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From the frequency table given below the mean value is computed as Iclass interval IFrequency /12-19/20-27128-35136-43144-51[52-59Select one; a.None6.5048.1d.30.8e....

Question

From the frequency table given below the mean value is computed as Iclass interval IFrequency /12-19/20-27128-35136-43144-51[52-59Select one; a.None6.5048.1d.30.8e.42.38

From the frequency table given below the mean value is computed as Iclass interval IFrequency /12-19 /20-27 128-35 136-43 144-51 [52-59 Select one; a.None 6.50 48.1 d.30.8 e.42.38



Answers

find the mean of the data summarized in the given frequency distribution. Also, compare the computed means to theactual means obtained by using the original list of data values, which are as follows: (Exercise 29) 35.9 years; (Exercise 30) 44.1 years;(Exercise 31) 84.4; (Exercise 32) 15.0 years.

Mhm. So to calculate the mean we just add up all our values and divide them by the number of values we have. However we are giving a frequency table so it's gonna be a little bit different to calculate the mean. To calculate our sums were given a couple of mid points and the number of times that happens. So what we can do is multiply the all of our mid points times the frequencies. And then we can just simply divide by the number of frequencies they are. So just sum up all the frequencies and what this is gonna turn out to be Is well that's a big equal sign. We're gonna get 3649 Out of our total observed 82. And dividing this further outs we will get 44 2.5. Yeah. Are estimated value of Or exact value specific of 44.1 Is actually pretty close to this. 44.5 that we have. So it's pretty good.

So the mean is the sum of all our values divided by the number of our observations. In this case, the number of observations is going to be the sum of our frequencies this part right here and then for our values, all we have to do is multiply our frequency times the midpoint values that were given. So once we do that we will end up with a some on the numerator of 10,244 and a half. And we'll divide that by a number of observations. So just some of that frequency column and will get 121. And this divided out is equal to 84.665 soul go 67. And our exact value that we got was 84.4. So just the difference of .27. So we got pretty close. Oh mm.

All right. So one last time we're gonna multiply our frequencies by our mid points to get the sum of all of our values. So that's gonna be the some of the frequencies times the mid points a couple different intervals to look at. So therefore a couple of frequencies. And we'll divide that by the sum of all the frequencies to get our total number of observations. And that's how we'll get that end. Mhm. So these values will come out to be um oh let me adjust that here. So we'll have 496 for some total and divide that by the number of observations which is 33. And this comes out to be approximately 15.03. And we'll compare that to the exact of 15.0 years. We are only 0.03 off and maybe we weren't supposed to round up or not. But this is very close.

For this problem. You're trying to find the means of the group data sets. So let's look at part A. We've got the grade and then we have the frequency and we're going from 49.5 to 59.5 and then 59.5 to 69.5, 69 5 to 79.5, 79.5 to 89.5 and 89.5 to 99.5. And our frequencies of each of those classes were 238 12 and five. Now, if you add up the frequencies, you are going to find out that there were 30 pieces of data in this data set. Now, in order to find the mean of grouped data, we kind of have to know what the data is. So in this particular instance, we have two numbers somewhere between these values. But we don't know if those two numbers were on the low end of that class. The upper end of that class, if they were the same, if they were different. So what we're going to do is we're going to find the mid point of each class, and that's going to be referred to as X, and that's going to be the representative number of that class now, in order to find the midpoint, what you do is you take the lower boundary and you add it to the upper boundary and you divide by two. So, in essence, you're averaging the lower boundary with the upper boundary. And in doing so, we would get 54.5 for this class and again if we take the lower boundary added to the upper boundary and divide by two. Because there's two pieces of data, we would get 64.5. We'll do it for the third class and you're gonna get 74.5, the fourth class 84.5 and the final class in this part will be 94.5. So now let's go back to our formula to find the average of group data you are going to sum up X times F divided by N. Well, we just got done saying that the number of pieces of data is what we get when we add up the frequencies so we know we're going to be dividing by 30 in this particular problem. So we're gonna create another column and we're going to call it X Times F. And for that we're going to multiply the frequency times its corresponding midpoint. And in doing so, for the first class, we get 109. Then we're gonna take the frequency times the corresponding midpoint, and we'll get 1 93.5 and will continue to do that for each of the classes. So we have 5, 96 1014 and 472.5. So now let's go to this formula. This here indicates that we need to add up this column. So when we add up that column, we're going to get a total of 2385. So we'll use that in our formula. And for this first part, we're going to get a mean of 79.5. So the mean of the grade in this instance was a 79.5. So we're gonna do the same thing for part B for Part B. We're going to have our chart again, But this time our classes are the daily low temperature, and we happen to have the same class boundaries 49.5 to 59.5, 59.5 to 69.5 69 5 to 79.5, 79.5 to 89.5 and 89.5 to 99.5. We were also given our frequency, and the frequency of each class was 53 32 15, 1 and zero. When we add up the frequencies, we're going to get that there was 101 pieces of data. Our mid points are going to be the same as they were in part a. So we're going to have 54.5 64.5 74.5 84.5 and 94.5 to find the average or the mean we need to sum up the X F column. That means we have to create an X F column. Mid points will be our X frequencies are your F, and then we have to divide it by en which is the same thing as saying the sum of our frequencies. And we've already determined that there were 100 pieces of data in this data set. So we're ready now to multiply the X times, the f of each of the classes. So we're gonna get 2888.5, 2000 64 1117.5, 84.5 and zero. And when we sum up the x f column, you get a total of 6154 0.5. So we'll use that in our formula. And we come to find out that the mean daily low temperature was approximately 69 3 six. And we're gonna do that for one more set of data. So we're now moving on to part C more in part C. We're going to set up our chart this time. Our boundaries are describing points per game. We happen to have the same class boundaries 49.5 to 59.5, 59.5 to 69.5, 69.5 to 79.5. 79 5 to 89.5 an 89.5 to 99 0.5. We just happen to have different frequencies. The frequencies of each were 14 32 15 23. And to when we add up the frequencies, we find out that there were 86 pieces of data. We will need the mid points so that we have a representative point for each of the classes, and they happen to be the same as part A and part B. So we had 54.5 64.5 74.5 84.5 and 94 0.5. Again, we're finding the mean and the meanness found by taking the sum of X times f and dividing it by n, which is the same as the sum of X times f divided by the sum of F. We've already declared 86 pieces of data, so we need to create that x f column. X will be our midpoint. F will be our frequency. And when we multiply the two of them together. 763 2064 1117.5, 1943.5 and 189. When we total up the x f column, you are going to get a result of 6077. So we're gonna use that value as the numerator in our formula, and we find that the mean points per game is approximately 70.66 three. So just to recap very quickly here, the first mean was a grade of 79.5. The second mean was a daily low temperature of approximately 69 36 And for part C, the mean points per game, it was approximately 70.663


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