Chapter 7: Random Variables 7.2 – Means and Variance of Random Variables

The Tri-State Pick 3 You choose a 3 digit number State chooses a 3 digit number Win $500 if they match Let X be the amount your ticket pays you. Create a probability distribution of this game

The Tri-State Pick 3 What is your average payoff from many tickets? 0(0.999) + 500(0.001) = $0.50 Since it costs $1 to play, how much does the state keep? half the money you wager in the long run

μ X = mean of X μ X does not have to be a value of X The mean of X is called the expected value Don’t necessarily expect one observation on X to be close to the expected value of X Can be thought of as a weighed average

General Definition of μ X The mean of a continuous random variable will be the center of the Normal curve for that distribution.

Benford’s Law Calculating the expected first digit If the first digits were truly random, then they would all have the same probability. The mean of this distribution is 45/9 = 5 If they follow Benford’s Law, the distribution of the first digit V is

Benford’s Law Calculating the expected first digit What is the mean of V? μ V = 1(0.301) + 2(0.176) + 3(0.125) + 4(0.097) + 5(0.079) + 6(0.067) + 7(0.058) + 8(0.051) + 9(0.046) = The mean of a uniform distribution is at its center. For skewed data, further calculations are needed

Locating the mean of a discrete random variable

Variance of a Random Variable Variance,, of a random variable is an average of the squared deviation (X – μ x ) 2 of the variable X from its mean μ x

Linda sells cars Linda is a sales associate at a large auto dealership. She motivates herself by using probability estimates of her sales. She estimates her car sales follow: Find her mean and her variance.

Linda sells cars μ X = 0(0.3) + 1(0.4) + 2(0.2) + 3(0.1) = 1.1 = (0 – 1.1) 2 (0.3) + (1 – 1.1) 2 (0.4) + (2 – 1.1) 2 (0.2) + (3 – 1.1) 2 (0.1) = When given a large data set your graphing calculator can be used. Let L 1 = x i, L 2 = p i. Use 1-var stats to find μ X = x and use σ x to find

Kids and toy Past experience with many subjects have shown that the probability distribution of the number X of toys played with is a follows: Find μ X = 0(0.03) + 1(0.16) + 2(0.30) + 3(0.23) + 4(0.17) + 5(0.11) = 2.68 Find σ X = (0 – 2.68) 2 (0.03) + (1 – 2.68) 2 (0.16) + … + (5 – 2.68) 2 (0.11) = 1.31

Some math humor HW: pg. 486 #7.24,

Statistical Estimation If we wanted to estimate the population mean height μ of all American women aged years we could SRS to obtain a sample mean, x. Statistics obtained from probability samples are random variables because their value would vary in repeated sampling. An SRS should fairly represent the population and the mean x should be somewhere near μ. If we keep adding observations to our sample, then x is guaranteed to get as close to μ as we wish and then stay that close.

Law of Large Numbers The sample mean always approaches the mean of the population

Law of Large Numbers The mean μ of a random variable is the average value of the variable in two senses. μ is the average of the possible values, weighted by their probability of occurring μ is also the long-run average of many independent observations on the variable The average result of many independent observations are stable and predictable Casinos will always win in the long run

Law of Small Numbers People think that every segment of a random sequence should reflect the true proportion Write down a sequence of head and tails you think imitates 10 tosses of a balanced coin. Recall that the probability of a run of 3 or more consecutive heads or tails is Belief in the law of small numbers influences behavior

How Large is Large Enough? It really depends on the variability of the random outcomes The more variability the more trials are needed to ensure x is close to μ Casinos understand this: The outcomes of games of chance are variable enough to hold the interest of the gamblers Only the casino plays often enough to rely on the law of large numbers

HW: pg. 491 #

Activity 7C: Rules for Means & Variances Find the mean and variance for X and Y μ X = 1(0.2) + 2(0.5) + 5(0.3) = 2.7 μ Y = 2(0.7) + 4(0.3) = 2.6 = ( ) 2 (0.2) + ( ) 2 (0.5) + ( ) 2 (0.3) = 2.41 = (2 – 2.6) 2 (0.7) + (4 – 2.6) 2 (0.3) = 0.84

Find the probability distribution for X + Y Hint: the smallest value of X + Y is 3 The P(X + Y = 3) = P(X = 1 and Y = 2) = P(X = 1)P(Y = 2) = (0.2)(0.7) = 0.14 (X + Y) P(X + Y)

Find μ X+Y and μ X+Y = 3(.14) + 4(.35) + 5(.06) + 6(.15) + 7(.21) + 9(.09) = 5.3 = (3 – 5.3) 2 (.14) + (4 – 5.3) 2 (0.35) + (5 – 5.3) 2 (.06) + (6 – 5.3) 2 (.15) + (7 – 5.3) 2 (.21) + (9 – 5.3) 2 (.09) = 3.25

Conclusion! ( = 5.3) ( = 3.25)

Rules for Means Rule 1: If X is a random variable and a and b are fixed numbers, then μ a+bX = a + bμ X Rule 2: If X and Y are random variables, then μ X+Y = μ X + μ Y Likewise: μ X-Y = μ X - μ Y

Linda sells cars and trucks The number X of cars that Linda hopes to sell has distribution Linda’s estimate of her truck and SUV sales is

Linda sell cars and trucks At her commission rate of 25% of gross profit on each vehicle she sells, Linda expects to earn $350 for each car and $400 for each truck/SUV sold. Her earnings are Z = 350X + 400Y What is Linda’s best estimate of her earnings for the day? Z = 350μ X + 400μ Y Z = 350(1.1) + 400(0.7) Z = $665 for the day

Rules for Variances Rule 1: If X is a random variable and a and b are fixed numbers, then Rule 2: If X and Y are independent random variables, then Note: we always add variances Note: a only affects position; b is squared because variance is squared

Regardless of adding or subtracting we are always increasing the spread Consider X: Y: (X + Y) = {6, 7, 8, …12} (X – Y) = {1, 2, 3, …7} Cannot add standard deviations because they are the square roots of variance Larger spread than you started with

SAT scores A college uses SAT scores as one criterion for admission. Experience has shown that the distribution of SAT scores among its entire population of applicants is such that SAT Math score Xμ X = 519σ X = 115 SAT Verbal score Yμ Y = 507σ Y = 111 What are the mean and st. dev. of the total score X + Y? μ X+Y = μ X + μ Y = = 1026 σ X+Y = N/A scores are not independent

Combining Normal Random Variables Any linear combination of independent Normal random variables will also be Normal Tom and George are playing in a golf tournament. Tom’s score X has the N(110, 10) and George’s score Y has the N(100, 8). They play independently of each other. What is the probability that Tom will score lower than George thus doing better in the tournament?

Tom and George play golf If Tom’s score is better then X < Y or X – Y < 0 Need to find P(X – Y < 0) We need to know μ X-Y and σ X-Y μ X-Y = μ X – μ Y = 110 – 100 = 10 σ X-Y =

Tom and George play golf P(X – Y < 0) = P(Z < -0.78) = Tom will beat George in about 1 of every 5 matches. HW: 499 # , 42 Convert to z-score