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**Random Variables Chapter 8**

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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: A continuous random variable can take any value in an interval or collection of intervals. A discrete random variable can take one of a countable list of distinct values. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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 ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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 interest” occurs in three independent tries. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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 ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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. Probability distribution function (pdf) X is a table or rule that assigns probabilities to possible values of X. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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 ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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 ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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 ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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 ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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 ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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 with probabilities p1, p2, p3, , then the expected value of X is calculated as Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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 ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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) = m, then Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**Example 8.7 Stability or Excitement**

Two plans for investing $ – 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 ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**Example 8.7 Stability or Excitement (cont)**

Variability for each plan: Plan 1: V(X ) = $29, and s = $172.92 Plan 2: V(X ) = $48.00 and s = $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 ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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 ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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 ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**Finding Binomial Probabilities**

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)? Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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 ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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 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 ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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 ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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 so total area under the curve is (0.10)(10) = 1 This is an example of a Uniform random variable Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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 ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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, m, and standard deviation, s. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**Example 8.14 College Women’s Heights**

Data suggest the distribution of heights of college women modeled by a normal curve with mean m = 65 inches and standard deviation s = 2.7 inches. Note: Tick marks given at the mean and at 1, 2, 3 standard deviations above and below the mean. Empirical Rule values are exact characteristics of a normal curve model Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**Standard Scores 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 ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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Example Height (cont) 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 ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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 ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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) = (from in the extreme) Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**Example 8.15 Probability Z > 1.31**

P(Z > 1.31) = 1 – P(Z 1.31) = 1 – = .0951 Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**Example 8.16 Probability Z is between –2.59 and 1.31**

P(-2.59 Z 1.31) = P(Z 1.31) – P(Z -2.59) = – = .9001 Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**Use z-scores to Solve General Problems**

Example Height (cont) 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 ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**Use z-scores to Solve General Problems**

Example Height (cont) What proportion of college woman are taller than 68 inches? About 13% of college women are taller than 68 inches. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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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*s + m. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**Example 8.17 75th Percentile of Systolic Blood Pressure**

Blood Pressures are normal with mean 120 and standard deviation 10. What is the 75th percentile? Step 1: Find closest z* with area of in body of Table A.1. z = 0.67 Step 2: Calculate x = z*s + m x = (0.67)(10) = or about 127. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**8.7 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 n(1 – p) are at least 10. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**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 Not unlikely to see 48% or less, even if 50% in population favor. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**8.8 Sums, Differences, Combinations of R.V.s**

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 + Y Difference = X – Y Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**Means of Linear Combinations**

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 ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**Variances of Linear Combinations**

If X, Y, are independent random variables, then Variance(L) = a2 Variance(X) + b2 Variance(Y) + … Most common: Variance( X + Y) = Variance(X) + Variance(Y) Variance(X – Y) = Variance(X) + Variance(Y) Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**Combining Independent Normal Random Variables**

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 Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**Example 8.20 Will Meg Miss Her Flight?**

X = Meg’s travel time is normal with mean 25 minutes and std dev 3 minutes. Y = Airport time is normal with mean 15 minutes and std dev 2 minutes. X and Y independent so 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 ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**Combining Independent Binomial Random Variables**

If X, Y, are independent binomial random variables with nX, nY, etc trials and all have same success probability p, then the sum X + Y + … is a binomial random variable with n = nX + nY + … and success probability p. If the success probabilities differ, the sum is not a binomial random variable. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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**Example 8.22 Donations Add Up**

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? 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 Using calculator or computer: Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

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