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Chapter 8 Random Variables Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 8.1 What is a Random Variable? Random Variable: assigns g 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 continuous random variable can take any 1. value in an interval or collection of intervals. 2 A discrete random variable can take one of a 2. countable list of distinct values. Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 2 Example 8.1 Random Variables at an Outdoor Graduation or Wedding Random factors that will determine how enjoyable the event is: Temperature: continuous random variable Number of airplanes that fly overhead: discrete random variable Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 3 Example 8.2 Probability an Event Occurs Three Times in Three Tries • What is the probability that three tosses of a fair coin will ill result l in i three h heads? h d? • 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 g of a population, what is the probability that 3 randomly selected individuals will all be taller than the median? Answer to all three questions = 1/8. Discrete Random Variable X = number of times the “ t “outcome off interest” i t t” occurs in i three th independent i d d t tries. t i Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 4 8.2 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. Discrete random variable: can only result in a countable set of possibilities – often a finite number of outcomes, but can be infinite. Example 8.3 It’s Possible to Toss Forever Repeatedly p y tossingg a fair coin,, and define: X = number of tosses until the first head occurs Any number of flips is a possible outcome. P(X = k) = (1/2)k Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 5 Probability Distribution of a Discrete R.V. Using the sample space to find probabilities: Step p 1: List all simple p events in sample p space. p Step 2: Find probability for each simple event. Step 3: List possible values for random variable X and identify the value for each simple event. Step 4: Find all simple events for which X = k, for each h possible ibl value l k. k Step 5: P(X = k) is the sum of the probabilities for all simple events for which X = k. k Probability distribution function (pdf) X is a table or rule l that th t assigns i probabilities b biliti to t possible ibl values l off X. X Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 6 Example 8.4 How Many Girls are Likely? 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 g possible p arrangements g 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. births For each simple event, the value of X is the number of G’s listed. Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 7 Example 8.4 How Many Girls? (cont) Value of X for each simple event: Probability y distribution function for Number of Girls X: Graph of the pdf of X: Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 8 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 pprobabilityy of anyy specific p outcome for a discrete random variable must be between 0 and 1. Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 9 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 p ( ≤ k)) for anyy real number k. Cumulative probability = probability that X is less than or equal to a particular value. Example 8.4 Cumulative Distribution Function for the Number of Girls (cont) Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 10 Finding Probabilities for Complex Events Example 8.4 A Mixture of Children 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 pdf for Number of Girls X: Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 11 8.3 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 x1, x2, x3, . . . , occurring i with ith probabilities b biliti p1, p2, p3, . . . , then the expected value of X is calculated as μ = E ( X ) = ∑ xi pi Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 12 Example 8.6 California Decco Lottery Player chooses one card from each of four suits. Winning card drawn from each suit. If one or more matches the winning cards => > prize prize. It costs $1.00 $1 00 for each play. play How much would you win/lose per ticket over long run? => Lose an average of 35 cents per play. Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 13 Standard Deviation for a Discrete Random Variable The standard deviation of a random variable is essentially the average distance the random variable falls from its mean over the long run. If X is a random variable with possible values x1, x2, x3, . . . , occurring with probabilities p1, p2, p3, . . . , and expected value E(X) = μ, then Variance of X = V ( X ) = σ 2 = ∑ ( xi − μ ) pi 2 Standard Deviation of X = σ = Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. ∑ (x − μ ) 2 i pi 14 Example 8.7 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 ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 15 Example 8.7 Stability or Excitement (cont) Variability for each plan: Plan 1: V(X ) = $29,900.00 and σ = $172.92 Plan 2: V(X ) = $48.00 σ = $6.93 and 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 ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 16 8.3 Binomial Random Variables Class of discrete random variables = Binomial -- results from a binomial experiment experiment. Conditions for a binomial experiment: p 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. 3. next 4. Probability of a “success”, denoted by p, remains same from one trial to the next. Probability of “failure” is 1 – p. Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 17 Examples of Binomial Random Variables A binomial random variable is defined as X=number of successes in the n trials of a binomial experiment. experiment Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 18 Finding Binomial Probabilities n! n−k k P( X = k ) = p (1 − p ) for k = 0, 1, 2, …, n k!(n − k )! Example 8.9 Probability of Two Wins in Three Plays p = pprobabilityy win = 0.2; plays p y of game g are independent. p X = number of wins in three plays. What is P(X = 2)? 3! 3− 2 P( X = 2) = .2 2 (1 − .2 ) 2!(3 − 2 )! = 3(.2) 2 (.8)1 = 0.096 Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 19 Expected Value and Standard Deviation for a Binomial Random Variable For a binomial random variable X based on n trials and success probability p, p Mean μ = E ( X ) = np Standard deviation σ = np (1 − p ) Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 20 Example 8.12 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 i l a binomial bi i l random d variable. i bl Mean μ = E ( X ) = 100((.5) = 50 Standard deviation σ = 100(.5)(.5) = 5 In repeated samples of n=100, on average 50 people would say “yes”. The amount by which that number would ld differ diff from f sample l to t sample l is i about b t 5. 5 Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 21 8.5 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 d density d i curve over the h interval between the values a and b. Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 22 Example 8.13 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 U o random do vvariable be Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 23 Example 8.13 Waiting for Bus (cont) What is the probability the waiting time X was in the interval from 5 to 7 minutes? Probability = area under curve between 5 and 7 )( g ) = (2)(.1) ( )( ) = .2 = ((base)(height) Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 24 8.6 Normal Random Variables If a population l ti off measurements t ffollows ll a normall curve, and if X is the measurement for a randomly selected individual from that population, population then X is said to be a normal random variable X is also said to have a normal distribution Any normal random variable can be completely characterized by its mean, μ, and standard deviation, σ. Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 25 Example 8.14 College Women’s Heights Data suggest the distribution of heights of college women modeled by a normal curve with mean μ = 65 inches and standard deviation σ = 2.7 inches. Note: Tick marks given at the mean and at 1, 2, 3 standard deviations above andd bbelow l the th mean. Empirical p Rule values are exact characteristics of a normal curve model Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 26 Standard Scores The formula for converting any value x to a z-score is Value − Mean x−μ z= = σ Standard deviation A z-score measures the number of standard deviations that a value falls from the mean. Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 27 Example 8.14 Height (cont) For a population of college women, the z-score corresponding to a height of 62 inches is Value − Mean 62 − 65 z= = = −1.11 2 .7 Standard deviation This z-score tells us that 62 inches is 1.11 standard deviations below the mean height for this population. Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 28 Finding Probabilities for z-scores Table A.1 = Standard Normal (z) Probabilities • Body of table contains P(Z ≤ z*). z*) • Left-most column of table shows algebraic sign, digit before the decimal place, place the first decimal place for z*. • Second S d ddecimal i l place l off z** is i in i column l heading. h di Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 29 More Finding Probabilities for z-scores Table A.1 = Standard Normal (z) Probabilities P(Z ( ≤ -3.00)) =.0013 ((see in pportion above)) P(Z ≤ −2.59) = .0048 P(Z ≤ 1.31) 1 31) = .9049 9049 P(Z ≤ 2.00) = .9772 P(Z ≤ -4.75) 4 75) = .000001 000001 (from in the extreme) Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 30 Example 8.15 Probability Z > 1.31 P(Z > 1.31) = 1 – P(Z ≤ 1.31) = 1 – .9049 = .0951 Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 31 Example 8.16 Probability Z is b between –2.59 2 59 andd 1.31 1 31 P(-2.59 ≤ Z ≤ 1.31) = P(Z ≤ 1.31) – P(Z ≤ -2.59) = .9049 9049 – .0048 0048 = .9001 9001 Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 32 Use z-scores to Solve General Problems Example 8.14 Height (cont) What is the probability that a randomly selected college woman is 62 inches or shorter? 62 − 65 ⎞ ⎛ P ( X ≤ 62 ) = P⎜ Z ≤ ⎟ 2 .7 ⎠ ⎝ = P (Z ≤ −1.11) = .1335 About 13% of college women are 62 inches or shorter. Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 33 Use z-scores to Solve General Problems Example 8.14 Height (cont) What proportion of college woman are taller than 68 inches? 68 − 65 ⎞ ⎛ P ( X > 68) = P⎜ Z > ⎟ = P (Z > 1.11) = 1 − P (Z ≤ 1.11) 2 .7 ⎠ ⎝ = 1 − .8665 = .1335 About 13% of college women are taller than 68 inches. Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 34 Finding Percentiles If 25th percentile of pulse rates is p , then 25% of pulse p rates 64 bpm, are below 64 and 75% are above 64. The percentile is 64 bpm, and the percentile ranking is 25%. Step 1: Find z-score z score that has specified cumulative probability. Using Table A.1, find percentile rank in body of table and read off the z-score. z-score Step 2: Calculate the value of variable that has the z score found in step 1: x = zz*σ + μ. z-score Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 35 Example 8.17 75th Percentile of S Systolic li Blood Bl d Pressure P Blood Pressures are normal with mean 120 and standard deviation 10. What is the 75th percentile? Step 1: Find closest z* with area of 0.7500 in body of Table A.1. z = 0.67 Step p 2: Calculate x = z*σ + μ x = (0.67)(10) + 120 = 126.7 or about 127. Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 36 8.7 Approximating Binomial Distribution Probabilities If X is i a binomial bi i l random d variable i bl based b d on n trials ti l with success probability p, and n is large, then the random variable X is also approximately normal, with mean and standard deviation given as: Mean μ = E ( X ) = np Standard deviation σ = np (1 − p ) Conditions: Both np and n(1 – p) are at least 10. Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 37 Example 8.18 Number of H in 30 Flips X = number of heads in n = 30 flips of fair coin Binomial distribution with n = 30 and p = .5. Distribution is bell-shaped p and can be approximated by a normal curve. Mean μ = E ( X ) = 30(.5) = 15 Standard deviation σ = 30(. ( 5)(.5) = 2.74 Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 38 Example 8.19 Political Woes Poll: n = 500 adults; 240 supported position pp pposition,, what is the If 50% of all adults support probability that a random sample of 500 would find 240 or fewer holding this position? P(X ≤ 240) X is approximately normal with Mean μ = E ( X ) = 500(.5) = 250 Standard deviation σ = 100(.5)(.5) = 11.2 240 − 250 ⎞ ⎛ P ( X ≤ 240 ) ≈ P⎜ Z ≤ ⎟ = P (Z ≤ −.89 ) = .1867 11.2 ⎠ ⎝ Not unlikelyy to see 48% or less,, even if 50% in ppopulation p favor. Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc. 39