2. Probability

3. The definition – probability of an Event Applies only to the special case when 1.The sample space has a finite no.of outcomes, and 2.Each outcome is equi-probable If this is not true a more general definition of probability is required.

4. Summary of the Rules of Probability

5. The additive rule P[A  B] = P[A] + P[B] – P[A  B] and if P[A  B] =  P[A  B] = P[A] + P[B]

6. The Rule for complements for any event E

7. Conditional probability

8. The multiplicative rule of probability and if A and B are independent. This is the definition of independent

9. Counting techniques

10. Summary of counting results Rule 1 n(A 1  A 2  A 3  …. ) = n(A 1 ) + n(A 2 ) + n(A 3 ) + … if the sets A 1, A 2, A 3, … are pairwise mutually exclusive (i.e. A i  A j =  ) Rule 2 n 1 = the number of ways the first operation can be performed n 2 = the number of ways the second operation can be performed once the first operation has been completed. N = n 1 n 2 = the number of ways that two operations can be performed in sequence if

11. Rule 3 n 1 = the number of ways the first operation can be performed n i = the number of ways the i th operation can be performed once the first (i - 1) operations have been completed. i = 2, 3, …, k N = n 1 n 2 … n k = the number of ways the k operations can be performed in sequence if

12. Basic counting formulae 1.Orderings 2.Permutations The number of ways that you can choose k objects from n in a specific order 3.Combinations The number of ways that you can choose k objects from n (order of selection irrelevant)

13. Applications to some counting problems The trick is to use the basic counting formulae together with the Rules We will illustrate this with examples Counting problems are not easy. The more practice better the techniques

14. Random Variables Numerical Quantities whose values are determine by the outcome of a random experiment

15. Random variables are either Discrete –Integer valued –The set of possible values for X are integers Continuous –The set of possible values for X are all real numbers –Range over a continuum.

16. Examples Discrete –A die is rolled and X = number of spots showing on the upper face. –Two dice are rolled and X = Total number of spots showing on the two upper faces. –A coin is tossed n = 100 times and X = number of times the coin toss resulted in a head. –We observe X, the number of hurricanes in the Carribean from April 1 to September 30 for a given year

17. Examples Continuous –A person is selected at random from a population and X = weight of that individual. –A patient who has received who has revieved a kidney transplant is measured for his serum creatinine level, X, 7 days after transplant. –A sample of n = 100 individuals are selected at random from a population (i.e. all samples of n = 100 have the same probability of being selected). X = the average weight of the 100 individuals.

18. The Probability distribution of A random variable A Mathematical description of the possible values of the random variable together with the probabilities of those values

19. The probability distribution of a discrete random variable is describe by its : probability function p(x). p(x) = the probability that X takes on the value x. This can be given in either a tabular form or in the form of an equation. It can also be displayed in a graph.

20. Example 1 Discrete –A die is rolled and X = number of spots showing on the upper face. x123456 p(x)1/6 formula –p(x) = 1/6 if x = 1, 2, 3, 4, 5, 6

21. Graphs To plot a graph of p(x), draw bars of height p(x) above each value of x. Rolling a die

22. Example 2 –Two dice are rolled and X = Total number of spots showing on the two upper faces. x 23456789101112 p(x) 1/362/363/364/365/366/365/364/363/362/361/36 Formula:

23. Rolling two dice

24. 36 possible outcome for rolling two dice

25. Comments: Every probability function must satisfy: 1.The probability assigned to each value of the random variable must be between 0 and 1, inclusive: 2.The sum of the probabilities assigned to all the values of the random variable must equal 1: 3.

26. Example In baseball the number of individuals, X, on base when a home run is hit ranges in value from 0 to 3. The probability distribution is known and is given below: PX()the random variable equals 2 p()  2 3 14 Note: This chart implies the only values x takes on are 0, 1, 2, and 3. If the random variable X is observed repeatedly the probabilities, p(x), represents the proportion times the value x appears in that sequence.

27. A Bar Graph

28. Discrete Random Variables Discrete Random Variable: A random variable usually assuming an integer value. a discrete random variable assumes values that are isolated points along the real line. That is neighbouring values are not “possible values” for a discrete random variable Note: Usually associated with counting The number of times a head occurs in 10 tosses of a coin The number of auto accidents occurring on a weekend The size of a family

29. Continuous Random Variables Continuous Random Variable: A quantitative random variable that can vary over a continuum A continuous random variable can assume any value along a line interval, including every possible value between any two points on the line Note: Usually associated with a measurement Blood Pressure Weight gain Height

30. Probability Distributions of Continuous Random Variables

31. Probability Density Function The probability distribution of a continuous random variable is describe by probability density curve f(x).

32. Notes: The Total Area under the probability density curve is 1. The Area under the probability density curve is from a to b is P[a < X < b].

33. Normal Probability Distributions (Bell shaped curve)

34. Mean and Variance (standard deviation) of a Discrete Probability Distribution Describe the center and spread of a probability distribution The mean (denoted by greek letter  (mu)), measures the centre of the distribution. The variance (  2 ) and the standard deviation (  ) measure the spread of the distribution.  is the greek letter for s.

35. Mean of a Discrete Random Variable The mean, , of a discrete random variable x is found by multiplying each possible value of x by its own probability and then adding all the products together: Notes: The mean is a weighted average of the values of X. The mean is the long-run average value of the random variable. The mean is centre of gravity of the probability distribution of the random variable

37.  2 Variance and Standard Deviation Variance of a Discrete Random Variable: Variance,  2, of a discrete random variable x is found by multiplying each possible value of the squared deviation from the mean, (x   ) 2, by its own probability and then adding all the products together: Standard Deviation of a Discrete Random Variable: The positive square root of the variance:

38. Example The number of individuals, X, on base when a home run is hit ranges in value from 0 to 3.

39. Computing the mean: Note: 0.929 is the long-run average value of the random variable 0.929 is the centre of gravity value of the probability distribution of the random variable

40. Computing the variance: Computing the standard deviation:

41. Random Variables Numerical Quantities whose values are determine by the outcome of a random experiment

42. Random variables are either Discrete –Integer valued –The set of possible values for X are integers Continuous –The set of possible values for X are all real numbers –Range over a continuum.

43. The Probability distribution of A random variable A Mathematical description of the possible values of the random variable together with the probabilities of those values

44. The probability distribution of a discrete random variable is describe by its : probability function p(x). p(x) = the probability that X takes on the value x. This can be given in either a tabular form or in the form of an equation. It can also be displayed in a graph.

45. Example In baseball the number of individuals, X, on base when a home run is hit ranges in value from 0 to 3. The probability distribution is known and is given below: PX()the random variable equals 2 p()  2 3 14 Note: This chart implies the only values x takes on are 0, 1, 2, and 3. If the random variable X is observed repeatedly the probabilities, p(x), represents the proportion times the value x appears in that sequence.

46. A Bar Graph

47. Probability Distributions of Continuous Random Variables

48. Probability Density Function The probability distribution of a continuous random variable is describe by probability density curve f(x).

49. Notes: The Total Area under the probability density curve is 1. The Area under the probability density curve is from a to b is P[a < X < b].

50. Mean, Variance and standard deviation of Random Variables Numerical descriptors of the distribution of a Random Variable

51. Mean of a Discrete Random Variable The mean, , of a discrete random variable x is found by multiplying each possible value of x by its own probability and then adding all the products together: Notes: The mean is a weighted average of the values of X. The mean is the long-run average value of the random variable. The mean is centre of gravity of the probability distribution of the random variable

53.  2 Variance and Standard Deviation Variance of a Discrete Random Variable: Variance,  2, of a discrete random variable x is found by multiplying each possible value of the squared deviation from the mean, (x   ) 2, by its own probability and then adding all the products together: Standard Deviation of a Discrete Random Variable: The positive square root of the variance:

54. Example The number of individuals, X, on base when a home run is hit ranges in value from 0 to 3.

55. Computing the mean: Note: 0.929 is the long-run average value of the random variable 0.929 is the centre of gravity value of the probability distribution of the random variable

56. Computing the variance: Computing the standard deviation:

57. The Binomial distribution An important discrete distribution

58. Situation - in which the binomial distribution arises We have a random experiment that has two outcomes –Success (S) and failure (F) –p = P[S], q = 1 - p = P[F], The random experiment is repeated n times independently X = the number of times S occurs in the n repititions Then X has a binomial distribution

59. Example A coin is tosses n = 20 times –X = the number of heads –Success (S) = {head}, failure (F) = {tail –p = P[S] = 0.50, q = 1 - p = P[F]= 0.50 An eye operation has %85 chance of success. It is performed n =100 times –X = the number of Sucesses (S) –p = P[S] = 0.85, q = 1 - p = P[F]= 0.15 In a large population %30 support the death penalty. A sample n =50 indiviuals are selected at random –X = the number who support the death penalty (S) –p = P[S] = 0.30, q = 1 - p = P[F]= 0.70

60. The Binomial distribution 1.We have an experiment with two outcomes – Success(S) and Failure(F). 2.Let p denote the probability of S (Success). 3.In this case q=1-p denotes the probability of Failure(F). 4.This experiment is repeated n times independently. 5.X denote the number of successes occuring in the n repititions.

61. The possible values of X are 0, 1, 2, 3, 4, …, (n – 2), (n – 1), n and p(x) for any of the above values of x is given by: X is said to have the Binomial distribution with parameters n and p.

62. Summary: X is said to have the Binomial distribution with parameters n and p. 1.X is the number of successes occurring in the n repetitions of a Success-Failure Experiment. 2.The probability of success is p. 3. The probability function

63. Example: 1.A coin is tossed n = 5 times. X is the number of heads occurring in the 5 tosses of the coin. In this case p = ½ and x012345 p(x)

64. Note:

66. Computing the summary parameters for the distribution – ,  2, 

67. Computing the mean: Computing the variance: Computing the standard deviation:

68. Example: A surgeon performs a difficult operation n = 10 times. X is the number of times that the operation is a success. The success rate for the operation is 80%. In this case p = 0.80 and X has a Binomial distribution with n = 10 and p = 0.80.

69. Computing p(x) for x = 0, 1, 2, 3, …, 10

70. The Graph

71. Computing the summary parameters for the distribution – ,  2, 

72. Computing the mean: Computing the variance: Computing the standard deviation:

73. Notes The value of many binomial probabilities are found in Tables posted on the Stats 245 site. The value that is tabulated for n = 1, 2, 3, …,20; 25 and various values of p is: Hence The other table, tabulates p(x). Thus when using this table you will have to sum up the values

74. Example Suppose n = 8 and p = 0.70 and we want to compute P[X = 5] = p(5) Table value for n = 8, p = 0.70 and c =5 is 0.448 = P[X ≤ 5] P[X = 5] = p(5) = P[X ≤ 5] - P[X ≤ 4] = 0.448 – 0.194 =.254

75. We can also compute Binomial probabilities using Excel =BINOMDIST(x, n, p, FALSE) The function will compute p(x). =BINOMDIST(c, n, p, TRUE) The function will compute

76. Mean, Variance and standard deviation of Binomial Random Variables

77. Mean of a Discrete Random Variable The mean, , of a discrete random variable x Notes: The mean is a weighted average of the values of X. The mean is the long-run average value of the random variable. The mean is centre of gravity of the probability distribution of the random variable

78.  2 Variance and Standard Deviation Variance of a Discrete Random Variable: Variance,  2, of a discrete random variable x Standard Deviation of a Discrete Random Variable: The positive square root of the variance:

79. The Binomial ditribution X is said to have the Binomial distribution with parameters n and p. 1.X is the number of successes occurring in the n repetitions of a Success-Failure Experiment. 2.The probability of success is p. 3. The probability function

80. Mean,Variance & Standard Deviation of the Binomial Ditribution The mean, variance and standard deviation of the binomial distribution can be found by using the following three formulas:

81. Solutions: 1)n = 20, p = 0.75, q = 1 - 0.75 = 0.25  np()(0.)207515  npq()(0. )..2075253751936 Example: Find the mean and standard deviation of the binomial distribution when n = 20 and p = 0.75 px x x xx ()(0.) )          20 7525 20 for0, 1, 2,..., 20 2)These values can also be calculated using the probability function:

82. Table of probabilities

83. Computing the mean: Computing the variance: Computing the standard deviation:

84. Histogram  

85. Probability Distributions of Continuous Random Variables

86. Probability Density Function The probability distribution of a continuous random variable is describe by probability density curve f(x).

87. Notes: The Total Area under the probability density curve is 1. The Area under the probability density curve is from a to b is P[a < X < b].

88. Normal Probability Distributions

89. The normal probability distribution is the most important distribution in all of statistics Many continuous random variables have normal or approximately normal distributions

90. The Normal Probability Distribution Points of Inflection

91. Main characteristics of the Normal Distribution Bell Shaped, symmetric Points of inflection on the bell shaped curve are at  –  and  +  That is one standard deviation from the mean Area under the bell shaped curve between  –  and  +  is approximately 2/3. Area under the bell shaped curve between  – 2  and  + 2  is approximately 95%.

92. There are many Normal distributions depending on by  and  Normal  = 100,  = 40Normal  = 140,  =20 Normal  = 100,  =20

93. The Standard Normal Distribution  = 0,  = 1

94. There are infinitely many normal probability distributions (differing in  and  ) Area under the Normal distribution with mean  and standard deviation  can be converted to area under the standard normal distribution If X has a Normal distribution with mean  and standard deviation  than has a standard normal distribution. z is called the standard score (z-score) of X.

95. Converting Area under the Normal distribution with mean  and standard deviation  to Area under the standard normal distribution

96. Perform the z-transformation then Area under the Normal distribution with mean  and standard deviation  Area under the standard normal distribution

97. Area under the Normal distribution with mean  and standard deviation   

98. Area under the standard normal distribution 0 1

99. Using the tables for the Standard Normal distribution

100. Table, Posted on stats 245 web site The table contains the area under the standard normal curve between -∞ and a specific value of z

101. Example Find the area under the standard normal curve between z = -∞ and z = 1.45 A portion of Table 3: z0.000.010.020.030.040.050.06 1.40.9265............

102. Pz(0.)  98.01635 Example Find the area to the left of -0.98; P(z < -0.98)

103. Example Find the area under the normal curve to the right of z = 1.45; P(z > 1.45)

104. Example Find the area to the between z = 0 and of z = 1.45; P(0 < z < 1.45) Area between two points = differences in two tabled areas

105. Notes Use the fact that the area above zero and the area below zero is 0.5000 the area above zero is 0.5000 When finding normal distribution probabilities, a sketch is always helpful

106. Example: Find the area between the mean (z = 0) and z = -1.26

107. Example: Find the area between z = -2.30 and z = 1.80

108. Example: Find the area between z = -1.40 and z = -0.50

109. Computing Areas under the general Normal Distributions (mean , standard deviation  ) 1.Convert the random variable, X, to its z-score. Approach: 3.Convert area under the distribution of X to area under the standard normal distribution. 2.Convert the limits on random variable, X, to their z-scores.

110. Example 1: Suppose a man aged 40-45 is selected at random from a population. X is the Blood Pressure of the man. Assume that X has a Normal distribution with mean  =180 and a standard deviation  = 15. X is random variable.

111. The probability density of X is plotted in the graph below. Suppose that we are interested in the probability that X between 170 and 210.

112. Let Hence

115. Example 2 A bottling machine is adjusted to fill bottles with a mean of 32.0 oz of soda and standard deviation of 0.02. Assume the amount of fill is normally distributed and a bottle is selected at random: 1)Find the probability the bottle contains between 32.00 oz and 32.025 oz 2)Find the probability the bottle contains more than 31.97 oz

116. When x = 32.00 Solution part 1) When x = 32.025

117. PXP X Pz (.) 0.. (.). 32.032025 32.0 02 32.0 02 3202532.0 02 012503944              Graphical Illustration:

118. PxP x Pz(.). (....             3197 32.0 0.02 319732.0 0.02 150) 100000066809332 Example 2, Part 2)

119. Summary Random Variables Numerical Quantities whose values are determine by the outcome of a random experiment

120. Types of Random Variables Discrete Possible values integers Continuous Possible values vary over a continuum

121. The Probability distribution of a random variable A Mathematical description of the possible values of the random variable together with the probabilities of those values

122. The probability distribution of a discrete random variable is describe by its : probability function p(x). p(x) = the probability that X takes on the value x.

124. The Binomial distribution X is said to have the Binomial distribution with parameters n and p. 1.X is the number of successes occurring in the n repetitions of a Success-Failure Experiment. 2.The probability of success is p. 3. The probability function

125. Probability Distributions of Continuous Random Variables

126. Probability Density Function The probability distribution of a continuous random variable is describe by probability density curve f(x).

127. Notes: The Total Area under the probability density curve is 1. The Area under the probability density curve is from a to b is P[a < X < b].

128. The Normal Probability Distribution Points of Inflection

129. Normal approximation to the Binomial distribution Using the Normal distribution to calculate Binomial probabilities

130. Binomial distribution Approximating Normal distribution Binomial distribution n = 20, p = 0.70

131. Normal Approximation to the Binomial distribution X has a Binomial distribution with parameters n and p Y has a Normal distribution

132. Binomial distribution Approximating Normal distribution P[X = a]

135. Example X has a Binomial distribution with parameters n = 20 and p = 0.70

136. Using the Normal approximation to the Binomial distribution Where Y has a Normal distribution with:

137. Hence = 0.4052 - 0.2327 = 0.1725 Compare with 0.1643

138. Normal Approximation to the Binomial distribution X has a Binomial distribution with parameters n and p Y has a Normal distribution

141. Example X has a Binomial distribution with parameters n = 20 and p = 0.70

142. Using the Normal approximation to the Binomial distribution Where Y has a Normal distribution with:

143. Hence = 0.5948 - 0.0436 = 0.5512 Compare with 0.5357

144. Comment: The accuracy of the normal appoximation to the binomial increases with increasing values of n

145. Normal Approximation to the Binomial distribution X has a Binomial distribution with parameters n and p Y has a Normal distribution

146. Example The success rate for an Eye operation is 85% The operation is performed n = 2000 times Find the probability that 1.The number of successful operations is between 1650 and 1750. 2. The number of successful operations is at most 1800.

147. Solution X has a Binomial distribution with parameters n = 2000 and p = 0.85 where Y has a Normal distribution with:

148. = 0.9004 - 0.0436 = 0.8008

149. Solution – part 2. = 1.000

150. Next topic: Sampling TheorySampling Theory