Suppose that the handedness of the last fifteen U.S. presidents is as follows:40% were left-handed (L)47% were Democrats (D)If a president is left-handed, there is a 13% chance that the president is a Democrat.Based on this information on the last fifteen U.S. presidents, is “being left-handed” independent of “being a Democrat”? No, since 0.47 is not equal to 0.13. No, since 0.40 is not equal to 0.13. Yes, since 0.40 * 0.47 is not equal to 0.13. Yes, since 0.47 is not equal to 0.13.

Suppose that the handedness of the last 15 U.S. presidents is as follows:(i) 40% were left-handed (L)(ii) 47% were democrats (D)(iii) If a president is left-handed, there is a 13% chance that the president is a Democrat.What is the probability that a randomly chosen U.S. president is left-handed and a democrat? 0.40 * 0.47 = 0.1880 0.40 * 0.13 = 0.0520 0.47 * 0.13 = 0.0611 0.40/0.47 = 0.8510 0.40/0.13 = 3.0769 0.47/0.13 = 3.6154Question 2Select one answer.10 pointsTwo methods, A and B, are available for teaching Spanish. There is a 70% chance of successfully learning Spanish if method A is used, and a 85% chance of success if method B is used. However, method B is substantially more time consuming and is therefore used only 20% of the time (method A is used the other 80% of the time). The following notations are suggested:A—Method A is used.B—Method B is used.L—Spanish was learned successfully.Which of the following is the correct probability tree for this problem?Question 3Select one answer.10 pointsTwo teaching methods, A and B, are implemented for learning Spanish. There is a 70% chance of successfully learning Spanish if method A is used, and a 85% chance of success if method B is used. However, method B is substantially more time consuming and is therefore used only 20% of the time (method A is used the other 80% of the time). The following notations are suggested:A—Method A is used.B—Method B is used.L—Spanish was learned successfully.What is the probability that a randomly chosen person will learn Spanish successfully? P(L) = 0.20 * 0.85 = 0.1700 P(L) = 0.80 * 0.70 = 0.5600 P(L) = 0.80 * 0.70 + 0.20 * 0.85 = 0.7300 P(L) = 0.80 * 0.85 + 0.20 * 0.70 = 0.8200 P(L) = 0.80 * 0.20 + 0.70 * 0.85 = 0.7550Question 4Select one answer.10 pointsTwo methods, A and B, are available for teaching a certain industrial skill. There is an 80% chance of successfully learning the skill if method A is used, and a 95% chance of success if method B is used. However, method B is substantially more expensive and is therefore used only 25% of the time (method A is used the other 75% of the time). The following notations are suggested:A—Method A is used.B—Method B is used.L—The skill was learned successfully.Which of the following is the correct representation of the information that is provided to us? P(A) = 0.75, P(B) = 0.25, P(L | A) = 0.80, P(L | B) = 0.95 P(A) = 0.75, P(B) = 0.25, P(A | L) = 0.80, P(B | L) = 0.95 P(A) = 0.75, P(B) = 0.25, P(A and L) = 0.80, P(B and L) = 0.95 P(A | L) = 0.75, P(B | L) = 0.25, P(L | A) = 0.80, P(L | B) = 0.95 P(A and L) = 0.75, P(B and L) = 0.25, P(L | A) = 0.80, P(L | B) = 0.95

Left-handedness occurs in about 12% of all Americans. Males are slightly more likely than females to be left-handed, with 13% of males and 11% of females being left-handed. Suppose a random sample of 80 females and 100 males is chosen.Let X be the number of males (out of the 100) who are left-handed.Let Y be the number of females (out of the 80) who are left-handed.Let Z be the total number of left-handed individuals in the sample (males and females together).Which of the following is true regarding the random variables X and Y? Both X and Y can be well approximated by normal random variables. Only X can be well approximated by a normal random variable. Only Y can be well approximated by a normal random variable. Neither X nor Y can be well approximated by a normal random variable.Question 2Type numbers in the boxes.10 pointsSuppose the time to complete a 200-meter backstroke swim for female competitive swimmers is normally distributed with a mean μ = 141 seconds and a standard deviation σ = 7 seconds. Suppose that to qualify for the Nationals, a woman must complete the 200-meter backstroke in less than 128 seconds. What proportion of competitive female swimmers will qualify for the Nationals? Give your answer to four (4) decimal places.Click here to download the normal table.Question 3Select one answer.10 pointsSuppose the scores on an exam are normally distributed with a mean μ = 75 points and standard deviation σ = 8 points.What is the exam score for an exam whose z-score is 1.25? 65 75 85 0.8944 0.1056

According to previous studies, 16% of the U.S. population is left-handed. Not knowing this, a high school student claims that the percentage of left-handed people in the U.S. is 15%.The student is going to take a random sample of 2100 people in the U.S. to try to gather evidence to support the claim. Let p be the proportion of left-handed people in the sample.Answer the following. (If necessary, consult a list of formulas.)(a)Find the mean of p.(b)Find the standard deviation of p.(c)Compute an approximation for P≥p0.15, which is the probability that there will be 15% or more left-handed people in the sample. Round your answer to four decimal places.

Assume that 25% of people are left-handed. If we select 10 people at random, find the probability that exactly 3 are lefties.

Which of the following statements best describes the party identification of U.S. adults?A.47 percent favor the Democratic party; 40 percent favor the Republican party; 13 percent claim to be independent or favor some other partyB.27 percent favor the Democratic party; 60 percent favor the Republican party; 13 percent claim to be independent or favor some other partyC.17 percent favor the Democratic party; 17 percent favor the Republican party; 65 percent claim to be independent or favor some other partyD.67 percent favor the Democratic party; 20 percent favor the Republican party; 13 percent claim to be independent or favor some other party