Suppose a student takes a multiple-choice quiz. There are 4 choices for each question. The student does not knowthe answers to the last 5 questions, so she guesses the answers. How many trials of the simulation wereconducted? What was the average number of correct answers for these trials? Use the simulation results below,where 1 represents a correct answer and 2, 3, and 4 represent incorrect answers.31242 41211 34141 41212 23342 23242 1341211313 42433 32334 31234 13314 41432 23413How many trials of the simulation were conducted?(Type a whole number.)What was the average number of correct answers for these trials?Next

Simulation is the type of probability model. It used to consider the likelihood of observation…

Answered: Suppose a student takes a…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A sample of ten commuters were asked for the distance and the time required commuting to their jobs on a given day. The data is shown in the table along with some calculations from the data. Which of the following statements are true in reference to this data? Distance x I 3 717 18 1011012 Time =y 5 10 15 2015 25 2025 35 35 From the data one obtains: = 276 SS = 103.6 SS SS = 872.5 EX =68 2Y = 205 I) X and Y show a positive correlation with r = .918 II) X and Y show a positive correlation with r = .843 III) Approximately 84.3% of the variation in time is explained by distance O l only O Il only O III only O I and III only Il and III only O I, II, and lII are all true
Consider a medical test with a false positive rate of 1% and a false negative rate of 2%. The frequency of the condition in people who take this test is 3% Q4. If this test was given to 100,000 healthy people, how many false diagnoses would there be? Q5. What is the overall percentage of people who test "positive" who are actually negative for the condition? Q6. What is the overall percentage of people who test "negative" who are actually positive for the condition? # wrong diag. = ??? % pos. = ??? % % pos. = ??? %
A certain affects virus 0.4% of the population (in a population of 100,000 people, 400 will be infected with the virus). A test used to detect the virus in a person is positive 87% of the time if the person has the virus (true positive) and it is positive 14% of the time if the person does not have the virus (false positive). Fill in the remainder of the following table and use it to answer the questions below. Infected Not Infected Total Positive Test Negative Test Total 400 99,600 100,000
A survey was conducted on how many texts (in average) students receive during an in-person class time. A survey data in a class in a particular day is given in the table. Number students Number of texts 1 20 2 14 12 4 5 If the average number texts that students received during this class time is 3, then determine the number of students (the value of x) that received 5 texts during the class time on that particular day.
PLEASE ANSWER THIS USING YOUR OWN ANSWERS. DO NOT COPY ON INTERNET. IF YOU SEE THIS QUESTION AGAIN, DO NOT ANSWER. IF YOU ANSWER THIS IN TYPEWRITTEN, COMPLETE, CORRECT, AND IN ORIGINAL WAY, I WILL UPVOTE. THANK YOU!
We are interested in estimating the proportion of people that can identify Nebraska but not Vermont. In a class of 55 students, each student was requested to select a sample of 6 people, show each person a map of the US and ask the person to identify the states of Vermont and Nebraska. The following numbers are the counts that each student got, that is, each number is the count of people in the student sample of 6, that could identify Nebraska but not Vermont.3344010110020030002020001100002010000001000200000001010Note: You can cut and past these numbers into an excel spreadsheet to help you calculate the mean and the standard deviation and compute the confidence interval. If you have trouble pasting them, paste them to a text file first and then copy from there.When we collect all the students' data, we have a sample size of 6x55=330.Of the sample of 330 people, what proportion were able to locate Nebraska but not Vermont? (round answer to 3 decimals) Find a 95% confidence interval for…
Can a person train to become better at holding their breath? An experiment was designed to find out. Twelve volunteers were randomly assigned to 1 of 2 groups. The 6 volunteers assigned to group 1 were given breath-holding exercises to perform for 2 weeks. The other group was not given any information about the experiment. At the end of the 2 weeks, all 12 volunteers were individually tested to determine how long they could hold their breath. Here are the data (in seconds). Group 1: 90, 88, 70, 110, 75, 105 Group 2: 40, 48, 35, 50, 55, 62 The researcher would like to determine if these data provide convincing evidence that the true mean amount of time volunteers who were given training held their breath is greater than volunteers without training. Let ₁ = the true mean amount of time that volunteers who were given training held their breath and ₂ = the true mean amount of time that volunteers without training held their breath. What are the appropriate hypotheses? Ho: U1 - U2 = 0, Ha:…
If the mean of 5 values equals 34, then ΣΧ=
PLEASE HELP SOLVE THE PROBLEM BELOW A candy manufacturer selects mints at random from the production line and weighs them. For one week, the day shift weighed n1 = 194 mints and the night shift weighed n2 = 162 mints. The numbers of these mints that weighed at most 21 grams was y1 = 28 for the day shift and y2 = 11 for the night shift. Let p1 and p2 denote the proportions of mints that weigh at most 21 grams for the day and night shifts, respectively. Give a point estimate of p1 Give the 95% confidence interval for p1 Give a point estimate of p1 − p2 Find the 95% confidence interval for p1 − p2
Two varieties of corn are grown in 4 different locations. Each location provides a value in pounds for each variety.The table below lists the results. Find the standard error for the difference in the production of the two corn varieties to two decimal places. Variety 1 Variety 2 difference location 1 32.1 34.5 -2.4 location 2 30.6 32.6 -2.0 location 3 33.7 34.6 -0.9 location 4 29.7 31.6 -1.3 mean value 31.53 33.18 -1.65 standard deviation 1.76 1.72 0.68 1.23 0.75 0.34 0.98

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