4.1 4.2 Probability Distribution for Discrete Random Variables Key concepts: discrete random variable, probability distribution, expected value, variance, and standard deviation of a discrete random variable. A random variable is a function that to each outcome (sample point) of an experiment assigns a numerical value associated with this outcome. In other words, a random variable is numerical description of outcomes of a random experiment. Radom variables are classified as Continuous set of possible values is an interval Discrete set of possible values is countable (= one can list all possible values of a random variable) Examples 1. A coin is tossed and we assign x = 1 if it turns heads and x = 0 if it turns tails (discrete, possible values are 0,1) 2. A die is rolled and we assign x = number of dots shown on the face of the die (discrete, possible values 1, 2, 3, 4, 5, 6) 3. Two dice are rolled and we define x = sum of dots on the two dice (discrete, possible values 2, 3, 4,, 12) 4. A light bulb is randomly selected and let t = lifetime of a randomly chosen light bulb (continuous, set of possible values ion an interval (0, )) 5. A student is selected and x = his/her GPA is recorded (continuous, set of possible values ion an interval [0,4])
The probability distribution of a discrete random variable is a graph, table, or formula that specifies the probability associated with each possible value that the random variable can assume. Notation: For a discrete random variable x, p(x) will stand for the probability that the observed value of the random variable is a number x, for example p(2) = P(x = 2) = probability that x =2 Requirements for the probability distribution of a discrete random variable 1. p(x) 0 for all values of x 2. p(x) = 1, where the summation is over all possible values of x. Example. (Example 4.4, p. 188) Two fair coins are tossed and the number of heads x is observed. The sample space : S = {HH, HT, TH, TT} Assigned values x: 2 1 1 0 Possible values of x are 0, 1, 2. Distribution: p(0)=1/4, p(1) = 1/2, p(2) = 1/4 Distribution is often given in the form of a table or a graph: x 0 1 2 p(x) 1/4 1/2 1/4
The mean, or expected value, of a discrete random variable x is defined as µ = E (x) = x p(x) Interpretation: µ is a measure of a center of the distribution of x µ is the average value of x observed in a very large (more precisely as n ) number of repetitions of the experiment For this reason is often referred as the population mean. The variance of a discrete random variable x is defined as σ 2 = E[ (x-µ) 2 ] = (x-µ) 2 p(x) The standard deviation of a discrete random variable x is equal to the square root of the variance, i.e. σ = σ 2 Interpretation: σ 2 and σ are measures of variability of x σ 2 is the average of the squared distance of x from µ in a very large number of observations of the experiment (n ) For this reason is often referred as the population variance.
Example 1. Two fair coins are tossed and the number of heads x is observed. The probability distribution of x is in the table. Compute the expected value µ = E (x), variance σ 2, and the standard deviation σ of x µ = 0 (1/4)+ 1 (1/2) + 2 (1/4) =1 σ 2 = (0-1) 2 (1/4)+ (1-1) 2 (1/2) + (2-1) 2 (1/4) =1/2 x 0 1 2 p(x) 1/4 1/2 1/4 σ = 0.5 = 0.7071 TI-83: Enter the values of x in L 1 and corresponding values of p(x) in L 2, then use CALC 1-Var Stats L 1, L 2 ENTER
Example 2. An insurance policy costs $100 and will pay policyholder $10,000 if they suffer a major injury or $3,000 if they suffer a minor injury. The company estimates that each year 1 in every 2,000 policyholders may have a major injury, and 1 in 500 may have a minor injury a. Create a model for x = profit of the company on one policy b. What is the company expected profit on one policy? What is its standard deviation? c. If the company sells 900 these policies, what is the company expected profit? SOLUTION a. x 100-9,900-2,900 p(x) 0.9975 0.0005 0.0020 b. Recall that µ = x p(x) and σ 2 = (x-µ) 2 p(x) x p(x) x p(x) (x- ) (x - ) 2 (x - ) 2 p(x) 100 0.9975 99.75 11 121 120.70-9,900 0.0005-4.95-9,989 99,780,121 49,890.06-2,900 0.0020-5.8-2,989 8,934,212 17,868.24 Total 1 89 67,879.00 µ = E(X) = 89, σ 2 = 67879, σ = = 260.54 [do it using TI-83] c. Expected profit on 900 policies = 900 89 = $80,100
Exercise 1. Consider the given discrete probability distribution. Find the probability that x equals Exercise 2. (Ex. 4.8, p. 193) Consider the probability distribution for the random variable x shown below: a. What is the probability that x is less than 2 b. What is the probability that 1< x 4 c. Find µ, σ 2, and σ by hand, and then use TI-83 d. Graph p(x). e. Locate µ and the interval µ ± 2σ on your graph. What is the probability that x will fall in this interval. Compare this result with estimates obtained using Chebyshev s and Empirical Rules. Exercise 2 [based on 4.28, p. 197] The USDA reports that one in every 100 slaughtered chickens has fecal contamination. Consider a random sample of three slaughtered chickens. Let x equal the number of chickens in the sample that have fecal contamination. a. Find the probability distribution p(x) of x b. Find the probability P(x 1) c. What is the probability that at least one of the three have fecal contamination d. Find the mean and the standard deviation of x