Learning Outcomes
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The idea of a random variable can be confusing. In this video we help you learn what a random variable is, and the difference between discrete and continuous random variables. A discrete probability distribution function has two characteristics:
ExampleA child psychologist is interested in the number of times a newborn baby’s crying wakes its mother after midnight. For a random sample of 50 mothers, the following information was obtained. Let [latex]X=[/latex] the number of times per week a newborn baby’s crying wakes its mother after midnight. For this example, [latex]x = 0, 1, 2, 3, 4, 5[/latex]. [latex]P(x) =[/latex] probability that [latex]X[/latex] takes on a value [latex]x[/latex].
[latex]X[/latex] takes on the values [latex]0, 1, 2, 3, 4, 5.[/latex] This is a discrete PDF because:
[latex](\frac{2}{50})+(\frac{11}{50})+(\frac{23}{50})+(\frac{9}{50})+(\frac{4}{50})+(\frac{1}{50})=1[/latex] Try itSuppose Nancy has classes three days a week. She attends classes three days a week [latex]80[/latex]% of the time, two days [latex]15[/latex]% of the time, one day [latex]4[/latex]% of the time, and no days [latex]1[/latex]% of the time. Suppose one week is randomly selected. a. Let [latex]X[/latex] = the number of days Nancy ____________________. b. [latex]X[/latex] takes on what values? c. Suppose one week is randomly chosen. Construct a probability distribution table (called a PDF table) like the one in Example 1. The table should have two columns labeled [latex]x[/latex] and [latex]P(x)[/latex]. What does the [latex]P(x)[/latex] column sum to?
ExampleJeremiah has basketball practice two days a week. Ninety percent of the time, he attends both practices. Eight percent of the time, he attends one practice. Two percent of the time, he does not attend either practice. What is X and what values does it take on? Solution: [latex]X[/latex] is the number of days Jeremiah attends basketball practice per week. [latex]X[/latex] takes on the values [latex]0, 1,[/latex] and [latex]2.[/latex] The characteristics of a probability distribution function (PDF) for a discrete random variable are as follows: Solution: a. Let [latex]X[/latex] = the number of days Nancy attends class per week. b. [latex]0, 1, 2,[/latex] and [latex]3[/latex] c.
Learning 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 111213141516212223 2425263132333435364142434445 46515253545556616263646566
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−212 3.5P(x)0.210.340.240.21 Solution: The formula in the definition gives μ=ΣxP(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 σ 2=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:
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Additional ExercisesAnswers
How do you find the probability density function of a discrete random variable?The probabilities in the probability distribution of a random variable X must satisfy the following two conditions: Each probability P(x) must be between 0 and 1: 0≤P(x)≤1. The sum of all the possible probabilities is 1: ∑P(x)=1.
What is probability density function of a random variable?In probability theory, a probability density function (PDF) is used to define the random variable's probability coming within a distinct range of values, as opposed to taking on any one value. The function explains the probability density function of normal distribution and how mean and deviation exists.
Is probability density function continuous or discrete?Hence its difficult to sum these uncountable values like discrete random variables and therefore integral over those set of values is done. Probability distribution of continuous random variable is called as Probability Density function or PDF.
Do discrete random variables have a PDF?The probability density function (PDF) of a random variable is a function describing the probabilities of each particular event occurring. For instance, a random variable describing the result of a single dice roll has the p.d.f.
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