1. Chapter 6 1

2. Chebychev’s Theorem The portion of any data set lying within k standard deviations (k > 1) of the mean is at least: 2 k = 2: In any data set, at least of the data lie within 2 standard deviations of the mean. k = 3: In any data set, at least of the data lie within 3 standard deviations of the mean.

3. Example: Using Chebychev’s Theorem The age distribution for Florida is shown in the histogram. Apply Chebychev’s Theorem to the data using k = 2. What can you conclude? 3

4. Solution: Using Chebychev’s Theorem k = 2: μ – 2σ = 39.2 – 2(24.8) = -10.4 (use 0 since age can’t be negative) μ + 2σ = 39.2 + 2(24.8) = 88.8 4 At least 75% of the population of Florida is between 0 and 88.8 years old.

5. Standard Deviation for Grouped Data Sample standard deviation for a frequency distribution When a frequency distribution has classes, estimate the sample mean and standard deviation by using the midpoint of each class. 5 where n= Σf (the number of entries in the data set)

6. Example: Finding the Standard Deviation for Grouped Data 6 You collect a random sample of the number of children per household in a region. Find the sample mean and the sample standard deviation of the data set. Number of Children in 50 Households 13111 12210 11000 15036 30311 11601 36612 23011 41122 03024

7. xfxf 0100(10) = 0 1191(19) = 19 272(7) = 14 373(7) =21 424(2) = 8 515(1) = 5 646(4) = 24 Solution: Finding the Standard Deviation for Grouped Data First construct a frequency distribution. Find the mean of the frequency distribution. 7 Σf = 50 Σ(xf )= 91 The sample mean is about 1.8 children.

8. Solution: Finding the Standard Deviation for Grouped Data Determine the sum of squares. 8 xf 0100 – 1.8 = –1.8(–1.8) 2 = 3.243.24(10) = 32.40 1191 – 1.8 = –0.8(–0.8) 2 = 0.640.64(19) = 12.16 272 – 1.8 = 0.2(0.2) 2 = 0.040.04(7) = 0.28 373 – 1.8 = 1.2(1.2) 2 = 1.441.44(7) = 10.08 424 – 1.8 = 2.2(2.2) 2 = 4.844.84(2) = 9.68 515 – 1.8 = 3.2(3.2) 2 = 10.2410.24(1) = 10.24 646 – 1.8 = 4.2(4.2) 2 = 17.6417.64(4) = 70.56

9. Solution: Finding the Standard Deviation for Grouped Data Find the sample standard deviation. 9 The standard deviation is about 1.7 children.

10. Quartiles Fractiles are numbers that partition (divide) an ordered data set into equal parts. Quartiles approximately divide an ordered data set into four equal parts.  First quartile, Q 1 : About one quarter of the data fall on or below Q 1.  Second quartile, Q 2 : About one half of the data fall on or below Q 2 (median).  Third quartile, Q 3 : About three quarters of the data fall on or below Q 3. 10

11. Example: Finding Quartiles The test scores of 15 employees enrolled in a CPR training course are listed. Find the first, second, and third quartiles of the test scores. 13 9 18 15 14 21 7 10 11 20 5 18 37 16 17 11 Solution: Q 2 divides the data set into two halves. 5 7 9 10 11 13 14 15 16 17 18 18 20 21 37 Q2Q2 Lower half Upper half

12. Solution: Finding Quartiles The first and third quartiles are the medians of the lower and upper halves of the data set. 5 7 9 10 11 13 14 15 16 17 18 18 20 21 37 12 Q2Q2 Lower half Upper half Q1Q1 Q3Q3 About one fourth of the employees scored 10 or less, about one half scored 15 or less; and about three fourths scored 18 or less.

13. Interquartile Range Interquartile Range (IQR) The difference between the third and first quartiles. IQR = Q 3 – Q 1 13

14. Example: Finding the Interquartile Range Find the interquartile range of the test scores. Recall Q 1 = 10, Q 2 = 15, and Q 3 = 18 14 Solution: IQR = Q 3 – Q 1 = 18 – 10 = 8 The test scores in the middle portion of the data set vary by at most 8 points.

15. Box-and-Whisker Plot Box-and-whisker plot Exploratory data analysis tool. Highlights important features of a data set. Requires (five-number summary):  Minimum entry  First quartile Q 1  Median Q 2  Third quartile Q 3  Maximum entry 15

16. Drawing a Box-and-Whisker Plot 1.Find the five-number summary of the data set. 2.Construct a horizontal scale that spans the range of the data. 3.Plot the five numbers above the horizontal scale. 4.Draw a box above the horizontal scale from Q 1 to Q 3 and draw a vertical line in the box at Q 2. 5.Draw whiskers from the box to the minimum and maximum entries. 16 Whisker Maximum entry Minimum entry Box Median, Q 2 Q3Q3 Q1Q1

17. Example: Drawing a Box-and-Whisker Plot Draw a box-and-whisker plot that represents the 15 test scores. Recall Min = 5 Q 1 = 10 Q 2 = 15 Q 3 = 18 Max = 37 17 510151837 Solution: About half the scores are between 10 and 18. By looking at the length of the right whisker, you can conclude 37 is a possible outlier.

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19. The Standard Score Standard Score (z-score) Represents the number of standard deviations a given value x falls from the mean μ. 19

20. Example: Comparing z-Scores from Different Data Sets In 2007, Forest Whitaker won the Best Actor Oscar at age 45 for his role in the movie The Last King of Scotland. Helen Mirren won the Best Actress Oscar at age 61 for her role in The Queen. The mean age of all best actor winners is 43.7, with a standard deviation of 8.8. The mean age of all best actress winners is 36, with a standard deviation of 11.5. Find the z-score that corresponds to the age for each actor or actress. Then compare your results. 20

21. Solution: Comparing z-Scores from Different Data Sets 21 Forest Whitaker Helen Mirren 0.15 standard deviations above the mean 2.17 standard deviations above the mean

22. Solution: Comparing z-Scores from Different Data Sets 22 The z-score corresponding to the age of Helen Mirren is more than two standard deviations from the mean, so it is considered unusual. Compared to other Best Actress winners, she is relatively older, whereas the age of Forest Whitaker is only slightly higher than the average age of other Best Actor winners. z = 0.15z = 2.17

23. Chapter 6 Summary Determined the quartiles of a data set Determined the interquartile range of a data set Created a box-and-whisker plot Interpreted other fractiles such as percentiles Determined and interpreted the standard score (z-score) 23