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The probabillty is 15% that a student taking the test scoros higher Ihan (Round Io Ihe nearest integer a5 needed ) If the professor grades on curve ((or example the...

Question

The probabillty is 15% that a student taking the test scoros higher Ihan (Round Io Ihe nearest integer a5 needed ) If the professor grades on curve ((or example the prolossor _ could give A'5 Io Ihe top 10% of the class. regardless 0f the score_ ol 89 on the exam with . mean ol 73 and standard deviallon of Or a grade 0l 70 On difforent exam, where Ihe mean is 66 and your answer statistically and explainA student is tor Iho grade (Round to twwith grade 0l 89 on the exam with mean ol 73 and

The probabillty is 15% that a student taking the test scoros higher Ihan (Round Io Ihe nearest integer a5 needed ) If the professor grades on curve ((or example the prolossor _ could give A'5 Io Ihe top 10% of the class. regardless 0f the score_ ol 89 on the exam with . mean ol 73 and standard deviallon of Or a grade 0l 70 On difforent exam, where Ihe mean is 66 and your answer statistically and explain A student is tor Iho grade (Round to tw with grade 0l 89 on the exam with mean ol 73 and slandard devlation of because the Z value for thi Jnl uxam bolter oli needed ) worse oli



Answers

The grades on an examination whose mean is 525 and whose standard deviation is 80 are normally distributed. a. Anyone who scores below 350 will be retested. What percentage does this represent? b. The top $12 \%$ are to receive a special commendation. What score must be surpassed to receive this special commendation? c. The interquartile range of a distribution is the difference between $Q_{1}$ and $Q_{3}, Q_{3}-Q_{1}$. Find the interquartile range for the grades on this examination. d. Find the grade such that only 1 out of 500 will score above it.

In this problem, We have a statistics class with a mean of 72 And a standard deviation of nine. And the scores on this final exam are normally distributed, and we are trying to determine where the lowest score for an A is for A B and so forth. So we're gonna separate our curve here is gonna be the A's, The top 10%,, the peas Are the next 20%,, The seas are the Middle 40%,, The DS are the next 20%,, And the EFS are the bottom 10%. So we've got to determine The cut offs that transition from one grade to the next. So I'm going to start this problem by doing The F two D cut off, so I'm just going to draw it right there 10% here. And what we need to do is we need to find the Z score associated with that cut off. And the best way to handle that is to use an in versed norm function on your graphing calculator. And when you use inverse norm, you have to provide the area in the left tail, the average and the standard deviation. So in our problem, the area in the left tail is 10 of the curve. Since we are trying to find a Z score, we are working on the standard normal curve and on the standard normal curve, the mean is always zero and the standard deviation is always one. So if I bring in my graphing calculator, I'm going to find inverse norm under distributions. So on this calculator, you'd hit the second key and the bears button And then it's number three in this menu. So we'll type in the area in the left tail, followed by the mean and the standard deviation of the standard normal curve. And we are getting a Z score of negative one 0.28 Now, we have to transition that back into a raw score. So we have a formula to find our X. Now you have a Z score formula and the z score formula reads x minus mu over sigma. So if we were to solve that for X by finding our cross products, we have x minus mu equals Z times sigma. And if I add you to both sides, I can generate this raw score form. Uh huh. So we're going to use that formula. So our average arm, you was 72. The Z score we just found was negative 1-8, And our standard deviation was nine. So if you were to multiply this all out, You would end up with an x. score of 60 0.48 Now, let's talk about the barrier between the D and the sea. Can I draw my picture and here's between the D and the sea. But keep in mind there's 20% here and 10% here. So when we use our inverse norm to find our Z score, We're going to say that there is 30 in the left tail. And again I can bring in my graphing calculator and I can do in verse norm And this time we've got 30 in that left tail again. Still standard deviation. An average of the standard normal curve. zero and one. And we get -52. So the Z score here Is -52. So our X score or are raw score Would be found by taking 72 Plus that Z score -52, multiplied by the standard deviation of nine. And you are going to get a value of 67 point 32. We want to do the same thing now for the barrier between the sea and the B. So I'm going to draw a bell shaped curve For the c. And the B. We're kind of over here but keep in mind there's still this and this over here, so we've got 10%,, 20% and 40%. So this time when I do my Z as the inverse norm, my area is going to be those three values added together. So I'm gonna have 30.40 plus 0.20 plus 0.10 or how I have 70% of the curve in the left tail. And I'm going to get a Z. Score For this as positive .52. So my ex or my raw data will be X equals the average plus the Z. Score times the standard deviation. And my ex score for that is going to be a 76.68. And then finally I want to find the barrier between the B. And the A. So I'm going to again draw a bell curve, This is where the B. And the A. Is. But keep in mind there's these other little sections in there, so we've got 20% here, 40% 20% And 10%. So the Z score finding here is going to be in verse norm of 90 Which then is going to yield a Z score of a 1.28. So a raw score will be X equals 72 plus 1 to 8, Multiplied by the standard deviation of nine. So there for that cut off is going to be an 83 0.52. So let's take all of these values and we are going to put them back on our original picture. So between the F and the D Was a 60 48. Between the D. And the sea Was a 67 32. Between the sea and the B Was a 76 68. And between the B. And the A. Would be in 83 point 50 two. So this would be our distribution That meets the criteria. So the lowest you can get was that 83 52 for an a. The B is between the 76.68 and the 835 to the sea is between 6732 and 76.68. The d. is between 60.48 and 67.32. And an f would be anything lower than a 60.48.

Problem. 35. We have the scores of a test which are normally distributed with a mean of 60 and standard deviation of 12. Then we have the mean equals 60 and the standard deviation equals 12. The professor of this test plans to curve the scores for 40 if she curves by adding 15 to each grid. What is the new mean and standard deviation? To visualize this we have. This curve is normally distributed about six, and Standard Division equals 12. Two. Curve the scores, she adds. 15 to old grids. This will make the curve is the same. It's a curved, it's the same, but it will be distributed about the means of 16 plus 15 only, which means then you mean will be 60 plus 15. We can make it new A. As for both eight equals 75. And but the standard vision well still the same remains the same. And it will be 12. This which will be 12. The curve will not be flattened. This is for birth. A for birth. We Is it fair to care by adding 15 to each grade? Of course not. It's not fair because this will change only the mean but not the standard division, and it will be more fair bye, changing the mean and the standard division. Then it's not fair because we should count for Variation or the Central Division, because it's not fair to add 15 to a student which take the grade of the and we will add 15 to a student who took he. Of course, it's not fair for Board C. We will curve the grades so that the great B will be from the range of 70 bottom 70% and below the top 10. By visualizing this in the normal distribution, it's below 70% then above the bottom 70%. We have 70%. Almost here. It's above this value and below the top. Then we have the top 10 here and we have here the bottom 70%. The in between 20% is great B for this strange. We want to get the grades of the you want to get this value. We can make it as X one, and we get we want to get next to thanks to X one has area to the left of it, which equals 70% or it's all 0.7. Using the standard normal distribution tables, we can get the equivalent that the score, then they'd want equals by entering the tables. We want to get the value of 4.7. It's here. It's adjacent to a point five three. Then the score is 4.53 and we know the score equals X minus mu divided by six. We can make it X one. Then x one equals 4.53 multiplied by sigma, which is 12 plus 60 which equals 66.3, 66 0.36 We can do the same to get there's a score of X two. The area to the left of X two is one minus 4.1. Then we can get the tube as the area of 4.9. We can search the table, the standard normal distribution tables to the value of all point mind. We can find that a 0.9 is here 4.9 this value or this value we can find. This value is more close. It's one point to it. I'm going to eat then so I'm going to eat equals next to minus mu, divided by Sigma we have x two equals 1.28 multiplied by sigma plus 60 equals 1.28 75 0.36 Then grade B will be from 66 point 36 to 75.36 Finally, for both d, which method of curving the grades is fairer, adding 15 old bird. See, we can see that by doing so in birth. See, we account four the mean and the variation for each grid has arranged and the change in each range differs from grade to grade. Then we account for the mean and the operation and the scheme of birth. See, then it's more fear.

The question here basically wants us to talk about a normal distribution here. So just to recall a normal distribution is going to be a bell shaped curb where we're gonna have a particular mean here, which is gonna be the average. And the variants here is going to show us how spread it is in one particular direction here. So in this scenario, or given that scores on examination are it seemed to be normally distributed with a mean of 78. So if I just draw that, that is gonna be at 78 year and our variance is gonna be 36 here, so it's gonna essentially be plus or minus in both directions here. So for a here asked us to find the probability that a person taking the examination is going to be higher than 72. So in order to do this, we're essentially going Thio split this. So we know that variants here is going to be the standard deviation square. So 36 the square of 36 year is going to be six years, so we know that one standard deviation is going to be six units in terms of great here. So in this case, we're taking one standard deviation to the left. So therefore, if we just check our standard deviation chart, we know that the probability that the student will score greater than 72 year so X is greater than 72 is gonna be 0.8413 or 84 point. Um, 13% here, be here. It asks us, um, that we want to suppose that the students during the top 10% are to receive an A grade. So what's the minimum score that they must achieve? So if we know that our standard Armenia is gonna be the 50th percent here, we just need to go a couple of standard deviations achieved that highest. Great. So in this case, it's gonna be 85.68 Um, for see here. It asked us what must be the cut off point for passing examinations if the Examiner only want to talk 28.1%. So we need to take the center deviations that is going to include that. So that's gonna be 81.48 percent or the grade that they're going to get is 81.48 for D here, it asks us approximately, what's the proportion of students that have a score of five or more? Um, points above the score. That cuts off the lowest 25%. So essentially wants us to find the proportion of students. So this proportion when we calculated it's gonna be equal to 43.7% here, Um, so lastly, it asks us if it is known that a student's score exceeds 72. What's the probability that, um, his or her score exceeds 84 here? So essentially we're going instead of going one standard deviation till up, we're gonna go one standard deviation, um, to the right here. And essentially we're going to take the opposite of this previous a portion that we found, which is gonna be a 0.1886 or 18.86% is are gonna be our probability of scoring greater than 84 here. This is as we're going to essentially take the center deviation. We're gonna try to find everything outside of that or above. In this case,

Okay for this problem were asked Teoh show our normal curve and Z score understanding to calculate a few different things. So the general procedure is going to be Ah, draw our home. See table are I'm sorry? A normal distribution and Ah, and look at the percentiles percents they want us to figure out. And that will give you use an apple. It too. Figure that out. So per day. Um, we're talking was referring back to example 6.11 and so we want to figure out the top 15%. Get a So we're looking at the top 15% of our data. So, uh, it's gonna restate that as a probability that X is greater than 0.85. So that means that's why I like to draw my diagrams out here. So just for a little throwback here because we need the information, you look back at the problem. It's said that the mean for that test was 72 and the standard deviation for that waas 13. So for considering 85 72 plus 13. Yeah. So we're looking at a above one standard deviation above here. Such of the right basically is what's looking at. You want the 10.85 percentile or the top? 15% is what's represented in that diagram. Let's go over here and use our Creeley available Apple it. I, like stop wits, have a lot of different applications, and we're kind of normal problem. Normal distributions. So our stop list and give this a little nicer diagram, and we know the meaning of standard deviation for this problem, we can see it. Um, And if you want to skip spit out Hagen positive video and see the difference Dinner deviations. Um, but we care about values to the right of a certain point. In this case, we care about values to the right of no area. Second with him, score at wording, working backwards. Okay, I think out loud. So working backwards from that, we're still using the same, uh, diagram, but reaction I can find in the area under the normal curve. We're going to work backwards to find the value that goes with that area. So really already winners, we're basically looking. If the area is the 15% of here, we're looking for the value that goes with that, so and we still have the same mean standard deviation. But I've seen just to the area calculation. So generally thinking we had to do there waas knowing that top 15% working backwards, this 150.85. So that's going to give us back Thea Value. That goes with this very point, um, and so it looks like 85.4 with our answer. We want for that. So to get that top score, you had to have at least a score of 85.4 seven, so we'll save 85.5. So that's part a two days. Do you have a table to or work from the graphing calculator and do something called Inverse Normal? You could do that there as well. So let's look at Part B is a different set of scores that refers to example 6 10 So this one was the 25th percentile. So again, the same process so we're looking at are diagrams, and we let's represent with the 25th percentile. Looks like for a que ah, 25th percentile would be kind of looking over here. What's make it green 25th percentiles here. So we're looking for us if the 25th percentile, we want to know the value that goes with that. Um, so we want to know Is Thea score? If you're the 25th percentile, it's the same thing working backwards. Now we have different. Ah, different normal distributions. So we do want a name that what we have. Look, looking back for the problem, I Q scores are centered at 100 and plus or minus. Ah, standard deviation is 16. So we could still do the same thing. Work backwards here. We need to change our mean. It's Jenner deviation to match the idea of set for I Q 116. We're gonna leave this alone. We have the area. We want the value. Um, it's gonna be a left tail area again. So, uh, let's work backwards. But we want the 25th percentile. So to get that value, I keep score. That goes with the 25th percentile. It is 89.2. Okay. And look apart. See another distribution. That's normal. But as a t scores so that s a T scores and me and 500 standard deviation 100. I want to be in a college. It takes the top 7%. Okay, so let's go over here. Top some percents. Gonna be a very slim appear. Must be a predictive February picky college. Okay, so they told us in this problem that 500 is the mean in the there s a T scores and the standard deviation is closer. Riotous 100. Yes. We want to know this four that goes with the top 7% top 7% score. So really, what I want to think about is the the 93rd percentile. So let's go over here and ah, give me the change or information because we have a different distribution of blue on our experience. So we know the mean for these s a T scores versus the A. C T scores is 500. Move and hold. That 100 is the standard deviation of scores, and we want to be the top 7% which means that you're the 93rd percentile or above the 93rd percentile. So you calculate that still gonna be left tail. So you it means you need an S a t of 6 47 0.57 Okay, so for that? I mean, the S A t not sure they have. Ah, decimal values for S a T scores. So let's call it the um 6 47 point 37 Hollywood is best wants. It's constant. 48. Okay, since you 6 47.5


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