Copyright ©2011 Brooks/Cole, Cengage Learning Random Variables Chapter 8 1

Copyright ©2011 Brooks/Cole, Cengage Learning What is a Random Variable? Random Variable: assigns a number to each outcome of a random circumstance, or, equivalently, to each unit in a population. Two different broad classes of random variables: 1.A discrete random variable can take one of a countable list of distinct values. 2.A continuous random variable can take any value in an interval or collection of intervals.

Copyright ©2011 Brooks/Cole, Cengage Learning 3 Random factors that will determine how enjoyable the event is: Temperature: continuous random variable Number of airplanes that fly overhead: discrete random variable Example 8.1 Random Variables at an Outdoor Graduation or Wedding

Copyright ©2011 Brooks/Cole, Cengage Learning 4 What is the probability that three tosses of a fair coin will result in three heads? Assuming boys and girls are equally likely, what is the probability that 3 births will result in 3 girls? Assuming probability is 1/2 that a randomly selected individual will be taller than median height of a population, what is the probability that 3 randomly selected individuals will all be taller than the median? Example 8.2 Probability an Event Occurs Three Times in Three Tries Answer to all three questions = 1/8. Discrete Random Variable X = number of times the “outcome of interest” occurs in three independent tries.

Copyright ©2011 Brooks/Cole, Cengage Learning Discrete Random Variables X the random variable. k = a number the discrete r.v. could assume. P(X = k) is the probability that X equals k. Example 8.5 Number of Courses 35% of students taking four courses, 45% taking five, and remaining 20% are taking six courses. X = number of courses a randomly selected student is taking The probability distribution function of X can be given by: Probability distribution function (pdf) X is a table or rule that assigns probabilities to possible values of X.

Copyright ©2011 Brooks/Cole, Cengage Learning 6 Conditions for Probabilities for Discrete Random Variables Condition 1 The sum of the probabilities over all possible values of a discrete random variable must equal 1. Condition 2 The probability of any specific outcome for a discrete random variable must be between 0 and 1.

Copyright ©2011 Brooks/Cole, Cengage Learning 7 Probability Distribution of a Discrete R.V. Using the sample space to find probabilities: Step 1: List all simple events in sample space. Step 2: Identify the value of the random variable X for each simple event. Step 3: Find the probability for each simple event (often equally likely). Step 4: Find P(X = k) as the sum of the probabilities for all simple events where X = k. Probability distribution function (pdf) X is a table or rule that assigns probabilities to possible values of X.

Copyright ©2011 Brooks/Cole, Cengage Learning 8 Example 8.6 PDF for Number of Girls Family has 3 children. Probability of a girl is ½. What are the probabilities of having 0, 1, 2, or 3 girls? Sample Space: For each birth, write either B or G. There are eight possible arrangements of B and G for three births. These are the simple events. Sample Space and Probabilities: The eight simple events are equally likely. Random Variable X: number of girls in three births. For each simple event, the value of X is the number of G’s listed.

Copyright ©2011 Brooks/Cole, Cengage Learning 9 Example 8.6 & 8.7 Number of Girls Probability Distribution Function for Number of Girls X: Value of X for each simple event: Graph of the pdf of X:

Copyright ©2011 Brooks/Cole, Cengage Learning 10 Cumulative Distribution Function of a Discrete Random Variable Cumulative distribution function (cdf) for a random variable X is a rule or table that provides the probabilities P(X ≤ k) for any real number k. Cumulative probability = probability that X is less than or equal to a particular value. Example 8.8 Cumulative Distribution Function for the Number of Girls

Copyright ©2011 Brooks/Cole, Cengage Learning 11 Finding Probabilities for Complex Events Example 8.9 A Mixture of Children pdf for Number of Girls X: What is the probability that a family with 3 children will have at least one child of each sex? If X = Number of Girls then either family has one girl and two boys (X = 1) or two girls and one boy (X = 2). P(X = 1 or X = 2) = P(X = 1) + P(X = 2) = 3/8 + 3/8 = 6/8 = 3/4

Copyright ©2011 Brooks/Cole, Cengage Learning Expectations for Random Variables The expected value of a random variable is the mean value of the variable X in the sample space or population of possible outcomes. If X is a random variable with possible values x 1, x 2, x 3,..., occurring with probabilities p 1, p 2, p 3,..., then the expected value of X is calculated as Expected value = Sum of “value × probability”

Copyright ©2011 Brooks/Cole, Cengage Learning 13 Example 8.12 California Decco Lottery How much would you win/lose per ticket over long run? Player chooses one card from each of four suits. Winning card drawn from each suit. If one or more matches the winning cards  prize. It costs $1.00 for each play.  Lose an average of 35 cents per play.

Copyright ©2011 Brooks/Cole, Cengage Learning 14 Standard Deviation for a Discrete Random Variable The standard deviation of a random variable is roughly the average distance the random variable falls from its mean, or expected value, over the long run. If X is a random variable with possible values x 1, x 2, x 3,..., occurring with probabilities p 1, p 2, p 3,..., and expected value E(X) = , then

Copyright ©2011 Brooks/Cole, Cengage Learning 15 Example 8.13 Stability or Excitement Two plans for investing $100 – which would you choose? Expected Value for each plan: Plan 1: E(X ) = $5,000  (.001) + $1,000  (.005) + $0  (.994) = $10.00 Plan 2: E(Y ) = $20  (.3) + $10  (.2) + $4  (.5) = $10.00

Copyright ©2011 Brooks/Cole, Cengage Learning 16 Example 8.13 Stability or Excitement Variability for each plan: Plan 1:V(X ) = $29,900.00and  = $ Plan 2: V(X ) = $48.00and  = $6.93 The possible outcomes for Plan 1 are much more variable. If you wanted to invest cautiously, you would choose Plan 2, but if you wanted to have the chance to gain a large amount of money, you would choose Plan 1.

Copyright ©2011 Brooks/Cole, Cengage Learning 17 Expected Value (Mean) and Standard Deviation for a Population

Copyright ©2011 Brooks/Cole, Cengage Learning Binomial Random Variables 1. There are n “trials” where n is determined in advance and is not a random value. 2. Two possible outcomes on each trial, called “success” and “failure” and denoted S and F. 3. Outcomes are independent from one trial to the next. 4. Probability of a “success”, denoted by p, remains same from one trial to the next. Probability of “failure” is 1 – p. Class of discrete random variables = Binomial -- results from a binomial experiment. Conditions for a binomial experiment:

Copyright ©2011 Brooks/Cole, Cengage Learning 19 Examples of Binomial Random Variables A binomial random variable is defined as X=number of successes in the n trials of a binomial experiment.

Copyright ©2011 Brooks/Cole, Cengage Learning 20 Finding Binomial Probabilities p = probability win = 0.2; plays of game are independent. X = number of wins in three plays. What is P(X = 2)? for k = 0, 1, 2, …, n Example 8.15 Probability of Two Wins in Three Plays

Copyright ©2011 Brooks/Cole, Cengage Learning 21 Expected Value and Standard Deviation for a Binomial Random Variable For a binomial random variable X based on n trials and success probability p,

Copyright ©2011 Brooks/Cole, Cengage Learning 22 Example 8.18 Extraterrestrial Life? 50% of large population would say “yes” if asked, “Do you believe there is extraterrestrial life?” Sample of n = 100 is taken. X = number in the sample who say “yes” is approximately a binomial random variable. In repeated samples of n=100, on average 50 people would say “yes”. The amount by which that number would differ from sample to sample is about 5.

Copyright ©2011 Brooks/Cole, Cengage Learning Continuous Random Variables Continuous random variable: the outcome can be any value in an interval or collection of intervals. Probability density function for a continuous random variable X is a curve such that the area under the curve over an interval equals the probability that X is in that interval. P(a  X  b) =area under density curve over the interval between the values a and b.

Copyright ©2011 Brooks/Cole, Cengage Learning 24 Example 8.19 Time Spent Waiting for Bus Bus arrives at stop every 10 minutes. Person arrives at stop at a random time, how long will s/he have to wait? X = waiting time until next bus arrives. X is a continuous random variable over 0 to 10 minutes. Note: Height is 0.10 so total area under the curve is (0.10)(10) = 1 This is an example of a Uniform random variable

Copyright ©2011 Brooks/Cole, Cengage Learning 25 Example 8.20 Probability about Wait Time What is the probability the waiting time X is in the interval from 5 to 7 minutes? Probability = area under curve between 5 and 7 = (base)(height) = (2)(.1) =.2

Copyright ©2011 Brooks/Cole, Cengage Learning Normal Random Variables Most commonly encountered type of continuous random variable is the normal random variable, which has a specific form of a bell-shaped probability density curve called a normal curve. A normal random variable is also said to have a normal distribution Any normal random variable can be completely characterized by its mean, , and standard deviation, .

Copyright ©2011 Brooks/Cole, Cengage Learning 27 Example 8.21 College Women’s Heights Data suggest the distribution of heights of college women described well by a normal curve with mean  = 65 inches and standard deviation  = 2.7 inches. Note: Tick marks given at the mean and at 1, 2, 3 standard deviations above and below the mean. Empirical Rule are exact characteristics of a normal curve model.

Copyright ©2011 Brooks/Cole, Cengage Learning 28 Useful Probability Relationships

Copyright ©2011 Brooks/Cole, Cengage Learning 29 Useful Probability Relationships

Copyright ©2011 Brooks/Cole, Cengage Learning 30 Example 8.22 Prob for Math SAT Scores Math SAT scores have a normal distribution with mean  = 515 and standard deviation  = 100. Q: Probability a score is less than or equal to 600? A: (using Minitab).8023.

Copyright ©2011 Brooks/Cole, Cengage Learning 31 Example 8.22 Prob for Math SAT Scores Math SAT scores have a normal distribution with mean  = 515 and standard deviation  = 100. Q: Probability a score is greater than 600? A: By complement, 1 –.8023 =.1977

Copyright ©2011 Brooks/Cole, Cengage Learning 32 Example 8.22 Prob for Math SAT Scores Math SAT scores have a normal distribution with mean  = 515 and standard deviation  = 100. Q: Probability a score is between 515 and 600? A: (since 50% of scores are larger than the mean of 515),.8023 –.5000 =.3023

Copyright ©2011 Brooks/Cole, Cengage Learning 33 Example 8.22 Prob for Math SAT Scores Math SAT scores have a normal distribution with mean  = 515 and standard deviation  = 100. Q: Probability a score is more than 85 points away from the mean in either direction? A: (since 600 is 85 points above the mean) =.3954.

Copyright ©2011 Brooks/Cole, Cengage Learning 34 Using a Table to Find Probabilities Table A.1 = Standard Normal (z) Probabilities Body of table gives cumulative probabilities P(Z  z). First column gives digit before the decimal place, the first decimal place for z. Second decimal place of z is in column heading. P(Z  1.82) =.9656

Copyright ©2011 Brooks/Cole, Cengage Learning 35 More Finding Probabilities for z-scores Table A.1 = Standard Normal (z) Probabilities P(Z  -2.5) =.0048 (see in portion above) Verify on own: P(Z  1.31) =.9049 P(Z  -2.00) =.0228

Copyright ©2011 Brooks/Cole, Cengage Learning 36 Calculating a Standard Score The formula for converting any value x to a z-score is A z-score measures the number of standard deviations that a value falls from the mean.

Copyright ©2011 Brooks/Cole, Cengage Learning 37 Example 8.23 z-Score for Height For a population of college women, the z-score corresponding to a height of 62 inches is This z-score tells us that 62 inches is 1.11 standard deviations below the mean height for this population.

Copyright ©2011 Brooks/Cole, Cengage Learning 38 Use z-scores to Solve General Problems Example 8.24 Probability of Smaller Height What is the probability that a randomly selected college woman is 62 inches or shorter? About 13% of college women are 62 inches or shorter.

Copyright ©2011 Brooks/Cole, Cengage Learning 39 Finding Percentiles If 25th percentile of pulse rates is 64 bpm, then 25% of pulse rates are below 64 and 75% are above 64. The percentile is 64 bpm, and the percentile ranking is 25%. Step 1: Find z-score that has specified cumulative probability. Using Table A.1, find percentile rank in body of table and read off the z-score. Step 2: Compute x = z*  + . This is desired percentile.

Copyright ©2011 Brooks/Cole, Cengage Learning 40 Example th Percentile of Systolic Blood Pressure Blood Pressures are normal with mean 120 and standard deviation 10. What is the 75 th percentile? Step 1: Find closest z* with area of in body of Table A.1. z* = 0.67 Step 2: Compute x = z*  +  x = (0.67)(10) = or about 127.

Copyright ©2011 Brooks/Cole, Cengage Learning Approximating Binomial Distribution Probabilities If X is a binomial random variable based on n trials with success probability p, and n is large, then the random variable X is also approximately normal, with mean and standard deviation given as: Conditions: Both np and np(1 – p) are at least 10.

Copyright ©2011 Brooks/Cole, Cengage Learning 42 Example 8.27 Number of Heads in 60 Flips X = number of heads in n = 60 flips of fair coin Binomial distribution with n = 60 and p =.5. Distribution is bell-shaped and can be approximated by a normal curve.

Copyright ©2011 Brooks/Cole, Cengage Learning 43 Example 8.28 Another Normal Approx Binomial distribution with n = 300 and p =.3 Normal curve closely approximates the binomial distribution

Copyright ©2011 Brooks/Cole, Cengage Learning 44 Example 8.29 Political Woes Poll: n = 500 adults; 240 supported position If 50% of all adults support position, what is the probability that a random sample of 500 would find 240 or fewer holding this position? P(X  240) X is approximately normal with Not unlikely to see 48% or less, even if 50% in population favor.

Copyright ©2011 Brooks/Cole, Cengage Learning Sums, Differences, Combinations of r.v. A linear combination of random variables, X, Y,... is a combination of the form: L = aX + bY + … where a, b, etc. are numbers – positive or negative. Most common: Sum = X + YDifference = X – Y

Copyright ©2011 Brooks/Cole, Cengage Learning 46 Means of Linear Combinations L = aX + bY + … The mean of L is: Mean(L) = a Mean(X) + b Mean(Y) + … Most common: Mean( X + Y) = Mean(X) + Mean(Y) Mean(X – Y) = Mean(X) – Mean(Y)

Copyright ©2011 Brooks/Cole, Cengage Learning 47 Variances of Linear Combinations If X, Y,... are independent random variables, then Variance(L) = a 2 Variance(X) + b 2 Variance(Y) + … Most common: Variance( X + Y) = Variance(X) + Variance(Y) Variance(X – Y) = Variance(X) + Variance(Y)

Copyright ©2011 Brooks/Cole, Cengage Learning 48 If X, Y,... are independent normal random variables, then L = aX + bY + … is normally distributed. In particular: X + Y is normal with X – Y is normal with Combining Independent Normal Random Variables

Copyright ©2011 Brooks/Cole, Cengage Learning 49 Example 8.31 Will Meg Miss Her Flight? X = driving time, normal, mean 25 min, std dev 3 min Y = airport time, normal, mean 15 min, std dev 2 min X and Y independent T = X + Y = total time is normal with If Meg leaves with 45 minutes before flight takes off, what is probability she will miss her flight?

Copyright ©2011 Brooks/Cole, Cengage Learning 50 If X, Y,... are independent binomial random variables with n X, n Y, etc trials and all have same success probability p, then the sum X + Y + … is a binomial random variable with n = n X + n Y + … and success probability p. If the success probabilities differ, the sum is not a binomial random variable. Combining Independent Binomial Random Variables

Copyright ©2011 Brooks/Cole, Cengage Learning 51 Example 8.33 Donations Add Up X = # donors by volunteer 1; binomial n = 10, p =.2 Y = # donors by volunteer 2; binomial n = 12, p =.2 W = # donors by volunteer 3; binomial n = 18, p =.2 T = X + Y + W is binomial n = 40, p =.2 Fund drive: Three volunteers call potential donors. About 20% called make a donation, independent of who called. If volunteers make 10, 12 and 18 calls, respectively, what is the probability that they get at least ten donors? (Verify it!)