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Three students are comparing their scores on their first test inthe Intro to Comm class and arguing about who did better.Student A took the class in 2018 and scored...

Question

Three students are comparing their scores on their first test inthe Intro to Comm class and arguing about who did better.Student A took the class in 2018 and scored a 85 (mean= 90,sd=5)Student B took the class in 2019 and scored a 85 (mean=83, sd=5)Student C took the class in 2020 and scored a 90 (mean = 90sd=3)Determine who fared the best and who the worst among the threeof them.

Three students are comparing their scores on their first test in the Intro to Comm class and arguing about who did better. Student A took the class in 2018 and scored a 85 (mean= 90, sd=5) Student B took the class in 2019 and scored a 85 (mean=83, sd= 5) Student C took the class in 2020 and scored a 90 (mean = 90 sd=3) Determine who fared the best and who the worst among the three of them.



Answers

An instructor announces to the class that the scores on a recent exam had a bell-shaped distribution with mean 75 and standard deviation $5 .$ a. What is the median score? b. Approximately what proportion of students in the class scored between 70 and $80 ?$ c. Approximately what proportion of students in the class scored above $85 ?$ d. What is the percentile rank of the score $85 ?$

So here we're told that the teacher is teaching to AP Stats classes and on the final exam, the 20 students in the first class. So class 1 20 students averaged 92 while the 25 students in the second class average only 83. And if the teacher combines the classes, what will the average final exam score be? So the some of the scores in one class is 20 times 90 to 18 40? Well, some of the other is 25 times 83 2000 and 75. So the total sum So adding these together gives us 3915 and there are 45 students total. Yeah, it's just 20 plus 25. And so our average will just be the mean, um, which is going to be the sun for data points over the number of data points that we have. So in this case, that would be 3915 over 45 which will give us an average score of 87 if she if she takes the two scores from her two stats classes and combines them, that will be the average final exam score

We have done this very same question in this exercise. The correct answer here is a we have been given to means we're going to calculate means for before and after. These will be two means. So the answer correct answer. Over here is a

In the situation. We have 30 papers that have been grated, that result in an average score of 81.3, a median score of 80.5. But we're supposing that something happened with the papers, and the maximum grade is increased by not nine points. We wonder, then what will be the new mean and the new media first, if we make only one change that infects only the maximum score than the median will still be at exactly 80.5. The reason being is that the median is a measure of position, the central value of the data set, and it doesn't care how big or how small. The maximum or minimum values happen to be the mean, however, will change. Let's write out that the observations are x one x two x three all the way over to accept 30 and suppose except 30 is the maxes score for this exam. That how we calculate X bar is we take X bar, set it equal to x one plus x two plus x three all the way over to X up 30. Then we divide the entire total by the number of observations, which is 30. Let's do a small change in this computation for X bar. Recall that we're going to increase the maximum score by nine and accept 30 is that Mexico score. So it's called this X Bar NU to denote as the new mean and right plus nine. Let's do a bit of algebra on this arrangement. We can. If we like Associate X one through X 30 then 30 can divide into this group as well as into the plus nine. So X Bar Lou will be equal to x one plus x two plus x three All the way through X sub 30 divided by 30 plus the quantity nine. Divide by 30 At this stage, we recognize that this quantity here is exactly the expert quantity. Before any change was being made, that means we can write that X Bar new is going to be equal to 81.3. The old value for the mean plus 9 30 It's so altogether the new mean expire. New is going to be equal to 81 0.6. So after changing the maximum score to the maximum score plus nine points, this will be our medium for no change and we have a very small change on the new sample mean and for a process on what happens when you change a maximum score by a certain amount. This is the formula. Essentially, that helps us get to the right value. We take the old mean plus the increase or decrease divide by the sample size, which was 30 in this case.

Okay for this problem, We're gonna compare some test scores. So we've been told that two different students equal not a points for their test force, but we're gonna try to use these scores. Our plan is to use, uh, Z scores to see who did better relative to the rest of the class and just did better overall. So test scores. We had some information that test one had a meaning of 65 standard deviation of 10. And for test two, that had a mean of Okay. Okay. 80 and a standard deviation of five. So we're gonna do is we're gonna look at each student's test scores and do this the Squire, both of those. So the students names are Derek and Julie. Get Dirks thescore here, and we'll get Julie easy score and see who the winner is. So because there's so many calculations and when he does my calculator over here. So, um, wait. Told Derek that eighties on both tests, So we're gonna take a score. It's backed away that mean for that first test and divided by the standard deviation of further first test. So that means he has of these squirt 1.5 on the first test. And for the second test, you got an 80 again, Mr Consistency. And the the extended deviation for that one. It's taking away 80 and actually zero since he got the meat zero. Well, before my calculation outside. Got a zero. These four on that for Julie. She, uh, did well in the first test, so she had 70 minus 65. First test, maybe a little bit harder. Um and so she had an overall there. A 0.5. The score for test one and four. Test to shed much better. She got a 90 but no test, too, with a little bit heart with a little bit easier. So she 10 points better than the mean in the variations Little assets to so putting all the pieces together, we can get that. Answer the question. So for Derek, he had a Z score of the first test of 1.5, and on the second test, he had a Z score of zero. He hit the mean Julie had a Z score of 0.5 in the second test. She had a Z score of two and we can actually add this together. It's their combines. The score. So overall is the scores 2.5, whereas his overall Z score was only 1.5 z for combined. So even though they have the same amount of points of the problems stated relative to the rest of the class and standardized, Julie actually did better. Eso Julie will answer the question. Julie had better test results. Overall, Ask the purpose of these Forced gives an idea relative to the rest of the class.


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