Chapter 4 Discrete Random VariablesIt is often the case that a number is naturally associated to the outcome of a random experiment: the number of boys in a three-child family, the number of defective light bulbs in a case of 100 bulbs, the length of time until the next customer arrives at the drive-through window at a bank. Such a number varies from trial to trial of the corresponding experiment, and does so in a way that cannot be predicted with certainty; hence, it is called a random variable. In this chapter and the next we study such variables. Show
Learning Objectives
DefinitionA random variableA numerical value generated by a random experiment. is a numerical quantity that is generated by a random experiment. We will denote random variables by capital letters, such as X or Z, and the actual values that they can take by lowercase letters, such as x and z. Table 4.1 "Four Random Variables" gives four examples of random variables. In the second example, the three dots indicates that every counting number is a possible value for X. Although it is highly unlikely, for example, that it would take 50 tosses of the coin to observe heads for the first time, nevertheless it is conceivable, hence the number 50 is a possible value. The set of possible values is infinite, but is still at least countable, in the sense that all possible values can be listed one after another. In the last two examples, by way of contrast, the possible values cannot be individually listed, but take up a whole interval of numbers. In the fourth example, since the light bulb could conceivably continue to shine indefinitely, there is no natural greatest value for its lifetime, so we simply place the symbol ∞ for infinity as the right endpoint of the interval of possible values. Table 4.1 Four Random Variables
DefinitionA random variable is called discreteA random variable with a finite or countable number of possible values. if it has either a finite or a countable number of possible values. A random variable is called continuousA random variable whose possible values contain an interval of decimal numbers. if its possible values contain a whole interval of numbers. The examples in the table are typical in that discrete random variables typically arise from a counting process, whereas continuous random variables typically arise from a measurement. Key Takeaways
Exercises
BasicAnswers
4.2 Probability Distributions for Discrete Random VariablesLearning Objectives
Probability DistributionsAssociated to each possible value x of a discrete random variable X is the probability P(x) that X will take the value x in one trial of the experiment. DefinitionThe probability distributionA list of each possible value and its probability. of a discrete random variable X is a list of each possible value of X together with the probability that X takes that value in one trial of the experiment. The probabilities in the probability distribution of a random variable X must satisfy the following two conditions:
Example 1A fair coin is tossed twice. Let X be the number of heads that are observed.
Solution:
Figure 4.1 Probability Distribution for Tossing a Fair Coin Twice Example 2A pair of fair dice is rolled. Let X denote the sum of the number of dots on the top faces.
Solution: The sample space of equally likely outcomes is 11 1213141516212223242526313233 34353641424344454651525354 5556616263646566
Figure 4.2 Probability Distribution for Tossing Two Fair Dice The Mean and Standard Deviation of a Discrete Random VariableDefinitionThe meanThe number ΣxP(x) , measuring its average upon repeated trials. (also called the expected valueIts mean.) of a discrete random variable X is the number μ=E(X)=ΣxP(x) The mean of a random variable may be interpreted as the average of the values assumed by the random variable in repeated trials of the experiment. Example 3Find the mean of the discrete random variable X whose probability distribution is x−2123.5P(x)0.21 0.340.240.21 Solution: The formula in the definition gives μ=Σ xP(x)=(−2)·0.21+(1)·0.34+(2)·0.24+(3.5)·0.21 =1.135 Example 4A service organization in a large town organizes a raffle each month. One thousand raffle tickets are sold for $1 each. Each has an equal chance of winning. First prize is $300, second prize is $200, and third prize is $100. Let X denote the net gain from the purchase of one ticket.
Solution:
The concept of expected value is also basic to the insurance industry, as the following simplified example illustrates. Example 5A life insurance company will sell a $200,000 one-year term life insurance policy to an individual in a particular risk group for a premium of $195. Find the expected value to the company of a single policy if a person in this risk group has a 99.97% chance of surviving one year. Solution: Let X denote the net gain to the company from the sale of one such policy. There are two possibilities: the insured person lives the whole year or the insured person dies before the year is up. Applying the “income minus outgo” principle, in the former case the value of X is 195 − 0; in the latter case it is 195−200,000=−199,805. Since the probability in the first case is 0.9997 and in the second case is 1−0.9997=0.0003, the probability distribution for X is: x195 −199,805P(x)0.99970.0003 Therefore E(X)=ΣxP(x)=195·0.9997+(−199,805)·0.0003=135 Occasionally (in fact, 3 times in 10,000) the company loses a large amount of money on a policy, but typically it gains $195, which by our computation of E(X) works out to a net gain of $135 per policy sold, on average. DefinitionThe variance, σ2, of a discrete random variable X is the number σ2=Σ(x−μ) 2P(x) which by algebra is equivalent to the formula σ2= Σx2P(x)−μ2 DefinitionThe standard deviationThe number Σ(x−μ)2P(x) (also computed using [Σx2P(x)]−μ2), measuring its variability under repeated trials., σ, of a discrete random variable X is the square root of its variance, hence is given by the formulas σ=Σ(x−μ)2P (x)=Σx2P(x)−μ2 The variance and standard deviation of a discrete random variable X may be interpreted as measures of the variability of the values assumed by the random variable in repeated trials of the experiment. The units on the standard deviation match those of X. Example 6A discrete random variable X has the following probability distribution: x−1014P(x )0.20.5a0.1 A histogram that graphically illustrates the probability distribution is given in Figure 4.3 "Probability Distribution of a Discrete Random Variable". Figure 4.3 Probability Distribution of a Discrete Random Variable Compute each of the following quantities.
Solution:
Key Takeaways
Exercises
Basic
Applications
Additional ExercisesAnswers
4.3 The Binomial DistributionLearning Objectives
The experiment of tossing a fair coin three times and the experiment of observing the genders according to birth order of the children in a randomly selected three-child family are completely different, but the random variables that count the number of heads in the coin toss and the number of boys in the family (assuming the two genders are equally likely) are the same random variable, the one with probability distribution x0123P(x)0.1250.3750.3750.125 A histogram that graphically illustrates this probability distribution is given in Figure 4.4 "Probability Distribution for Three Coins and Three Children". What is common to the two experiments is that we perform three identical and independent trials of the same action, each trial has only two outcomes (heads or tails, boy or girl), and the probability of success is the same number, 0.5, on every trial. The random variable that is generated is called the binomial random variableA random variable that counts successes in a fixed number of independent, identical trials of a success/failure experiment. with parameters n = 3 and p = 0.5. This is just one case of a general situation. Figure 4.4 Probability Distribution for Three Coins and Three Children DefinitionSuppose a random experiment has the following characteristics.
Then the discrete random variable X that counts the number of successes in the n trials is the binomial random variable with parameters n and p. We also say that X has a binomial distribution with parameters n and p. The following four examples illustrate the definition. Note how in every case “success” is the outcome that is counted, not the outcome that we prefer or think is better in some sense.
Probability Formula for a Binomial Random VariableOften the most difficult aspect of working a problem that involves the binomial random variable is recognizing that the random variable in question has a binomial distribution. Once that is known, probabilities can be computed using the following formula. If X is a binomial random variable with parameters n and p, then P(x)=n!x!(n−x)! pxqn−x where q=1−p and where for any counting number m, m! (read “m factorial”) is defined by 0!=1,1!=1,2!=1·2,3!=1· 2·3 and in general m!=1·2 · · · (m−1)·m Example 7Seventeen percent of victims of financial fraud know the perpetrator of the fraud personally.
Solution:
Special Formulas for the Mean and Standard Deviation of a Binomial Random VariableSince a binomial random variable is a discrete random variable, the formulas for its mean, variance, and standard deviation given in the previous section apply to it, as we just saw in Note 4.29 "Example 7" in the case of the mean. However, for the binomial random variable there are much simpler formulas. If X is a binomial random variable with parameters n and p, then μ=npσ2=npqσ=npq where q=1−p Example 8Find the mean and standard deviation of the random variable X of Note 4.29 "Example 7". Solution: The random variable X is binomial with parameters n = 5 and p = 0.17, and q=1−p =0.83. Thus its mean and standard deviation are μ=np=5·0.17=0.85 (exactly) and σ=npq=5·0.17·0.83=.7055≈0.8399 The Cumulative Probability Distribution of a Binomial Random VariableIn order to allow a broader range of more realistic problems Chapter 12 "Appendix" contains probability tables for binomial random variables for various choices of the parameters n and p. These tables are not the probability distributions that we have seen so far, but are cumulative probability distributions. In the place of the probability P(x) the table contains the probability P(X≤x)=P (0)+P(1)+ · · · +P(x) This is illustrated in Figure 4.6 "Cumulative Probabilities". The probability entered in the table corresponds to the area of the shaded region. The reason for providing a cumulative table is that in practical problems that involve a binomial random variable typically the probability that is sought is of the form P(X≤x) or P(X≥x). The cumulative table is much easier to use for computing P(X≤x) since all the individual probabilities have already been computed and added. The one table suffices for both P(X≤x) or P(X≥x) and can be used to readily obtain probabilities of the form P(x), too, because of the following formulas. The first is just the Probability Rule for Complements. Figure 4.6 Cumulative Probabilities If X is a discrete random variable, then P(X≥x)=1−P( X≤x−1) andP(x)=P(X≤x)−P(X≤x−1) Example 9A student takes a ten-question true/false exam.
Solution: Let X denote the number of questions that the student guesses correctly. Then X is a binomial random variable with parameters n = 10 and p = 0.50.
Example 10An appliance repairman services five washing machines on site each day. One-third of the service calls require installation of a particular part.
Solution: Let X denote the number of service calls today on which the part is required. Then X is a binomial random variable with parameters n = 5 and p= 1∕3=0.3-.
Key Takeaways
Exercises
Basic
Applications
Additional ExercisesAnswers
What is the probability of obtaining 2 heads from a coin that was tossed 5 times?∴ Five coins are tossed the probability of getting two heads is 5/16.
What is the probability of getting at least one head in 5 coin toss?With 5 coins to flip you just times 16 by 2 and then minus 1, so it would result with a 31 in 32 chance of getting at least one heads.
What are the chances of flipping heads 5 times in a row?Your proposed answer of 13/32 is correct. If there are four or five heads in the sequence of five coin tosses, at least two heads must be consecutive.
Is tossed what are the possible values of the random variable for the number of heads?If the random variable Y is the number of heads we get from tossing two coins, then Y could be 0, 1, or 2.
|