Chapter 8. Variables. Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.

Chapter 8 Random Variables Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.

8.1 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 continuous random variable can take any value in an interval or collection of intervals. 2. A discrete random variable can take one of a 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 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? Answer to all three questions = 1/8. Discrete Random Variable X = number of times the outcome of finterest t occurs in three independent d tries. 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 tossing 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 1: List all simple events in sample space. 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 possible value 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 that t assigns probabilities biliti to possible values of 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 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 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 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 probability of any specific 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 ( k) ) for any 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 =1orX 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 x 1, x 2, x 3,..., occurring with probabilities biliti p 1, p 2, p 3,..., then the expected value of X is calculated as μ ( X ) = = E x i 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. It costs $1.00 for each 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 x 1, x 2, x 3,..., occurring with probabilities p 1, p 2, p 3,..., and expected value E(X) = μ, then Variance Standard of X 2 ( X ) = σ = ( μ ) 2 = V x i pi Deviation of X ( μ ) = σ = x i pi 2 Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc. 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 and σ = $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 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. Conditions for a binomial experiment: 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. 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. Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc. 18

Finding Binomial Probabilities P ( X k ) k = = p 1 k! n! ( n k )! ( ) n p k for k = 0, 1, 2,, n Example 8.9 Probability of Two Wins in Three Plays p = probability win = 0.2; plays of game are independent. X = number of wins in three plays. What is P(X = 2)? P ( X = 2) = = 2! 3!.2 2! ( 3 ) 3(.2) 2 (.8) 1 2 = ( 1.2) 0.096 3 2 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, Mean μ = E ( X ) = np Standard deviation ( p) σ = np 1 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 a binomial i random variable. E( ( X ) = 100(.5) 50 Mean μ = = 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 differ from sample to sample is about 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 density curve over the 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 random variable 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 = (base)(height) ) = (2)(.1) ) =.2 Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc. 24

8.6 Normal Random Variables If a population of measurements follows a normal curve, and if X is the measurement for a randomly selected individual from that 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 and dbelow the mean. Empirical 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 z = Value Mean Standard deviation = x μ σ 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 z = Value Mean Standard deviation = 62 65 2.7 = 1.11 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*). Left-most column of table shows algebraic sign, digit before the decimal place, the first decimal place for z*. Second decimal place of z* *is in column heading. 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 portion above) P(Z 2.59) =.0048 P(Z 1.31) =.9049 P(Z 2.00) =.9772 P(Z -4.75) =.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 between 2.59 259and d131 1.31 P(-2.59 Z 1.31) = P(Z 1.31) P(Z -2.59) =.9049.0048 =.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? P ( X 62) 62 65 = P Z 2.7 = P Z 1 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? P 68 65 2.7 ( X > 68) = P Z > = P( Z > 1.11) = 1 P( Z 1.11) = 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 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: Calculate the value of variable that has the z-score found in step 1: x=z*σ σ + μ. μ Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc. 35

Example 8.17 75 th Percentile of Systolic Blood dpressure 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 0.7500 in body of Table A.1. z = 0.67 Step 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 a binomial i random variable based on n til 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: 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 and can be approximated by a normal curve. Mean μ = ( X ) = 30(.5) = 15 (.5 ) = 2. 74 Standard deviation σ = 30(.5) = E Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc. 38

Example 8.19 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 Mean μ = E = ( X ) = 500(.5) 250 (.5) 11. 2 Standard deviation σ = 100(.5) = 240 250 11.2 P( X 240 ) P Z = P ( Z.89 ) =. 1867 Not unlikely to see 48% or less, even if 50% in population p favor. Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc. 39