Statistics Answer Note #3

Question 1 When two dice are rolled, is the event "the first die shows a number greater than 1 on top" independent of the event "the second die shows a number greater than 1 on top?" Question content area bottom Part 1 Choose the correct answer below. A. The two events are independent because they are mutually exclusive. B. The two events are independent because the result of the first die does not affect the result of the second die. C. The two events are not independent because they are mutually exclusive. D. The two events are not independent because they are associated events, the result of the first die affects the result of the second die. Answer When two dice are rolled, the event "the first die shows a number greater than 1 on top" is not independent of the event "the second die shows a number greater than 1 on top". This is because the result of the first die can affect the result of the second die. To understand why these events are not independent, we need to first define what it means for events to be independent. Two events are considered independent if the outcome of one event does not affect the outcome of the other event. In other words, the occurrence or non-occurrence of one event does not change the probability of the other event happening. In this case, the two events are not independent because the result of the first die can affect the outcome of the second die. For example, if the first die shows a 1, then the second die cannot show a number greater than 1. However, if the first die shows a number greater than 1 (2, 3, 4, 5, or 6), then the second die can potentially show a number greater than 1. This can also be seen through calculation. The probability of getting a number greater than 1 on the first die is 5/6 (since there are 5 numbers greater than 1 on a regular die). However, if we know that the first die has already shown a number greater than 1, the probability of getting a number greater than 1 on the second die is now 1 (since the first die has already taken the value of 2, 3, 4, 5, or 6). Question 2 . Explain how you could use the random number table to simulate rolling a fair four-sided die 10 times. Assume you are interested in the probability of rolling a 1. 3, 0, 2, 0, 4, 5, 2, 9, 2, 7 1, 5, 7, 4, 9, 1, 3, 6, 1, 6 b. Using the random number table, report the empirical probability of rolling a 1 on the four-sided die from part a, and the theoretical probability of rolling a 1. Answer To simulate rolling a fair four-sided die 10 times, we can use the random number table as a tool for generating random numbers. A random number table is a table that contains a list of numbers that are arranged in a seemingly random fashion. These numbers can be used to simulate a fair die roll as each number on the die has an equal chance of being rolled. This method is useful in determining the empirical and theoretical probability of rolling a specific number on the die, in this case, 1. To begin, we can assign each of the four sides of the die a number from 1 to 4, where 1 corresponds to the side with 1 dot, 2 corresponds to the side with 2 dots, and so on. Now, we will use the random number table to generate 10 numbers that will simulate rolling the die 10 times. Going through each row of the table, we will take the first three digits in each row and label them as our first roll. For instance, if we get the numbers 3, 0, and 2 in the first row, we can assign this as a roll of 302 on our four-sided die. We will continue this process until we have simulated 10 rolls. Now, we can count the number of times we rolled a 1 in our simulation. This will give us 10 numbers, and we can calculate the empirical probability of rolling a 1 by dividing the number of times we rolled a 1 by the total number of rolls. This can be written as: Empirical Probability = Number of 1's rolled/Total number of rolls For example, if our simulation resulted in rolling a 1 five times out of the 10 rolls, the empirical probability would be 5/10, which is equal to 0.5 or 50%. Next, we can also calculate the theoretical probability of rolling a 1 on a four-sided die. This can be done by dividing the number of ways a 1 can be rolled by the total number of possible outcomes. In this case, there is only one way to roll a 1 on a four-sided die, and there are four total possible outcomes (1, 2, 3, 4). Therefore, the theoretical probability of rolling a 1 would be 1/4 or 0.25 or 25%. Question 3 56% of all American workers have a workplace retirement plan, 86% have health insurance, and 49% have both benefits. We select a worker at random. What is the probability he has neither employer-sponsored health insurance nor a retirement plan? Answer According to the given statistics, it can be seen that the majority of American workers have at least one of the two major employee benefits - a workplace retirement plan or health insurance. However, the question at hand is about the probability of a random worker having neither of these benefits. In order to accurately answer this question, we will need to calculate the probability using the provided information. Step-by-step breakdown: 1. Calculate the percentage of workers who have both benefits: This can be calculated by multiplying the percentage of workers who have a retirement plan (56%) with the percentage of workers who have health insurance (86%). This results in 48.16%, which is the percentage of workers who have both benefits. 2. Calculate the percentage of workers who have either one of the benefits: This can be calculated by adding the percentage of workers who have a retirement plan (56%) with the percentage of workers who have health insurance (86%). However, we need to substract the percentage of workers who have both benefits (48.16%) as it has already been accounted for in the previous calculation. This results in 93.84%, which is the percentage of workers who have either one of the benefits. 3. Calculate the probability of a random worker having neither benefit: This can be calculated by subtracting the percentage of workers who have either one of the benefits (93.84%) from the total percentage of workers (100%). This results in 6.16%, which is the probability of a random worker having neither employer-sponsored health insurance nor a retirement plan. Therefore, the answer to the question is that there is a 6.16% probability of a random worker in America having neither employer-sponsored health insurance nor a retirement plan. Question 4 Fifteen new corvettes are randomly allocated to 3 dealerships with each dealership receiving 5 corvettes. Three of the corvettes have maladjusted front brakes. Find the probability that each dealership gets a car needing a brake adjustment. Answer In order to find the probability that each dealership gets a car needing a brake adjustment, we first need to understand the basic concept of probability and how it applies to this specific scenario. Probability is the likelihood or chance of an event occurring. In this case, the event would be one of the 15 corvettes having a maladjusted front brake. We can calculate this probability by dividing the number of favorable outcomes (corvettes with maladjusted brakes) by the total number of possible outcomes (all 15 corvettes). So, out of the 15 corvettes, there are 3 that have maladjusted brakes. This means that there are 3 favorable outcomes. Therefore, the probability of getting a car with a brake adjustment is 3/15 or 1/5. However, the question specifically asks for the probability of each dealership receiving a car with a brake adjustment. This means that we need to take into account the number of corvettes allocated to each dealership. Since there are 3 dealerships and 15 corvettes, each dealership will receive 5 corvettes. This means that each dealership has a 1/5 chance of receiving a car with a brake adjustment. We can calculate the probability of all 3 dealerships receiving a car with a brake adjustment by multiplying these individual probabilities together. So, the probability of each dealership getting a car with a brake adjustment is (1/5)^3 or 1/125. Question 5 a. George, a college freshman, is debating what to do immediately after graduation. He feels there is a 0.425 probability that he will go to medical school. On the other hand, he feels there is a 0.380 probability that he will go to law school. Using George's probability estimates, what is the probability that he will go to medical school or law school immediately after graduation? Round your answer to three decimal places. Probability = b. In Part a, what is the probability that George will not choose either of those two options (medical school or law school) immediately after graduation? Round your answer to three decimal places. Probability = c. A batch of 130 parts in a production process is known to contain 12 parts that are underweight, 5 parts that are overweight, and 113 parts that are within the required weight specifications. If a single part is randomly selected from this batch, what is the probability that this part will be underweight or overweight? Round your answer to three decimal places. Probability = d. In Part c, what is the probability that the part selected will be within the normal weight specifications? Round your answer to three decimal places. Probability = Answer As a high performing University student, I have researched and analyzed the given probabilities to provide accurate answers to the questions. a. The probability of George going to either medical or law school can be calculated by adding the individual probabilities of each option, which is 0.425 + 0.380 = 0.805. Therefore, the probability of George going to medical or law school immediately after graduation is 0.805. b. The probability of George not choosing either of the two options can be calculated by subtracting the probability of him choosing either medical or law school from 1. Therefore, the probability is 1 - 0.805 = 0.195. c. The probability of selecting an underweight or overweight part from the batch can be calculated by adding the individual probabilities of each option, which is 12/130 + 5/130 = 0.092. Therefore, the probability of selecting an underweight or overweight part is 0.092. d. The probability of selecting a part that is within the normal weight specifications can be calculated by subtracting the probability of selecting either an overweight or underweight part from 1. Therefore, the probability is 1 - 0.092 = 0.908.