Use the frequency distribution in Exercise 4 to estimate the s...

Question

Use the frequency distribution in Exercise 4 to estimate the sample mean and sample standard deviation of the data. Do the formulas for grouped data give results that are as accurate as the individual entry formulas? Explain.

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Use the frequency distribution, estimate the data sample mean, and sample standard deviation. Are the mean and standard deviation calculated from grouped data as precise as those from ungrouped individual data? Explain. Awesome. So it appears for this particular prompt we're asked to use frequency distribution to estimate the data sample mean and the sample standard deviation. So those are our first two answers we're trying to solve for. And finally, the last answer we're trying to solve for. So ultimately we're trying to solve for 3 separate answers. So our third and final answer we're trying to solve for is we're trying to determine if the mean and standard deviation calculated from group data as precise as those from ungrouped or individual data, and we're asked to explain. Awesome. So now that we know what we're ultimately trying to solve for, let's take a moment to examine our data table that is provided to us by the prom itself. As you can see, we have two columns. Our left column is the score range and our right column is the frequency. So going down our column for the frequency, we have 48, 1510, 6, and 7, and for our score ranges going down our column, we have 10 to 1920 to 29, 30 to 39, 40 to 49. 50 to 59 and 60 to 69. Awesome. So with that in mind, our first step to solve for this particular problem is we need to calculate the midpoints of each score we're trying to calculate the midpoints for the class. Or rather the score range using our score range and frequency, we're trying to determine the midpoint. So let us recall and use our midpoint equation. So I'm going to say M is our midpoint and the midpoint M is equal to the lower limit, which I'll call L L plus the upper limit, which I'll call UL upper limit. Divided by 2. So the midpoint is equal to the lower limit, plus the upper limit, all divided by 2. Fantastic. So to keep our data organized, I've gone ahead and collected all of our midpoints for each of our score ranges, and I've put it into a table for us to refer to. And instead of repeating our score range again and our frequency again, I'll read going down the column starting from our 10 to 19. Ending with 60 to 69 for our midpoint values which we're denoting as X in our table. So we have 14.5, 24.5, 34.5, 44.5, 54.5, and 64.5. Fantastic. So once again, that's when we plug in our known information for each score range into our midpoint equation to solve for our midpoint, and then we will be able to fill in our table as such. So, moving right along here, our next step now, our second step. is we need to go ahead and calculate the sample mean using the grouped data formula. So as we should recall, X bar. is equal to The sum of F I or F subscript I multiplied by X subscript I, all divided by the sum. Of F subscript I. So, Using our table, once again, we can go ahead. And determine Our values for for FI and XI. So when we do that, We will determine That using our table, we can, we can go ahead and easily write that the sum of FI multiplied by XI using our table. will be equal to 58.0 plus 196.0, so 196.0 plus 517.5 plus 445.0 plus 327.0. Plus 451.5, which when we add all those values together, we'll get 1,995. 0.0. Fantastic. And our value for N will be equal to the sum of F which is equal to 4 + 8 + 15 + 10 + 6 + 7, which is going to be all equal to 50 when we plug in those values into our calculator. So that will mean, without a doubt, our value for X bar will be equal to 1,995.0 divided by 50, which will give us an answer of 39.9. Hooray, we did it, and this is one of our final answers. Hooray, we did it. So moving right along here, our third step now, moving on to our third step. Would be we need to calculate the sample standard deviation using the grouped data formula, which states that S is equal to the square root of The sum of F subscript I multiplied by X I 2 minus The sum So it's gonna be minus the sum. Of F I multiplied by XI alter the power of 2. All divided by N minus 1. And note that this large expression is all under a square root. Sign, fantastic. So, with that in mind, Our next step now is we need to compose a new data table. So using our data table. That we need to, so we're gonna need to use a new data table to help us calculate the sample standard deviation using our group data formula. So when we do that, we will obtain the following. So we to go down our columns, so for our values for FI, we have 48, 1510, 6, and 7, and for Xi, we got 14.5, 24.5, 34.5, 44.5, 54.5, and 64.5. And when we square our XI values, we will get 210.25, 600.25, 1,190.25, 1,980.25. 2,970.25 and finally 4,160.25. And then when we multiply our FI value multiplied by our XI squared value, we'll get 841.0, 4,802.0, 17,853.75. 19,802.5, 17,821.5 and 29,120.75. Fantastic. So, with that in mind, we can go ahead and easily write using our table that the sum of FI multiplied by XI squared is equal to. 9 90,000. 342.5. So that will mean that S is equal to the square root of 90,342.5 minus 1,995, so 1,995.0 squared divided by 50, all divided by 50 minus 1. And don't forget that this is all under a square root symbol. So when we plug that into our calculator and we round to two decimal places, we will find that it's equal to 14.81, and that is our final answer. Hooray, we did it. And that is our 2nd final answer, but we still need to solve for our 3rd final answer. So, that means we need to determine are the mean and standard deviation calculated from grouped data as precise as those from ungrouped or individual data? So, as we should recall a note, the group data calculations use midpoints to represent all values in a class interval. This approximation may lead to slight inaccuracies compared to calculations done from the exact raw data. And if the data, as we should recall, are evenly distributed within intervals, the grouped data estimates or estimates will be quite close. However, for skewed or unevenly distributed data within intervals, grouped data formulas are less accurate. Therefore, as we should recall, grouped data formulas give a good estimate, but generally are less precise than individual entry formulas using raw data. So, with that in mind, in conclusion, We can safely say that grouped data gives approximate results because midpoints are used to represent a range of values. The mean and standard deviation are estimates and may differ slightly from those calculated using raw data. So that will mean when we go to look at our original problem, we can safely say that when we Combine all of our knowledge and we write down our final answer, we can safely say that our final answers. Without a doubt, have to be the following. That X bar is equal to 39.9, S is equal to 14.81, and grouped data gives approximate results because midpoints are used to represent a range of values. The mean and standard deviation are estimates are estimates and may differ slightly from those calculated using raw data. Awesome, and that's it. That's our final set of answers. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Use the frequency distribution in Exercise 4 to estimate the sample mean and sample standard deviation of the data. Do the formulas for grouped data give results that are as accurate as the individual entry formulas? Explain.

Key Concepts

Sample Mean

Sample Standard Deviation

Grouped Data vs. Individual Data

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