1. Chapter 5: Exploring Data: Distributions Lesson Plan Exploring Data Displaying Distributions: Histograms Interpreting Histograms Displaying Distributions: Stemplots Describing Center: Mean and Median Describing the Spread: The Quartiles The Five-Number Summary and Boxplots Describing Spread: The Standard Deviation Normal Distributions The 68-95-99.7 Rule 1 Mathematical Literacy in Today’s World, 8th ed. For All Practical Purposes © 2009, W.H. Freeman and Company

2. Chapter 5: Exploring Data: Distributions Exploring Data 2 Statistics is the science of collecting, organizing, and interpreting data. Data  Numerical facts that are essential for making decisions in almost every area of life and work.  Spreadsheet programs are used to organize data by rows and columns. Exploratory data analysis 1.Examine each variable by itself and then the relationship among them. 2.Begin with a graph or graphs, then add numerical summaries of specific aspects of the data. Individual – The objects described by a set of data. May be people or may also be animals or things. Variable – Any characteristic of an individual. A variable can take different values for different individuals.

3. Chapter 5: Exploring Data: Distributions Displaying Distributions: Histograms 3 Distribution – The pattern of outcomes of a variable; it tells us what values the variable takes and how often it takes these values. Histogram  The graph of the distribution of outcomes (often divided into classes) for a single variable. Steps in Making a Histogram 1.Choose the classes by dividing the range of data into classes of equal width (individuals fit into one class). 2.Count the individuals in each class (this is the height of the bar). 3.Draw the histogram:  The horizontal axis is marked off into equal class widths.  The vertical axis contains the scale of counts (frequency of occurrences) for each class. Histogram of the percent of Hispanics among the adult residents of the states

4. Example Construct a histogram given the following data. Value Count 12 3 142 16 5 18 4 20 2 4

5. 5

6. Example Given the following 18 quiz scores (out of 30 points), construct a histogram. 12 16 13 9 28 10 22 25 29 20 24 27 28 25 24 26 19 30 6

7. 7 Examining a Distribution  Overall Pattern What does the histogram graph look like? Shape –  Single peak (either symmetric or skewed distribution)  Symmetric – The right and left sides are mirror images.  Skewed to the right – The right side extends much farther out.  Skewed to the left – the left side extends much farther out.  Irregular distribution of data may appear clustered and may not show a single peak (due to more than one individual being graphed). Center – Estimated center or midpoint of the data. Spread – The range of data outcomes (minimum to maximum).  Deviation Are there any striking differences from the pattern? Outlier – An individual value that clearly falls outside the overall pattern; possibly an error or some logical explanation. Chapter 5: Exploring Data: Distributions Interpreting Histograms

8. 8 Examples of Distribution Patterns and Deviations  Regular Single-Peak Distributions Chapter 5: Exploring Data: Distributions Interpreting Histograms Histogram of the tuition and fees charged by four-year colleges in Massachusetts Two separate distributions, graphing two individuals (state and private schools)  Histogram of the percent of Hispanics among the adult residents of the states Single Peak Skewed to Right with Outlier   Irregular Clustered Distributions Histogram of Iowa Test of Basic Skills vocabulary scores for 947 seventh-grade students Single Peak Symmetric 

9. Example Given the following data regarding exam scores, construct a histogram. Describe its overall shape and identify any outliers. Class Count Class Count 0 – 9 1 50 – 59 6 10 – 19 0 60 – 69 7 20 – 29 2 70 – 79 7 30 – 39 3 80 – 89 2 40 – 49 4 90 – 99 1 9

10. Shape appears to be skewed to the left. The score in the class 0-9 could be considered an outlier 10

11. 11 Stemplot  A display of the distribution of a variable that attaches the final digits of the observation as leaves on stems made up of all but the final digit, usually for small sets of data only. Stemplots look like histograms on the side. How to Make a Stemplot 1.Separate each observation into a stem (all but the final rightmost digit) and a leaf (the final rightmost digit). 2.Write the stems in a vertical column, smallest at top, sequentially down to the largest value. Draw a vertical line to the right of this column. 3.Write each leaf in the row to the right of its stem, in increasing order out from the stem. Chapter 5: Exploring Data: Distributions Displaying Distributions: Stemplots Stemplot of the percent of Hispanics among the adult residents of the states

12. Example Recall the 18 quiz scores (out of 30 points) stated earlier. Each score has been converted to a percentage (rounded to the nearest tenth of a percent). Construct a stemplot. 20.0% 53.3% 43.3% 30.0% 93.3% 33.3% 73.3% 83.3% 96.7% 66.7% 80.0% 90.0% 93.3% 83.3% 80.0% 86.7% 63.3% 100.0% 12

13. Solution Given the format of the converted scores, we need to further round to the nearest whole percent. The stemplot would not be meaningful with the tenth of a percent being the leaf. 20% 53% 43% 30% 93% 33% 73% 83% 97% 67% 80% 90% 93% 83% 80% 87% 63% 100% In the stemplot, the ones digit will be the leaf. 13 2 0 303 4 3 5 3 6 37 73 8 0337 937 100

14. 14 Two Most Common Ways to Describe the Center: Mean and Median  Mean “average value” Ordinary arithmetic average of a set of observations, average value. To find mean of a set of observations, add their values, and divide by the number of observations, n. x-bar, = (x 1 + x 2 + … +x n )/n  Median “middle value” The midpoint or center of an ordered list; middle value of a set of observations; half fall below the median and half fall above. Arrange observations in increasing order (smallest to largest). If the number of observations is odd, the median M is the center observation in the ordered list. If the number of observations is even, the median M is the average of the two center observations in the ordered list. Chapter 5: Exploring Data: Distributions Describing Center: Mean and Median

15. 15 Chapter 5: Exploring Data: Distributions Describing Center: Mean and Median Finding the Mean and Median Mean average value, ¯ {x-bar} Mean, ¯ = ( x 1 + x 2 + … x n ) /n The mean city mileage for the 13 cars in Table 5.2: Median middle value, M Arrange observations in order, then choose the middle value: 15 16 18 19 20 20 21 21 21 22 24 27 48 The median city mileage for the 13 cars in Table 5.2: For 13 cars (odd): (n + 1)/2 = (13 + 1) /2 = 7 The 7th observation is 21 (in red above), the median. Note: If the Toyota Prius is removed there are 12 observations (even): (n + 1)/2 = (12 + 1)/2 = 6.5 Median = Average of 6th and 7th value (20 + 21)/2 = 20.5 x x

16. Example Calculate the mean of each following data set. a) 13, 6, 8, 12, 15, 14, 26, 12, 10, 11 b) 20, 61, 3, 2, 4, 5, 10, 7, 2 16

17. A) The mean is =12.7 B) The mean is ≈12.7 17

18. Example Calculate the median of each data set. a) 13, 6, 8, 12, 15, 14, 26, 12, 10, 11 b) 20, 61, 3, 2, 4, 5, 10, 7, 2 18

19. Solution For each of the data sets, the first step is to place the data in order from smallest to largest. a) 6, 8, 10, 11, 12, 12, 13, 14, 15, 26 Since there are 10 pieces of data, the mean of the 10/2 = 5th and 6th pieces of data will be the median. Thus, the median is (12 +12) / 2 = 12. Notice, if you use the general formula (n+ 1)/ 2, you would be looking for a value (10 + 1) / 2 = 11 / 2 = 5.5 “observations” from the bottom. This would imply halfway between the actual 5th observation and the 6th observation. Notice since the 5th observation and the 6th observation were the same, we didn’t really need to calculate the median. b) 2, 2, 3, 4, 5, 7, 10, 20, 61 Since there are 9 pieces of data, the (9 + 1) / 2 = 10 / 2 = 5th piece of data, namely 5, is the median. 19

20. Example Given the following stemplot, determine the median. 11 029 12 3478 13 034679 14 012359 15 01359 16 09 17 1 18 0 20

21. Solution Since there are 28 pieces of data, the mean of the 28 / 2 =14th and 15th pieces of data will be the median. Thus, the median is (140 + 141) / 2 = 281 / 2 =140.5. Notice, if you use the general formula (n+ 1) / 2, you would be looking for the value (28 + 1) / 2 = 29 / 2 = 14.5 “observations” from the bottom (or top). 21

22. Chapter 5: Exploring Data: Distributions Describing Center: Mode Mode, most frequent value  Since 21 appears 3 times and no other mileage appears in the list of city mileages more than twice, then the mode of the data set would be 21.  If there is a tie for the most occurrences in a data set, then there may be multiple modes.  Example: For highway mileage in the table, 27, 29, 30, and 33 all appear twice and no mileage appears more times than twice. Hence there are several modes.

23. 23 Include Spread and Center to Better Describe a Distribution  Range – Measures the spread of the set of observations. Subtract the smallest observation from the largest observation  Quartiles – The center and the middle of the top and bottom halves. Calculating the Quartiles 1.Arrange the observations in increasing order and locate the median M in the ordered list of observations.  If n = even, split group in half and use all the numbers.  If n = odd, circle the median and do not use it in finding quartiles. 2.The first quartile, Q 1 is the median of the observations whose position in the ordered list is to the left of the overall median (midpoint of lower half). 3.The third quartile, Q 3 is the median of the observations whose position in the ordered list is to the right of the overall median (midpoint of upper half).  First quartile, Q 1 is larger than 25% of the observation.  Third quartile, Q 3 is larger than 75% of the observations.  Second quartile, Q 2 is the median, and larger than 50% of observations. Chapter 5: Exploring Data: Distributions Describing Spread: The Quartiles

24. Example Find the range for following data set: 12, 14, 14, 14, 16, 16, 18. Solution: 12-18 = 6. So there is a range of 6 from this data set 24

25. 25 The Five-Number Summary  A summary of a distribution that gives the median, the first and third quartiles, and the largest and smallest observations.  These five numbers offer a reasonably complete description of center and spread.  In symbols, the five-number summary is: Minimum Q 1 M Q 3 Maximum Examples Five-number summary for the fuel economies in Table 5.2: For the city mileage: 15 18.5 21 23 48 For the highway mileage: 23 27 29 32 45 Chapter 5: Exploring Data: Distributions The Five-Number Summary and Boxplots

26. 26 Boxplots  A boxplot is a graph of the five-number summary.  Boxplots are often used for side-by-side comparison of one or more distributions (they show less detail than histograms or stemplots). A box spans the quartiles, with an interior line marking the median. Lines extend out from this box to the extreme high and low observations (maximum and minimum). A box plot may be drawn vertically or horizontally. Chapter 5: Exploring Data: Distributions The Five-Number Summary and Boxplots Boxplots of the highway and city gas mileages for cars classified as midsized by the Environmental Protection Agency.

27. Example: Draw a boxplot for the following data set. 21, 25, 10, 40, 43, 19, 12 27

28. Solution The first step is to place the data in order from smallest to largest: 10, 12, 19, 21, 25, 40, 43 Since there are 7 pieces of data, the median isthe (7 +1) / 2 = 8 / 2 = 4 th piece of data, namely 21. There are 3 pieces of data below the median, M. Thus, the (3 +1) / 2 = 4 / 2 = 2 nd piece of data is the first quartile. Thus, 1 Q = 12. Now since there are 3 pieces of data above M, 3 Q will be the 2nd piece of data to the right of M. Thus, 3 Q = 40. The smallest piece of data is 10 and the largest is 43. Thus, the five-number summary is 10, 12, 21, 40, 43. 28

29. So the box plot is: Max = 43 3Q = 40 2Q (median)= 21 1Q = 12 Min = 10 29

30. Example Draw a boxplot for the following data set: 31, 16, 11, 18, 10, 9, 12, 15, 15, 17, 20, 25 30

31. Solution The first step is to place the data in order from smallest to largest. 9, 10, 12, 11, 15, 15, 16, 17, 18, 20, 25, 31 Since there are 12 pieces of data, the median is between the 6th and 7th pieces of data. 9, 10, 11, 12, 15, 15, (median) 16, 17, 18, 20, 25, 31 Thus the median, M, is (15 +16) / 2 = 31 / 2 = 15.5 There are 6 pieces of data below M. Since (6 +1) / 2 = 7 / 2 = 3.5. Q1 will be the mean of 3rd and 4th pieces of data, namely (11 +12) / 2 = 23 / 2 = 11.5. Now since there are 6 pieces of data above M, 3 Q will be the mean of the 3rd and 4th pieces of data to the right of M. Thus, (18 + 20) / 2 = 38 / 2 = 19. 9, 10, 11, (1Q = 11.5) 12, 15, 15, (Median = 15.5), 16, 17, 18, (3Q = 19), 20, 25, 31 The smallest piece of data is 9, and the largest is 31. Thus, the five-number summary is: 9, 11.5, 15.5,19, 31. 31

32. The boxplot is as follows: 32

33. 33 Standard Deviation s  “Standard” or average amount that the observed data values deviate from the mean  Calculated by taking the square root of the mean of the squared deviations except dividing by n-1 instead of n  The standard deviation of n observations Chapter 5: Exploring Data: Distributions Describing Spread: The Standard Deviation

34. Standard deviation example  7 purchase prices for Radiohead “In Rainbows” download: 3 4 5 7 10 12 15 (in dollars) The mean is The standard deviation is

35. Chapter 5: Exploring Data: Distributions Describing Spread: The Standard Deviation Properties of the standard deviation s:  s measures spread about the mean  s=0 only when there is no spread, otherwise s>0  s has the same units of measurement as the original observations  s is sensitive to extreme observations or outliers Choosing a Summary The five-number summary is usually better than the mean and standard deviation for describing a skewed distribution or a distribution with outliers. Use the mean and standard deviation only for reasonably symmetric distributions with no outliers. Many calculators and computer programs can easily calculate the standard deviation.

36. 36 Normal Distributions  When the overall pattern of a large number of observations is so regular, we can describe it as a smooth curve.  A family of distributions that describe how often a variable takes its values by areas under a curve. Chapter 5: Exploring Data: Distributions Normal Distributions  Normal curves are symmetric and bell-shaped, smoothed-out histograms.  The total area under the Normal curve is exactly 1 (specific areas under the curve actually are proportions of the observations). Histogram of the vocabulary scores of all seventh-grade students. The smooth curve shows the overall shape of the distribution.

37. 37 Calculating Quartiles  The first quartile of any Normal distribution is located 0.67 standard deviation below the mean. Q 1 = Mean − (0.67)(Stand. dev.)  The third quartile is 0.67 standard deviation above the mean. Q 3 = Mean + (0.67)(Stand. dev.) Chapter 5: Exploring Data: Distributions Normal Distributions Standard Deviation of a Normal Curve  The shape of a Normal distribution is completely described by two numbers, the mean and its standard deviation. The mean is at the center of symmetry of the Normal curve. The standard deviation is the distance from the center to the change-of-curvature points on either side. Example: Mean = 64.5, Stand. dev.= 2.5 Q 3 = 64.5 + 0.67(2.5) = 64.5 + 1.7 = 66.2

38. Example The scores on a marketing exam were normally distributed with a mean of 68 and a standard deviation of 4.5. a) Find the first and third quartile for the exam scores. b) Find a range containing exactly 50% of the students’ scores. 38

39. Solution a) The quartiles are: μ + 0.67σ = 68 + 0.67(4.5) = 68 + 3 = 71 μ - 0.67σ = 68 - 0.67(4.5) = 68 - 3 = 65 1 Q = 65 and 3 Q = 71. b) Since 25% of the data lie below the first quartile and 25% of the data fall above the third quartile, 50% of the data would fall between the first and third quartiles. We would say an interval would be [65, 71]. So 50% of the students got between a 65 and a 71 on the marketing exam. 39

40. 40 Chapter 5: Exploring Data: Distributions The 68-95-99.7 Rule SAT scores have Normal distribution Normal Distributions 68-95-99.7 Rule  68% of the observations fall within 1 standard deviation of the mean.  95% of the observations fall within 2 standard deviations of the mean.  99.7% of the observations fall within 3 standard deviations of the mean. Example  SAT scores are close to a Normal distribution, with a mean = 500 and a standard deviation = 100. What percent of scores are above 700? Answer: Score of 700 is +2 stand. dev. Since 95% of data is between +2 and −2 stand. dev., then above 700 is in top 2.5%.

41. Example: The scores on a marketing exam were normally distributed with a mean of 71.3 and a standard deviation of 5.5. a) Almost all (99.7%) scores fall within what range? b) What percent of scores are more than 82? c) What percent of scores fall in the interval [66, 82]? 41

42. 42 a) Since 99.7% of all scores fall within 3 standard deviations of the mean, we find the following. μ ± 3σ = 71.3± 3(5.5) = 71.3±16.5 71.3−16.5 = 54.8 and 71.3+16.5 = 87.8 Thus, the range of scores is 54.8 to 87.8. If scores on the exam are understood To be whole numbers, then the range of Scores would be the interval [55, 87]. b) Scores above 82, such as 83 or more are two σ above μ; 95% are within 2σ of μ. 5% lie farther than 2σ. Thus, half of these, or 2.5%, lie above 82. c) 34% (half of 68%) of the scores would be between 66 and the mean. 47.5% (half of 95%) of the scores would be between the mean and 82. Thus, 34% + 47.5% = 81.5% of scores fall in the interval [66, 82].