CHAPTER 5 DISCRETE RANDOM VARIABLES AND THEIR PROBABILITY

CHAPTER 5 DISCRETE RANDOM VARIABLES AND THEIR PROBABILITY DISTRIBUTIONS 1

5. 1 PROBABILITY DISTRIBUTION (Random Variable) Tila College is a small private school with 2, 500 students. According to the registrar’s office, the frequency and relative frequency distributions of the number of courses taken by all the students in Fall 2010 are shown in the table below. Table 1: Frequency and Relative Frequency Distribution of the number of Courses Taken by Students Number of Courses Frequency Relative Frequency 1 150 0. 06 2 500 0. 20 3 1000 0. 40 4 600 0. 24 5 250 0. 10 f = 2500 = 1. 00 2

Random Variables Now, suppose we want to conduct an experiment by randomly selecting students from the population of 2, 500 students. Definition 1. The process of randomly selecting a student from the population is called a random or a chance experiment. 2. Let x = the number of courses taken by the randomly selected student. Then x depends on the outcome of the experiment and can assume any values (1, 2, 3, 4, or 5). x is also called a random variable or a chance variable. 3. A random variable is a variable whose value is the outcome of a random or a chance experiment. 4. A random variable x can be either discrete random variable or continuous random variable. 3

Discrete Random Variable Definition p A discrete random variable is a random variable that assumes countable values. p In our example, the number of courses taken by students in Fall 2010 is a discrete random variable because it can assume any value, 1, 2, 3, 4, or 5. p Other examples of discrete random variables are: Ø Ø Ø The number of smokers in STA 120 class The number of cars owned by families in your city The number of fish caught on a fishing trip 4

Continuous Random Variable Definition q A continuous random variable is a random variable that assumes any values within one or more intervals. q Suppose the volume of water used per month per household in Victorville ranges from 2, 000 to 4, 000 gallons. Ø Ø Let’s randomly select a household in the city of Victorville. Let x =the volume of water used per month by the randomly selected household. Since there are infinitely many values between 2, 000 and 4, 000 gallons, the random variable, x, is a continuous random variable and can assume infinite number of values. Graphically, 2, 000 x can assume any value on this number line 4, 000 5

Discrete and Continuous Random Variables Example #1 Classify each of the following random variables as discrete or continuous. a. b. c. d. e. f. The time left on a parking meter The number of bats broken by a major league baseball team in a season The number of cars in a parking lot The total pounds of fish caught on a fishing trip The number of cars crossing a bridge on a given day The time spent by a physician examining a patient Solution a. b. Continuous Discrete d. Discrete Continuous e. Discrete f. Continuous c. 6

PROBABLITY DISTRIBUTION OF A DISCRETE RANDOM VARIABLE Definition A probability distribution of a discrete random variable lists all possible values that the random variable can assume with their corresponding probabilities. q q q As discussed in Section 5. 1, let x = number of courses taken by a randomly selected student. Then, we can write the probability distribution of x as shown to the right. Note that the probabilities are actual since the relative frequencies represent population. As stated in Chapter 4, if the relative frequencies were for a sample, then the probabilities would have been approximate probabilities. 7

Probability Distribution of a Discrete Random Variable q q q A probability distribution of a discrete random variable must satisfy the following conditions: a. The probability of each value of x is from 0 to 1. b. The sum of the probabilities for all values of x is 1. We can read the probability for any value of x as P(x = 2) = P(2) = 0. 20 P(x > 3) = P(4)+ P(5) =0. 24 + 0. 10 = 0. 34 P(x =< 3) = P(1) + P(2) + P(3) =. 06 +. 20 +. 40 = 0. 66 A probability distribution of a discrete random variable can be represented by, a. b. c. Graph Formula Table 8

Probability Distribution of a Discrete Random Variable Example #2 Each of the following tables lists certain values of x and their probabilities. Determine whether or not each one satisfies the two conditions required for a valid probability distribution. (1) (2) (3) x P(x) 5 -. 36 1 . 27 0 . 15 6 . 48 2 . 24 1 . 08 7 . 62 3 . 49 2 . 20 8 . 26 3 . 50 Solution 1. Although the sum of probabilities for all values of x is equal to 1, the probability of x = 5 is negative. Thus, this table does not represent a valid probability distribution. 2. Each probability ranges from 0 to 1, and the sum of the probabilities is 1. 00. Thus, this table represents a valid probability distribution. 3. Although each probability lies between 0 and 1, the sum of the probabilities is less than 1. 0. Thus, this table does not represent a valid probability 9 distribution.

Probability Distribution of a Discrete Random Variable Solution Example #3 The following table gives the probability distribution of a discrete random variable x. x 0 1 2 3 4 5 P(x) . 03 . 17 . 22 . 31 . 15 . 12 From the given table, Find the following probabilities: 10

Probability Distribution of a Discrete Random Variable Example #4 The following table shows the probability distribution for the sum, denoted by x, of the faces on a pair of Nathan’s dice: x 2 3 4 5 6 7 8 9 10 11 12 P(x) . 065 . 08 . 095 . 11 . 17 . 11 . 095 . 08 . 065 a. Draw a bar graph for this probability b. Determine the probability that the sum of the faces on a single roll of Nathan’s dice is: (i) an even number (ii) 7 or 11 (iii) 4 to 6 (iv) no less than 9 Solution 11

Probability Distribution of a Discrete Random Variable Example #5 According to a survey, 30% of adults are against using animals for research. Assume that this result holds true for the current population of all adults. Let x be the number of adults who are against using animals for research in a random sample of two adults. Obtain the probability distribution for x. Draw a tree diagram for this problem. Let A = F= against animal research for animal research Solution x can assume 0, 1, or 2. Thus, the probability distribution is x P(x) 0 P(F)=. 7 (. 7) = 0. 49 1 P(F)P(A)+P(A)P(F)=. 7(. 3)+. 3(. 7) =. 42 2 P(A)=. 3(. 3) =. 09 A. 30 F. 70 P(AA) =. 09 P(AF) =. 21 A. 30 F P(FA) =. 21 . 70 P(FF) =. 49 12

5. 2 MEAN, VARIANCE, STANDARD DEVIAITON AND EXPECTATION The mean of a discrete variable is denoted by µ and is define as the sum of the product of each value of the discrete variable and the corresponding probabilities. It is also called an expected value and denoted by E(x) Formal Definition: The mean of a discrete random variable x is the value that is expected to occur per repetition, on average, if an experiment is repeated a large number of times. It is calculated as µ = E(x) = Σ x P(x) 13

EXPECTED VALUE p The mean of a probability distribution is also the expected value of the discrete random variable. The expected value is expressed as E(x). p We can also interpret this result by saying that if infinitely many families with three children were observed, we would expect an average of 1. 5 girls per family. p µ = E(x) = Σ x P(x) 14

Mean of a Discrete Random Variable Example #6 The following table gives the probability distribution of the number of camcorder sold on a given day at an electronic store Camcorder sold 0 1 2 3 4 5 6 Probability . 05 . 12 . 19 . 30 . 20 . 10 . 04 Calculate the mean for this probability distribution. Give a brief interpretation of the value of the mean. Solution X 0 1 2 3 4 5 6 P(x) . 05 . 12 . 19 . 30 . 20 . 10 . 04 x. P(x) 0 . 12 . 38 . 90 . 80 . 50 . 24 The electronic store could sell 3 camcorders one day, 4 camcorders the next day, 6 camcorders on another day, and none on some other day, but on an average over a certain period, the electronic store is expected 15 to sell about 3 camcorders a day.

STANDARD DEVIATION Definition The standard deviation of a discrete random variable is denoted by and is define as the measure of the spread of its probability distribution. It is computed as, Basic Formula Short-Cut Formula Variance Standard deviation Note p A higher value of about the mean p signifies that x can assume values over a larger range A smaller value of signifies that x can assume values that are closely clustered about the mean. 16

Standard Deviation of a Discrete Random Variable Example #7 The following table gives the probability distribution of the number of camcorder sold on a given day at an electronic store Camcorder sold 0 1 2 3 4 5 6 Probability . 05 . 12 . 19 . 30 . 20 . 10 . 04 Calculate the standard deviation for this probability distribution. Solution X 0 1 2 3 4 5 6 P(x) . 05 . 12 . 19 . 30 . 20 . 10 . 04 x. P(x) 0 . 12 . 38 . 90 . 80 . 50 . 24 X 2 0 1 4 9 16 25 36 x 2 P(x) 0 . 12 . 76 2. 7 3. 2 2. 5 1. 44 17

Standard Deviation of a Discrete Random Variable Example #8 An instant lottery ticket costs $2. Out of a total of 10, 000 tickets printed for this lottery, 1000 tickets contain a price of $5 each, 100 tickets have a price of $10 each, and 5 tickets have a price of $1000 each, and 1 ticket has a price of $5000. Let x be the random variable that denotes the net amount a player wins by playing this lottery. Write the probability distribution of x. Determine the mean and standard deviation of x. How will you interpret the values of the mean and standard deviation. Solution x f P(x) x. P(x) X 2 x 2 P(x) -2 8894 1. 7788 4 3. 5576 3 1000 0. 3000 9 0. 9000 8 100 0. 0800 64 0. 6400 998 5 0. 0005 0. 4990 996004 498. 002 4998 1 0. 0001 0. 4998 24980004 2498. 0004 Players are expected to lose, on an average, 40 cents per ticket with a standard 18 deviation of $54. 78.

5. 3 THE BINOMIAL PROBABILITY DISTRIBUTION Suppose we want to find the probability that head will turn up 6 times when a coin is tossed 10 times. The binomial probability distribution is best suited for this type of question. To use the binomial probability distribution, the experiment must meet the following conditions: 1. There are n identical trials. In other words, the experiment is repeated n times and each repetition is performed under identical conditions. Note that each repetition of the experiment is called a trial or a Bernoulli trial. 2. Each trial has only two possible outcomes. That is, the outcome is either a “success” or a “failure. ” 3. The probability of each outcome remains constant. The probability of success is denoted by p and probability of failure is denoted by q. Therefore, p + q = 1. 4. The trials are independent. In other words, the occurrence of one trial does not affect the probability of the other. Note: a. Success refers to the outcome to which the question is stated. Then failure refers to the outcome that the question does not stated. b. Success does not mean that the outcome is favorable. Likewise, 19 failure does not mean that the outcome is unfavorable.

The Binomial Probability Distribution Example #14 Which is the following are binomial experiment? Explain why. a. Rolling a die many times and observing the number of spots. b. Rolling a die many times and observing whether the number obtained is even or odd? c. Selecting a few voters from a very large population of voters and observing whether or not each of them favors a certain proposition in an election when 54% of all voters are known to be in favor of this proposition. Solution a. Rolling a die many times and observing the number of spots. This experiment has n identical trials, and each repetition is performed under identical conditions. ii. The experiment, however, has six rather than two possible outcomes. iii. Therefore, this experiment is not a Binomial experiment. i. 20

The Binomial Probability Distribution Solution b. Rolling a die many times and observing whether the number obtained is even or odd i. iii. iv. v. c. This experiment has n identical trials, and each repetition is performed under identical conditions. The experiment has only two possible outcomes, an “even” and an “odd”. The probability of even is ½ and probability of odd is ½. The probabilities of the two outcomes remain constant throughout the experiment. The events are independent. Therefore, this experiment is a Binomial experiment. Selecting a few voters from a very large population of voters and observing whether or not each of them favors a certain proposition in an election when 54% of all voters are known to be in favor of this proposition i. iii. iv. This experiment has n identical trials. The experiment has only two possible outcomes, “favor” and “not favor”. The probability of favor, p, is 0. 54 and “not favor” , q, is 0. 46, p + q =1, and both p and q are constant for each trial. The events are independent. Therefore, experiment is Binomial 21

The Binomial Probability Distribution and Binomial Formula So, knowing that an experiment is a binomial experiment, then we can apply the Binomial probability distribution to find the probability of exactly x successes in n trials by using the binomial formula where n = total number of trials p = probability of success q = 1 – p = probability of failure x = number of successes in n trials n - x = number of failures in n trials 22

The Binomial Probability Distribution and Binomial Formula Example #15 Solution According to a Harris Interactive poll, 52% of American college graduates have Facebook accounts. Suppose that this result is true for the current population of American college graduates. a. b. Let x be a binomial random variable that denotes the number of American college graduates in a random sample of 15 who have Facebook accounts. What are the possible values that x can assume? Find the probability that exactly 9 American college graduates in a sample of 15 have Facebook accounts. 23

The Binomial Probability Distribution and Binomial Formula Example #16 According to an October 27, article in Newsweek, 65% of Americans said they take expired medicines. Suppose this result is true of the current population of Americans. Find the probability that the number of Americans in a random sample of 22 who take expired medicines is a. exactly 17 b. none c. exactly 9 Solution 24

Using the Table of Binomial Probabilities So, rather than calculate the probability of x success in n trials by using the Binomial formula, we can use Table I of Appendix C – Tables of Binomial Probabilities. Lets do this by solving the following problem. Example #17 According to a March 25, 2007 Pittsburgh Post-Gazette article, 30% to 40% of U. S. taxpayers cheat on their returns. Suppose that 30% of all current U. S. taxpayers cheat on their returns. Use the binomial probabilities table (Table I of Appendix C) or technology to find the probability that the number of U. S. taxpayers in a random sample of 14 who cheat on their taxes is a. at least 8 b. at most 3 c. 3 to 7 25

Using the Table of Binomial Probabilities Solution 26

Probability of Success and the Shape of the Binomial Distribution The shape of a binomial distribution depends on the value of the probability of success. If p = 0. 50, then the binomial probability distribution is symmetric. If p < 0. 50, then binomial probability distribution is right skewed. If p > 0. 50, then binomial probability distribution is left skewed. 27

Mean and Standard Deviation of the Binomial Distribution For a binomial distribution, we can still use the formulas in Sections 5. 3 and 5. 4 to compute its mean and standard deviation. µ = E(x) = Σ x P(x) However a simpler methods are: where, n = total number of trials p = probability of success q = probability of failure. 28

Mean and Standard Deviation of the Binomial Distribution Example #18 A fast food chain store conducted a taste survey before marketing a new hamburger. The results of the survey showed that 70% of the people who tried hamburger liked it. Encouraged by this result, the company decided to market the new hamburger. Assume that 70% of all people like this hamburger. On a certain day, eight customers bought it for the first time. a. Let x denote the number of customers in this sample of eight who will like the hamburger. Using the binomial probabilities table, obtain the probability distribution of x and draw a graph of the probability distribution. Determine the mean and standard deviation of x. a. Using the probability distribution of part a, find the probability that exactly three of the eight customers will like this hamburger. 29

Mean and Standard Deviation of the Binomial Distribution Solution a. Using Table I of Appendix C, the probability distribution of x is, x P(x) 0 0. 0001 1 0. 0012 2 0. 0100 3 0. 0467 4 0. 1361 5 0. 2541 6 0. 2965 7 0. 1977 8 0. 0576 30

5. 4 OTHER TYPES OF DISTRIBUTION THE POISSON PROBABILITY DISTRIBUTION Suppose we know that 4 Toyota Corollas breakdown on an average every month. Based on this information, we would like to know the probability that exactly 3 corollas will breakdown in the next month. The binomial probability distribution technique may not fit well for this scenario especially because: 1. Each breakdown is probably under non-identical conditions, and the number of breakdown per month is both random and independent. 2. Each breakdown does not follow any pattern. 3. The breakdowns are considered with respect to an interval. This type of scenario is better addressed by using a Poisson probability distribution, in which each breakdown is called an occurrence. To apply a Poisson probability distribution, an experiment must meet the following conditions: 1. x is a discrete random variable. 2. The occurrences are random 3. The occurrences are independent. 31

The Poisson Probability Distribution Examples include: • The number of patients in an emergency room per week • The number of customers at a post office per day. So, knowing that an experiment follows a Poisson distribution, then we can apply the Poisson probability distribution to find the probability of x occurrences in an interval by using where = average number of occurrences in a given interval x = the actual number of occurrences in an interval e = 2. 71828 Note: 1. The intervals for and x must be equal. If not, we need to adjust to make them equal. 2. We could apply normal distribution for Poisson probability distribution in 32 cases where n>25 and

The Poisson Probability Distribution Example #19 A household receives an average of 1. 7 pieces of junk mail per day. Find the probability that this household will receive exactly 3 pieces of junk mail on a certain day. Use the Poisson probability distribution formula. Solution 33

Using the Table of Poisson Probabilities We can also use Table III in Appendix C to find the probabilities for Poisson distribution. Example #20 A university police department receives an average of 3. 7 reports per week of lost student ID cards. a. Find the probability that at most 1 such report will be received during a given week by this police department. Use the Poisson probability distribution formula. b. Using the Poisson probabilities table, find the probability that during a given week the number of such reports received by this police department is: (i) 1 to 4 (ii) at least 6 (iii) at most 3 34

Using the Table of Poisson Probabilities Example #20 – Solution 35

Mean and Standard Deviation of the Poisson Probability Distribution For a Poisson distribution, the formulas to compute its mean and standard deviation are 36

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