use the frequency polygon to identify the class with the great...

Question

use the frequency polygon to identify the class with the greatest, and the class with the least, frequency.

Line graph showing frequency of commuting distances from 1 to 10 miles, peaking at 4 miles and lowest at 10 miles.

Accepted Answer

Welcome back, everyone. In this problem, the following graph represents the weight of apples in grams. Identify the class with the greatest and the class with the least frequency. That is not zero. Here we have a frequency polygon that has the weight of apples against their frequency. And for our answer choices A says the highest class is 159.5 to 179.5 g, while the least is 199.5 to 219.5 g. B says they are 139.5 to 159.5 and 179.5 to 199.5 respectively. CE says they are 139.5 to 159.5 and 99.5 to 119.5 respectively, and the D says they are 139.5 to 159.5 and 199.5 to 219.5. Now let's go back to our frequency polygon to help us figure this out. To find the greatest frequency, for starters, we would locate the highest point or the peak of the polygon. At that point, we can figure out which X-axis value it corresponds to. Now, by looking at our graph, we can tell that the peak, OK, is at the point where our frequency is 30, OK. And at that peak, OK, if we compare it to our X-axis values, notice that it is directly above 149.5 g. OK? So let me just, let me, there's a dotted line here. This is the corresponding midpoint because on our X-axis we have the midpoints of the classes of the apple weights. Now notice here our midpoints go from 89.5 to 109.5, 109.5 to 129.5. They're increasing by 20 or they're spaced by 20 g, and in that case, that means that since they're spaced by 20, the class width is 20. So to find the interval, the class, sorry, to find our bones of our class, remember that the interval is the midpoint plus or minus the class with divided by 2. So let me put this in red for the greatest. Or for the highest frequency, OK. For the highest frequency. The lower bound OK, it is going to be equal to or midpoint 149.5 minus or class with 20 divided by 2, and that would be equal to 149 plus 0.5 minus 10, which is 139.5. On the other hand, or upper bound. Is going to be equal to 149.5 plus 20 divided by 2, and that's gonna be 149.5 plus 10, which is 159.5. Thus The class is going to be 139.5 to 159.5 g. This is the class with the highest frequency. We can apply the same logic to find the least frequency, the class with the least frequency. Now, if we go to our diagram here, remember it's non-zero, so uh one with the midpoint 89.5 wouldn't count because here is where or Y value is zero. However, right above that where Y is greater than 0. Notice that. We have a point here at 209.5. And even though we don't know exactly what valid this is, we know that it is the least frequency. And it's right above 209.5. So that means this midpoint, or this is the midpoint of the class with the least frequency. So what is the class going to be? Well, Let's write that here for the least frequency. If we apply the same logic, then to get the lower bound, all we need to do is to subtract 10. From 209.5, which is 199.5. And for the upper bound, all we'd have to do is to add a thin. That's 209.5 plus 10, which is 219.5. Therefore, The class with the least frequency is going to be 199.5 to 219.5 g. Therefore, we have our frequency, or sorry, our class with the highest frequency and our class with the least frequency, and if we compare them to our answer choices, this means that D is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

use the frequency polygon to identify the class with the greatest, and the class with the least, frequency.

Key Concepts

Frequency Polygon

Frequency

Class Interval

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