Ch 7.1 Discrete and Continuous Random Variables Introduction A random variable is a variable whose value is a numerical outcome of a random phenomenon. If an experiment or sample survey is repeated, different results will be obtained. The probability theory for random variables tells us that there is a certain type of regularity or predictability in these results. A random variable can be either discrete or continuous. Discrete Random Variables A discrete random variable X has a countable number of possible values. The probability distribution of X lists the values and their probabilities. Values of X x 1 x 2 x 3 x k Probability p 1 p 2 p 3 p k The probability distribution assigns each value of X a probability between 0 and 1 such that the sum of all the probabilities is exactly 1. The probability of any event is the sum of the probabilities of all the values that make up the event. Use a probability histogram to picture a probability distribution. page 1
Example Car Ownership o Choose an American household at random and let the random variable X be the number of cars they own. Here is the probability model if we ignore the few households that own more than 5 cars. Number of Cars X 0 1 2 3 4 5 Probability.09.36.35.13.05.02 o Verify that this is a legitimate discrete distribution. Display the distribution in a probability histogram. o Say in words what the event {X 1} is. Find P(X 1). o Say in words what the event {X > 1} is. Find P(X > 1). o A housing company builds houses with two-car garages. What percent of households have more cars than the garage can hold? page 2
Continuous Random Variables A continuous random variable X takes all values in an interval of numbers. The probability distribution of X is described by a density curve. The probability of any event is the area under the density curve and above the values of X that make up the event. Any density curve has area exactly 1 underneath it, corresponding to total probability 1. The probability model for a continuous random variable assigns probabilities to intervals of outcomes rather than to individual outcomes. In fact, all continuous probability distributions assign probability 0 to every individual outcome. Example The Sum of Two Random Decimals o Generate two random numbers between 0 and 1 and take Y to be their sum. Then Y is a continuous random variable that can take any value between 0 and 2. The density curve of Y is the triangle shown in the figure. o Verify that the area under this curve is 1. o What is the probability that Y is less than 1? o What is the probability that Y is less than 0.5? page 3
o Use the following commands to simulate the last two answers. rand(200) L 1 rand(200) L 2 L 1 +L 2 L 3 SortA(L 3 ) Generates 200 random numbers and stores them in List 1 Generates 200 random numbers and stores them in List 2 Adds the first number in L 1 and the first number in L 2 and stores the sum in L 3, and so forth Sorts the sums in L 3 in ascending order o Count the number of sums in L 3 that are less than 1 and determine the relative frequency. sum(l 3 < 1) o Count the number of sums in L 3 that are less than 0.5 and determine the relative frequency. sum(l 3 <.5) o Picture the distribution as follows: Define Plot1 to be a histogram using L 3. Set the window dimensions as follows: X[0, 2] 0.1 and Y[-6, 25] 5. Does the resulting histogram resemble the triangle? Of course, some bars will be too short and others will be too long, but this is due to chance variation. Overlay the triangle: Y1 = 25X, Y2 = -25X + 50. How well does this curve fit your histogram? page 4
Normal Distributions as Probability Distributions Normal distributions are one type of continuous probability distribution. Recall that N(μ,σ) is the shorthand notation for the normal distribution having mean μ and standard deviation σ. In the language of random variables, if X has the N(μ,σ) distribution, then the standardized variable Z is the standard normal random variable having the distribution N(0,1). Z = X μ σ Example Joggers o An opinion poll asks an SRS of 1500 adults, Do you happen to jog? Suppose that the population proportion who jog is p = 0.15. To estimate p, we use the proportion p ˆ in the sample who answer Yes. The statistic p ˆ is a random variable that is approximately normally distributed with mean μ = 0.15 and standard deviation σ = 0.0092. What is the probability that the poll results differ from the truth about the population by more than one percentage point? o Find P( ˆ p 0.14) o Find P( ˆ p 0.16) o Find P(0.14 ˆ p 0.16) o Find P( ˆ p 0.14 or ˆ p 0.16) page 5
o Carry out a simulation to approximate the answers to the above questions. Don t forget to change the counts to relative frequencies by dividing by 500. randnorm (.15,.0092, 500) L 1 SortA(L 1 ) sum(l 1.14) sum(l1.16) o Find P( ˆ p 0.14) o Find P( ˆ p 0.16) o Find P(0.14 ˆ p 0.16) o Find P( ˆ p 0.14 or ˆ p 0.16) o How does your simulation compare to the calculations? page 6
Ch 7.2 Means and Variances of Random Variables The Mean and Variance of a Random Variable The probability distribution of a random variable X has a mean μ x and a standard deviation σ x. The mean of a random variable X is also called the expected value of X. If X is discrete, the mean is the average of the values of X, each weighted by its probability. μ x = x 1 p 1 + x 2 p 2 +...+ x k p k =Σx i p i The variance σ 2 x is the average squared deviation of the values of the variable from their mean. For a discrete random variable: σ 2 x = (x 1 μ) 2 p 1 + (x 2 μ) 2 p 2 +...+ (x k μ) 2 p k =Σ(x i μ) 2 p i The standard deviation σ x is the square root of the variance. The standard deviation measures the variability of the distribution about the mean. The mean and variance of a continuous random variable can be computed from the density curve, but to do so requires more advanced mathematics. For a symmetric density curve, the mean lies at the center. Example Grade Distribution o The following is a grade distribution in a large class. Grade 0 1 2 3 4 Probability.10.15.30.30.15 o Find the expected value of the distribution of grades. o Find the standard deviation of the distribution of grades. page 1
Law of Large Numbers The law of large numbers says that the actual mean outcome of many trials x gets close to the distribution mean μ as more trials are made. Example Law of Large Numbers Simulation o The distribution of heights of all young women is close to the normal distribution with mean 64.5 inches and standard deviation 2.5 inches. Simulate picking a young woman at random from the population and plotting the values of the mean height x as you add women to the sample. o Simulation: seq(x,x,1,200) L 1 Enters the positive integers 1 to 200 randnorm(64.5,2.5,200) L 2 Generates 200 random heights cumsum(l 2 ) L 3 Provides a cumulative sum L 3 /L 1 L 4 Calculates the average heights o Plot an xyline with L1 vs L4 and Y1 = 64.5 o Set window to X[1,100] 10 and Y[60,69] 1 o What happened to the average heights in the short run? o What happened to the average heights in the long run? page 2
Rules for Means and Variances If X is a random variable and a and b are fixed numbers, then μ a +bx = a + bμ X If X and Y are random variables, then μ X +Y = μ X + μ Y If X is a random variable and a and b are fixed numbers, then 2 σ a +bx = b 2 2 σ X If X and Y are independent random variables, then 2 σ X +Y = σ 2 2 X + σ Y 2 σ X Y = σ 2 2 X + σ Y For sums or differences of independent random variables, variances add. Notice that the formula looks very similar to the Pythagorean theorem. Just as the Pythagorean theorem applies only to right triangles, this relationship applies only to independent random variables. Why add even for the difference of the variables? o We buy some cereal. The box says "16 ounces." We know that's not precisely the weight of the cereal in the box, just close. Weights of such boxes of cereal vary somewhat, and our uncertainty about the exact weight is expressed by the variance (or standard deviation) of those weights. o Next we get out a bowl that holds 3 ounces of cereal and pour it full. Our pouring skill certainly is not very precise, so the bowl now contains about 3 ounces with some variability (uncertainty). o How much cereal is left in the box? Well, we'd assume about 13 ounces. But notice that we're less certain about this remaining weight than we were about the weight before we poured out the bowlful. The variability of the weight in the box has increased even though we subtracted cereal. o Moral: Every time something happens at random, whether it adds to the pile or subtracts from it, uncertainty (read "variance") increases. Example Heights o Men s heights (X) are normally distributed with a mean of 68 inches and a standard deviation of 3 inches. Women s heights (Y) are normally distributed with a mean of 63 inches and a standard deviation of 2 inches. o Let Z be the sum of men s and women s heights, Z = X+Y. Find the mean and standard deviation of Z. o Let D be the difference of men s and women s heights, D = X-Y. Find the mean and standard deviation of D. page 3
Example Ferry Prices o Consider a large ferry that can accommodate cars and buses. The toll for cars is $3, and the toll for buses is $10. Let X = # of cars and Y = # of buses carried on a single trip. Cars and buses are accommodated on a different levels of the ferry, so the number of buses accommodated on any trip is independent of the number of cars on the trip. o Probability distributions for X = # of cars and Y = # of buses X 0 1 2 3 4 5 P(X).05.10.25.30.20.10 Y 0 1 2 P(Y).50.30.20 o Compute the mean and standard deviation of X. o Compute the mean and standard deviation of Y. o Compute the mean and variance of the total amount of money collected in tolls from cars. o Compute the mean and variance of the total amount of money collected in tolls from buses. o Compute the mean and variance of Z = total number of vehicles. o Compute the mean and variance of W = total amount of money collected. page 4
Combining Normal Random Variables Any linear combination of independent normal random variables is also normally distributed. Example Capping Machine o A machine fastens plastic screw-on caps onto containers of motor oil. If the machine applies more torque than the cap can withstand, the cap will break. Both the torque applied and the strength of the caps vary. The cappingmachine torque has the normal distribution with mean 7 inch-pounds and standard deviation 0.9 inch-pounds. The cap strength has the normal distribution with mean 10 inch-pounds and standard deviation 1.2 inchpounds. o Explain why it is reasonable to assume that the cap strength and the torque applied by the machine are independent. o What is the probability that a cap will break while being fastened by the capping machine? page 5