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1 Overview This chapter will deal with the construction of probability distributions by combining the methods of Chapter 2 with the those of Chapter 4. Probability Distributions will describe what will probably happen instead of what actually did happen.

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2 Combining Descriptive Statistics Methods and Probabilities to Form a Theoretical Model of Behavior 4 5

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3 Definitions Random Variable a variable (typically represented by x ) that has a single numerical value, determined by chance, for each outcome of a procedure Probability Distribution a graph, table, or formula that gives the probability for each value of the random variable

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4 Probability Distribution Number of Girls Among Fourteen Newborn Babies 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0.000 0.001 0.006 0.022 0.061 0.122 0.183 0.209 0.183 0.122 0.061 0.022 0.006 0.001 0.000 x P(x)P(x)

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5 Probability Histogram

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6 Definitions Discrete random variable has either a finite number of values or countable number of values, where ‘countable’ refers to the fact that there might be infinitely many values, but they result from a counting process. Continuous random variable has infinitely many values, and those values can be associated with measurements on a continuous scale with no gaps or interruptions.

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7 Requirements for Probability Distribution P ( x ) = 1 where x assumes all possible values 0 P ( x ) 1 for every value of x

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8 µ = [x P (x)] 2 = [ (x - µ) 2 P (x )] 2 = [ x 2 P (x )] - µ 2 (shortcut) = [ x 2 P (x )] - µ 2 Mean, Variance and Standard Deviation of a Probability Distribution

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9 Roundoff Rule for µ, 2, and Round results by carrying one more decimal place than the number of decimal places used for the random variable x. If the values of x are integers, round µ, 2, and to one decimal place.

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10 Definition Expected Value The average value of outcomes E = [ x P( x )]

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11 E = [x P(x)] Event Win Lose x $499 - $1 P(x) 0.001 0.999 x P(x) 0.499 - 0.999 E = -$.50

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13 Definitions Binomial Probability Distribution 1.The experiment must have a fixed number of trials. 2.The trials must be independent. (The outcome of any individual trial doesn’t affect the probabilities in the other trials.) 3.Each trial must have all outcomes classified into two categories. 4.The probabilities must remain constant for each trial.

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14 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0.206 0.343 0.267 0.129 0.043 0.010 0.002 0.0+ P( x ) n x 15 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0.206 0.343 0.267 0.129 0.043 0.010 0.002 0.000 P( x ) x Table B Binomial Probability Distribution For n = 15 and p = 0.10

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15 Example: Using Table B for n = 5 and p = 0.90, find the following: a) The probability of exactly 3 successes b) The probability of at least 3 successes a) P(3) = 0.073 b) P(at least 3) = P(3 or 4 or 5) = P(3) or P(4) or P(5) = 0.073 + 0.328 + 0.590 = 0.991

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16 This is a binomial experiment where: n = 5 x = 3 p = 0.90 q = 0.10 Using the binomial probability chart to solve: P(3)= 0.0.0729 Example : Find the probability of getting exactly 3 correct responses among 5 different requests from AT&T directory assistance. Assume in general, AT&T is correct 90% of the time.

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17 Notation for Binomial Probability Distributions n = fixed number of trials x = specific number of successes in n trials p = probability of success in one of n trials q = probability of failure in one of n trials ( q = 1 - p ) P( x )= probability of getting exactly x success among n trials Be sure that x and p both refer to the same category being called a success.

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18 P(x) = p x q n-x n !n ! ( n - x ) ! x ! Number of outcomes with exactly x successes among n trials Probability of x successes among n trials for any one particular order Binomial Probability Formula

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19 P(x) = p x q n-x ( n - x ) ! x ! Binomial Probability Formula n !n ! Method 1 P(x) = n C x p x q n-x for calculators with n C r key, where r = x

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20 This is a binomial experiment where: n = 5 x = 3 p = 0.90 q = 0.10 Using the binomial probability formula to solve: P(3) = 5 C 3 0.9 01 = 0.0.0729 Example : Find the probability of getting exactly 3 correct responses among 5 different requests from AT&T directory assistance. Assume in general, AT&T is correct 90% of the time. 3 2

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22 = n p q For Binomial Distributions: µ = n p 2 = n p q

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23 We previously discovered that this scenario could be considered a binomial experiment where: n = 14 p = 0.5 q = 0.5 Using the binomial distribution formulas: µ = (14)(0.5) = 7 girls = (14)(0.5)(0.5) = 1.9 girls (rounded) Example: Find the mean and standard deviation for the number of girls in groups of 14 births.

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24 For this binomial distribution, µ = 50 girls = 5 girls µ + 2 = 50 + 2(5) = 60 µ - 2 = 50 - 2(5) = 40 The usual number girls among 100 births would be from 40 to 60. So 68 girls in 100 births is an unusual result. Example: Determine whether 68 girls among 100 babies could easily occur by chance.

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