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Unit 8, Lesson 6

Areas in Histograms

Algebra 2

6.1

Warm-up

Standards Alignment

Building On

Addressing

Building Toward

Instructional Routines

None

Materials

None

Activity Narrative

The mathematical purpose of this activity is for students to find the area of 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.

Launch

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

Activity

None

Find the shaded area between the function, the -axis, and the boundaries and . Explain or show your reasoning.
What proportion of the area between the function, the -axis, and the boundaries and is shaded? Explain or show your reasoning.

<p>A bell curve, made up of smaller line segments.</p>

Student Response

to access Student Response.

Building on Student Thinking

Activity Synthesis

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 , , , and 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 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 is a line of symmetry.)

6.2

Activity

Instructional Routines

5 Practices Analyze It

Materials

To Copy (from Blackline Masters)

Story Submission Handout

Activity Narrative

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. Students are asked to determine the proportion of submitted stories that are within certain ranges.

Monitor for students who use these methods to find the area. The strategies are listed in order from most straightforward to more use of tools:

Return to the original data set for each question and count the values in each interval.
Create a frequency table to break the data into natural intervals.
Create a histogram to visualize the areas in question.
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.

In the digital version of the activity, students use a spreadsheet to find the proportions in the activity. The spreadsheet allows students to count the number of values in an interval more quickly. Use the digital version if students could benefit from extra time to understand the proportions rather than to count values from the data set.

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

6.3

Activity

Instructional Routines

None

Materials

None

Activity Narrative

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.

Launch

Keep students in groups of 2.

Activity

None

Lesson Synthesis

Tell students, “A biologist measures the stride lengths of a population of emus, the second-tallest birds in the world, and the stride lengths of a population of ostriches, the tallest birds in the world. The biologist found that the stride lengths of both populations are approximately normally distributed.” Display the information for all to see.

The mean stride length of the population of emus is 3 meters with a standard deviation of 0.5 meter.
The mean stride length of the population of ostriches is 4.5 meters with a standard deviation of 0.75 meter.

Here are some questions for discussion:

“Approximately 34% of the ostriches have stride lengths between 4.5 and 5.25 meters. Describe these values in terms of the mean and standard deviation only. What interval would represent a similar percentage of emus?” (4.5 is the mean, and 5.25 is one standard deviation above the mean. For emus, a similar percentage can be found between 3 and 3.5 meters since it is between the mean and one standard deviation above the mean.)
“How can this percentage be seen using a graph of the normal curve?” (There is a vertical line down the middle, then another vertical line to the right. The area between the two lines is shaded.)

“Approximately 95% of the emus have a stride length between 2 and 4 meters. What interval for the ostriches has a similar percentage?" (3 to 6 meters, since these values are within 2 standard deviations of the mean)
“How can this percentage be seen using a graph of the normal curve?” (The area under the normal curve is shaded two standard deviations to the left and right of the mean. The shading is symmetric around the center of the curve.)

to access Cool-down.

Student Lesson Summary

There is an important connection between areas in histograms and the data represented by the histogram. In particular, the proportion of the total area in the histogram that is represented by a single bar in the histogram is equivalent to the proportion of all the data that is included in that interval. This is made more interesting by the fact that, for normally distributed data, the proportion of values in an interval whose endpoints are described by the mean and standard deviation is always the same.

For example, a woodshop produces boards of various lengths. During a certain week, 5,000 boards are produced and measured. The mean length is 6 feet, and the standard deviation length is 1 foot. The table and histogram show a summary of the board lengths.

board length	3.5–4	4–4.5	4.5–5	5–5.5	5.5–6	6–6.5	6.5–7	7–7.5	7.5–8	8–8.5
frequency	113	220	460	747	960	955	753	460	220	112

The total area of all the rectangles in the histogram is 2,500 since we could stack all the bars on top of one another and have a rectangle that is 5,000 tall and 0.5 wide. If we look at just the rectangles representing boards between 5.5 and 6 feet wide, the area is 480, which is 19.2% of the total area, since . Similarly, we can see from the data that 19.2% of the data is in this same interval since . It is not a coincidence that these values are the same! The proportion of the total area that is in one of the rectangles is always equivalent to the proportion of all the data values that are in the same interval.

When the data is normally distributed, the proportions of certain regions are always the same. For example, there is always about 68% of the data within 1 standard deviation of the mean. Since the boards produced by the woodshop are approximately normal, we can test this information.

The boards within one standard deviation of the mean are between 5 and 7 feet long. Using the table, we can see that 3,415 boards are in this range () and those represent 68.3% () of the boards produced in the woodshop.

Let’s say that, another week, the woodshop produces 5,000 boards again, but this time, the mean is 6.5 feet and the standard deviation is 0.75 foot. As long as the board lengths continue to be approximately normal, we can expect about 68% of the boards to be within 1 standard deviation of the mean. For that week, it means that about 68% of the boards are between 5.75 and 7.25 feet long.

In fact, as long as the interval can be described using only the mean and standard deviation, and the data is normally distributed, the proportion of data values in the interval can be found. In general, about 68% of the data is within 1 standard deviation of the mean, about 95% of the data is within 2 standard deviations of the mean, and more than 99% of the data is within 3 standard deviations of the mean.

Activity Synthesis

The goal of this discussion is to get students thinking about histograms as an area model for a distribution.

Invite previously selected groups to share how they computed the area. Sequence the discussion of the strategies in the order listed in the Activity Narrative. If possible, record and display the students’ work for all to see.

Connect the different responses to the learning goals by asking questions, such as:

“What are the advantages and disadvantages of each strategy?”
“Which strategy is most efficient with this many values in the data set?”
“How could you check whether your answer is reasonable using each strategy?”

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 Warm-up?” (In the Warm-up, 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?” (The are the same since the area came from , which is equivalent to .)
“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?” (It can be useful to know that distributions that can be modeled with a normal distribution have certain percentages of the data clustered near the center of the distribution.)

A company collects data from 10,000 websites about how long it takes to load the site. The number of seconds it takes to fully load the website is summarized in the relative frequency table.

seconds to load	relative frequency
1.4–1.6	0.0003
1.6–1.8	0.0012
1.8–2.0	0.0053
2.0–2.2	0.0181
2.2–2.4	0.0442
2.4–2.6	0.0910
2.6–2.8	0.1555
2.8–3.0	0.1861
3.0–3.2	0.1938
3.2–3.4	0.1447
3.4–3.6	0.0923
3.6–3.8	0.0447
3.8–4.0	0.0166
4.0–4.2	0.0048
4.2–4.4	0.0012
4.4–4.6	0.0002

The relative frequency histogram summarizes the same data.

The mean time to load a website is 3 seconds, and the standard deviation is 0.4 second.

Would a normal distribution be a good model for this distribution? Explain your reasoning.
What proportion of websites loaded within 1 standard deviation of the mean?
What proportion of websites loaded within 2 standard deviations of the mean?
What proportion of websites loaded within 1 standard deviation of 2.8 seconds?
Compare the proportion of websites within 1 standard deviation of the mean to the proportion of stories in the submissions that are within 1 standard deviation of the mean number of words from the previous task. Do the same for the proportion within 2 standard deviations.

Activity Synthesis

Invite 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 68% 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 approximately 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.

<p>Area under a curve of website load times.</p>

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 problem. The shaded regions represent 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 we can compare data sets with different means and different standard deviations.)

Engagement: Provide Access by Recruiting Interest. Revisit math community norms to prepare students for the whole-class discussion.
Supports accessibility for: Social-Emotional Functioning, Attention

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Lesson 6

Areas in Histograms

Warm-up

Standards Alignment

Building On

Addressing

HSG-GPE.B.7

Building Toward

HSS-ID.A.4

Instructional Routines

Materials

Activity Narrative

Launch

Activity

Student Response

Building on Student Thinking

Activity Synthesis

Activity

Instructional Routines

Materials

To Copy (from Blackline Masters)

Activity Narrative

Activity

Instructional Routines

Materials

Activity Narrative

Launch

Activity

Lesson Synthesis

Student Lesson Summary

Launch

Activity

Student Response

Activity Synthesis

Student Response

Building on Student Thinking

Activity Synthesis

Standards Alignment

Building On

Addressing

HSS-ID.A.1

Building Toward

HSS-ID.A.4

Standards Alignment

Building On

Addressing

HSS-ID.A.1

Building Toward

HSS-ID.A.4

Building on Student Thinking