Example1:Find the cumulative area that corresponds to a z-score of 2.71 Find the area by finding 2.7 in the left hand column, and then moving across the row to the column under 0.01 The area to the left of z = 2.71 is 0.9966

110. Example: Find the cumulative area that corresponds to a z-score of -0.25. Find the area by finding -0.2 in the left hand column, and then moving across the row to the column under 0.05. The area to the left of z = -0.25 is 0.4013

111. Guidelines for Finding Areas Finding Areas Under the Standard Normal Curve 1. Sketch the standard normal curve and shade the appropriate area under the curve. 2. Find the area by following the directions for each case shown a. To find the area to the left of z, find the area that corresponds to z in the Standard Normal Table.

112. Find the area that corresponds to a z-score of 1 .23

113. b. To find the area to the right of z, use the Standard Normal Table to find the area that corresponds to z. Then subtract the area from 1

114. Examples a. What is the probability that z < ‐1.96? Sketch a normal curve Draw a perpendicular line for z = ‐1.9 Find the area in the table The answer is the area to the left of the line P(z < ‐ 1.96) = 0.0250 use the standard normal table. 114

115. b. What is the probability that -1.96 < z < 1.96? The area between the values P(-1.96 < z < 1.96) = .9750 - .0250 = .9500 Or =1- 2*(0.0250) = 0.9500 115

116. Th e ND is u se d as a mode l t o study many different variables. The ND can be used to answer probability questions about continuous random variables. Following the model of the ND, a given value of x must be converted to a z score before it can be looked up in the z table. 116

117. a. Suppose x is distributed normally with mea n 8. an d standar d deviatio n 5.0 . Find P(8 < x < 8.6) 117

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120. Suppos e x i s n o r m a l w it h me a n 8. 0 and standard deviation 5.0. Now Find P(7.4 < x < 8) 120

121. Example: The DBP of males 35–44 years of age are normally distributed with µ = 80 mm Hg and σ 2 = 144 mm Hg2 σ = 12 mm Hg Let individuals with BP above 95 mm Hg are considered to be hypertensive 121

122. a. What is the probability that a randomly selected male has a BP above 95 mm Hg? Approximately 10.6% of this population would be classified as hypertensive 122

123. b. What is the probability that a randomly selected male has a DBP above 110 mm Hg? Z = 110 – 80 = 2.50 12 P (Z > 2.50) = 0.0062 Approximately 0.6% of the population has a DBP above 110 mm Hg 123

124. c. What is the probability that a randomly selected male has a DBP below 60 mm Hg? Z = 60 – 80 = ‐1.67 12 P (Z < ‐1.67) = 0.0475 Approximately 4.8% of the population has a DBP below 60 mm Hg 124

125. d. What value of DBP cuts off the upper 5% of this population? Looking at the table, the value Z = 1.645 cuts off an area of 0.05 in the upper tail We want the value of X that corresponds to Z = 1.645 X = 99.7 Approximately 5% of the men in this population have a DBP greater than 99.7 mm Hg   125

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128. Compute P(-1 ≤ Z ≤ 1.5) Ans: _______ Fin d th e a re a unde r th e S N D fro m t o 1.45 Ans: ______ 3. Compute P(-1.66 < Z < 2.85) Ans: ______ 128 Exercise

129. 11/28/2022 Answer 1) 0.7745 , 2) 0.4265 3) 0.9493