In Exercises 5 and 6, use the data set, which represents the n...

Question

In Exercises 5 and 6, use the data set, which represents the number of rooms reserved during one night’s business at a sample of hotels.

153 104 118 166 89 104 100 79 93 96 116

94 140 84 81 96 108 111 87 126 101 111

122 108 126 93 108 87 103 95 129 93 124

Construct a frequency distribution for the data set with six classes and draw a frequency polygon.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says the numbers below represent daily customer accounts from a sample of coffee shops, and we're given the data set of 92, 75, 81, 97, 102, 85, 94, 76, 88, 90, 100, 83, 105, 95, 89, 110, 7887. 91 99 84101 106 93 82 86 97104 92 88 80 107 and 96. struct a frequency distribution for the data set using 6 classes and draw a frequency polygon for the frequency distribution. And below the text we're given an empty graph on which to create our frequency polygon. So, the first thing we need to do is to construct our frequency distribution using 6 classes. And so the first thing we'll actually need to really do is determine the range of our data set, since we'll need to use that in order to determine our class intervals. So, we call that our range. is equal to our maximum value minus our minimum value. So if you look at our data set, we see that our maximum value is going to be 110, and our minimum value is going to be 75. So range is going to be equal to 110 minus 75. Which one we evaluate that is going to be equal to 35. So now that we have the range, we can use it to calculate our class width. So this is going to be the size of each of our class intervals. So we call that our class width is going to be equal to our range, which is 35. Divided by the number of classes, which in this case, we're told the problem that we'll need 6 classes, so we'll have 35 divided by 6. And evaluating this, this is going to be equal to 5.83, and we need to round this up to the next whole number. So that means our class width is going to be equal to 6. So, we'll now use this class width, you calculate our class intervals. And this is actually best done in a table, since we'll need to calculate other quantities. As well. So, if we look at our table here, In the first column, we have our class interval. And the way we calculate our class intervals is we start with our minimum value, which in this case is 75, and then we make each one of our intervals have 6 values. So in our first class interval, we have 75 to 80, since we'll have the values of 75, 76, 77, 78, 79, and 80. So those are 6 different values. And we'll have to repeat this for the next several class intervals until we've covered all of our six intervals. So, for the next interval, we have 81 to 86. The next interval is 87 to 92. The next interval is 93 to 98, the next interval is 99 to 104, and the next interval is 105 to 110. Now, in order to to find our um calculate and draw our frequency polygon, we'll need the class midpoints. And so, to calculate the midpoint, we just take the upper end of our interval, add to at the lower end of our interval, and then divide by 2. So, the midpoints for each of our class intervals are going to be 77.5, 83.5, 89.5, 95.5, 101.5, and 107.5. And then we need to find the frequency for each of our class intervals. So this means looking through our data and seeing how many times. A quantity, in this case, the first class interval from 75 to 80, shows up in our data set. And so, our frequencies for the 75 to 80 class interval is going to be 4, or 81 to 86 class interval, we have a frequency of 6. Or the class interval of 87 to 92. We have a frequency of 8. For the class interval of 93 to 98, we have a frequency of 6. For the class interval of 99 to 104, we have a frequency of 5, and then for the class interval from 105 to 110, we have a frequency of 4. So this here is the frequency distribution that the first part of the problem was asking for. So now we need to use this information to actually draw our frequency polygon. And so we're going to Come up to a graph, and we're actually going to plot the frequency. On our graph. So, on the x-axis, we have our class midpoints, and on the Y axis, we have our frequency. So, for the class interval, going from 75 to 80, we're actually going to use the class midpoint. So, at 77.5, we're going to have a frequency of 4. We'll place a point there on the graph. For the next class interval from 81 to 86, we have a class midpoint of 83.5, and this has a frequency of 6, so we'll place. A dot at 6 With the class midpoint of 83.5. The next class interval is the 872 92, which has a class midpoint of 89.5, and here we have a frequency of 8, so we'll plot that point on our graph. For the next class interval, we have 93 to 98, which has a class midpoint of 95.5, and here we have a frequency of 6, so we'll place a frequency of 6 at the 95.5 mark on our graph. The next class interval is the 99 to 104, which has a class midpoint of 101.5, and here we have a frequency of 5. So here we replace another dot, a frequency of 5. When we have 101.5 on our X axis. And then our final class interval is going to be 105 to 110. And here we have a class midpoint of 107.5, and here we have a frequency of 4. So here we will place a point. At a frequency of 4 above our 107.5 midpoint mark on the X-axis. And so now what we need to do is to connect these points with a straight line, and we're actually going to be starting from a frequency of zero at 71.5, which would be the midpoint of the class interval before the 75 to 80 class interval. So we'll start there at 0. And go up to our first point at 77.5. Then we'll go from there up to the next point. At 83.5, and then we'll go up again with a straight line, connecting to the point at 89.5, which gives us a frequency of 8, so that's gonna be our maximum point. Then from that point, we're going to come down. To a frequency of 6 at 95.5. And then we will connect. From that point to our next point on the graph, which is at a frequency of 5, at 101.5, and then finally, a frequency of 4. At 107.5. Then we will actually. Come back down to the zero frequency line, back down to our X-axis, from that point of a frequency of 4 and an X value of 107.5, so we'll go down to 0 and And with added X value of 113.5. So we just basically closed off our polygon towards the X axis. And so the graph that we have on the screen here is the frequency polygon that the problem was asking for. So, I hope this explanation has been useful, and I'll see you in the next video.

Key Concepts

Frequency Distribution

Classes and Class Width

Frequency Polygon

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