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5) For this next problem; let p1 0.5 and p2 0.75 . Let the fraction of students at Boston University that took Astronomy be given by p1- Let the fraction of student...

Question

5) For this next problem; let p1 0.5 and p2 0.75 . Let the fraction of students at Boston University that took Astronomy be given by p1- Let the fraction of students at Oxford University that took Astronomy be given by p2. Suppose you are not sure how many students to survey; SO YOU continue to survey indefinitely:(a) What is the probability that none of the first 3 people surveyed took As- tronomy? (b) What is the probability that the first person that answers that they have taken Astronomy is

5) For this next problem; let p1 0.5 and p2 0.75 . Let the fraction of students at Boston University that took Astronomy be given by p1- Let the fraction of students at Oxford University that took Astronomy be given by p2. Suppose you are not sure how many students to survey; SO YOU continue to survey indefinitely: (a) What is the probability that none of the first 3 people surveyed took As- tronomy? (b) What is the probability that the first person that answers that they have taken Astronomy is the Sth person asked? (c) What is the probability that the first person that answers that they have taken Astronomy is the Sth person asked GIVEN that none of the first 3 people have taken Astronomy?



Answers

Group Activity Investigating an Athletic Program A university widely known for its track and field program claims that 75$\%$ of its track athletes get degrees. A journalist investigates what happened to the 32 athletes who began the program over a 6-year period that ended 7 years ago. Of these athletes, 17 have graduated and the remaining 15 are no longer attending any college. If the university’s claim is true, the number of athletes who graduate among the 32 examined should have been governed by binomial probability with $p=0.75$.
(a) What is the probability that exactly 17 athletes should have graduated?
(b) What is the probability that 17 or fewer athletes should have graduated?
(c) If you were the journalist, what would you say in your story on the investigation?

For this problem. We're told that 2% of two million high school students would take the S A T every year, requires special accommodations and then were asked to consider a random sample of 25 students who have recently taken the test. And we're asking probabilities so weakened we can define X as the number of successes being the number of students who required special accommodation and because we're sampling without replacement from the population. The samples are not strictly independent, but because 25 is far less than 5% of the total population, X can be pretty reliably estimated as a binomial random variable based on 25 size of 25 and the probability of success of 0.2 part A were asked, What is the probability that exactly one required special accommodation? What is the probability that we have exactly one success? So that's the probability mass function for the binomial, and this comes out to approximately zero point 308 now for B were asked, what is the probability that at least one received special accommodation? So what is the probability that we have at least one success we can rewrite this as one minus the probability of getting exactly zero successes, and this comes out to approximately zero point 397 or C. We're looking for the probability of at least two successes, so this could be rewritten as one minus probability. That X is less than or equal to one, and this equals one minus 0.911 which equals 0.89 for a party, you were asked, what is the probability that the number of successes is within two standard deviations of the mean number of successes or the expected number of successes? So to calculate this, we must calculate the mean number of successes where this distribution and the standard deviation So for a binomial random variable, the mean or expected value for the number of successes is he going to end terms? P, which comes out to you 1/2 and the standard deviation of a binomial random variable is given by the square root of N times. P Times Q, which comes out to 0.7, so within two standard deviations of the expected value. We're within this range, so the mean value is 0.5 and so plus or minus two times 0.7, which is the standard deviation, and that gives us a range from minus 0.9 to 1.9. So that means we're looking for the probability of a certain number of successes. So we for outcomes we can only have integer numbers of students, and that can't be negative. So this would be replaced by zero Can't be any lesson zero, and the number two would be more than two standard deviations greater than the expected number. So therefore, we must look for outcomes that are less than or equal to one. So we're looking for the probability that the number of successes is between zero and one, which is the same as saying we're looking for the probability of at most one success, which comes out to 0.911 and finally for party. It's explained to us that students who receive special accommodation are allowed 4.5 hours for the S A T, and students who do not are allowed three hours to rate the S A T. So we're asked, what do we expect? The average time allowed for the 25 students to be, so we can begin by writing a function that defines the average time spent by the 25 students. So for a student who is given special accommodation to get 4.5 hours, so we have 4.5 hours times number of students who get the special accommodation and then the students who don't get three hours and the number of them will be 25 minus the number of successes. And to make it an average, we must divide by the number of students, divide by 25. So this function defines the average number of hours that the students are given for the S A. T out of a sample of 25 and this can be simplified as 75 plus 1.5 x over 25. So for the question, we're asked, what is the number of hours allocated to the students on average, that we expect. So what is their expected number for the average number of hours allocated to the students? So what is asking for us? The expected value of the function h of X. So we can say that is equal to the expected value of 75 plus 1.5 x over 25. And because of the linearity of expectation, we can factory with the 1/25 and rewrite the rest of it like this. Now we have 1/25 so the expected value of a constant is just that constant. So we have 75 plus 1.5 times the expected value for X. And we've already calculated that previously, as 0.5. So this comes out to 3.3 So are we expect that the average number of hours allocated to the students to the 25 students writing S A T will be 3.3 hours.

And this problem, we need to find the probability of a certain event run. This question is that that 60% of students who take the statistics exam, score four or 5 25% of these students score three. And the remaining percentage of students, or one or two Out of the, uh, 90 out of those scoring four or 5, 95 of those students receive college credit. And out of those students scoring 3, 50 of them score the credit and four of the students scoring one or to receive credit. Now, what we need to do is find the probability that if a student is chosen at random from among those taking the exam and that student receives college credit, then what is the probability that students scored three on the exam? So first of all, we can arrange the information given in the question Using tree Diagram. So out of the total number of students, um, 60 of them receive a four or five. So 60 means 0.0 60 They receive four or 5 and 25 for three. So that means the score of three is a plane by 25 or 0.25. And the remaining percentage gets a one or two. And the remaining percentage there is .60 and .25 already, so that's .85. So the remaining percentage will be 1 -15, which is 0.15. Now, amongst those scoring four or 5, 95 of them receive credit. So that means that 0.95 of them receive credit. And for those who received the three, only 50 receive credits, those 0.50 Reervive credit. And for those getting a one or two only four receive credit. So that will be 0.04. So from here we can obtain that. The probability of getting four or five and also getting a credit can be obtained from this first branch of the tree diagrams will be zero points 60 times 0.95 We can obtain that following the branches. So that would be 0.60 times 0.95. The next one is the probability of getting A three and also getting a credit. So we're here as well, we follow the branches and it will be the product of these two numbers. So The probability is 0.25 times 0.50. This is a plane by using the multiplication rule of probability. And we can use that rule again to and the probability that the student gets a one or two and also gets credit again, we just want to play These numbers by following the branches so that 0.15 times 0.04. So from here we can obtain that the probability of getting a credit will be the sum of all of this. Because these are the only options. These three probabilities, you will either get a four or five and a credit or student will get a three and a credit or one or two and a credit. So these are the options. So by the additional rule of probability, the probability that a student receives credit is the sum of these obtained probabilities. So just be 0.60 Time 0.95 plus 0.25 Time 0.50 plus 0.15. I'm 0.04. And what we need to do is uh find the probability that the students called a three if it is given that the student of paint credit. So that means we need to find the probability that the students got a three given that the student received credit. So by the definition of conditional probability this will be the probability of three intersection getting a credit divided by the probability of getting a credit. So the probability of three intersection credit. That will be this. And the probability of a credit is this expression. So if we just substitute those expressions, what we will get is the probability of getting A three and a credit Is 0.25 times 0.50. And the probability of getting a credit will be the sum that we obtained. So 0.60 times 0.950 25 times 0.50 plus 0.15 times 0.4 So this is the required probability and this will be option number eight. So the correct option for this problem is option E.

26 found the probability that a physics major will do postgraduate research for at most three years, or the probability that X is less than or equal to three. In order to figure this out, we want to find all of the years that represent less than or equal to three years. So we're gonna find the probability of one year. Add that to the probability of two years. And because three is included in this, we're gonna add the probability of three to this as well. Those three probabilities air in the chart, we have 0.35 plus 0.20 0.15 these three all some 2.7. So the probability that a physics major would do postgraduate research for at most three years is about 70%.

Question number 16. It's from a given in equally then the financial binomial probability Be off X equal key equal in CK the people key the one minus b power in minus K equal factorial in over factorial key Multiply boy factorial in minus key The people k dot one minus p or in minus key. One off, one out off. Five answers is correct. The probability the number off favorable outcomes divided by the number off possible outcomes be equal 1/5 equally point to below it at K equal thin. So probability off X equal 10 equal Factorial 10 over. Factorial 10 Multiply Boyton minus 10 story In that 0.2. Parton the one minus point to Burton minus 10 equal. Almost zero question number we given an equal. Then we would use the above formula. So 11 out off five answers is correct. The probability is a number off favorable outcome divided by the number off possible outcomes The equal 1/5 equal point to value it AT T equals zero probability Off X equals zero equal factorial 10 over. Factorial zero boy didn't minus zero factorial that 0.2 point zero that one minus point to for them minus zero almost equal 0.17 four equal 10 point 10.74% Question Number. C Result. Port P Prosperity of X equal zero equal 00.1074 equals 10.4 at a 10.74 person A p equal 4.2 addition rule for multiple exclusive events. The probability off E or me equal probability off e plus probability Off be did remind the probability using this rule and table toe in the appendix. So probability off is bigger than or equal one equal probability of X plus X equals one plus the probability of X equal to and so on toe Probability of X equal 10 equals coin 268 plus 2680.302 plus 0.20 On plus 0.8 eight plus 80.26 plus 0.6 plus 0.1 plus zero plus zero plus zero equal 0.892 Equal 89.2 percent Compliment rule Peak off. Not a equal one minus B off A. Did you mind the probability you think the compliment rule? So probability Off X is bigger than or equal one equal one minus probability off X equal zero equals one minus 10.1 074 equal 0.89 to 6 equal 89.26%. Thes probability should be equal. They aren't equal, but the difference is most likely due to rounding errors in the use deep.


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