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The following data represents the age of 30 lottery winners22 30 31 31 33 40 40 41 41 42 " 47 49 52 53 53 56 58 58 64 66 68 69 70 71 72 75 82Complete the frequ...

Question

The following data represents the age of 30 lottery winners22 30 31 31 33 40 40 41 41 42 " 47 49 52 53 53 56 58 58 64 66 68 69 70 71 72 75 82Complete the frequency distribution for the data. Age Frequency 20-2930-3940-4950-5960-6970-7980-89

The following data represents the age of 30 lottery winners 22 30 31 31 33 40 40 41 41 42 " 47 49 52 53 53 56 58 58 64 66 68 69 70 71 72 75 82 Complete the frequency distribution for the data. Age Frequency 20-29 30-39 40-49 50-59 60-69 70-79 80-89



Answers

Charity Donations A random sample of 30 large companies in the United States shows the amount, in millions of dollars, that each company donated to charity for a specific year. Construct a frequency distribution for the data, using 9 classes. $$ \begin{array}{lllll}{26} & {25} & {19} & {31} & {14} \\ {48} & {35} & {43} & {25} & {46} \\ {17} & {21} & {57} & {58} & {34} \\ {41} & {12} & {27} & {15} & {53} \\ {16} & {63} & {82} & {23} & {52} \\ {56} & {75} & {19} & {26} & {88}\end{array} $$

Hell in this problem we have to calculate the mean deviation about meeting each for the age distribution of 100 persons by using the given data. We have given the following table. Now, first of all, we will convert the given data into continuous frequency distribution by subtracting 0.5 from the global limit and adding 0.2 to the upper limit of each class interval. So we get here, we have the following modified class. Now we will find the mid value. That is exciting. It would be 18, 23 28 33 28 43 48 and 53. Now, in order to find me in division about meeting, first of all, we will find a cumulative frequency, it would be 55 plus six is 11, 11 plus 12 is 23 23 plus 14 is 37, 37 plus 26 is 63, 63 plus 12 is 75, 75 plus 16 is 91 and 91 plus nine is 100. Here we have an is equals 200, so we get and upon two is equal to 100 upon to Which is equals to 15. Therefore, medium class would be 35.5 to 40.5, so we get median is equal to 35.5 plus 50 -37 are born 26, It would be equals two, 35.5 Plus 2.5, Which is equals to 38. Now we will find the absolute value of deviation from the median, it would be AB -28 is 20 as we're finding the absolute value, so we will ignore the minus sign, 23 -38 would be 15, 28 -38 is then 33 -38 is by you, 38 -38 is zero, 43 -38 is five, 48 -38 is Dane, And 53 -38 is 15. Now we will find a fly into absolute value of deviation from median, it would be five into 20 years, 100 16 to 15 is 90 12 and detain is 120 14 into 5, 70, 26 20 is zero, 12 into five is 60 16 into 10 is 160, and 19 to 15 is 135 And the submission is 735. Now, main division about meeting as equals two, one upon N sigma, F I into absolute value of division from me. Well, I goes from one to and by substituting the known values, we get 735 upon 100 It would be equals to 7.35. Hence we get main deviation about meeting Is equal to 7.35.

Before we can make our frequency distribution, we'll first need to find the width of each of our classes. To do this, we'll start by finding the minimum and maximum in our data. Our minimum is 0.8 and our maximum is 51.7. Technically, we could treat this 51.7 as an outlier because it is one. But it's not an extreme outlier. So I'm going to treat this as a normal problem. If you were used as an outlier, just like in the following problems after this one, then that would still be fine. Now we confined our range which will be 51.7 minus 0.8 and that equals 50 0.9. Now we want to divide our 50.9 range into five equal groups, which equals 10.18. And we'll want to just round this up to the nearest whole number. Our data goes into the tens place, so we want to round to the nearest one. So are with here will be 11. In general, you want to go one place value lower than your highest place value. Alright. So back to our table, we will start with zero and go up to a number that will find out in a second. Since our with is 11 will just add 11 to get to our next class is starting point. And we can just keep counting by Elevens for these five, and I'll go ahead and make a note of what the next one would be. So we know where to end off, all right. And before this 11 will end at 10 for our first class, 21 will come right for the 22 32 43 and then finally, 54 comes before this 55. Right now we will count up our frequencies and we should end up with seven six to zero and one. Even though one of our frequencies is zero, we do still need to include it because we don't ever want to leave any numbers out from our rage again. We really should start with zero. Starting at one obviously does not work, since we have a data point lower than one and we don't want to start in the negative numbers. Zero is really the only good starting place for this one, so your table should look exactly like this

Okay. So for this problem, we're going to use a frequency table to find out the average age at which the an actress wins the best actress Oscar. Yeah. So we have different classes, which are a drain show. 29 to 2022 29 years old, 30 to 39 years old, etcetera. And we're gonna and we have different frequencies. 29 actress have won this award when they were between 20 and 29 years old. Have 30 fours. 34 actresses have won between 30 and 39 years old, etcetera. And we're gonna take inspiration from the example seven in your book. And the first figure out was the class midpoint for these values. Now there are something a bit odd, which is our road plus should have been adjusted a dash. Really? It's not the minus or anything. Okay, So what's the class midpoint in 20 on 29 years old. Well, if you do 20 plus 29 divided by two, you would get 24.5. So you just take what's straight in the middle of those values. Same here. 30 to 39 is 34.5 44.5, 50.54 point five before 0.5. 74.5 and 84.5. Now we can assume that no one has no one who is younger than 21 it and no one who was older than 81. Now we're going to multiply every, um, frequency by the class midpoint. So 29 times 24.5 is 710 plus five. 34 times 34.5 is 1173 14 times 44.5 is 623 three times 54 uh, 0.5 is 163.5. Five times 64.5 is equal to 322.5. Yeah, one time 74.5 is 74.5 and one times 84.5 is 84.5. Good. Now we're gonna add all of our frequency here. So 29 plus 34 plus 14 plus three plus five plus one plus one that gives us 87. So we have 87 actresses who want the Academy Award for best actress. This sum here is irrelevant. It's not important we don't need it. And we're gonna have every f ah, every frequency times midpoint. So that's gonna be 710.5 plus 1073 101 173 623 163 0.5 plus 322.5 plus 74.5 plus a 4.5, which gives us 3100 51 0.2. And now what we're gonna do to find the mean we're simply gonna, uh we're gonna take the sum of all f times X divide by the sum of F. So 3151.2 divided by 87 which give us 36.2. And if you look in the appendix three appendix of your book, you will see that if we use the raw data without putting anything in a frequency table, we would have noticed that this result is the same as if we used every single entry in the data. So which is the same which is the same as that mean is we take yeah, quote unquote ra data in the appendix

Okay, So in this problem, we once more have ah, frequency table for the age at which the some actors have won the best actor at the Academy Awards. Um, it's very similar to problem 30 29. So so similar. In fact, I forgot to change title, but this is problem 30 and s. So it's the same age ranges we had in problem. 29 different frequencies and therefore a different mean at the end. Once again, we take expiration from example seven in the book, and we're gonna find the mean using the stable here. So first thing we need to do is find the mid point of each class, so each age range. So what's 20 plus 29 divided by two is 24.5, 30 plus 39 divided by two is 34.5 and so forth. So 44.5 54.5, 64.5 74.5. And here I use the same table as in problem 39 no actor in the age range of 82 89 won the Academy Award for best actor. Uh, the mid boy point is 84 but the zero here. We'll make sure we don't use that information later on. So now we need to Because remember, when we compute the mean using frequency table, use the sum of the frequency and the value divided by the total frequency. Uh, so we need to compute every frequency times Classmate Point. So one times 24.5 is 24.5, 28 times 34.5 is 966 966. 36 times 44.5 is 1602 15 times 64.5 kids, UH, 108 117.5. Six times 64.5 is 387 and one times 74.58 74.5 and zero times 84.50 We compute the total frequency. So one plus 28 plus 36 plus 15 plus six plus one, which has got well plus six plus one, which is 87. So same thing if you did the problems 29 with actresses, uh, that there's the same amount of Academy Award winners for best actor. And now we need to compute the sum of f of X 24 f times X so 24.5 plus 966 plus 1602 plus 817.5 plus 387 plus a 74.5 plus zero, which is 3000 871 0.5. Wow. And of course, the class, the some of the class midpoint is irrelevant. We don't care for it. So to find the mean using the frequency table, we simply divide 300 eighties 3000, 871.5 divided by 87. And so we have a mean of 44.5. If we had taken the raw data in our ah, using the appendix table, we would have found that mean is 44 years old. So this is not the same this mean isn't the same as the one we find using the ah rah data. So raw, of course, is quote unquote hero data and the appendix And remember that mean is 44 so it's slightly, uh, it's still very similar so 440.5 years of six months. So still, uh, totally adequate in that situation, to use the frequency at table and using age range to find the mean. But it's just slightly different than if we had taken the age of every single actor divided by 87 yeah.


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