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.