Descriptive Statistics LESSON 11–1

Lesson Menu Five-Minute Check (over Chapter 10) Then/Now New Vocabulary Key Concept: Symmetric and Skewed Distributions Key Concept: Choosing Summary Statistics Example 1:Skewed Distribution Example 2:Bimodal Distribution Key Concept: Symmetric and Skewed Box Plots Example 3:Describe a Distribution Using a Box Plot Example 4:Compare Position Using Box Plots Example 5:Construct and Use a Percentile Graph

Over Chapter 10 5–Minute Check 1 Find the fifth term of the sequence A. B. C. D.

Over Chapter 10 5–Minute Check 2 Write an explicit formula and a recursive formula for finding the nth term of the arithmetic sequence −2, 5, 12, 19, …. A.a n = 7n – 3; a 1 = –2, a n = a n – B.a n = –2n + 7; a 1 = –2, a n = 7a n – 1 – 2 C.a n = –2 + 7n; a 1 = –2, a n = –2 + 7n D.a n = 7n – 9; a 1 = –2, a n = a n – 1 + 7

Over Chapter 10 5–Minute Check 3 A.181,398,527 B.60,455,176 C.177,146 D.118,098 Evaluate

Over Chapter 10 5–Minute Check 4 Use Pascal’s triangle to expand (x – 2y) 4. A.x 4 – 8x 3 y + 24x 2 y 2 – 32xy y 4 B.x 4 + 4x 3 y + 6x 2 y 2 + 4xy C.x 4 – 8x 3 y + 12x 2 y 2 – 8xy 3 + 2y 4 D.x 4 + 8x 3 y + 24x 2 y + 32xy 3 – 16y 4

Over Chapter 10 5–Minute Check 5 Which of the following represents −4 − 4i in exponential form? A. B. C. D.

Then/Now You found measures of central tendency and standard deviations. (Lesson 0-8) Identify the shapes of distributions in order to select more appropriate statistics. Use measures of position to compare two sets of data.

Vocabulary univariate negatively skewed distribution symmetrical distribution positively skewed distribution resistant statistic cluster bimodal distribution percentile percentile graph

Key Concept 1

Example 1 Skewed Distribution A. GOLF Twenty members of a golf team each hit a golf ball as far as they could. The table shows the distance each team member drove his or her ball. Construct a histogram and use it to describe the shape of the distribution.

Example 1 Skewed Distribution The graph shown has three equal bars. Using the TRACE feature, you can determine that these peaks represent distances from 185 to 215 yards. On a graphing calculator, press STAT EDIT and input the data into L1. Then turn on Plot1 under the STAT PLOT menu and choose. Graph the histogram by pressing ZoomStat or by pressing Graph and adjusting the window manually.

Answer: The graph is positively skewed. Most golf shots range from 185 to 215 yd, but a few shots were much longer, so the tail of the distribution trails off to the right. Example 1 Skewed Distribution

Example 1 Skewed Distribution B. GOLF Twenty members of a golf team each hit a golf ball as far as they could. The table shows the distance each team member drove his or her ball. Summarize the center and spread of the data using either the mean and standard deviation or the five-number summary. Justify your choice.

Example 1 Answer: five-number summary; The median is 205 yd, and the interquartile range is from 195 to yd. Skewed Distribution Since the distribution is skewed, use the five-number summary instead of the mean and standard deviation to summarize the center and spread of the data. To display this summary, press STAT, select 1-Var Stats under the Calc submenu, and scroll down.

Example 1 TRAVEL The distance from Reemi's lake cottage to 20 towns are given in the table. Summarize the center and spread of the data using either the mean and standard deviation or the five-number summary. Justify your choice. A.mean 62.8, standard deviation 23.35; because the graph is symmetrical B.mean 62.8, standard deviation 23.35; because the graph is skewed C.median 67, interquartile range from 50.5 to 82.5; because the graph is positively skewed D.median 67, interquartile range from 50.5 to 82.5; because the graph is negatively skewed

Example 2 Bimodal Distribution A. CELL PHONES People from two professions were asked how much they paid for their work cell phone. Construct a histogram and use it to describe the shape of the distribution.

Example 2 Bimodal Distribution The histogram of the data has not one but two major peaks. Therefore, the distribution is neither symmetrical nor skewed but bimodal. The two separate clusters suggest that two types of cell phones are mixed in the data set. It is likely that there are several relatively inexpensive cell phones and some much more expensive ones.

Example 2 Bimodal Distribution Answer: The distribution is bimodal.

Example 2 Bimodal Distribution B. CELL PHONES People from two professions were asked how much they paid for their work cell phone. Summarize the center and spread of the data using either the mean and standard deviation or the five-number summary. Justify your choice.

Example 2 Bimodal Distribution Since the distribution is bimodal, an overall summary of center and spread would give an inaccurate depiction of the data. Instead, summarize the center and spread of each cluster. Since the lower cluster of the data is positively skewed, the five-number summary shows that the prices ranged from $50 to $70, with a median of $60, and half of the prices were between $55 and $60. Since the upper cluster is symmetrical, the mean of $100 and standard deviation of $4.50 can be used to describe the center and spread, respectively.

Example 2 Bimodal Distribution

Example 2 Bimodal Distribution Answer: Sample answer: The lower cluster of the data is positively skewed, so the five- number summary shows that the prices ranged from $50 to $70, with a median of $60, and half of the prices were between $55 and $60. Since the upper cluster is symmetrical, the mean of $100 and standard deviation of $4.50 can be used to describe the center and spread, respectively.

Example 2 HOTEL PRICES Summarize the center and spread of the data using either the mean and standard deviation or the five-number summary. Justify your choice.

Example 2 A.The distribution is bimodal. Since the lower cluster is positively skewed, the five-number summary indicates that the prices range from $49 to $99, with a median of $70.5, and half are between $56 and $82. The upper cluster is symmetrical; the mean price is $121.7 with standard deviation $16.6. B.The distribution is bimodal. Since the lower cluster is symmetrical, the mean price is $71 with standard deviation $15.5. The upper cluster is positively skewed so the five- number summary indicates that the prices range from $100 to $150, with a median of $118, and half are between $108 and $135. C.The distribution is symmetrical. The mean is $93.32 with standard deviation $29.8. D.The distribution is negatively skewed. The five-number summary shows that the prices ranged from $49 to $150, with a median of $95, and half the prices between $69 and $115.

Key Concept 3

Example 3 Describe a Distribution Using a Box Plot A. HOME SALES The table shows the prices of homes recently sold by a realtor. Construct a box plot and use it to describe the shape of the distribution.

Example 3 Describe a Distribution Using a Box Plot Input the data into L1 on a graphing calculator. Then turn on Plot1 under the STAT PLOT menu and choose. Graph the box plot by pressing ZoomStat or by pressing Window and adjusting the window manually.

Example 3 Describe a Distribution Using a Box Plot Answer: The distribution is negatively skewed. Since the left whisker is longer than the right whisker and the line representing the median is closer to Q 3 than to Q 1, the distribution is negatively skewed.

Example 3 Describe a Distribution Using a Box Plot B. HOME SALES The table shows the prices of homes recently sold by a realtor. Summarize the center and spread of the data using either the mean and standard deviation or the five-number summary. Justify your choice.

Example 3 Describe a Distribution Using a Box Plot Since the distribution is skewed, use the five-number summary. This summary indicates that while the home sale prices ranged from $95,000 to $184,000, the median price was $165,000. Prices in the middle half of the data varied by $170,000 – $150,000 or $20,000, which is the interquartile range.

Example 3 Describe a Distribution Using a Box Plot Answer: Since the distribution is skewed, use the five-number summary. The median is $165,000. The interquartile range is $20,000. The range is $89,000.

Example 3 A. BOOKS The number of pages in the books that Cecelia read this year is shown in the table. Construct a box plot and use it to describe the shape of the distribution.

Example 3 A.The distribution is symmetrical. B.The distribution is bimodal. C.The distribution is negatively skewed. D.The distribution is positively skewed.

Example 4 Compare Position Using Box Plots GAMES Jessica and Miguel earned these scores on a new computer game. Construct side-by-side box plots of the data sets. Then use this display to compare the distributions.

Example 4 Compare Position Using Box Plots Input the data into L1 and L2. Then turn on Plot1 and Plot2 under the STAT PLOT menu, chose, and graph the box plots by pressing ZoomStat or by pressing Graph and adjusting the window manually.

Example 4 Answer:Miguel's median score of 84 was much higher than Jessica's median score of 35. The first quartile in Miguel's distribution of scores (81) is much higher than Jessica's third quartile of 38, making the difference substantial. Miguel's distribution is negatively skewed. Jessica's is positively skewed. Compare Position Using Box Plots Miguel's median score of 84 was much higher than Jessica's median score of 35. The first quartile in Miguel's distribution of scores (81) is much higher than Jessica's third quartile of 38, making the difference substantial. Miguel's distribution is negatively skewed. Jessica's is positively skewed.

Example 4 GROCERIES The amount the Dentons and the Sherwoods spent on groceries each week are shown in the tables. Construct side-by-side box plots of the data sets to compare the distributions.

Example 4 A.The Sherwood's median is about equal to the Denton's top expenditure. Half of the Sherwood's amounts are more than any of the Denton's. B.The Sherwood's median amount is greater than the Denton's top amount. More than half of the Sherwood's amounts are greater than all those of the Dentons. C.The Sherwood's median amount is in the top quartile of the Denton's amounts. Three-fourths of the Denton's amounts are less than all of those of the Sherwoods. D.The Sherwood's least amount is greater than the third quartile of the Denton's amounts. More than three-fourths of the Denton's amounts are less than all those of the Sherwoods.

Example 5 Construct and Use a Percentile Graph A. FOOTBALL The table gives the frequency distribution of weights of football players from 11 teams. Construct a percentile graph of the data.

Example 5 Construct and Use a Percentile Graph First, find the cumulative frequencies. Then find the cumulative percentages by expressing the cumulative frequencies as percents. The calculations for the first two players are shown.

Example 5 Construct and Use a Percentile Graph Finally, graph the data with the class boundaries along the x-axis and the cumulative percentages along the y-axis, as shown. Answer:

Example 5 Construct and Use a Percentile Graph B. FOOTBALL The table gives the frequency distribution of weights of football players from 11 teams. Estimate the percentile rank a player weighing 210 pounds would have in this distribution and interpret its meaning.

Example 5 Construct and Use a Percentile Graph Find 210 on the x-axis and draw a vertical line to the graph. This point on the graph corresponds to approximately the 79th percentile. Therefore, a football player weighing 210 pounds is heavier than about 79% of the other players included in the distribution.

Example 5 Answer:About 79th percentile; the football player is heavier than about 79% of the other players included in the distribution. Construct and Use a Percentile Graph

Example 5 MOVIES The table gives the frequency distribution of running time for 25 movies. Estimate the percentile rank of a movie that runs 110 minutes. A.55% B.50% C.35% D.30%

Descriptive Statistics LESSON 11–1