Lesson 6

The mathematical purpose of this activity is for students to find the area of a different regions under a curve. Monitor for students who divide the shaded area into a square and triangle and those who treat the entire shape as a trapezoid. Later in this lesson, students will find areas under a normal curve to approximate the proportion of values in an interval.

Tell students that the phrase “the area under the curve” means the area between the \(x\)-axis and the curve.

The purpose of this discussion is to find strategies for comparing the area of a shaded region to the total area under a curve. Select previously identified students to share how they determined the area of the shaded region. Here are some questions for discussion.

• “How did you determine the area of the shaded region?” (I see that the region is composed of a unit square and a triangle with base 1 and height 2. I know the formula for a triangle is one-half base times height so 2 square units is the area of the shaded region.)
• “How did you find the area under the curve?” (I used the vertical lines \(x = 3\), \(x = 4\), \(x = 5\), and \(x = 6\), to divide the region into triangles and rectangles, then I found the area of each of the regions and added them together.)
• “Andre said he drew the vertical line \(x = 3.5\) and found the area to the left of the line and then doubled it. Will Andre’s method work? Explain your reasoning.” (Yes, Andre’s method will work because \(x = 3.5\) is a line of symmetry.)

The mathematical purpose of this activity is for students to understand that approximately normal distributions have a similar proportion of area in regions described by the mean and standard deviations. Students are given a data set including the number of words for 200 submitted short stories. They are asked to determine the proportion of submitted stories that are within certain ranges. 

Monitor for students who:

1. return to the original data set for each question and count the values in each interval
2. create a frequency table to break the data into natural intervals
3. create a histogram to visualize the areas in question
4. use technology to count values in each region

The problem mentions that a page holds approximately 200 words to help students identify natural intervals to summarize the data into a frequency table or histogram.

Making statistical technology available gives students an opportunity to choose appropriate tools strategically (MP5).

Some students may have trouble understanding the phrase “within 1 standard deviation of the mean.” Ask students to draw a number line containing the mean value and mark on the number line 1 standard deviation below and 1 standard deviation above the mean.

The goal of this discussion is to get students thinking about histograms as an area model for a distribution. Select previously identified students to share in this order:

1. return to the original data set for each question and count the values in each interval
2. create a frequency table to break the data into natural intervals
3. create a histogram to visualize the areas in question
4. use technology to sort or count values in each interval

Connect the methods used by asking students about the advantages and disadvantages of the methods they used. (The summary methods, such as a frequency table or histogram, require additional time to set up, but are quicker to find the useful values for questions such as the ones included in the activity. Using technology, such as spreadsheets, to sort or count the data is also quick and accurate, but may take some time to enter the data if it is not already on the computer.) 

If possible, display a histogram created by the students. If none of the students created a histogram, display the one here.

Here are some questions for discussion.

• “How is this activity related to the previous one?” (In the previous activity, we figured out the area of a shaded region and divided it by the total area under the curve. In this activity, we can think of the bars as having an area.)
• “Why would a normal distribution be a good model for the data summarized in the histogram?” (The distribution is approximately symmetric and bell-shaped.)
• “Why do you think the area for a bar representing 2,000 to 2,200 words has the same proportion of the whole as the proportion of stories in the same region to the whole?” (Since the area came from \(\frac{18 \boldcdot 200}{200 \boldcdot 200}\), which is equivalent to \(\frac{18}{200}\).)
• “Was it surprising to you that more than 95% of the data was within two standard deviations of the mean or that almost 70% of the data was within one standard deviation of the mean? Explain your reasoning.” (Both were surprising to me. I had never really thought about using the standard deviation to create an interval around the mean, so I had never thought about using the standard deviation to estimate the proportion of the data that was within a specific interval. But since most of the data is near the center, I can understand why it is this way.)
• “Why do you think it could be useful to look at the proportion of values that are within one standard deviation of the mean?” (In the last lesson, we were introduced to normal curves that were based on the mean and standard deviation. If we modeled a histogram with a standard curve, it would be interesting to know what proportion of values fall within a given interval on normal curves.)

The mathematical purpose of this activity is for students to recognize that, with approximately normally distributed data, regions described completely by the mean and standard deviation represent the same proportion of the whole. For example, there should always be approximately 68% of the data within one standard deviation of the mean. This activity provides a second context for approximately normal data so that comparisons can be made and to highlight the similar proportions for regions described only in terms of the mean and standard deviation.

Tell students that they are going to do an activity very similar to the previous one involving finding proportions using a histogram.

Select students to share their comparisons.

Students may comment on the fact that the two activities had some questions that are very much alike, and their answers are very much alike. If no students mention this, point it out. For example, each task asked about the proportion of stories or websites that were within 1 standard deviation of a value that was half of a standard deviation below the mean (within 1 standard deviation of 2,400 words or within 1 standard deviation of 2.8 seconds), and they both had about 62% of the values in that region.

Display the images for the normal distributions that model each situation. Tell students that since both questions ask for the proportion of participants within one standard deviation of a value that is the same number of standard deviations from the mean, and we know that these distributions are normal, we would expect the region in each image to include approximately the same proportion of values from the data set. That is what it means to say that a normal distribution is defined only by the mean value and the standard deviation.

Here are some questions for discussion:

• “Describe what is happening in the two graphs.” (Each graph represents a normal curve fit to the data in the words or the website problem. Each of the shaded regions represents the proportion of the data that is within one standard deviation of 2,400 words and 2.8 seconds, respectively.)
• “Describe a story that includes 2,400 words in terms of the mean and standard deviation only. Then do the same for a website that loads in 2.8 seconds.” (Both values are half of a standard deviation below the mean.)
• “How would this graph change if the shaded region was within one standard deviation of the mean?” (The shaded region would be shifted to the right and it would take up about 68% or 69% of the area under the curve. It would be symmetric around the normal curve’s vertical line of symmetry.)
• “Why do you think it is helpful to model bell-shaped symmetric data using a normal curve?” (It is helpful because then you can compare data sets with different means and different standard deviations.)