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YMS Chapter 7 Random Variables Q1. A random variable is a variable whose value is a ________ of a random phenomenon. A1. Numerical outcome Q2. A random variable with a countable number of possible values is a _____ random variable. A2. Discrete. Q3. What is a probability distribution of a discrete random variable? A3. A list of the values the variable can take on, and the probability for each value. Q4. For the probability distribution of a discrete random variable, every probability is between ___ and ___, and the sum of all the probabilities is equal to ___. A4. 0 and 1, 1 Q5. In a probability histogram, what quantity do the horizontal and vertical axes represent, respectively? A5. The horizontal axis represents the possible values the random variable can take on, and the vertical axis represents the probability of that value. Q6. A continuous random variable can take on how many values for a certain interval in its domain? A6. An infinite number Q7. A continuous random variable’s probability distribution is described by a graph called the ___. A7. Density curve, or probability density curve.(This is the graph of the probability density function, or pdf.) Q8. Events, for continuous random variables, are described by the random variable’s taking on a value within a certain interval. The probability of that event is represented by what aspect of the density curve? A8. The area under the curve, between the two points that bound the interval, or the area under the curve, over the values (on the x-axis) that make up the event. Q9. Suppose you have a continuous random variable X. What is the probability that X=10? A9. Zero. Continuous probability distributions assign probability 0 to every individual outcome. Q10. In a continuous probability distribution, what is the relationship between the probability that X<10 and the probability that X<=10? A10. The two are equal, because the probability that X=10 is 0. Q11. True or false: the normal distribution is an example of a continuous probability distribution. A11. True. Q12. The mean of a discrete random variable is the sum of the products of all the possible values and the __________. A12. Probabilities of those values. Q13. Suppose there are two possible outcomes for a certain random variable, 0 and 100. The probability of getting 0 is .99 and the probability of getting 100 is .01. What is the mean of the random variable? A13. 1. Q14. The mean of a random variable is often called the e_____ v_____ of the variable. A14. Expected value Q15. Someone is offered a gambling game where there is a .25 chance of her losing $100, and a .75 chance of her winning $60. If she plays many times, what would her average winnings be? A15. $20. (Because this number is positive, it is unlike any gambling games anyone is likely to be offered.) Q16. Someone is invited to send in a contest entry in which the chances are 1 in 50 million of winning a million dollars, and one in a million of willing a thousand dollars. How do the expected earnings (or the mean earnings) from this contest compare with the price of a first-class stamp? A16. The expected earnings are 1/50million * 1million + 1/1million*1000 + the rest of the probability*0. This comes out to one fiftieth of a dollar plus one thousandth of a dollar, or 2.1 cents. At the time of this writing, a first class stamp costs 37 cents. So the cost is 34.9 cents more than the expected earnings. Q17. The mean of symmetric continuous probability distributions lies at the ____ of the curves. A17. Center Q18. The variance of a discrete random variable is the sum of the products of the squared deviation of each possible value from the mean of the distribution and the _____ for that value. A18. Probability Q19. Suppose there is a distribution with possible values 0, 1, and 2, each with probability 1/3. What is the variance, i.e. the sigma-squared, of this distribution? (This is also known as the variance of the population.) A19. 1*1/3 +0*1/3 +1*1/3, or 2/3. Q20. Think back to the definition of the variance of a sample. Suppose you had a sample consisting of 0, 1, and 2, with mean 1. Is the variance of this sample the same as the variance of the population? A20. No. The variance of the sample is the sum of the squared deviations over n-1. So the variance of the sample would be (1+0+1)/(3-1) or 1, rather than 2/3. Q21. Please take a few seconds to enter 0,1,and 2 in a list on your calculator. (On the TI83 or 84, stat >edit.) Then please compute 1 variable stats on these. (Stat>calc>1varstats, listname). Look at sx and sigma x results. What are they, and why do they make sense vis a vis the different definitions of population and sample variance and standard deviation? A21. They are 1 and .81650, respectively, and they do make sense because they are the square roots of 1 and 2/3, respectively. Q22. What is the law of large numbers, in your own words? A22. One way of putting it is that as the sample size approaches infinity, the sample mean approaches the population mean. Another is that you can make the sample mean get as close as you want to the population mean by getting a large enough sample. Q23. True or false: If by chance, you flip a coin and get 10 heads in a row, the law of large numbers tells us that if we flip many more times, we will get just a tiny bit under 50% heads in the remaining tosses, to compensate for the first 10 heads and make the long-range probability equal 50%. A23. False. The definition of independent trials implies that the coin “doesn’t remember” the first 10 flips and the subsequent results are not influenced by the initial ones. Q24. True or false. When you are picking a “large number” so as to make the sample mean get within a certain distance of the population mean, you need a larger number the greater the variability of the random outcomes. A24. True. In other words, with very small variability, the sample mean gives a close approximation to the population mean with much smaller sample size than with very great variability. Q25. True or false: the mean of a linear function of a random variable is that same linear function of the mean of the random variable. In other words, the mean of a + bX is a +b*the mean of X. A25. True. Q26. The mean of the sum of two random variables equals what? A26. The sum of the means of the two variables. Q27. If the mean amount that Linda makes at her lemonade stand per day is $10 and the mean amount her brother Tom makes is $9, what’s the average of their total daily receipts? A27. $19. Q28. Suppose someone tells you that the standard deviation of single scores on the SAT is 100 points. Suppose that there are two people who take the SAT independently of one another. How would you find the sd of the sum of their scores? A28. You’d first square their scores to get the variances of their individual scores – 10000 apiece. Then you add those variances, to get the total variance – 20000. Then you take the square root of that to get the sd of the sum, 100*the square root of 2, or 141.42. Q29. If the standard deviation of the SAT math and critical reading are both about 100, is the standard deviation of the sum of these two scores for an individual more or less than 141.42? Why? A29. More, because the two scores are correlated with each other. The variance of the sum is the sum of the variances plus twice the correlation between the variables times the product of their standard deviations. Q30. Can you give an intuitive explanation for why the variance of the sum of two random variables is increased, the more highly they are correlated with each other? A30. The variance is increased the more extreme values you have. If the two variables are independent, then a high value of one variable tends to be balanced by a lower value of the other. For example, if you get a 6 on rolling one die, you’ll on the average get a sum of 9.5 once you add in the value of the other die – not a sum of 12. The same thing goes in the other direction: low values of one variable are on the average, balanced by higher values of the other. But when the values are highly correlated, then high values of one variable predict high values of the other, and you sums that are higher, and low values of one variable predict low values of the other, and you get sums that are lower. So with sums that are higher and lower, the variance of the sum is greater with correlated variables. Q31. A linear combination of two independent normally distributed random variables is distributed how? A31. Normally. YMS Chapter 8 The Binomial and Geometric Distributions Q1. Suppose someone looks at the numbers of 1’s, 2’s, 3’s, 4’s, 5’s, and 6’s that result from 600 die rolls. Is this situation an example of the “binomial setting”? A1. Almost, but not quite. For there to be a binomial setting, you have to have each observation fall into only two categories, rather than the 6 categories described here. However, if you defined a 1 as a “success” and anything else as a “failure,” then you would have a binomial setting, and you could then do the same thing separately with 2, 3, 4, 5, and 6. Q2. What are the four requirements for the binomial setting? A2. 1. Two categories 2. Fixed n of observations 3. Independence 4. p of success same for all observations. Q3. The distribution of the number of successes out of n trials (with probability of success p on each trial) is the ______ _______. A3. binomial distribution Q4. If someone has 51 socks in a drawer, with 1/3 red and 2/3 black, and the person grabs a handful of 5 of them, and counts the number of black, will the results of such a trial follow the binomial distribution? Why or why not? A4. Not quite, because grabbing a handful of 5 is equivalent to sampling without replacement. The probability of a black’s being included in the handful is altered some depending on what other socks are also in the handful. If you picked a sock one at a time, replaced the sock and mixed them thoroughly, and then picked again, the binomial distribution would apply. Q5. Suppose you roll a die 1000 times and count “1” as success and “anything else” as failure. Is this an example of the binomial setting, and does the count have the binomial distribution? Why or why not? A5. Yes and yes, because the conditions of 2 categories, fixed n of observations, independence, and p of success constant all hold. Q6. In Chapter 1, the word distribution was defined as what values the variable takes and how often it takes these values. Let’s say you roll a die 1000 times and count the number of 1’s. The count of successes comes out to 165. Someone asks you, “What does this have to do with a distribution? We just got one number from this experiment. How’s anybody going to plot a histogram or any other representation of a distribution with this?” What would be your answer? A6. The number of 1’s that you would get in such an experiment is a random variable. If you did the experiment many times, you would get a distribution that could be plotted with a histogram, and which would take the approximate shape of the theoretically derived binomial distribution for this situation. Q7. When we say a certain random variable has a B(100, .7) distribution, what do we mean? A7. That there is a binomial distribution with 100 observations and probability of success on each observation .7. Q8. If there is a discrete random variable (such as a binomial), and you want to find the probability of any given value of X, what function do you use – the cumulative distribution function or the probability distribution function? (cdf or pdf?) A8. pdf Q9. Suppose you want to know the probability that a binomial random variable B(100, .7) takes on a value less than or equal to 60. One way would be to use the binomial pdf for values 0, 1, .. 60 and then add them all up. A much less laborious way would be to do what? A9. Use the binomial cdf function. On the TI-83 or TI-84 the command would be binomcdf(100, .7, 60). Q10. Suppose you roll a die six times, and you want the probability of getting exactly 3 1’s. What would be the appropriate expression of the binomial formula that would give the answer to this? A10. (6 choose 3) * (1/6^3 * 5/6^3). Q11. Can you please explain why the binomial probability formula is as it is, using this example of rolling a die six times? Please give an explanation for why each of the three factors is what it is. A11. You have 6 independent rolls. The probability of getting 1 on the first three and something other than 1 on the next three is 1/6^3 * 5/6^3 because of the multiplication rule for independent events. The number of different ways you can get three ones is the number of ways you can select 3 dice to be ones out of the 6 different rolls: i.e. roll 1, 2, and 3, roll 1, 2, and 4, etc. This is the number of combinations of six things taken 3 at a time, or 6 choose 3. Because each of these ways of getting 3 1’s is mutually exclusive of the others, you can use the addition rule to add the probability of each of the 6 choose 3 ways of getting the outcome of 3 1’s, and since these probabilities are all the same, a shorter way is to multiply the probability of any one of them time 6 choose 3. Q12. Can you please explain why, to obtain the binomial coefficient, you use the number of combinations rather than the number of permutations, in calculating n choose k? A12. In our die roll example, using the number of permutations would for example count the event of getting 1’s on roll number 1, 3, and 5 as a different event from getting 1’s on rolls number 5, 3, and 1. Since these events are the same, order does not make a difference in enumerating subsets, and therefore you want combinations rather than permutations. Q13. In chapter 7 we learned that means and variances are additive when you want to know the mean and variance of sums of independent random variables. How are these facts crucial in figuring out the formulas for the mean and variance of a binomial random variable? A13. In the case of both the mean and the variance, we consider a certain random variable to be the outcome of any one trial, giving a success numerical value 1 and failure value 0. We calculate the mean and variance of this variable from the defining formulas. Then we define another random variable to be the sum of n of these random variables, which is the number of successes in n trials. We use the additive properties of the mean and the variance that we learned in chapter 7 to move us from an expression for the mean and variance of any one trial, to the mean and variance of the sum of n trials. Q14. What are the formulas for the mean, variance, and standard deviation of a binomial random variable in terms of n and p, and, if you want, q (or 1-p). A14. The mean mu=np. The variance sigma squared = npq. The standard deviation is the square root of npq or (np(1-p))^.5. Q15. When n is “large,” the binomial distribution with n trials and success probability p can be approximated by what? A15. The normal distribution with mean np and standard deviation (npq)^.5. Q16. As a rule of thumb, the normal distribution may be used as an approximation to the binomial when both np and nq (expected successes, expected failures) equal or exceed what number? A16. 10 Q17. Please describe how to have your calculator simulate a binomial experiment. What are the keys that you press? A17. For the TI-83 or 84, You press math>prb>randbin( and then enter 1, p, and n in parentheses. You press the sto> key to store the results, and 2nd L1 to get the results stored in L1; then you hit the colon which is done by alpha and then the decimal point key; then you do the sum function which is under list>math>. Then in parentheses you enter L1. So the command is randbin(1, p, n)->L1:sum(L1). What this does is to generate n numbers that are either 0 or 1, with a p probability of being 1, and adds them to get the number of successes in the n trials, and displays that number. Each time you press the enter key, this will be repeated. So you take your number 2 pencil and write down each of these numbers, and then you count how many of them had the result you were looking for, and you express that result as a fraction of the number of trials you did. Q18. For a binomial setting, the number of trials is fixed, and the random variable is the number of successes in that trial. For a geometric setting, the random variable is the number of ____ necessary to achieve the first _____. A18. trials, success Q19. True or false: in the geometric setting, as in the binomial setting, you have 1. two categories, 2. with the same probability for each observation, and 3. independent observations. A19. True. Q20. In a geometric setting, with probability of success p, what is the probability that the first success will occur on the nth trial? A20. P(X=n) = (1-p)^n-1 * p Q21. True or false: the probabilities of success on the first, second, third, etc. trial in a geometric setting, when arranged in order, form a geometric series where p is the first term and each successive term being (1-p) (or q) times the previous one? A21. True Q22. True or false: if you apply the formula a/(1-r) for the sum of the terms of an infinite geometric series, where a is the first term and r is the ratio of each term to the previous one, for the geometric setting p is the first term and (1-p) is the ratio, so the sum becomes p/(1-(1-p)) or 1. Thus even though there are infinitely many possibilities for the outcome of the experiment in the geometric setting, the probabilities of each outcome sum to 1. A22. True Q23. If your chances of rolling a 1 on a die roll are one in 6, what is the expected or average number of times that you would have to roll the die before getting a 1? A23. 6 times. Q24. If your chances of getting a success at anything in the geometric setting is p, what is the average or expected number of trials that you would have to conduct before getting a success? A24. 1/p trials. Q25. What is the variance in the geometric random variable? A25. q/p^2, or (1-p)/p^2 Q26. In the geometric setting, if q=1-p, what is the probability that it takes more than n trials to see the first success? A26. P(X>N)= q^n. Q27. On page 470 of YMS there is a derivation of the formula for the probability that it takes more than n trials to see the first success. Can you think of a really simple way to arrive at the same formula? A27. If, and only if, the first n trials are failures, it will take more than n trials to get the first success. The probability of the first n trials being failures, using the multiplication rule for independent events, is q^n. Q28. For a geometric distribution, would you say that it is approximately true that 34% of the observations would fall between the mean and 1 standard deviation above the mean, and 34% would fall between the mean and 1 standard deviation below the mean? Why or why not? A28. No, because the geometric distribution is always strongly skewed to the right, and its shape doesn’t resemble the normal distribution (for which the above statement is true). Q29. Suppose that some experts estimate that the probability of a major nuclear war in any given year is 1%. You think that you will live another 65 years. You are wondering what the chance is that you will and will not see a nuclear war. Please fit some of the concepts of this chapter to this situation, and calculate the probability. A29. This is the geometric setting, where “success” is defined as nuclear war, and “failure” is defined as no nuclear war! What you are being asked is the probability that it will take more than 65 trials to see a “success.” You use the formula P(X>n)=q^n, to get the P(X>65)=.99^65. The probability of no nuclear war comes out to .52. So would that be comforting, or what?