5

Using 5 clesses of width 50 starting at 100, construct a frequency distribution for the following numbemt [4138 286 345 211 117 321 4119 1179 297 244 168 145 2251 1...

Question

Using 5 clesses of width 50 starting at 100, construct a frequency distribution for the following numbemt [4138 286 345 211 117 321 4119 1179 297 244 168 145 2251 176 ] Find out the frequency counts for all 5 classes, starting with class 100 to 149, for your answer 043,3,2,2 4,4,2.2,2 0511,5,2 0 5,3,2,3,1

Using 5 clesses of width 50 starting at 100, construct a frequency distribution for the following numbemt [4138 286 345 211 117 321 4119 1179 297 244 168 145 2251 176 ] Find out the frequency counts for all 5 classes, starting with class 100 to 149, for your answer 043,3,2,2 4,4,2.2,2 0511,5,2 0 5,3,2,3,1



Answers

Reading a Frequency Distribution ,Use the frequency distribution to find the (a) class width, (b) class midpoints, and (c) class boundaries. Travel Time to Work (in minutes) $$\begin{array}{|c|c|}\hline \text { Class } & \text { Frequency, } \boldsymbol{f} \\\hline 0-9 & 188 \\ 10-19 & 372 \\20-29 & 264 \\30-39 & 205 \\40-49 & 83 \\50-59 & 76 \\60-69 & 32 \\\hline\end{array}$$

So for this question, were pretty much asked to determine the total number of data. And so to do this, we're really just gonna focus on the frequency portion because the frequency portion tells is like the number of you know, each value, right, the number of data pieces that show up in a certain class interval. So this is a table that we were given. So all we're really going to do is just go in and add all the frequencies together. So we're gonna have 29 plus 12 about 29 plus 12 plus 12 plus six plus 22 plus 56 plus 60. So we know that 12.6. So I'm gonna go home. 12 plus six is 18 18 plus 22 is 40 and 40 plus 60 is 100. So we have 100 plus 56 right? Plus 29 56 post 29 right. Is just going to be 85 66 76 then we have the nine. So that's gonna be 85 85 close. 100 is 1 85 So the answer is gonna be 1 85

So in this problem we have a data set with whole numbers with a low value of 10 and a high value of 20. And we want to find the class within class limits for frequency table with five classes. So our range is 10, 220 and our table, we'll have five classes. So the way to find the class with let's just take our range. So 1 20 minus 10 over the number of desired classes. So in this case, it would be 1. 29 is 10/5, which will give us a class with of 22. So that would mean that our class limits it would be 10, 32 33 2, 55 etcetera. So until you get to the upper end of your range, So you would just want to make sure that you're not overlapping values and that your with is matching the range, so you should end up with five classes all falling within this range with a width of 22. That's how you would find the appropriate limits for this frequency table. Given our data set. Yeah. Mhm. Mm hmm.

At least problem off frequency distribution. We have to use this frequency distribution table to find the class with classmate points. Okay, class boundaries. Given frequencies, the high temperatures frequencies for the high temperature ranges. Do we have to find class? Good points? Yeah. New class. Don't things? Yeah. First, we we can't believe that class wit class with will be. By the definition, it is a distance between us. The poor limits upper or lower limits off consecrated classes. So if we see save the day global limits of the conservative pluses we see the difference is 11th, therefore is going to be 31 minus 20. Say, which is going to be 11 or we can just calculated by the upper, not some data entry. There is 96 minus 20. That is a minimum data's data entry. Do I buy the number of classes? The seventh to the standpoint date. The next convenient number will be 11. So that the class with this level now we have to find a midpoint. The midpoint can be calculated by adding the lower class limit. That is my point equation. Is the point of the film too lower class limit Yeah. Place Popular's limit. Delighted bite. Therefore, here it is. Lower classes 20 and thirties upper classes 30 Developed by 2025. Similarly, that even place 40 to 41. Divided by two. That is 72 by two 36 94 by two, which is 47 53 less, 63 by two, 116. By truth, that is. 58 60 focus and D four bedroom with a 69 75 85 close to just 80. Finally, just 91. You see the difference. The consequent values. We're also going to be there class with now. To find the class boundaries, we have to subtract the lower class limit with this lower class boundary. Zeke Winter Lower class limit minus 0.5. Similarly, after class boundary is equal to after plans limit. Let's zero point I. So if we do that, we'll get 20 minus 200.5, 19.5 2 30 less 0.5 30 points Fight. Similarly, we can do for the rest, that is 30.5, and it is complete with there is no gap to 30.5 to 41.5 41 points. Siding speaking point, died escaping from inside. 63.5, 63.5 to 74.5. We'd 74.5 until 85 on five. 85.5. 96 point. Yeah. Are you so

Yeah, this problem is a problem about turning raw data into something more visual so that you can actually see patterns or things that are unusual, things that you might not be able to see if you don't have your data in some kind of graph. So you're being given a bunch of times uh 57 actually to count them up 57 different times. So I'm just gonna make a note here that uh we have 57 different pieces of data and is equal to 57 that represent the time that it takes for bobsleds to finish a race. Okay, so the first thing that you're asked to do is to divide this into five classes. So five categories to break this down. So for the class with what we're going to do is just take the highest value of 3, 60 minus the lowest value of 2 36. And we're going to divide that into five equal parts. All right, well it turns out that that is 24.8 which is kind of an awkward number to use. So we're going to round that up to 25. Um You could round it up to 30 but the more you round the less accurate your graph is going to be. So we know that we have class widths of 25. Now what you'll notice about the data that I have at the top is that I already put all of the data into numeric order. I happened to do it with Excel. Um It's a bit tedious if you don't import all the data into Excel but it's certainly doable. And so now we're ready to figure out what the categories or the classes look like. So the lowest category is going to start at 2 36 and then I'm going to add 25. So I end up getting, yeah uh sorry to 61 I'm going to add 25 again and I get to 86 add 25 again, I get 3 11 and one more time I get 3 36. Okay, so that's those are the lower limits of each class. Now the upper limit is easy to figure out. This is going to go from 2 36 to 2 60 just before I hit the next class to 61 to 2 85 to 86 to 3 10. 3 11 to 3 35. And then the next class after 3 36 would actually be 3 61 although I'm not going to end up using that, but it does tell me that the high end is 3 60 so you might just want to take one second and make sure that your classes will fit all the data that you've done it correctly, So they all fit and we don't have to go any higher than 3 60. So we're all set all right class boundaries. So I'm going to go across just one class for right now and this is the class that I have in green up above. So just four pieces of data. So actually I'll go ahead and put that in under the frequency. There's four pieces of data. All right, the class boundaries um, the rule for the class boundaries is that you're just going to go from 0.5 below the low boundary. So that would be to 35 5 and you're going to go to 0.5 above the higher boundary. So we're going to say that the boundaries go from to 35.5 to 65. And these boundaries are what you're going to use when you make your graphs. Okay, so these are going to be used for the purpose of making your graphs. You'll see that in a little bit. Alright, midpoint should be fairly straightforward. We're just going to take the average of 2 36 and 2 60. So to 36 plus 2 60 divided by two is going to give me 2 48. Okay, the relative frequency is just the frequency divided by the total. And then you're going to want to turn that into a percent or a decimal if it's approximate, that's completely fine. So this would be if you do that on your calculator, it's about 7% but I'll leave it as a decimal. And then the cumulative frequency, that refers to the number of pieces of data you have so far. So by the time you get to to 60 how many pieces of data have you accumulated? And we've accumulated only four. Right okay I'll do one more row kind of talk through it and then I'll show you the rest without going through all of them. Okay So here's the next rope. So notice that um the boundaries are 0.5 below and 05 above the class limits. The midpoint comes from adding to 61 plus 2 85. Taking that entire quantity and dividing by two. Okay. The frequencies come from going back to looking at the raw data and just counting the number of pieces of data that fall between 2 61 and 2 85. And if I count those up 123456789 Okay. The relative frequency is just nine divided by the total which I'm going to round to about 16%. And now by the time I've gotten 2 to 85 I've accumulated a total of 13 pieces of data. So it's really just my original four plus the nine that were in this category. Mhm. Alright filling out the rest of the table. I'm just gonna go ahead and put this up there. Um The one thing that I think might bear looking at are the cumulative frequencies. So by the time I get to 3 10 I have I have accumulated the original four plus the 13 and now I've accumulated 25 more. So if I do I'm sorry the four plus the nine plus the 13. So let me go back just for a second here. So it was my original four. Yeah which was here. Then the nine that I got from the second category and then finally the 25 that I got from the third category and that gives me a cumulative of 38. The other way you can do it is just take the previous category and add on the amount in that category. So we started with 13 from the previous two. We added 25. Okay. We started with for the next one. We started with 38 we added another 16 in that category. Okay. And then finally we started with 54 added another three in this category. Yeah. To get a total of 57 it is worth checking. By the time you get to the end here you should have the total of the number of pieces of data. So if you if you ended up for example with 56 you would know that you added something wrong along the way. All right. That's kind of the tedious part. Now putting this into a picture. That's the part. That's a little bit more interesting. So, first of all, when I make a history Graham, I want you to notice that the class boundaries are what I use on the X axis. Right. Another thing to notice is that you always want to label your axis, your both of your axes. So these are finished times to the nearest our and then along the y axis for a hissed a gram. I'm going to use the frequencies. So just how many pieces of data were in each category. So that was that would be this column right here. This is the column I'm going to use for hist a grams. Mhm. Yeah. All right. And then it's just a question of plotting the various points. So we know that from 2 35.52 to 60.5, we had four pieces of data. So, oh, sorry. And then the other thing that we need to do is label the y axis. So we know we're going all the way up to the highest frequency is 25. So I set this out so that it would be easily divisible and I'm just going to go 5 10 15 20 25. Okay so in the first class we had four so I'm just gonna kind of eyeball this. Yeah. Okay. And we draw a bar. Alright. Second when we had nine. So again this just can be pretty approximate right? Third category that was the big one. We had 25 so we're jumping all the way up to here and then drawing a bar all the way down. Okay. Fourth category we had 16. So that's that bar there. And last category we had three so we're back down to just a few. So put that at three and there's your instagram. Now, a relative frequency hissed A gram is very very similar. It's just that you have different values on the Y axis and these instead of being the actual number is going to be the percent. So that's gonna be the relative frequency. And for that we're going to use this column right here. Okay, that's going to be for a relative frequency, hissed a gram. So this one is the relative frequency hist a gram. All right. So if you look at those numbers, the highest is 44% and the lowest is 5%. Okay, so again, I kind of planned this out ahead of time so I'm going to go by uh yeah, 10%. So this is going to be 10% 20% 30% 40 and 50. And then you do exactly the same thing that you did for the hist a gram of above. So the shape should look exactly the same. So we're starting with let's see, 7%. So again, I'm going to kind of just eyeball this right. Next one was 16%. That's a little bit more than halfway between 10 and 20. Okay, next one is 44%. That was the big jump. And again, there will be a big jump here as well. Mhm. Mhm. All right. Next one is 28%. So put that roughly here and last is 5%. So again, that's going to be the lowest one of all the bars. So you should be able to tell that if you compared the history graham and the relative frequency hissed a gram and you didn't have anything on the Y axis, they would look exactly alike. All right, categorizing the shape. So our choices are a mound shaped, right skewed, left skewed. Um Got a couple other answers that we can go with but this one is clearly a mound. So we're going to say this is mound shaped. But since it's not perfectly symmetric there's a little less data over here and a little bit more data over here, we're going to call this slightly left skewed and that's the part that can be a little bit counterintuitive. The skew is in the direction of the tail. So where there is less data. So just as a reminder a left skewed like something that was really left skewed would look like that and something that was right skewed really right skewed would be like that. Okay so this this would be a left skew and the bottom one would be a right to skew. Mhm. All right. Last one got an O. Jive. I love this word. Okay, so an O jive is just a way to show how the data accumulates. So for an O jive you're going to end up using this last column right here, that will give you an O jive. Mhm. Alright, so notice that for all of these the X axis is always the same. It's just the class boundaries. So we have to get up to 57 on the Y axis. So this is gonna be the cumulative frequency. Yeah and we have to get all the way up to 57. So again I planned out the space so that I could get up to 60. So here's 10 20 30 40 50 and 60. Okay, so when we start at 2 35.5 we have no data yet. Okay, the lowest piece of data was 2 36. By the time we get up to 2 65 if you look at back up here, we have accumulated for pieces of data. Mhm. So come down here and we're going to graph this at four. All right. And we're gonna actually put a four by it. Just to be a little bit helpful to folks trying to read something like this. All right. By the time we get to the next boundary to the upper level of the next boundary, So by the time we get to to 85.5, we have accumulated a total of 13 pieces of data. Right next. The next accumulation, By the time we get to 3, 10.5 is 38. So that's here. Yeah. Alright. So notice that you can tell that there is the most data in this category right here because it's the biggest jump. All right, Next one. We have accumulated 54. So put this right about here and last. We have by the time we get to the end we have accumulated all 57. So this is 54 and this is 57 then all you have to do is just connect them with straight line segments and that is an O. Jive. Okay, So once again, I just want to point out this section right here is where you know you have the most data because it's the steepest jump. So the most data. Okay, is in that frequency or is in this class? Sorry? Okay. Okay. And then the last thing is just to kind of make a note of some things that the graph tells you and this is kind of up to you. But some of the things you could talk about our um the range of data. So in this case we have a range of data that is between from 3 62 to 36. So the range is equal to 3 60 minus 2 36 which is 124 hours. Okay. So you have 100 and 24 hours difference between the slowest team which would have taken 360 hours and the fastest team which took only 2 36. Okay. Another thing that you could notice is where did most of the times lie? So most times or least times you kind of want to think about the the highs and the lows that you might be able to talk about. So the most times were between 2 86 and 310 hours. Mhm. Right. And then the only other thing to think about is whether there was anything unusual. So were there any pieces of data that really really were? Um outliers that really lay far apart from the rest of the pack and in this case there wasn't and that's something to notice also. So no unusual data or no unusual times. Okay. Um You might think about also talking about the shape but we already did that up above. We said it was mound shaped but skewed slightly left and with that I think we've done everything that the problem asked us to do. Sure.


Similar Solved Questions

5 answers
10) Consider the differential equation y' with the initial condition y(1) = 3_ Use Euler' Method with dx = h =,0,1 to approximate Y(1.2). Please build . table work: Round EVERY of values which demonstrates your answer t0 the nearest decimal places.
10) Consider the differential equation y' with the initial condition y(1) = 3_ Use Euler' Method with dx = h =,0,1 to approximate Y(1.2). Please build . table work: Round EVERY of values which demonstrates your answer t0 the nearest decimal places....
5 answers
Bf (b): For the vector field F(I,y,2) = r*y2 j+y'2'6,find curlF and divergence of F.
bf (b): For the vector field F(I,y,2) = r*y2 j+y'2'6,find curlF and divergence of F....
5 answers
IL 4{32}= OFa H a21 (eat + at + 1)Ob 41 (eat at a2 I1) OLc 3 (eat at I)OFd; Hl (eat] 02 #iat H4
IL 4{32}= OFa H a21 (eat + at + 1) Ob 41 (eat at a2 I1) OLc 3 (eat at I) OFd; Hl (eat] 02 #iat H4...
4 answers
(4 points) Use tha simplex mathod t0 maximizeP = Sx1 + 9x2subject to3x1X2 < 3*2X1 20X2 2 0
(4 points) Use tha simplex mathod t0 maximize P = Sx1 + 9x2 subject to 3x1 X2 < 3*2 X1 20 X2 2 0...
5 answers
Frohlar Let A ard B bc subscts of 4 univcrsal si U Show thal 4 € 8 if and only if B € T
Frohlar Let A ard B bc subscts of 4 univcrsal si U Show thal 4 € 8 if and only if B € T...
5 answers
For each of the following molecules: give the name of the shape; state whether the molecule is polar or non-polar and give the hybridization of the central atomFormulaMolecular ShapePolar/ Non-polarHybridization of the Central Atom(teSzBCla
For each of the following molecules: give the name of the shape; state whether the molecule is polar or non-polar and give the hybridization of the central atom Formula Molecular Shape Polar/ Non-polar Hybridization of the Central Atom (teSz BCla...
5 answers
'720143Evaluate the Integral. (Remember tO use absolute values where approprate: Usecot50 sin"0 d0Need Help?71.10 polnt SCalcET7 72.019Evaluate the Integral . (Remember t0 use absolute valles where appropriate: Use € cosX + sin 2x sin *Need Help?Ce potnt SCalcET7 7.2020Evaluatc thc integral: (Use for the constant integration:)cos?, sin 2x dxNeed Help? HlabA{oAr 5C4k2777.2021.,Evaluate the Integral: (Use for the constant 0t Integration )
'720143 Evaluate the Integral. (Remember tO use absolute values where approprate: Use cot50 sin"0 d0 Need Help? 71.10 polnt SCalcET7 72.019 Evaluate the Integral . (Remember t0 use absolute valles where appropriate: Use € cosX + sin 2x sin * Need Help? Ce potnt SCalcET7 7.2020 Evalu...
5 answers
Assuming - x? Zy2 6xy = definesdifferentiable function of X, use the theoremto findat the point (1, -1)(Type an integer or simplified fraction )
Assuming - x? Zy2 6xy = defines differentiable function of X, use the theorem to find at the point (1, -1) (Type an integer or simplified fraction )...
5 answers
Calculate the percent ionization in each of the following solutions.a. $0.10 mathrm{M} mathrm{NH}_{3}$b. $0.010 mathrm{M} mathrm{NH}_{3}$
Calculate the percent ionization in each of the following solutions. a. $0.10 mathrm{M} mathrm{NH}_{3}$ b. $0.010 mathrm{M} mathrm{NH}_{3}$...
5 answers
Sample 3;2.(X 1027 0.9475MunalumnTnntAculitedTcanannnEol ALcilculcd ErnHumtnumHTATAmeaurcd0475Te(show vourwod bdox)Sample 4:atoms of silver2XT6 21 etomemol 48calculated moles of silvergrams Agcalculated grams of silvermessured grams of silver14(show your work below)Brams
Sample 3; 2.(X 1027 0.9475 Mun alumn TnntA culited Tcanannn Eol AL cilculcd Ern Humtnum HTATA meaurcd 0475 Te (show vourwod bdox) Sample 4: atoms of silver 2XT6 21 etome mol 48 calculated moles of silver grams Ag calculated grams of silver messured grams of silver 14 (show your work below) Brams...
5 answers
Balanze tnu IdtoAll) 5,()-+AL5,6) LlartanaTqyblloiHsAI(S) 2S (S)=9AL,SainWealnol ArutIncorroct; Iry Aqain; J attomata romainingpCL () #ou) +H,BO,(44) HClaa) che meal equntiar_Uelentlty all ol the Knatoa
Balanze tnu Idto All) 5,()-+AL5,6) Llartana Tqyblloi HsAI(S) 2S (S)=9AL,S ain Wealnol Arut Incorroct; Iry Aqain; J attomata romaining pCL () #ou) +H,BO,(44) HClaa) che meal equntiar_Uelentlty all ol the Knatoa...
5 answers
The principal is borrowed at simple interest rate for a period of time Find the simple interest owed for the use of the money: Assume 360 days in Lycar and round answer to the nearest centS110 49 years84,4091600S127.60517.60
The principal is borrowed at simple interest rate for a period of time Find the simple interest owed for the use of the money: Assume 360 days in Lycar and round answer to the nearest cent S110 49 years 84,40 91600 S127.60 517.60...
5 answers
Determine whether each graph is the graph of a function. If it is not, find two ordered pairs where more than one value of y corresponds to a single value of x. See Example 9.CAN'T COPY THE GRAPH
Determine whether each graph is the graph of a function. If it is not, find two ordered pairs where more than one value of y corresponds to a single value of x. See Example 9. CAN'T COPY THE GRAPH...
5 answers
A. A company decided to distribute 5 identical computers amongits employees. In how many different ways can be distributed amongthree distinct people if each person receives at least 1 computerand no more than 3 ? . b. Given the two generating functions f(x)=2/(1-x)^2 andg(x)=1/(1-x)^2 find the generating function for f(x)+g(x)?
a. A company decided to distribute 5 identical computers among its employees. In how many different ways can be distributed among three distinct people if each person receives at least 1 computer and no more than 3 ? . b. Given the two generating functions f(x)=2/(1-x)^2 and g(x)=1/(1-x)^2 find the...
5 answers
Height of Father; Xi 71.6 73.1 69.8 73.3 67.9 67.3 73.2 68.4 71.7 72.7 70.5 71.5 68.1Height of Son; Yi 76.7 76.6 72.3 75.1 69.0 67.8 73.3 67.8 70.5 70.9 68.0 68.0 63.2
Height of Father; Xi 71.6 73.1 69.8 73.3 67.9 67.3 73.2 68.4 71.7 72.7 70.5 71.5 68.1 Height of Son; Yi 76.7 76.6 72.3 75.1 69.0 67.8 73.3 67.8 70.5 70.9 68.0 68.0 63.2...
5 answers
Rurvorourchayeo518.937Eledleunne
Rurvor ourchayeo 518.937 Eledl eunne...

-- 0.019895--