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Ch5: Discrete Probability Distributions Section 5-1: Probability Distribution Ch5: Discrete Probability Distributions Section 5-1: Probability Distribution Recall: A variable is a characteristic or attribute that can assume different values. o Various letters of the alphabet (e.g. X, Y, Z) are used to represent variables. A random variable is a variable whose values are determined by chance. Discrete variables are countable. Example: Roll a die and let X represent the outcome so X = {1,2,3,4,5,6} Ch5: Discrete Probability Distributions Santorico - Page 147 Discrete probability distribution - the values a random variable can assume and the corresponding probabilities of the values. The probabilities may be determined theoretically or by observation. They can be displayed by a graph or a table. How does this connect to our frequency distributions, tables and graphs from Chapter 2? Ch5: Discrete Probability Distributions Santorico - Page 148 Example: Create a probability distribution for the number of girls out of 3 children. We previously used a tree diagram to construct the sample space which consisted of 8 possible outcomes: BBB X=0 BBG, BGB, GBB X=1 BGG, GBG, GGB X=2 GGG X=3 The corresponding (discrete) probability distribution is: Number of Girls X 0 1 2 3 Probability P(X) 1/8 3/8 3/8 1/8 Check of calculations in table: MUST SUM TO 1! Ch5: Discrete Probability Distributions Santorico - Page 149 Graph the probability distribution above. 0.35 0.30 0.25 Probability 0.20 0.15 0.0 0.5 1.0 1.5 2.0 2.5 3.0 Number of Girls Ch5: Discrete Probability Distributions Santorico - Page 150 Example: The World Series played by Major League Baseball is a 4 to 7 game series won by the team winning four games. The data shown consists of the number of games played in the World Series from 1965 through 2005. The number of games played is represented by the variable X. Ch5: Discrete Probability Distributions Santorico - Page 151 Construct the corresponding discrete probability distribution and graph the probability distribution above. Ch5: Discrete Probability Distributions Santorico - Page 152 Two Requirements for a Probability Distribution 1. The sum of the probabilities of all the outcomes in the sample space must be 1; that is P(X) 1. 2. The probability of each outcome in the sample space must be between or equal to 0 and 1; that is 0 P(X) 1. These are good checks for you to use after you have computed a discrete probability distribution! The “sums to 1” check will often find a calculation error! Ch5: Discrete Probability Distributions Santorico - Page 153 Example: Determine whether each distribution is a probability distribution. Explain. Ch5: Discrete Probability Distributions Santorico - Page 154 Section 5-2: Mean, Variance, Standard Deviation, and Expectation The mean, variance, and standard deviation for a probability distribution are computed differently from the mean, variance, and standard deviation for sample. Recall that a parameter is a numerical characteristic of a population. The mean of a probability distribution is denoted by the symbol, . The mean of a probability distribution for a discrete random variable is XPX() where the sum is taken over all possible values of X. Rounding Rule: Round to one more decimal place than the outcome X when finding the mean, variance, and standard deviation for variables of a probability distribution. Ch5: Discrete Probability Distributions Santorico - Page 155 Example: Find the mean number of girls in a family with two children using the probability distribution below. 1 1 1 XPX ( ) 0 1 2 1 4 2 4 X Ch5: Discrete Probability Distributions Santorico - Page 156 Example: Find the mean number of trips lasting five nights or longer that American adults take per year using the probability distribution below. X P(X) Ch5: Discrete Probability Distributions Santorico - Page 157 Variance and Standard Deviation The variance of a probability distribution, σ², for a discrete random variable is found by multiplying the square of each outcome, X, by its corresponding probability, summing those products, and subtracting the square of the mean. 2 X2 P(X) 2 The standard deviation, σ, of a probability distribution is: 2 Ch5: Discrete Probability Distributions Santorico - Page 158 Example: Calculate the variance and standard deviation for the number of girls in the previous example: 2[XPX 2 ( )] 2 [XPX2 ( )] 12 1 1 1 3 1 02 1 2 2 2 1 1 4 2 4 2 2 1 2 0.707 2 Ch5: Discrete Probability Distributions Santorico - Page 159 Example: Calculate the variance and standard deviation for the number of trips five nights or more in the previous example. 2 [X 2 P(X)]2 2 Ch5: Discrete Probability Distributions Santorico - Page 160 Expectation Another concept closely related to the mean of a probability distribution is the concept of expectation. Expected value has wide uses in the insurance industry, gambling, and other areas such as decision theory. The expected value of a discrete random variable of a probability distribution is the theoretical average of the variable. E(X) X P(X) Does this look familiar? Ch5: Discrete Probability Distributions Santorico - Page 161 Example: Suppose one thousand tickets are sold at $10 each to win a used car valued at $5,000. What is the expected value of the gain if a person purchases one ticket? The person will either win or lose. If they win which will happen with probability 1/1000, they have gained $5000-$10. If they lose, they have lost $10. Gain, X Probability, P(X) $4990 1/1000 -$10 999/1000 1 999 EX $4990 ( $10) $5 1000 1000 Ch5: Discrete Probability Distributions Santorico - Page 162 Example: Suppose one thousand tickets are sold at $1 each for 3 prizes of $150, $100, and $50. After each prize drawing, the winning ticket is then returned to the pool of tickets. What is the expected value if a person purchases 3 tickets? Gain, X Probability, P(X) E(X) = Ch5: Discrete Probability Distributions Santorico - Page 163 When gambling: If the expected value of the game is zero, the game is said to be fair. If the expected value of a game is positive, then the game is in the favor of the player. If the expected value of the game is negative, then the game is said to be in the favor of the house. o This means you lose should expect to lose money in the long run. o Every game in Las Vegas has a negative expected value!!! Ch5: Discrete Probability Distributions Santorico - Page 164 Section 5-3: The Binomial Distribution A binomial experiment is a probability experiment that satisfies the following four requirements: 1. Each of the n trials has two possible outcomes or can be reduced to two outcomes: “success” and “failure”. The outcome of interest is called a success and the other outcome is called a failure. 2. The outcomes of each trial must be independent of each other. 3. There must be a fixed number of trials. 4. Each trial has the same probability of success, denoted by p. Ch5: Discrete Probability Distributions Santorico - Page 165 The acronym BINS may help you remember the conditions: B – Binary outcomes I – Independent outcomes N – number of trials is fixed S – same probability of success Examples: Ch5: Discrete Probability Distributions Santorico - Page 166 Notation: P (S), probability of success P (F), probability of failure p, the numerical probability of success q, The numerical probability of failure P(S) p and P(F) 1P(S) 1 p q n, the number of trials. X, the number of successes in n trials. NOTE: 0 X n and Xn 0,1,2,3,..., Binomial distribution – the outcomes of a binomial experiment along with the probabilities of these outcomes. Ch5: Discrete Probability Distributions Santorico - Page 167 Probabilities for a Binomial Distribution In a binomial experiment, the probability of exactly X successes in n trials is n! X nX P(X) p (1 p) . X!(n X)! Note: x ! stands for x factorial where x is a non-negative integer. x! x ( x 1)( x 2)...(2)(1) when x > 0 0 !1 You can use your calculator or Table C at the back of the book to solve binomial probabilities for selected values of n and p. Ch5: Discrete Probability Distributions Santorico - Page 168 Examples: 5! = 5*4*3*2*1 = 120 8! = 5! = 5*4=20 3! 25! = Ch5: Discrete Probability Distributions Santorico - Page 169 Example: Dionne Warwick claims to possess ESP. An experiment is conducted to test her. A person in one room picks one of the integers 1, 2, 3, 4, 5 at random. In another room, Dionne identifies the number she believes was picked. The experiment is done with eight trials. Dionne gets the correct answer four times. If Dionne does not actually have ESP and is actually guessing the number, what is the probability that she’d make a correct guess in four of the eight trials? We have a Binomial experiment here since with each guess she will either be right (success) or wrong (failure). If she does not have ESP, then the probability of a correct guess is 1/5. Hence, we would like to know P(X=4) given we have a Binomial distribution with n=8 and p=1/5. 4 8 4 8! 1 4 PX( 4) 0.0459 4!8! 5 5 Ch5: Discrete Probability Distributions Santorico - Page 170 Example: Consider a family with six children and suppose there is a 25% chance that each child will be a carrier of a particular mutated gene, independent of the other children. What is the probability that exactly 2 of the children will carry the mutated gene? What is the probability that 2 or less children will carry the mutated gene? Ch5: Discrete Probability Distributions Santorico - Page 171 Binomial Mean and Standard Deviation The binomial probability distribution for n trials with probability p of success on each trial has mean , variance 2, and standard deviation given by: 2 np npq npq Ch5: Discrete Probability Distributions Santorico - Page 172 Example: You will take a 10 question multiple-choice test with 4 possible answers for each question.
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