Identifying the Shape of a Distribution In Ex...

Question

Identifying the Shape of a Distribution In Exercises 53–56, construct a frequency distribution and a frequency histogram for the data set using the indicated number of classes. Describe the shape of the histogram as symmetric, uniform, negatively skewed, positively skewed, or none of these.

Heights of Males

Number of classes: 5

Data set: The heights (to the nearest inch) of 30 males

67 76 69 68 72 68 65 63 75 69

66 72 67 66 69 73 64 62 71 73

68 72 71 65 69 66 74 72 68 69

Accepted Answer

Hello. In this video, we are told that a group of 29 students reported the number of books they read last year. The data set is given as the following. We want to construct a frequency distribution and a frequency histogram using 4 classes. We then want to describe the shape of the histogram as symmetric, uniform, negatively skewed, positively skewed, or none of these. So, let's go ahead and start by setting up a frequency distribution. Now, since we are going to be working with a frequency distribution, and we are going to work with a histogram, let's go ahead and reorder the data from least to greatest. This is going to give us the following. OK, so this is going to be the data set listed from least to greatest. Now, in order to create a frequency table, the first thing we need to do is we need to define the class width. Now, class width, It is defined as the following. It is the range of the data set divided by the number of classes we are working with. Now, the number of classes we want to work with is given to us. We want to work with a total of 4 classes, but we need to define the range for the class width. Now, in order to find the range, the range is defined as the maximum value of the data set minus the minimum value of the data set. Now, the maximum value is the largest number in the data set, and that is going to be 9. The minimum value is the smallest value in the data set, and that is going to be 3. So the range is going to be defined as 9 minus 3, which is going to give us a range of 6. So now that we have the range and we have the number of classes we are working with, we can define the class width. Class width is going to be defined as 6 divided by 4. And that is going to simplify to 1.5. Now, for classes, we want to round up to the nearest whole number, and that number is going to be 2. So the class width is approximately 2. Now that we have the class width, we want to go ahead and define the classes we are working with. Now we know that for our range, the total data set ranges from 3 to 9, and we want to go ahead and split this up into classes of width 2. Specifically, we want to divide this into 4 classes. So the first class we can define is 3 to 4. Because this has a width of 2. The 2nd class can be 5 to 6, the 3rd class can be 7 to 8, and the 4th class can be 9 to 10. So, these are going to be the classes that we are working with. And for the frequency table, we want to go ahead and define or figure out how many of each of the data points fall into these classes. So for our frequency. How many numbers in the data set do we have that land between 3 and 4? Well, that is going to be the 1st 5 numbers in the data set. So in total, we have a frequency of 5 of the values that land between 3 and 4. Now, what about the total amount of values that land between 5 and 6? Well, those total values are going to be 12. There are 12 total data points that land between 5 and 6. We're gonna repeat this process for 7 and 8. These are all the values that land between 7 and 8, and there is a total of 11 values. And for 9 and 10, well, that's just going to be the maximum value at the end of the data set, and that is going to be 1. So, what we have here is the frequency table for the total amount of 4 classes of width 2. And now that we have the frequency table, let's go ahead and set up a frequency histogram. Now, in order to set up a frequency histogram, the first thing we want to do is you want to go ahead and define the mid points for the histogram. In order to define the midpoints. What we want to do is we want to take the midpoints of each of the classes. So, for example, for our first class, which is between 3 and 4, we are going to add the values of those end points in the class, which is 3 + 4, then average those values, which means we divide by 2. So 3 + 4 divided by 2 is going to give us a value of 3.5, and that's going to be one midpoint that we are going to plot on the histogram. We are going to repeat this process for each of our classes. So, for the second class, we have 5 + 6 divided by 2, which gives us 5.5. For our third class, we have 7 + 8 divided by 2, which gives us 7.5. And for our last class, the fourth one, we have 9 + 10 divided by 2, which gives us 9.5. Now what we are going to do is we are going to set up the histogram. So, we are going to first label the midpoints on the histogram, 3.5, 5.5. 7.5. And 9.5. Then we are going to set up the values, all the possible values, which are the frequencies. Now, our maximum frequency is 12 from the frequency table. So let's go ahead and set up some heights in counts of 2. 1210, 864, and 2. Now, we are going to label the amount of numbers that fall within the class from 3 to 4. That total number is 5, so our height for the first rectangle of the histogram is going to land between 4 and 6, which is 5. Now, to draw the second rectangle in the histogram, we want to connect the rectangles together. Now, how many numbers fall between 5 and 6? Well, that's going to be 12. So, we're going to have a rectangle of a high 12 on our histogram. Next, we are going to label the height for all the values that fall between 7 and 8, and that's going to be 11, so just slightly less. Than this rectangle here. And finally, we are going to label the rectangle for 9.5, which is just 1. So it is the shortest rectangle on the histogram. Now, in terms of this histogram, since this histogram is right heavy, what this means is that the histogram is positively skewed. And since the histogram is positively skewed, this is going to be the solution for the histogram and for the frequency table. We also know that based off the histogram, the histogram is positively skewed because it is right heavy, and with that being said, the answer to this problem is going to be B. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

Key Concepts

Frequency Distribution

Histogram

Shape of Distribution

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