Section 5.1 Random Variables

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Section 5.1 Random Variables Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Section 5.1 Random Variables Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Objectives Distinguish between discrete and continuous random variables Determine a probability distribution for a discrete random variable Describe the connection between probability distributions and populations Construct a probability histogram for a discrete random variable Compute the mean of a discrete random variable Compute the variance and standard deviation of a discrete random variable Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Objective 1 Distinguish between discrete and continuous random variables Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Random Variable If we roll a fair die, the possible outcomes are the numbers 1, 2, 3, 4, 5, and 6, and each of these numbers has probability 1/6. Rolling a die is a probability experiment whose outcomes are numbers. The outcome of such an experiment is called a random variable. A random variable is a numerical outcome of a probability experiment. Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Discrete and Continuous Random Variables Discrete random variables are random variables whose possible values can be listed. Examples include: The number that comes up on the roll of a die. The number of siblings a randomly chosen person has. Continuous random variables are random variables that can take on any value in an interval. Examples include: The height of a randomly chosen college student. The amount of rain that falls within a certain period. Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Objective 2 Determine a probability distribution for a discrete random variable Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Probability Distribution A probability distribution for a discrete random variable specifies the probability for each possible value of the random variable. Properties: 1. 0 ≤ P(x) ≤ 1 for every possible value x. 2. ΣP(x) = 1 Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Example Decide if each of the following is a probability distribution: This is not a probability distribution. P(3) is not between 0 and 1. This is a probability distribution. All the probabilities are between 0 and 1, and they add up to 1. This is not a probability distribution. P(0) is not between 0 and 1. x 1 2 3 4 P(x) 0.25 0.65 –0.30 0.11 x 3 4 5 6 7 P(x) 0.17 0.25 0.31 0.22 0.05 x 1 2 3 P(x) 1.02 0.31 0.90 0.43 Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Example Four patients have made appointments to have their blood pressure checked at a clinic. Let X be the number of them that have high blood pressure. Based on data from the National Health and Examination Survey, the probability distribution of X is (a) Find P(2 or 3) (b) Find P(More than 1) (c) Find P(At least one) x 1 2 3 4 P(x) 0.23 0.41 0.27 0.08 0.01 Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Solution Solution: (a) Find P(2 or 3) The events “2” and “3” are mutually exclusive, since they cannot both happen. We use the Addition Rule for Mutually Exclusive events: P(2 or 3) = P(2) + P(3) = 0.27 + 0.08 = 0.35 (b) Find P(More than 1) “More than 1” means “2 or 3 or 4.” We the Addition Rule for Mutually Exclusive events: P(More than 1) = P(2 or 3 or 4) = 0.27 + 0.08 + 0.01 = 0.36 x 1 2 3 4 P(x) 0.23 0.41 0.27 0.08 0.01 Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Solution Solution: (c) Find P(At least one) We use the Rule of Complements: P(At least one) = 1 – P(0) = 1 – 0.23 = 0.77 x 1 2 3 4 P(x) 0.23 0.41 0.27 0.08 0.01 Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Objective 3 Describe the connection between probability distributions and populations Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Probability Distributions and Populations Statisticians are interested in studying samples drawn from populations. Random variables are important because when an item is drawn from a population, the value observed is the value of a random variable. The probability distribution of the random variable tells how frequently we can expect each of the possible values of the random variable to turn up in the sample. Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Example In a town with a population of 1000 households, 142 of the households have no car, 378 have one car, 423 have two cars, and 57 have three cars. A household is sampled at random. Let X represent the number of cars in the randomly sampled household. Find the probability distribution of X. Solution: To find the probability distribution, we must list the possible values of X and then find the probability of each of them. The possible values of X are 0,1,2,3. We find their probabilities. Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Solution In a town with a population of 1000 households, 142 of the households have no car, 378 have one car, 423 have two cars, and 57 have three cars. A household is sampled at random. Let X represent the number of cars in the randomly sampled household. Find the probability distribution of X. Solution (continued): The probability distribution is x 1 2 3 P(x) 0.142 0.378 0.423 0.057 Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Objective 4 Construct a probability histogram for a discrete random variable Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Probability Histograms In an earlier chapter we learned to summarize the data in a sample with a histogram. We can represent discrete probability distributions with histograms as well. A histogram that represents a discrete probability distribution is called a probability histogram. Constructing a probability histogram from a probability distribution is just like constructing a relative frequency histogram from a relative frequency distribution for discrete data. We draw a rectangle for each possible value of the random variable, whose height is equal to the probability of that value. Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Example The following presents the probability distribution and histogram for the number of boys in a family of five children, using the assumption that boys and girls are equally likely and that births are independent events. x P(x) 0.03125 1 0.15625 2 0.31250 3 4 5 Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Objective 5 Compute the mean of a discrete random variable Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Mean of a Random Variable The mean of a random variable provides a measure of center for the probability distribution of a random variable. To find the mean of a discrete random variable, multiply each possible value by its probability, then add the products. In symbols, µX = Σ[x·P(x)]. Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Example A computer monitor is composed of a very large number of points of light called pixels. It is not uncommon for a few of these pixels to be defective. Let X represent the number of defective pixels on a randomly chosen monitor. The probability distribution of X is as follows. Find the mean number of defective pixels. Solution: The mean is µX = 0(0.2) + 1(0.5) + 2(0.2) + 3(0.1) = 1.2. x 1 2 3 P(x) 0.2 0.5 0.1 Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Expected Value There are many occasions on which people want to predict how much they are likely to gain or lose if they make a certain decision or take a certain action. Often, this is done by computing the mean of a random variable. In such situations, the mean is sometimes called the “expected value” and is sometimes denoted by E(X). If the expected value is positive, it is an expected gain, and if it is negative, it is an expected loss. Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Example A mineral economist estimated that a particular mining venture had probability 0.4 of a $30 million loss, probability 0.5 of a $20 million profit, and probability 0.1 of a $40 million profit. Let X represent the profit, in millions of dollars. Find the probability distribution of the profit and the expected value of the profit. Does this venture represent an expected gain or an expected loss? Solution: The probability distribution of X is as follows: The expected value is the mean of X E(X) = µX = (–30)(0.4) + (20)(0.5) + (40)(0.1) = 2.0 The expected value is positive, so this is an expected gain of $2 million. x –30 20 40 P(x) 0.4 0.5 0.1 Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Objective 6 Compute the variance and standard deviation of a discrete random variable Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Variance and Standard Deviation of a Random Variable The variance of a random variable provides a measure of spread for the probability distribution of a random variable. The variance of a discrete random variable X is given by An equivalent expression that is often easier when computing by hand is The standard deviation of X is the square root of the variance: Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Example Let X represent the number of defective pixels on a randomly chosen computer monitor. The probability distribution of X is as follows. Find the variance and standard deviation of the number of defective pixels. Solution: In a previous example, we found that the mean µX = 1.2. Now, The standard deviation is the square root of the variance: x 1 2 3 P(x) 0.2 0.5 0.1 Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Mean/Standard Deviation on the TI-84 PLUS The mean and standard deviation can be found on the TI-84 PLUS Calculator. We use the following steps: Step 1: Enter the values of the random variable into L1 in the data editor and the associated probabilities into L2. Step 2. Press STAT and highlight the CALC menu and select 1-Var Stats. Step 3. If using Stat Wizards, enter L1 in the List field and L2 in the FreqList field. If not using Stat Wizards, enter L1, comma, L2 after the 1-Var Stats command on the home screen Step 4. Run the command. Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Mean/Standard Deviation on the TI-84 PLUS Compute the mean and standard deviation of the following probability distribution using the TI-84 PLUS. Solution: We first enter values of the random variable and the associated probabilities into the data editor and then run the 1-Var Stats command. We find µX = 1.2 and σX = 0.872. x 1 2 3 P(x) 0.2 0.5 0.1 Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Do You Know… The difference between discrete and continuous random variables? How to determine a probability distribution for a discrete random variable? How to construct a probability distribution for a population? How to construct a probability histogram? How to compute the mean of a discrete random variable? How to compute the variance and standard deviation of a discrete random variable? Copyright ©2014 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.