Introduction to the Practice of Statistics Fifth Edition Moore, McCabe

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1 Introduction to the Practice of Statistics Fifth Edition Moore, McCabe Section 4.2 Homework Answers 4.17 Choose a young adult (age 25 to 34 years) at random. The probability is 0.12 that the person chosen did not complete high school, 0.31 that the person has a high school diploma but no further education, and 0.29 that the person has at least a bachelor's degree. P(No H.S.) = 0.12 P(H.S. only) = 0.31 P(Bachelors or more) = 0.29 This is a type of problem that you need to play with the wording until it is clear that the statements made create disjoint statements. One group does not finish high school, while another only finishes high school, which is different from those that finished high school and attained at least a bachelors. (a) What must be the probability that a randomly chosen young adult has some education beyond high school but does not have a bachelor's degree? Draw the Venn diagram from the original set of information, and then see if you can shade in the area that is depicted by the statement. Notice the shaded area depicts someone that has a high school diploma but not a bachelors or higher degree, yet it is not only a high school degree. 1 ( ) = 0.28 (b) What is the probability that a randomly chosen young adult has at least a high school education? Notice that this statement includes everyone except those that did not attain a high school diploma. P(at least HS) = 1 P(no HS) = = 0.88

2 4.19 Dugout Lou thinks that the probabilities for the American League baseball champion are as follows. The Yankees have probability 0.6 of winning. The Red Sox and Angels have equal probabilities of winning. The Athletics and White Sox also have equal probabilities, but their probabilities are one-third that of the Red Sox and Angels. No other team has a chance. What is Lou's assignment of probabilities to teams? This problem just tests your ability to organize the information. Let us organize this information in a table. P(Yankees win) = 0.6 P(Red Sox win) = P(Angels win) P(Athletics win) = P(White Sox) = 1/3 P(Angels). P(other team wins) = 0 Let P(Angels win) = x Let us organize this information in a table. Team Yankees Red Sox Angels Athletics White Sox Other P(of winning) 0.6 x x (1/3)x (1/3)x 0 Now, you need to recall the one fact that sum of all the probabilities must equal 1. Solve 1 = x + x + (1/3)x + (1/3)x 4.21 There are many ways to produce crooked dice. To load a die so that 6 comes up too often and 1 (which is opposite 6) comes up too seldom, add a bit of lead to the filling of the spot on the 1 face. Because the spot is solid plastic, this works even with transparent dice. If a die is loaded so that 6 comes up with probability 0.2 and the probabilities of the 2, 3, 4, and 5 faces are not affected, what is the assignment of probabilities to the six faces? P( six) = 0.2 P(two) = 1/6 P(three) = 1/6, P(four) = 1/6, P(five) = 1/6. P(one) = 1 ( /6) = 2/15

3 4.22 The 2000 census allowed each person to choose from a long list of races. That is, in the eyes of the Census Bureau, you belong to whatever race you say you belong to. "Hispanic/Latino" is a separate category; Hispanics may be of any race. If we choose a resident of the United States at random, the 2000 census gives these probabilities: Hispanic Not Hispanic Total Asian Black White Other Total Let A be the event that a randomly chosen American is Hispanic, and let B be the event that the person chosen is white. (a) Verify that the table gives a legitimate assignment of probabilities All probabilities assigned are between 0 and 1. Also, the sum of all probabilities is 1. The assignment is legitimate. (b) What is P(A)? P(A) = (c) Describe B c in words and find P(B c ) by the complement rule. The event B c is a person is not white. P(B c ) = 1 P(B) = = (d) Express "the person chosen is a non-hispanic white" in terms of events A and B. What is the probability of this event? person is a non-hispanic White : A c and B. P(A c and B) =

4 4.23 Human blood is typed as O, A, B, or AB and also as Rh-positive or Rh-negative. ABO type and Rhfactor type are independent because they are governed by different genes. In the American population, 84% of people are Rh-positive. Use the information about ABO type in Exercise 4.13 to give the probability distribution of blood type (ABO and Rh) for a randomly chosen person. First I am going to write down the given probabilities using function notation. P(Rh-positive) = 0.84 thus, P(Rh-negative) = 0.16 Here is the table from problem 4.13 Blood Type O A B AB U.S. probability Next we go through all possible combinations. P(Rh-positive AND O) = P(Rh-positve)P(O) = 0.84(0.45) This calculation is correct because we know we have independence. We continue in the same manner to get the rest. P(Rh-positive AND A) = 0.84(0.4) P(Rh-positive AND B) = 0.84(0.11) P(Rh-positive AND AB) = 0.84(0.04) P(Rh-negative AND O) = 0.16(0.45) P(Rh-negative AND A) = 0.16(0.40) P(Rh-negative AND B) = 0.16(0.11) P(Rh-negative AND AB) = 0.16(0.04) A roulette wheel has 38 slots, numbered 0, 00, and 1 to 36. The slots 0 and 00 are colored green, 18 of the others are red, and 18 are black. The dealer spins the wheel and at the same time rolls a small ball along the wheel in the opposite direction. The wheel is carefully balanced so that the ball is equally likely to land in any slot when the wheel slows. Gamblers can bet on various combinations of numbers and colors. (a) What is the probability that the ball will land in any one slot? Since there are 38 slots available every slot has a 1/38 chance. (b) If you bet on "red," you win if the ball lands in a red slot. What is the probability of winning? P(red) = 18/38 (c) The slot numbers are laid out on a board on which gamblers place their bets. One column of numbers on the board contains all multiples of 3, that is, 3, 6, 9,..., 36. You place a "column bet" that wins if any of these numbers comes up. What is your probability of winning? I am going to assume that you can only bet on a single color. Since 12(3) = 36 there are 12 such multiples of 3. P(multiple of 3 single color) = 12/38.

5 4.29 Internet sites often vanish or move, so that references to them can't be followed. In fact, 13% of Internet sites referenced in major scientific journals are lost within two years after publication. 9 If a paper contains seven Internet references, what is the probability that all seven are still good two years later? What specific assumptions did you make in order to calculate this probability? P(link lost, within 2 yrs) = 0.13 P(link not lost, within 2 yrs) = 0.87 P(link not lost AND link not lost AND AND link not lost) = In order to calculate this I had to assume I had independence, which means the following. I am assuming that if a link is lost within two years, the probability of another link not being lost has not changed from the stated 13%. On a practical matter, suppose you knew that all seven links resided on the same server, then the independence assumption would not apply. Why? Because one can reason that if the links all reside on the same server, the manager of the server might decide to sever all links, or if the server goes down, then all the links go down. The fact that they all reside on the same server, would put in question the independence of their survival Government data show that 6% of the American population are at least 75 years of age and that about 51% are women. Explain why it is wrong to conclude that because (0.06)(0.51) = about 3% of the population are women aged 75 or over. If you calculate P(women AND 75 or greater) = 0.06(0.51) ONE is assuming independence between events. What exactly does that mean? Let start with P(woman) = 0.51 according to the stated figure. The population is the United States. In order to have independence between population of women and the population of people over 75 years of age, the following would have to be true: for the subpopulation (with respect to the U.S. population) of people over 75 years of age, the percentage of women in that subgroup would also need to be 51%. And here is the reason for the title of independence. If the above statement is true, it does not matter if I ask what is the percentage of women in the U.S. or what is the percentage of women in the U.S. over age 75, the response would come back 51%. Similarly, if we had independence between the two groups, we can the restate the subgroup the other way as well. The people aged over 75 in the U.S. is 6%. What if we looked at the subgroup of women in the U.S. With independence, the number of people age over 75 when we only consider the subgroup of women would continue to be 6%. To be able to calculate the AND statement as P(women AND 75 or greater) = 0.06(0.51), you need to know if one has independence between the groups the statements allude to.

6 4.33 The "random walk" theory of securities prices holds that price movements in disjoint time periods are independent of each other. Suppose that we record only whether the price is up or down each year, and that the probability that our portfolio rises in price in any one year is (This probability is approximately correct for a portfolio containing equal dollar amounts of all common. stocks listed on the New York Stock Exchange.) First I am going to organize the given information. P(price up) = 0.65 P(price down) = 0.35 (we will have to assume that if the price remains flat that we can place it in one of the two mentioned categories). Notice we have independence. (a) What is the probability that our portfolio goes up for 3 consecutive years? P(price up AND price up AND price up) = (b) If you know that the portfolio has risen in price 2 years in a row, what probability do you assign to the event that if will go down next year? P(price down) = 0.35 since I have independence (it does not matter what occurred the previous year). (c) What is the probability that the portfolio's value moves in. the same direction in both of the next 2 years? P(down And down OR up And up) = P(down And down) + P(up And up) =