In Exercises 5–8, identify the class width, class midpoints, a...

Question

In Exercises 5–8, identify the class width, class midpoints, and class boundaries for the given frequency distribution. Also identify the number of individuals included in the summary. The frequency distributions are based on real data from Appendix B.
8.

8. Table displaying blood platelet count ranges for females alongside their corresponding frequencies.

Accepted Answer

All right, hello, everyone. So this question says, the table below summarizes the time and minutes students spend on social media daily based on a survey. Determine the class width, class midpoints, and class boundaries for the given frequency distribution. Also, find the total number of students included in the summary. So here, in addition to the data that was provided, I went ahead and pre-made a table here with the information that we need to calculate, as well as the classes themselves from the given information. So we have the time spent in minutes, as well as the class width, the midpoints and the class boundaries. So first, let's begin with the class width. Now, the class width, if you recall, is found by subtracting the lower boundary of one class from the lower boundary of the next one. So looking at the first two classes, for example, the lower limits of these two are 0 and 30 respectively. So, 30 subtracted by 0 gives you the class width of 30. And that just so happens to be the case for all of the classes. So I'm just going to go ahead and write that out. And now for the midpoints. Now, to calculate a given class's midpoint, what you do is find the sum of the lower and upper boundaries, and then divide that sum by 2. So focusing on the first class, for example, ranging from 0 to 29. The sum of 29 and 0 gets subtracted by 2. And that equals the class midpoint of 14.5. So then you would repeat the steps for the remaining classes. For the second class, the midpoint would be 44.5. For the third that would be 74.5. For the 5th, or excuse me, for the 4th, that would be 104.5. And for the 5th, that would be 134.5. So here, for the last part of the table, we have the class boundaries. Now, recall the class boundaries and ensure that there are no gaps in between consecutive intervals. So here, For the lower boundary of each class, you take the lower limit and then subtract 0.5. For example, starting again with R. First class here, from 0 to 29, subtracting 0.5 from zero gives you the upper, excuse me, the lower class boundary of negative 0.5. And then for the upper class boundary, you take the upper limit and add 0.5. So in this case, that would be. 29.5, using the first class here as an example. And then you repeat those steps for the remaining classes. So for the second one The second class, rather, that would be from 29.5. 259.5. And then for the third class that would be 59.5. Up to 89.5. For the 4th, that would be 89.5. Up to 119.5. And then for the fifth class, that would be 119.5 up to 149.5. And now this leaves you with the total number of students. Now, because this is a frequency distribution, and to find the total number of students, you would add the frequencies of each class together. So in this case, you're adding together 1532, 48, 20, and 10. So in this case, the total number. Of students is equal to 125. And there you have it. So with that being said, thank you very much for watching and I hope you found this helpful.

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