Guided Exercise 3

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University of Delaware *

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200

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Statistics

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Apr 3, 2024

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1 STAT 200 Guided Exercise 3 For On-Line Students, be sure to:  Please submit your answers in a Word or PDF file to Canvas at the place you downloaded the file.  You can paste Excel/JMP output into a Word File. Please submit only one file for the assignment.  It is ok to do problems by hand. However, you will need to scan or take a picture of your work.  Guided Assignments are not graded but we check that you did the problems. Key Topics  Probability  Probability with Tables 3. An individual’s genetic makeup is determined by the genes obtained from each parent. For every genetic trait, each parent posses a gene pair; and each contributes ½ of the gene pair, with equal probabilities, to their offspring, forming a new gene pair. The offspring’s traits (eye color, baldness, etc.) come from this new gene pair, where each gene in this pair posses some characteristic. For the gene pair that determines eye color, each gene trait may be one of two types, dominant brown, (B) or recessive blue (b). A person possessing the gene pair BB or Bb has brown eyes, whereas the gene pair bb produces blue eyes. a. Suppose both parents of an individual are brown eyed, each with a gene pair Bb . What is the probability that a randomly selected child of this couple will have blue eyes? Only 1 out of 4 would have the bb combination for blue eyes if both parents were Bb. The probability would be ¼=.25 b. If one parent has brown eyes, type Bb , and the other has blue eyes, what is the probability that a randomly selected child of this couple will have blue eyes? If one parent was Bb while the other was bb, the probability of the child having blue eyes would be 2/4=.50 c. Suppose one parent is brown-eyed, type BB. What is the probability that a child has blue eyes? If one parent was bb and the other parent was BB there would be no chance of there being a child with blue eyes. This would be 0/4=0. 1

2. On a standard SAT test, a typical question has five possible answers; A, B, C, D, and E. Only one answer is correct. If you guess you have a 1 out of 5 or .20% chance of being correct. a. What is the probability of not being correct on a single question if you randomly guess? The probability of not being correct on a single question is equal to .8. This can be found by doing 1-.20=.80, which id the probability of being right. b. What is the probability of getting all three questions right on three questions if you randomly guess? P(3 right) = .2 x .2 x .2 = 0.008. This just means there is a small chance of getting all three questions correct if you are guessing. c. What is the probability of gettting at least one question right on three questions if you randomly guess? P(3 right) = .2 x .2 x .2 = 0.008 P(2 right) = .2 x .2 x .8 x 3 = 0.096 P(1right)=(.2*.8*.8)*3=.384 P(at least 1 right) = 0.008+0.096+0.348 = 0.488 d. What is the probability of getting all three questions wrong on three questions if you randomly guess? P(all 3 wrong) = .8 x .8 x .8 = .512 e. Training on how to do better on the SAT test advise that you should guess if you can eliminate possible answers. Suppose on a question you can eliminate two possible answers. What is the probability that you are right if you randomly guess your answer. This would be a Conditional Probability. The probability of a correct guess give that you would get rid of two answers. Without those two answers I would have a 1/3 chance of guessing correctly, which equals to .333 2

3. Across the U.S. in 2020 and 2021 there was increased attention to voting security. Some states began enacting more stringent voting requirements for national elections. These requirements came with controversy and debate. One requirement is identification, usually photo identification. According to the National Conference of State Legislatures (NSL), "a total of 36 states have laws requesting or requiring voters to show some form of identification at the polls, 35 of which are in force in 2020." (NSL, https://www.ncsl.org/research/elections-and-campaigns/voter-id.aspx#Laws%20in%20Effect ). The main form of photo identification for most adults in a driver’s license. However, not everyone has a driver's license, and the proportion with a license is decreasing over time. Whether you have a driver’s license depends upon a number of factors, including age, where you live, your race, and your income level, among others. I found data for those who have a license by age group in 2019. I applied this data to a pseudo survey of 1,000 adults. You can easily copy and paste this table directly into Excel and work on the questions there. Do You Have a Driver’s License? Age Yes No Total 18 to 29 170 41 211 30 to 54 377 34 411 55 to 69 220 15 235 70 and Over 117 26 143 Total 884 116 1000 a. What is the probability that a randomly selected adult in the U.S. has a driver’s license? P(yes0 =844/1000=0.840 b. What is the probability that a randomly selected adult in the U.S. is 70 and over? P(70 and over) = 143/1000 = .143 c. Given that a randomly selected adult is 18 to 29, what is the probability that he/she will not have a license? This is a Conditional Probability, and the conditions is at 18 to 29. P(no 18 to 29) = 41/211 =.1943 d. Given that a randomly selected adult is 55 to 69, what is the probability that he/she will not have a license? This is a Conditional Probability, and the conditions is at 55 to 69. P(no 55 to 69) = 15/235 = 0.0638 e. Are the events Age and Do you have a Driver’s License mutually exclusive? Explain. They are not mutually exclusive. There is nothing that says you must be 45 and you will have a license. This table shows that you can be whatever age and either have your license or not have your license. 3

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f. Are the events “55 to 69” and “Yes has a license” independent? Why? They are not independent because you do have to be a certain age to get your license, so age is kind of a dependent g. Calculate the following odds and odds ratio for Not Having a License. (1.) Odds of Not having a license (versus Yes) for persons 18 to 29. 18 to 29 No to Yes = 41/170 = .2412 (2.) Odds of Not having a license (versus Yes) for persons 55 to 69. 55 to 69 No to yes = 15/220 = 0.0682 (3.) Odds ratio of (1) versus (2). Explain this in words. Odds Ratio = .2412/.0682 = 3.5373 Adults 18 to 29 are 3.5 times more likely tnot have a license when compared to those 55 to 69. 4. A fast food restaurant has determined that the chance a customer will order a soft drink is .90. The chance that a customer will order a hamburger is .6, and the chance for ordering French Fries is .5. a. If a customer places an order, what is the probability that the order will include a soft drink and no fries if these two events are independent ? .9 x .5 = .45 b. The restaurant has also determined that if a customer orders a hamburger the chance the customer will also order fries is .8. Determine the probability that the order will include a hamburger and fries. The two events are no independent because the probability that the order will have a hamburger and fries is equal to .48. This is from a conditional probability. 5. No diagnostic tests are infallible, so imagine that the probability is 0.95 that a certain test will diagnose a diabetic correctly as being diabetic, and it is 0.05 that it will diagnose a person who is not diabetic as being diabetic. It is known that roughly 10% if the population is diabetic. 4

What is the probability that a person diagnosed as being diabetic actually is diabetic? Hint: This is a use Bayes’ theorem problem, which we did not cover in the lectures. There is another way to handle this problem – make a mock 2 by 2 table of the data based on the information you already know. Once the table is complete, you can solve for the conditional probability. Since some of the probabilities are small, I would sugget you make a table that is based on 100,000 people. I have started the table for you. Test Results Diabetes Status Diabetic Not Diabetic Diabetic 9500 500 10,000 Not Diabetic 4500 85500 90000 14000 86000 100,000 P(D Test days D) = 9500/14000 = .6786 5