Random Variables and Discrete probability Distributions Chapter 7 Random Variables and Discrete probability Distributions

Random Variables and Probability Distributions A random variable is a function or rule that assigns a numerical value to each simple event in a sample space. A random variable reflects the aspect of a random experiment that is of interest for us. There are two types of random variables: Discrete random variable Continuous random variable.

Discrete Probability Distribution A table, formula, or graph that lists all possible values a discrete random variable can assume, together with associated probabilities, is called a discrete probability distribution. To calculate the probability that the random variable X assumes the value x, P(X = x), add the probabilities of all the simple events for which X is equal to x, or Use probability calculation tools (tree diagram), Apply probability definitions

Developing Probability Distribution and Finding the Probability of an Event Example The number of cars a dealer is selling daily were recorded in the last 100 days. This data was summarized in the table below. Estimate the probability distribution, and determine the probability of selling more than 2 cars a day. Daily sales Frequency 0 5 1 15 2 35 3 25 4 20 100

Developing Probability Distribution and Finding the Probability of an Event Solution From the table of frequencies we can calculate the relative frequencies, which becomes our estimated probability distribution .05 .15 .35 .25 .20 Daily sales Relative Frequency 0 5/100=.05 1 15/100=.15 2 35/100=.35 3 25/100=.25 4 20/100=.20 1.00 0 1 2 3 4 X P(X>2) = P(X=3) + P(X=4) = .25 + .20 = .45

Describing the Population/ Probability Distribution The probability distribution represents a population We’re interested in describing the population by computing various parameters. Specifically, we calculate the population mean and population variance.

Population Mean (Expected Value) Given a discrete random variable X with values xi, that occur with probabilities p(xi), the population mean of X is.

Population Variance Let X be a discrete random variable with possible values xi that occur with probabilities p(xi), and let E(xi) = m. The variance of X is defined by

The Mean and the Variance Solution – continued The variance can also be calculated as follows:

Laws of Expected Value and Variance E(c) = c E(X + c) = E(X) + c E(cX) = cE(X) Laws of Variance V(c) = 0 V(X + c) = V(X) V(cX) = c2V(X)

Example 1 The random variable X has the following distribution: a. Find the mean and variance for the probability distribution b. Determine the probability distributions of Y where Y=5X c. Use the probability distribution in part (b) to compute the mean and variance of Y d. Use the laws of expected value and variance to find the expected value and variance of Y from the parameters of X x -4 5 7 8 P(x) .59 .15 .25 .01

Solution a. = E(X) = –4(.59) +5(.15) + 7(.25) +8(.01) = .22 V(X) = (–4–.22)^2(.59) + (5–.22)^2(.15) + (7–.22)^2(.25) + (8–.22)^2(.01) = 26.03 Also, you can use V(X)=E(X2)-{E(X)}2 =(-4)2(.59)+(5)2(.15)+(7)2(.25)+(8)2(.01)-(.22)2 =26.03

Solution b. x –4 5 7 8 y –20 25 35 40 P(y) .59 .15 .25 .01 c. E(Y) = –20(.59) + 25(.15) + 35(.25) + 40(.01) = 1.10 V(Y) = (–20–1.10)^2(.59) + (25–1.10)^2(.15) + (35–1.1)^2(.25) + (40–1.1)^2(.01) = 650.8

Solution d. E(Y) = E(5X) = 5E(X) = 5(.22) = 1.10 V(Y) = V(5X) = 25V(X) = 25(26.03) = 650.8

Bivariate Distributions The bivariate (or joint) distribution is used when the relationship between two random variables is studied. The probability that X assumes the value x, and Y assumes the value y is denoted p(x,y) = P(X=x and Y = y)

Bivariate Distributions

Bivariate Distributions Example 7.5 Xavier and Yvette are two real estate agents. Let X and Y denote the number of houses that Xavier and Yvette will sell next week, respectively. The bivariate probability distribution is presented next.

Bivariate Distributions 0.42 Example 7.5 – continued p(x,y) X Y 0 1 2 0 .12 .42 .06 1 .21 .06 .03 2 .07 .02 .01 0.21 0.12 0.06 X 0.06 y=0 0.07 0.03 0.02 y=1 0.01 Y y=2 X=0 X=1 X=2

Marginal Probabilities Example 7.5 – continued Sum across rows and down columns X Y 0 1 2 p(y) 0 .12 .42 .06 .60 1 .21 .06 .03 .30 2 .07 .02 .01 .10 p(x) .40 .50 .10 1.00 p(0,0) P(Y=1), the marginal probability. p(0,1) p(0,2) The marginal probability P(X=0)

Describing the Bivariate Distribution The joint distribution can be described by the mean, variance, and standard deviation of each variable. This is done using the marginal distributions. x p(x) y p(y) 0 .4 0 .6 1 .5 1 .3 2 .1 2 .1 E(X) = .7 E(Y) = .5 V(X) = .41 V(Y) = .45

Describing the Bivariate Distribution To describe the relationship between the two variables we compute the covariance and the coefficient of correlation Covariance: COV(X,Y) = S(X – mx)(Y- my)p(x,y) Coefficient of Correlation COV(X,Y) sxsy r =

Describing the Bivariate Distribution Example 7.6 Calculate the covariance and coefficient of correlation between the number of houses sold by the two agents in Example 7.5 Solution COV(X,Y) = S(x-mx)(y-my)p(x,y) = (0-.7)(0-.5)p(0,0)+…(2-.7)(2-.5)p(2,2) = -.15 r=COV(X,Y)/sxsy = - .15/(.64)(.67) = -.35

Sum of Two Variables The probability distribution of X + Y is determined by Determining all the possible values that X+Y can assume For every possible value C of X+Y, adding the probabilities of all the combinations of X and Y for which X+Y = C Example 7.5 - continued Find the probability distribution of the total number of houses sold per week by Xavier and Yvette. Solution X+Y is the total number of houses sold. X+Y can have the values 0, 1, 2, 3, 4.

The Probability Distribution of X+Y P(X+Y=0) = P(X=0 and Y=0) = .12 P(X+Y=1) = P(X=0 and Y=1)+ P(X=1 and Y=0) =.21 + .42 = .63 P(X+Y=2) = P(X=0 and Y=2)+ P(X=1 and Y=1)+ P(X=2 and Y=0) = .07 + .06 + .06 = .19 X Y 0 1 2 p(y) 0 .12 .42 .06 .60 1 .21 .06 .03 .30 2 .07 .02 .01 .10 p(x) .40 .50 .10 1.00 The probabilities P(X+Y)=3 and P(X+Y) =4 are calculated the same way. The distribution follows

The Expected Value and Variance of X+Y The distribution of X+Y The expected value and variance of X+Y can be calculated from the distribution of X+Y. E(X+Y)=0(.12)+ 1(63)+2(.19)+3(.05)+4(.01)=1.2 V(X+Y)=(0-1.2)2(.12)+(1-1.2)2(.63)+… =.56 x + y 0 1 2 3 4 p(x+y) .12 .63 .19 .05 .01

The Expected Value and Variance of X+Y The following relationship can assist in calculating E(X+Y) and V(X+Y) E(X+Y) =E(X) + E(Y); V(X+Y) = V(X) +V(Y) +2COV(X,Y) When X and Y are independent COV(X,Y) = 0, and V(X+Y) = V(X)+V(Y).

Example2 (Text 7.47&48) The bivariate distribution of X and Y is described here. a. Find the marginal probability distribution of X. b. Find the marginal probability distribution of Y c. Compute the mean and variance of X d. Compute the mean and variance of Y e. Compute the covariance and variance of Y x y 1 2 0.28 0.42 0.12 0.18

Solution a x P(x) 1 .4 2 .6 b y P(y) 1 .7 2 .3 1 .4 2 .6 b y P(y) 1 .7 2 .3 c E(X) = 1(.4) + 2(.6) = 1.6 V(X) = (1–1.6)^2*(.4) + (2–1.6)^2*(.6) = .24 or (1^2)*0.4+(2^2)*0.6-(1.6)^2=.24

Solution d E(Y) = 1(.7) + 2(.3) = 1.3 V(Y) = (1–1.3)^2(.7) + (2–1.3)^2(.3) = .21 e = (1)(1)(.28) + (1)(2)(.12) + (2)(1)(.42) + (2)(2)(.18) = 2.08 COV(X, Y) = – = 2.08 – (1.6)(1.3) = 0 = 0

The Binomial Distribution The binomial experiment can result in only one of two possible outcomes. Typical cases where the binomial experiment applies: A coin flipped results in heads or tails An election candidate wins or loses An employee is male or female A car uses 87octane gasoline, or another gasoline.

Binomial Experiment Binomial Random Variable There are n trials (n is finite and fixed). Each trial can result in a success or a failure. The probability p of success is the same for all the trials. All the trials of the experiment are independent. Binomial Random Variable The binomial random variable counts the number of successes in n trials of the binomial experiment. By definition, this is a discrete random variable.

Calculating the Binomial Probability In general, The binomial probability is calculated by:

Mean and Variance of Binomial Variable E(X) = m = np V(X) = s2 = np(1-p)

Example 3(Text 7.102) In the game of blackjack as played in casinos in Las Vegas, Atlantic City, Niagara Falls, as well as many other cities, the dealer has the advantages. Most players do not play very well. As a result, the probability that the average player wins a about 45%.Find the probability that an average player wins twice in 5 hands ten or more times in 25 hands

Solution a P(X = 2) = = .3369 b Excel with n = 25 and p = .45: P(X 10) = 1 – P(X 9) = 1 – .2424 = .7576

The Binomial Distribution [插入][函數][選取類別(統計)]BINOMDIST Number_s 表示二項分配中成功的次數 Trials 總共的實驗次數 Probability_s 成功的機率 Cumulative 打入true則為累加分配函數，打入false則為機率函數

Poisson Distribution The Poisson experiment typically fits cases of rare events that occur over a fixed amount of time or within a specified region Typical cases The number of errors a typist makes per page The number of customers entering a service station per hour The number of telephone calls received by a switchboard per hour.

Properties of the Poisson Experiment The number of successes (events) that occur in a certain time interval is independent of the number of successes that occur in another time interval. The probability of a success in a certain time interval is the same for all time intervals of the same size, proportional to the length of the interval. The probability that two or more successes will occur in an interval approaches zero as the interval becomes smaller.

The Poisson Variable and Distribution The Poisson Random Variable The Poisson variable indicates the number of successes that occur during a given time interval or in a specific region in a Poisson experiment Probability Distribution of the Poisson Random Variable.

Example 4 The number of students who seek assistance with their statistics assignments is Poisson distributed with a mean of three per day. What is the probability that no student seek assistance tomorrow? Find the probability that 10 students seek assistance in a week.

Solution a. P(X = 0 with = 3) = = = .0498 b. P(X = 10 with = 21) = = = .0035

The Poisson Distribution X 表示平均時間內發生的次數 Mean 平均值 Cumulative 打入true為累加分配函數，打入false則為機率函數