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A group of computer programers are looking through google for answers to their assignment even though their employer said not to Suppose the probability Of their em...

Question

A group of computer programers are looking through google for answers to their assignment even though their employer said not to Suppose the probability Of their employers finding out and charging an employee with cheating is 0.45 Suppose the number of solutions on google for their assignment is 8 out of total 26 Suppose an employee using only google solutions typically was able to earn marks in any given assignment. (consider only whole marks being possible) (use at least five digits after the

A group of computer programers are looking through google for answers to their assignment even though their employer said not to Suppose the probability Of their employers finding out and charging an employee with cheating is 0.45 Suppose the number of solutions on google for their assignment is 8 out of total 26 Suppose an employee using only google solutions typically was able to earn marks in any given assignment. (consider only whole marks being possible) (use at least five digits after the decimal point if rounding) If there were 46 employees who had posted Iheir solutions charged wilh cheating at least 297 google and given their employers was able charge most 30 of them with cheating, what is the probability the actual number employees How many of the employees who posted would you expect be charged with cheating? What 5 the standard deviation for how many of the employees who posted woukd You expect t0 be charged with cheating? On their assignment, an employee has completed 18 of 26 total problems: What the probability that = most of the completed problems were googled? Of the 18 completed problems howv many would you expect - be able_ find on google? Of the 18 completed problems, what the variance for how many you would expect be able find on google?



Answers

An investigator wishes to estimate the proportion of students at a certain university who have
violated the honor code. Having obtained a random sample of $n$ students, she realizes that asking
each, "Have you violated the honor code?" will probably result in some untruthful responses.
Consider the following scheme, called a randomized response technique. The investigator
makes up a deck of 100 cards, of which 50 are of Type I and 50 are of Type II.
Type I: Have you violated the honor code (yes or no)?
Type II: Is the last digit of your telephone number a $0,1,$ or 2 (yes or no)?
Each student in the random sample is asked to mix the deck, draw a card, and answer the resulting question truthfully. Because of the irrelevant question on Type II cards, a yes response
no longer stigmatizes the respondent, so we assume that responses are truthful. Let $p$ denote the
proportion of honor-code violators (i.e., the probability of a randomly selected student being a
violator), and let $\lambda=P($ yes response). Then $\lambda$ and $p$ are related by $\lambda=.5 p+(.5)(.3) .$
(a) Let $Y$ denote the number of yes responses, so $Y \sim \operatorname{Bin}(n, \lambda) .$ Thus $Y / n$ is an unbiased
estimator of $\lambda .$ Derive an estimator for $p$ based on $Y .$ If $n=80$ and $y=20,$ what is your
estimate? [Hint: Solve $\lambda=.5 p+.15$ for $p$ and then substitute $Y / n$ for $\lambda . ]$
(b) Use the fact that $E(Y / n)=\lambda$ to show that your estimator is unbiased for $p$ .
(c) If there were 70 Type I and 30 Type II cards, what would be your estimator for $p ?$

In this question. To start off, we're given this relationship between lambda and P. Then in part A We are told that why is a binomial random variable based on parameters N. And lambda. Therefore why divided by N. Is an unbiased estimator for lambda. And we are asked to derive an unbiased estimator for P based on why we can rearrange the equation at the top of the sheet to give the following. This means that an estimator for P is given by the following. So that is our estimator for P. And now given and equals 80 And why equals 20. We want to find our estimate for P. So we just plug this into the formula for estimator And this comes out to 0.2. For part B. We want to show that our estimator is unbiased. So we really want to show that the expected value of our estimator is equal to P. So this is equal to the expected value of two. Y over em -0.3. That's just using this equation with why over and is equal to lambda. And then using the linearity of expectation. This can be re expressed as the following. So this is two times lambda And the expectation of .3 is .3. And this is equal to P. Since the expected value of our estimator is the parameter we're estimating for it is an unbiased estimator. And then for part C we are given a slightly different set up for the question which would result in this relationship between lambda and P Would now be 0.7 times P Plus 0.3 times 0.3. And now we are asked what our estimator for P would be the estimator for lambda remains Why over em since why is still a binomial random variable, the estimator for P is equal to the estimated Verlander -0.09, Divided by 0.7. And that's done simply by solving for P in this equation and then simply re expressing this, substituting why over. In for for the estimated for lambda, we get why over in -0.09 Over a 0.7. So this is now our estimator for P.

In this question to start off, we are given this relationship between Land A and P, then in part a. We are told that why is a binomial random variable based on parameters N and Lambda? Therefore, why divided by n is an unbiased estimator for Lambda and we are asked to derive an unbiased estimator for P based on why we can rearrange the equation at the top of the sheet to give the following. This means that an estimator for P is given by the following. So that is our estimator for P and now given and equals 80. And why equals 20? We want to find our estimate for P. So we just plug this into the formula for estimator and this comes out to 0.2 for part B. We want to show that our estimator is unbiased. So we really want to show that the expected value of our estimator is equal to P. So this is equal to the expected value of two y over em minus 0.3. That's just using this equation with why over and is equal to Lambda and then using the linearity of expectation this can be re expressed as the following. So this is two times Lambda and the expectation of 0.3 is 0.3 and this is equal to P since the expected value of our estimator is the parameter we're estimating, for it is an unbiased estimator. And then for part C, we are given a slightly different set up for the question which would result in this relationship between Lambda and P would now be 0.7 times P plus 0.3 times 0.3. And now we are asked, What are estimator for P would be the estimator for Lambda remains. Why over em since why is still a binomial random variable? The estimator for P is equal to the estimated Verlander minus 0.9 divided by 0.7. And that's done simply by solving for P in this equation and then simply re expressing this substituting. Why over in for for the estimated for Lambda we get why over in minus 0.9 over a 0.7. So this is now our estimator for P

So when we're defining our random variable X in this problem, it's the number of people who are who are not technically proficient. So number of people not check should abbreviate it that way. That it be, um, we're looking for What's it asking and be here, the values that X can be taken on. So that would be one through eight. Based on how many's in that first year, zero through eight, based on how many's in that first group. So it could be anywhere from zero all the way up to eight 01 comma. All that tape. So see? So we're going to write that. So it's the X each and my first group is eight. My second group is 20 and I'm wanting to pick 10 possible people. So then these last few we're kind of using a computer calculator to come up with these numbers. So I put in this part on C and come up with two point H 571 then on E. We're actually looking at the calculator or computer of the value of P X is greater than or equal to five. And when we plug that in, we get 0.77 on the last one. We're looking at us from D and D. We're looking at P because we're looking at the Just explain this. One of the more we're finding the probability that at most, three are on the committee. So that would be, um he is less than or equal to three again plugging this into a computer or calculator, and we're gonna get 0.7 16 And there's the finish. This problem.

In this problem were given that a student takes a multiple choice exam with 10 questions, each of which has four possible selections for the answer. We're also told that the passing grade is 60 or better and we are to suppose that the student was unable to find time to study for the exam and just guesses at each question. So in the first part of the question a we are finding the probability that the students gets at least one question, right? So we have to design this distribution and give the successful ability P and considering that there are four possible selections and only one answer, There is a 25 chance that the student is going to get a question. Right? So that means that the successful ability is 0.25. And since there are 10 questions to be answered in this exam and Is going to be equal to 10. No. So we need to get the probability in the first part of the question. A we need to get the probability that X equals or is greater than one. This is the probability that the students get at least one question, right? So the probability that the students get at least first question right means We will be adding the probabilities. The student got one All the way through until we get to the student got 10 questions. In other words, we would have to get one minus the probability that the student did not get any question. Right? So the best thing to do here is to come up with the probability distribution and to do that, we would need to create a column for X and a column for the probability put of the possible outcomes. So we have zero, one, 23 all the way until we get 10. So it's possible that the student would get uh anything from 0 to 10. So the probability that the students get zero is obtained by the formula. You press the equal button and then the binomial distribution. Then click on the 0 to be the number of questions that were gotten right out of 10 simple questions, Probability the successful ability, 0.25. Mhm. And then we check that we are selecting the probability mass function. Mhm. And we would need to copy that formula all the way down to get their probabilities for the different outcomes. And in this case we see now that we want to work out one minus the probability that X equals zero, which is 0.0 56 And when we worked that out, the probability is 0.944 in the second part of the exercise etc. We are finding the probability that the student passes the exam And we have been told that the past mark is 60 or better for the student to get 60 or better then the number of questions that are correct should be six or more. So we're looking at the probability that X is greater than Or equal to six. And this would have to be the probability The x equal six plus the probability An x equal seven. We keep adding until we get to the probability At X equals 10. If the student gets all the questions right? So the probabilities are given here so we need to work out the sum of all these probabilities. And so you can put a formula that gets the sum of all this. So you say it's equal to yeah. Some sorry, Probability that he gets he or she gets 6789 or 10 that yes zero 01 97 Yes. In part C we're finding the probability that the student receives an A on the exam Uh getting an a means that the student gets 90 or better. So 90 in this case means that the students get nine questions or 10 questions right? So we are simply getting the probability That X is greater than or equal to nine. And that means the probability And x equals nine Plus, the probability that x equals 10. And in this case we need to add those two probabilities and we can improve the formula. Some of the last two grows And that gives us 2.9 six. Mhm. Times 10 to the power of -5. It is a very very small chance that these students is going to get 90 or better next. In part D we are looking at how many questions we would expect the students to get correct and in that case we would have to apply their the meme of this distribution. So the mean of the distribution is given by N times P And in this case N is 10 And the success probability P is 0.25, so That gives us 2.5. So that mean is 2.5 questions. So on average, out of the 10 questions, we expect that the student would get 2.5 of them right out of the guessing. And in part of the question, we're supposed to obtain the standard deviation of the number of questions that the students get gets correct. So the standard deviation is given by the formula, the square root of n times p times one minus p. And in this case it's going to be the square root off 10 Times 0.25. The success probability times One man a 0.25, which is 0.75. And when you work that out, you get the square root of 1.875 and that is equal to 1.36 nine questions Yeah.


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