1. Section 5.1 Random Variables
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Section 5.1 Random Variables
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2. 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
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3. Objective 1
Distinguish between discrete and continuous random variables
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4. 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.
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5. 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.
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6. Objective 2
Determine a probability distribution for a discrete random variable
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7. 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
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8. 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
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9. 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
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10. 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) = = 0.35 (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) = = 0.36 x 1 2 3 4
P(x)
0.23
0.41
0.27
0.08
0.01
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11. 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
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12. Objective 3
Describe the connection between probability distributions and populations
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13. 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.
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14. 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.
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15. 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
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16. Objective 4
Construct a probability histogram for a discrete random variable
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17. 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.
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18. 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) 1 2 3 4 5
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19. Objective 5 Compute the mean of a discrete random variable
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20. 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)].
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21. 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
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22. 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.
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23. 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
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24. Objective 6
Compute the variance and standard deviation of a discrete random variable
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25. 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:
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26. 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
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27. 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.
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28. 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 = x 1 2 3
P(x)
0.2
0.5
0.1
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29. 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?
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