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AP Statistics 2.1 Measures of Relative Standing Objectives: Explain what is meant by a Standardized Value. Compute the z-­‐score of an observation given the mean and standard deviation of a distribution. Compute the pth percentile of an observation. Define Chebyshev’s inequality, and give an example of its use. Intro Example: Here are the scores of all 25 students in Mr. Pryor’s Stat class on their chapter 1 test: 79, 81, 80, 77, 73, 83, 74, 93, 78, 80, 75, 67, 73, 77, 83, 86, 90, 79, 85, 83, 89, 84, 82, 77, 72 Jenny scored an 86, how did she do relative to her classmates? Where does Jenny fall relative to the center of this distribution? (Compute 5 number summary) N = 25 Mean = 80 Median = 80 StDev = 6.07 Min = 67 Max = 93 Q1 = 76 Q3 = 83.5 Since the mean and median are both 80, we can say that Jenny’s result is “above average”, but HOW MUCH “above average” is it???? One way to describe Jenny’s position within the distribution of test scores is to tell how many standard deviations above or below the mean her score is. Since the mean is 80 and standard deviation is about 6 Jenny’s score of 86 is about one standard deviation above the mean. Converting scores like this from original values to standard deviation units is known as standardizing. Measuring Relative Standing: z-­‐Scores: If x is an observation that has known mean and standard deviation, the standardized value or z-­‐
score of x is: !!!"#$
z=
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KEY CONCEPT: A z-­‐Score is directional!! That means, the x tells you how many standard deviations the score is from the mean. The SIGN of x tells you whether it is great than or less than the mean. The day after receiving her statistics test result of 86 from Mr. Pryor, Jenny earned an 82 on Mr. Goldstone’s Chem test. Mr. G told the class that the distribution of scores was fairly symmetric with a mean of 76 and a standard deviation of 4. On which test did Jenny do better relative to the class? Stat Chem z = !!!"
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= !"!!"
!.!"
= 0.99 z = !!!"
!
= !"!!"
!
= 1.5 Her 82 in chemistry was 1.5 standard deviations above the mean score for the class. Since she only scored 1 standard deviation about the mean on the stat test, Jenny actually did better in chem relative to the class. Percentiles: The number of observations at or below a given class (less than or equal to). ! Sometimes the pth percentile of a distribution is defined as the value with p percent strictly BELOW the observation. In this case, it is impossible to be in the 100th percentile. We can also describe Jenny’s performance on her first stat test using percentiles. Using the stemplot from page 116, we can see the Jenny’s 86 is the 22nd highest score on this test. Since 22 of the 25 observations (88%) are at or below her score; Jenny scored at the 88th percentile. (See pg.119-­‐ Example 2.3 for another example) Chebyshev’s Inequality: (This concept is not on the AP Syllabus, but you are free to use this appropriately on the open-­‐ended section of the exam) In ANY distribution (regardless of shape), the percent of observations falling within k standard deviations of the mean is AT LEAST: !
(100) (1 − ! ! ) Example: If k = 2 (100) (1 −
Read: k à at least (100) (1 −
!
!!
1
22
) = 75% à 75% of the data falls within 2 S.D of the mean. ) of observations within k standard deviations of mean. Example 2.5 (pg 121) FOR THIS HISTOGRAM! Begin at 2.25 Class intervals of .5 10 classes 2.25 to < 2.75 2.75 to < 3.25 3.25 to < 3.75 3.75 to < 4.25 4.25 to < 4.75 4.75 to < 5.25 5.25 to < 5.75 5.75 to < 6.25 6.25 to < 6.75 6.75 to < 7.25 b) The average unemployment rate is x-­‐bar = 4.896% and the standard deviation of the rates is s = 0.976%. The five-­‐number summary is: 2.7%, 4.1%, 4.8%, 5.5%, and 7.1%. The distribution is relatively symmetric with a center at 4.896%, a range of 4.4%, and no gaps or outliers. c) Since Illinois is the 42 number in the data set, you do 42/50 which equals 84%. This means Illinois is in the 84th percentile. 84% of the 50 states have unemployment rates at or below the unemployment rate of 84. Illinois has one of the highest unemployment rates in the country. d) Minesota’s unemployment rate is 4.3% (50 * 0.3) = 15 " place # in data. Z-­‐score is: z = (x – x-­‐bar)/ s z = (4.3 – 4.896)/ 0.976 = -­‐0.61 e) Example 2.16 (pg 131) a) Histogram b) Min = 316,000 Q1 = 800,000 Med =2,875,000 Q3= 7,000,000 Max = 22,000,000 x-­‐bar = 4410897 s = 4837406 S ! Skewed right O ! IQR = 6,200,000 x (1.5) = 9,300,000 so 7,000,000 + 9,300,000 = 16,300,000 so the maximum is an outlier. C ! is 2,875,000 S ! IQR is 6,200,000, Range = 21,684,000 c) z = (555,000 – 4,410,897)/ 4,837,406 = -­‐.08, Percentile = 4/28 = 14th percentile. HW: pg. 118; 2.1 – 2.4 pg. 121; 2.7, 2.8