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QuestonAracerain h Gh school 90% Ofali students carry a backpack and 674 of all students bring therr Junch Given that stubent carrles backpack; 702 ofthese backpack carriers Wll bring their lunch. Let B be the eventtha: a Student carnes . backpack Let [ bethe €vent that a squdent pnngs their lunch Summanize in symbols the probabilijies descr bed above PIB)PIL =Sdettanznwet0.7Find the probability that & randomly se ected student carries backpackand onngs lunchFind the probabilitythar &

Queston Aracerain h Gh school 90% Ofali students carry a backpack and 674 of all students bring therr Junch Given that stubent carrles backpack; 702 ofthese backpack carriers Wll bring their lunch. Let B be the eventtha: a Student carnes . backpack Let [ bethe €vent that a squdent pnngs their lunch Summanize in symbols the probabilijies descr bed above PIB) PIL = Sdettanznwet 0.7 Find the probability that & randomly se ected student carries backpackand onngs lunch Find the probabilitythar & randomy se ected student carries . Dackpack= or dringsa iunch, Fino the probablity thata randomy se ected Squden: does no:bringtheir lunch Engtheprobab ty that a ranoomly selected Student carries squjent bnngs tneir Iuncn Dackpack = eventharire LDeterrine f tne events, carryirg Explaln dackpock and bringing lunch Jre muua exciusive To petide WE check tre cquabon; (eean Compute te Icft Sioe and Tergntside 0f the equetion above: Lenside] ana Right side Condusion Bona Select Zc Determine # [he events, carrying backpack and bninging lunch; are injependenz Expiain To Dedidc wc check thc cquopon: [anDnswer Compuie the Icft side and the ghE side of tht equotion above Len *06 and Foght nJe Condusion: Band L ere Euaunthatr don pdf Type here to search



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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 for number one, we're looking at male and female students with long hair or with short hair. And we're just riding out the symbols for each of these problems and a through J you're gonna let f represent a female on film. A male s is gonna represent short hair, and l is gonna represent you want to go through these, eh? Probability that a student does not have long hair is just written probability parentheses. And because we're focusing on not having long hair, we're gonna right along here with an apostrophe. This is gonna indicate the compliment of long hair, which is, in essence, the same thing as the probability of short hair. Be the probability that a student is male or has short hair probability print the seas male. We're just gonna represent with the Capitol or s, which stands for short hair. You see the probability that a student is a female and has long hair? We're gonna do probability female. We can write the word and ill for long Here, Indy. The probability that a student is male given that student here, this looks a little bit differently in this case on D, the probability that a student his mail given we're gonna represent that what they line down the middle, that they have long hair. So what we're doing here is actually changing our interest from the entire population to Justus with here in e the probability that a student has long here given that the student is a mill, you're gonna flip those letters around long here. Given that we have a male, Yeah, all the female students probability that a student has short here, so it doesn't have the word given in this one. However, there indicating that that's who were interested in this female students. So of all female students, the probability that a student have short hair. So we're looking at the probability that a short hair out of all the females g of all the students with long hair. So that's what we're given the probability that a student is female. So we're looking at female given long here h the probability that a student this female or has long hair just gonna separate those with or but probability that I randomly slept. The student is a male student with short hair, So a male student with short hair indicates that he's a male and he has short hair. The probability that a student is female, it's just purely probability of.

For this problem were given the table pictured here above, in part a access to calculate the joint probability. To do this, we take the numbers in each column. The dry probability between meeting the proficiency standards being in the third grade is 47,401 divided by the total number of responses. Just 423,392. This gives a joint probability of 0.1120 Drunk probability between not meeting the proficiency standards in being in the third grade is 23,975 divided by 423,392 to give a joint probability of 0.566 We do this for all of the joint probabilities between Met proficiency standards and which grade the student is in to get the following table. Pictured here to calculate the marginal probability, we total up each column and row and divide that by the total number. So the marginal probability for the third grade is going to be number of yes and no responses added together, which totals to 71,376 divided by the total number of responses. Just 423,392 To give a marginal probability a 0.16 eat six part B s. What are the marginal probabilities? What do they say about the probabilities of a student meeting or not? Or reading the proficiency standards to do this, we look at the marginal probabilities for the columns of yes and no. The marginal probability for meeting the proficiency standards is 0.5586 While the marginal probability of not meeting the proficiency standards is 0.4414 These probabilities show that students are more likely to meet the proficiency standards, not meet the proficiency standards. Since the probability is higher for meeting the standards, Heart C s. A randomly selected student is 1/3 grader. What is the probability the student will meet the proficiency standards? This is the probability of meeting the standards. Given that the student is 1/3 grader, this is equal to the joint probability between meeting the standards and being 1/3 grader divided by the marginal probability of being 1/3 grader. This is equal toe, 0.1120 divided by 0.1686 to give a probability of 0.6643 Percy also asked What if the randomly such a student is 1/4 grader? What is the probability they will meet the proficiency standards? This is the probability of meeting the standards given the student is 1/4 grader. This is equal to the joint probability between meeting standards and being 1/4 grader about about the marginal probability of being 1/4 grader. This is equal 2.827 divided by 0.1648 to give a probability of 0.5018 Party of the problem asked. A random student is known to have met the standards. What is the probability the student is 1/3 grader? This is the probability of the student being 1/3 giver. Third Greer, Given that they did meet the standards, this is equal to the joint probability between being 1/3 grader in meeting the proficiency standards. Divided by the marginal probability of meeting the proficiency standards, this is equal to a 0.1120 divided by 0.5586 to give a probability of 0.2005 Her d also asked, What is it? A random student is known to have met the standards. What is the probability the student is 1/4 grader? This is the probability of being 1/4 grader, given that the student did meet the standards. This is equal to the joint probability between being 1/4 grader in meeting the standards divided by the marginal probability of meeting the standards, this equals 0.8 to 7, about about 0.5586 to give a probability of 0.4. What 0.1 for 80?


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