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Nalion; ed, Brookes'Physics examination was 73 and the The mean grade I a standard deviation was 8. Determine the following: grade The standard scores (z-value...

Question

Nalion; ed, Brookes'Physics examination was 73 and the The mean grade I a standard deviation was 8. Determine the following: grade The standard scores (z-value) corresponding to of 86. standard score is -0.8. The grade of the student whoseSolution:vasan. Series:

Nalion; ed, Brookes' Physics examination was 73 and the The mean grade I a standard deviation was 8. Determine the following: grade The standard scores (z-value) corresponding to of 86. standard score is -0.8. The grade of the student whose Solution: vasan. Series:



Answers

The scores on an economics examination are normally distributed with a mean of 72 and a standard deviation of $16 .$ If the instructor assigns a grade of $A$ to $10 \%$ of the class, what is the lowest score a student may have and still obtain an A?

In this question were given a mean and sample standard deviation for some test scores and asked to find the Z scores for several individual data values. So let's say we have the mean of 74.2 on a sample standard deviation of 11.5. If you recall the formula for finding a Z score is X minus X bar over s where X is the data value. X bar is the sample mean and s is the sample standard deviation. So for our first part of the question, we are given us test score of 54. So somebody who performed pretty well below average So to find there's you score will substitute all of our known information into the formula for the Z score and divide by our standard deviation. So subtraction first, following order of operations and then division gives us a Z score of negative 1.76 So somebody who scored between one and two standard deviations below the mean Our second test score is 68. So somebody who scored below the mean but not as far below the mean, so we would expect disease score below zero, but not as far below zero. So 68 minus 74.2 over 11.5. And we find that that Z score is in fact negative. But on Lee negative 0.54. What if somebody performs better than average? So what if somebody got a test score of 79 so a little bit above average. So we would again substitute all of our known information into the formula and divide by our standard deviation here. And we're going to expect a Z score above zero, but not a lot above zero, because this score is not very far above average. So that's a Z score of 0.42 and then finally parte de what if somebody did even better above average? What if they scored in 93? So we would expect again a positive Z score because the scores above average, but it's even further above average. So even higher number and this ups supposed to be a five, not a 11.2 11.5, and this sea score turns out to be 1.63

27 A set of 340 exam scores exhibiting a bell shaped relative frequency distribution has a mean of why bar of 72 and the standard deviation of S equals eight. Approximately how many of the scores would you expect to follow in the interval from 64 to 80 and then in the interval from 56 to 88? Well, so the first question is 64 to 80. The next question is 56 to 88. So let's draw a picture to see if we can approximate these answers. The mean is 72 we have a standard deviation of eight, so we have about three standard deviations on each side, we're gonna add eight, we have 80 than 88 in 96 on the left, we're gonna subtract eight, so we have 64 56 48. So for this first question, between 64 80 64 80 that's one standard deviation on each side of the mean. So according to the empirical rule, that would be approximately 68% for 56 to 88 those are each two standard deviations from the mean, and the empirical rule says approximately 95% of the data falls within two standard deviations. So between 64 80 I expect to see 80 68% of the data and between the interval 50 60 88 I would expect to see 95% of the data. So now we have to take these percentages and apply them to the number of exam scores we actually have, because we're looking for how many scores would you expect? So we take the number of exam scores we have of 3 40 and we're just going to multiply to figure out what is 68% of 3, 40 68% of 3 40 is 231.2, we'll do the same thing for the 95% 95% of 3 40 is 320 three

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.

So in this question, were given to test scores as well as the mean and standard deviation of the sample from which they're taken and were asked to find the Z scores. So the first person scored 92 the second person scored 63 were asked to find the Z scores. So let's review the formula for finding a Z score. Z is equal to X minus X bar over s where X bar is the mean of the sample and s is the standard deviation of the sample and then X is the individual data value. Mom, this is supposed to be extra. All right. So for the first test score, I'm gonna find the Z score by substituting this X value into the formula. And we're told that the meanness 72 and the standard deviation is 12. So is in simplifying this. You're going to this attraction on the numerator to follow order of operations and then divide by 12. And so you get 1.67 for the second test score. Gonna follow the same procedure 63 minus 72 over 12 and we find out that the Z score is negative 0.75


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