4

Fnos. 34 [0 36, please refer to the table below:ScoreFrequencCumulative Frequency 28Cumulative Percentage 10O40-44 35.39 30-34 25-2934.In solving for the 60th perce...

Question

Fnos. 34 [0 36, please refer to the table below:ScoreFrequencCumulative Frequency 28Cumulative Percentage 10O40-44 35.39 30-34 25-2934.In solving for the 60th percentile the lower boundary to be used is A.34 C.34.5 B. 395 D. 3935. What value Ol the cumulative frequency is to be used in solving for the 35"h percentile? C.12 B. D. 1836. The 45"h percentile is A.35.6 B.36.5C.33.5 D. 37.5

Fnos. 34 [0 36, please refer to the table below: Score Frequenc Cumulative Frequency 28 Cumulative Percentage 10O 40-44 35.39 30-34 25-29 34.In solving for the 60th percentile the lower boundary to be used is A.34 C.34.5 B. 395 D. 39 35. What value Ol the cumulative frequency is to be used in solving for the 35"h percentile? C.12 B. D. 18 36. The 45"h percentile is A.35.6 B.36.5 C.33.5 D. 37.5



Answers

Given here is the relative frequency histogram associated with grade point averages (GPAs) of a sample of 30 students:a. Which of the GPA categories identified on the horizontal axis are associated with the largest proportion of students? b. What proportion of students had GPAs in each of the categories that you identified? c. What proportion of the studen I GPAs less

For part A were asked to create a cumulative relative frequency distribution. So I'm gonna start by listing my AP scores. I am going to do this as a group AP score versus UN groups like we did in number 34 because we're going to construct an AWG. I've in part B and typically ogg lives are created for grouped distributions. So for AP, score minimalist 1 to 2. And remember what that means is a score up to two so below two, which would just be the scores of one in this case and then 2 to 3 would be scores of two all the way up to, but not including three. So in this case, it would just be scores of to. And then we keep going. So 3 to 4 would really be scores of three. 4 to 5 will be scores of four, which means we have this 5 to 6, which sounds weird because there's no score of six on the AP exam. But that's okay because this this class means scores of five up to not including six. So it just really includes the scores of five on the AP exam. But this is gonna help us to create our archive in part B. Now, we already found the relative frequencies for these and 34 part C. So we're going to go ahead and just list those from the previous problem. So the frequency of a score of one the relative frequency was 10.214 of two was 20.262 of the three was 30.310 of a 4.143 and of a five was 50.71 Now, to get the cumulative relative frequency for one, it's going to be the same point 214 But for a score for two, we add the frequency the relative frequency from a score of one to the relative frequency of a score of two, which is going to give us point for 76 for the key meals of relative frequency for 3 to 4. We're gonna add the relative frequency for that class 40.310 to the previous cumulus of relative frequency of 0.476 That's going to give us a value of 0.7 a six. Then for 4 to 5, we add 50.1 for three to the 30.1 point 786 which would give us 7860.9 to 9. And finally, when we add our relative frequency for 5 to 6 to the previous cumulative relative frequency 60.9 to 9, we get 1.0. Now we're going to greet our guys using the cumulative relative frequencies from Part A. So remember, we plot the cumulative relative frequency at the upper boundary for each class, so we start at zero for one, and we plot the 0.214 at two. And remember what that tells us is that 21.4% of students earned less than a two, meaning they earned a one on the AP exam, and we connect those points with a line. Then, at three, report the 30.476 That means 47.6% of students earned less than a three, meaning they earned a one or two, and we connect those points with the line. Then, at four, we plot 40.786 and connect with a line at five. We plot 50.9 to 9 and connect, and lastly, at six, we plot 1.0 I remember. That means 100% of students earned less than a six, which makes sense because they all students are between a one and a five. So this would be our archive for the A peace force. So looking at our ogg, I've above to find the cumulative relative frequency for two. We would actually go to the three on the archive and look at the cumulative relative frequency, which was 0.476 Because remember, that tells us that 47.6% of students earned less than a three, meaning they earned a one or two on the exam. For the next part, we want to know which students qualified to earn college credit, so that would mean they earn a three or above. Since we know the percentage for students earning a one or two, we can do 100% minus the 47.6% and that will give us 52.4%. And so 52.4% of students are going to earn college credit. They will earn a three or above. So when we're looking at part cnd together, the relationship here is that they add to 100% because together they include all the AP scores. And that's because the part C was a one or two, while Part D was three or above. So when you combine those, you have all of the AP scores or 100% of the scores earned. Lastly, looking at Part E, we're comparing our answer to what we found when we did our earlier problem. Number 34 and they are the exact same answer. They're both 52.4 person.

In this question, we're gonna be taking a look at the distribution of scores on achievement test using a relative frequency table and a percentile graph or a relative frequency graph. So we're going to start with our data and we're going to construct our team a lot of frequencies and then calculate a relative frequencies to make our graph. So relative frequency is just a fancy way of talking about percents. Cumulative frequency means you're just gonna add these numbers up. So we're going to bring the five over, and then we're going to keep adding them as we got so five plus 17 it's 20 to 22 plus 22 is 44 out on another 48 we get 92 add on another 22 or upto 1 14 and at on six, and that gives us a grand total of 20. So that means we are represented representing 120 pieces of data. So that is our end in this case, to find a relative frequency, we're gonna take each one of these cumulative frequencies and divide by 1 20 So, for our first interval five divided by 1 20 is approximately 200.4 So just to show you where that's coming from, I'm rounding these because we're going to be graphing on this graph where everything is estimated, so we don't have to be that precise. Eso 22 divided by 1 20 is 18% or 180.18 44 divided by 20th. About 440.37 or 37% 92 divided by 1 20 is about 77%. 1 14 divided by 20 is 95% and 1 20 divided by 1 20 is one. Now we're gonna graph this over here and we're starting with 1 96 But there are no achievement test scores below 1 96.5 So that's gonna get a zero. There's nothing before. And then as we go to the end of each interval, we're going to continue to build up our graphics. We move forward. So at the end of the next interval work, four percents about here at the end of the next interval were at 18% so just below 20% about here at the end of the next interval were at 37%. So about here, the end of the next interval were at 77%. So about here, the end of the next interval were 95% almost to the tall. And at the end of the last interval, we are at 100% now. One of the things that's important to know is that each of these intervals represents even other kind of knot standard ways with numb label, a scale they do represent consistently spaced out intervals. So I want to try to connect these with a straight line as much as I possibly can free so and this is what we make our cumulative frequency graph. Now, the reason we connect these with a straight line is because we are assuming that all of the achievement tests in the interval in between are evenly distributed. That's our best way of making a guess here. That may not be actually the case, but that's what we're going to use to make our estimation. Okay, so now we have our frequency graph curved, and we can use our frequency curve draft on. We can use that to make our estimates. So first thing that we want to estimate is what is the approximate percentile for somebody who scored 2 20 So to 17 to 2 38 to 59. What? Um, what we need to kind of know is how big that interval is. So we did say that they were evenly spaced on. That means that in between each of those intervals, we're looking at about 21 points. So each one of those lines is representing seven. So to 20 plus seven is to 24. So when I'm looking for to 20 I'm gonna be looking at the score of 2 20 It's gonna be about here, so I'm gonna follow to 20 up to the curve, and then I'm gonna follow that line over 2% all trying to draw this line a straight is that possibly can, and it looks like it's between 5% and 10% a little bit closer to 5%. So let's estimate that at about the sixth percentile, question B is asking for the percentile associated with a score of 2 45 So to 45 is going to be about here, Right? So about 7 to 38.7 about 2 45 about there. So again, we're gonna follow from a test score up to the graph and then follow from the curve over to the percent axis. So that looks to be at approximately 25. So about the 25th percentile. Her question. See, we're looking for the percentile associated with score of 2 76 So to 76 is going to be somewhere over here. So a little bit below to 80. I'm so I'm gonna estimate about here. You should not quite so straight there. Tried to even it out some of the end on. Then we'll move that over a little bit below 70%. So I'm gonna estimate that to be about the 68th percentile. These are just estimates. So if you're not exactly spot on, um, that's okay, that's that's a reasonable, uh, thing to happen with this kind of graph. So now what is the percentile associated with to 80? So, I mean, that's just like here, right? So we can kind of see that there's kind of a big gap that's happening there. So we're all the way up here now. All right, so we should really be drawing this line all the way down. This isn't much better done with a ruler on paper, then digitally, unless you were doing something like on Power Point or something. But we're above 75%. So maybe, like the 76 there, 77th percentile, and then the final score we're going to try to estimate the percentile for is 300. So three hundreds gonna be a little bit below here, and we're gonna follow that up a bit wonky. But it's a little bit before this line hears about their and then we're gonna follow that over, and we know that we're below 95% but not a lot below 95%. So maybe 94 93%. Ah, percent. All right, now we've got all those lines on there that we're gonna have to try to ignore a little bit, but we're gonna work the other direction, and we're going to try to, um, estimate the test score now given a percent. So if we have the 15th percentile, what is the approximate test for that goes with that? So we're gonna work in the opposite direction. 15th percentile. Follow that line over to the graph, and then we're gonna come down here and estimate. So we remember we said that each one of these lines is about seven. So this would be like to 24. This would be like to 31 2 38 So somewhere between 2 31 2 38 little closer, 2 to 31. So maybe, like 232 G. Um, we're gonna estimate that 29th percent up, so you can see, like, all this estimation, all we're going to be able to do is kind of guess right, the 29th percentile. Follow that line right below the 30th percentile over Teoh about here, maybe, and I will follow that down. And so this is to 38 to 45. The next one would be, what to 52. So we're a little bit before 2 50 like maybe 2 49 The important thing to kind of remember is that when you have the percentile, you work. You started to have relative frequency and work to the data. When you have the data, you started the data and you work to the relative frequency. Okay, so, H, we are looking at the 43rd percentile so the 43rd percent tells somewhere between 40 and 45 and we're going to come over to the graph over here and then follow that line down. And if this is to 59 this would be to 66. And so we're a little bit I don't know in the low in the low to sixties. So let's say like to 64 I were looking for the 65th percentile. So 65th percentile is here. Follow that over to the curve and then from the curve down. So it's gonna be a little bit above this line right here. So that was 2 59 to 66 to 70 three. So a little bit above to 73 I don't know. Let's say to 74 and then finally, the last one the 80th percentile. So we'll follow from the 80th percentile to the curve, but here, then we'll follow that down and like to 87 a little bit below 2 87 So let's say to 85. So, as you can see is the percentile was increasing. So where the test scores and that makes sense because we're accumulating more and more test scores as we go. Um, and the same was true here. The lower the test score, the lower the percentile that it was associated with.

And this question were given seven data values and asked to find the percentile rank of each. So we have the data values sorted in order from smallest to largest on. We need to use this formula that tells us what the percentile ISS so P equals the number of numbers that are lower than any other than whatever number we're looking to find. The percentile loves a number of numbers lower. We add 0.5 and divide by the total number of numbers that we have and then multiplied by 100. So in this particular problem, we have seven data values. So on is equal to seven. All right, so is we got through to find each percentile, we need to look at the number of numbers lower. So when I look at my smallest data value, there are zero data values lower than that. So zero plus 0.5 divided by seven and multiplied by 100 and that gives us 7.14 Since we don't have a seven point fourth per position, we go ahead and round that and this number would round down. So this would be approximately the seventh percentile and we continue on in that fashion for each. Each additional number that we find there is one more number that is lower than that. So it's sort of this pattern. So for 28 there is one number lower than 28 in our data set, and we're gonna divide that by seven, multiplied by 100 and that gives us 21 point for until again we round, and that's going to be the 21st percentile. And then 35 has two numbers smaller than it, so two plus zero point 5/7. Multiply that by 100 and we find out that's equal to 35.7, so that would round to the approximately 36 percentile. Now notice that the value of the number doesn't really have any relevance to it. It's just the position of the number that helps us identify percentile. So the 42nd are the value 42 is going to have three numbers less than it. When I add 30.5, I get 3.5 under by seven and multiply that by 100. That translates to be 50th percentile. It's the 40 said the data value 47 has four numbers when we add four numbers lower than it. When we add 40.5, divide by seven multiplied by 100 that becomes the 64th percentile, or we can see that that is the 64th. Percentile. 49 has five numbers lower than that. When we add fought 50.5 divide by seven multiplied by 100. That be we can see that that is the 79th percentile. And finally, the largest number has six numbers smaller than it. So 6.5, divided by seven multiplied by 100. And that is the 93rd percentile. So you may notice that there is a bit of a pattern, so seven plus 14 is 21 plus another. About 14 is 36 plus 14 plus 14 a little bit more than 14 plus 14. Like there's this pattern of adding 14 on. There is a reason for that. So if we take 100% of the numbers and we divide by seven, you'll see that that's a little over 14. It's 14.3, so each sing. Each one of those data values represents about 14.3% of the data, so That's why you see that pattern in the percentiles. Just a fun fact. All right. Next part of the question says, then, find the value of the number in the six or find the buy of the 60th percentile. So for finding the percentile, then we have to use this other formula. C is equal to end times p divided by 100. Where n is the number of numbers that we're looking for and P is the percentile that we're looking for, so C equals. We have seven numbers times. We're looking for the 60th percentile, divided by 100. So when we multiply about that that out, it comes out to 4.2. Well, we don't have a number in the 4.2 position because we only have whole numbers of positions the 1st 2nd 3rd etcetera. So we would round this and we always round up. So this is going to be the number that's in the fifth position. So if we go back over and look at our numbers that are in order, you've got 12345 This number is in the fifth position, so the 60th percentile is 47. Don't know where that line came from, so the 60th percentile is 47


Similar Solved Questions

5 answers
Use the WSEPR model to predict the bond angles about each numbered atomThe predicted angles about atom 2fedegreesThe predicted angles about atom 2 aredegreesThe predicted angles about atom 2fedegrees
Use the WSEPR model to predict the bond angles about each numbered atom The predicted angles about atom 2fe degrees The predicted angles about atom 2 are degrees The predicted angles about atom 2fe degrees...
5 answers
Exercise 3508 Show that the Fnk of A is fc dimemsion cf %{4}.Lemta 3.5.9 Le: A € R"*? Tags Vr & : €R aw % € R {Az,3} = (~,ATw
Exercise 3508 Show that the Fnk of A is fc dimemsion cf %{4}. Lemta 3.5.9 Le: A € R"*? Tags Vr & : €R aw % € R {Az,3} = (~,ATw...
3 answers
#1. each of the following cases. determine whether there is function defined neighborhood of (0,0) in R? so that f(0,0) = 0 and so that setting f(z,y) makes the corresponding relation hold identically: Tyz + sin(r + y | 2) =0 12 +y} + 22 + Vsin(z2 + y?) +32+7= 2, zy2(2 cos y cos -) + (-cos I COS !) = 0, I + y | - +g(I,y.-) = 0 where is any continuously differentiable function satisfying 9(0. 0.0 = 0 and 82 (0,0,0) > 0.
#1. each of the following cases. determine whether there is function defined neighborhood of (0,0) in R? so that f(0,0) = 0 and so that setting f(z,y) makes the corresponding relation hold identically: Tyz + sin(r + y | 2) =0 12 +y} + 22 + Vsin(z2 + y?) +32+7= 2, zy2(2 cos y cos -) + (-cos I COS !) ...
5 answers
HS(g) + ZHzO() ~3H2(g) + SO2(g)Using standard thermodynamic data at 298K , calculate the free energy change when 1.79 moles of H2S(g) react at standard conditions AG'ixn
HS(g) + ZHzO() ~3H2(g) + SO2(g) Using standard thermodynamic data at 298K , calculate the free energy change when 1.79 moles of H2S(g) react at standard conditions AG'ixn...
5 answers
3_ Consider the compound shown. Draw the product obtained if the compound is fully hydrolyzed under acidic water.H+, HzOCNThere are three carbons that are multiple bonded to different heteroatoms (labelled 1-3). Predict the position of the multiple bond stretch in an IR for these three carbon atoms_NCN
3_ Consider the compound shown. Draw the product obtained if the compound is fully hydrolyzed under acidic water. H+, HzO CN There are three carbons that are multiple bonded to different heteroatoms (labelled 1-3). Predict the position of the multiple bond stretch in an IR for these three carbon ato...
1 answers
Let $f(x)=\frac{e^{x}-1}{e^{x}+1}$. a. Plot the graph of $f$ using the viewing window $[-5,5] \times[-1,1] .$ b. Find the area of the region under the graph of $f$ over the interval $[0, \ln 3]$. c. Verify your answer to part (b) using a calculator or a computer.
Let $f(x)=\frac{e^{x}-1}{e^{x}+1}$. a. Plot the graph of $f$ using the viewing window $[-5,5] \times[-1,1] .$ b. Find the area of the region under the graph of $f$ over the interval $[0, \ln 3]$. c. Verify your answer to part (b) using a calculator or a computer....
5 answers
Let R be the region between y = x?,y = 4x. If R is revolved aboutx- axis, then the volume of the solid obtained is:Select one:TJ (16x? x)dx 21 J8 (16x? x")dxJG (x4 16x)dx21 JG (x16x2)dxNONE
Let R be the region between y = x?,y = 4x. If R is revolved aboutx- axis, then the volume of the solid obtained is: Select one: T J (16x? x)dx 21 J8 (16x? x")dx JG (x4 16x)dx 21 JG (x 16x2)dx NONE...
1 answers
Plot the curves of the given polar equations in polar coordinates. $$r+1=\sin \theta \quad \text { (cardioid) }$$
Plot the curves of the given polar equations in polar coordinates. $$r+1=\sin \theta \quad \text { (cardioid) }$$...
5 answers
Slide- Disc diffusion plate Escherichio coli (Gram- bacteria) with different plant extractsplants;YM= Yerba Mansa Mormon C= Creosote Junipe Yucer A= AspenSlide diffusion plate of Staphylococcus epidermis (Gram+ bacteria) wlth different plant extractsKey- pljals; YM= Yerba Mansa Te Mormon TeaCreosote Junipt Yuccz A= AspenSlide Disc diffusion platc of Serratia morescens Eleencle
Slide- Disc diffusion plate Escherichio coli (Gram- bacteria) with different plant extracts plants; YM= Yerba Mansa Mormon C= Creosote Junipe Yucer A= Aspen Slide diffusion plate of Staphylococcus epidermis (Gram+ bacteria) wlth different plant extracts Key- pljals; YM= Yerba Mansa Te Mormon Tea Cre...
5 answers
In Exercises 21-24, approximate the area of the region using the indicated number of rectangles of equal width.$$ f(x) = rac{1}{4} x^3 $$
In Exercises 21-24, approximate the area of the region using the indicated number of rectangles of equal width. $$ f(x) = \frac{1}{4} x^3 $$...
5 answers
Find the area of the shaded region. The graph depicts thestandard normal distribution with mean 0 and standard deviation1.Click to view page 1 of the table.LOADING...Click to view page 2 of the table.LOADING...z=0.51A graph with a bell-shaped curve, divided into 2 regions by a linefrom top to bottom on the right side. The region left of the lineis shaded. The z-axis below the line is labeled "z=0.51".The area of the shaded region is ______.
Find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. Click to view page 1 of the table. LOADING... Click to view page 2 of the table. LOADING... z=0.51 A graph with a bell-shaped curve, divided into 2 regions by a line from top ...
5 answers
What existing evidence was there that led the scientists tobelieve that Dengue virus modulates the interferon response inorder to replicate?
What existing evidence was there that led the scientists to believe that Dengue virus modulates the interferon response in order to replicate?...
5 answers
10. Find the indicated nth partial sum of the arithmetic sequence_ -31, ~17, -3, 11; n =15
10. Find the indicated nth partial sum of the arithmetic sequence_ -31, ~17, -3, 11; n =15...
5 answers
An electron moving with a speed of 3.5E6 m/s enters a region of space where a uniform magnetic field of 5.5 T is directed left to right: If the angle 0 in the image is 289,what are the magnitude and direction of the magnetic force that acts on the electron?B3.1E-12 N,into the screen3.1E-12 N,out of the screen1.4E-12N,into the screen1.4E-12 N,out of the screen27E-12,into the screen
An electron moving with a speed of 3.5E6 m/s enters a region of space where a uniform magnetic field of 5.5 T is directed left to right: If the angle 0 in the image is 289,what are the magnitude and direction of the magnetic force that acts on the electron? B 3.1E-12 N,into the screen 3.1E-12 N,out ...
5 answers
B) Let X and be discrete random variables with joint probability functionfor € 1,2 and y = 1,.2p(. !) =othierwiseCalculate E(YIX)
B) Let X and be discrete random variables with joint probability function for € 1,2 and y = 1,.2 p(. !) = othierwise Calculate E(YIX)...
5 answers
Which of the following statements is EALSE concerning solar radiation?Rays at the equator travel a shorter distance than those at the south pole: The sun's rays hit the Earth at a 90 degree angle at the equator: There is a 40% reduction in solar radiation at the poles compared to the tropics. Arayof sunlight at the equator covers the same surface area asone at the north pole:
Which of the following statements is EALSE concerning solar radiation? Rays at the equator travel a shorter distance than those at the south pole: The sun's rays hit the Earth at a 90 degree angle at the equator: There is a 40% reduction in solar radiation at the poles compared to the tropics. ...

-- 0.019193--