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Lecture 8 Andrew L. Zaeske Review 6A: Characterizing Data Lecture 8 6A, 6B and 6C 6B: Measures of Variation 6C: The Normal Distribution Andrew L. Zaeske Marian University 2/25/15 Test Review Lecture 8 Andrew L. Zaeske Review 6A: Characterizing Data 6B: Measures of Variation 6C: The Normal Distribution Distribution and Measures of Central Tendency Lecture 8 Andrew L. Zaeske Review 6A: Characterizing Data 6B: Measures of Variation 6C: The Normal Distribution Distribution of data : the way it is spread out Question: Where is the data centered? Mean (average) : Sum of all values divided by number of values Median : Middle value of ordered data (average if there are an even number) Mode : Most frequent value Shapes of Distributions Lecture 8 Andrew L. Zaeske Review 6A: Characterizing Data 6B: Measures of Variation 6C: The Normal Distribution Outlier : Number(s) that is far away from others Outliers can greatly change mean value, but not median or standard deviation Symmetry : mirror image around mean/median/mode Skewness : clustered to one side (left skew: spread out on left side, right skew: spread out on right side) Variation : how wide apart are values Simplest Measures: Range and 5 Number Summary Lecture 8 Andrew L. Zaeske Review 6A: Characterizing Data Range = maximum value - minimum value 6B: Measures of Variation 5 Number Summary : Minimum, Lower Quartile (Q1 ), Median, Upper Quartile (Q3 ), Maximum 6C: The Normal Distribution Lower Quartile is the median of the values between the Minimum and the Median Upper Quartile is the median of the values between the Median and the Maximum More complicated: Standard Deviation Lecture 8 Andrew L. Zaeske Review 6A: Characterizing Data 6B: Measures of Variation 6C: The Normal Distribution Deviations from mean = Value - mean Squared deviation = (Value - mean)2 Sum of squared deviations = square root of sum of squared deviations from mean, divided by number of values minus 1 p sumof (value − mean)2 countofvalues − 1 (1) Example #1 Lecture 8 Andrew L. Zaeske Review 6A: Characterizing Data Randomly generated integers: 10 2 2 8 2 10 3 1 8 5 6B: Measures of Variation Mean: 51/10=5.1, Median: (3+5)/2=4, Mode: 2 6C: The Normal Distribution 5 Number: 1, 2, 4, 8, 10 Standard deviation: 3.573 Example #2 Lecture 8 Andrew L. Zaeske Review 6A: Characterizing Data 6B: Measures of Variation 6C: The Normal Distribution Salaries of Brewer Starters: 550,000 11,000,000 11,250,000 12,500,000 3,325,000 Mean: 7,725,000 Median: 11,000,000 5 Number: 550,000 3,325,000 11,000,000 11,250,000 12,500,000 Standard deviation: 5,403,529 Example #3 Lecture 8 Andrew L. Zaeske Review 6A: Characterizing Data 6B: Measures of Variation 6C: The Normal Distribution Randomly Generated Numbers: 0.5188813 1.5681064 8.3100768 2.8875178 6.1874861 0.9655630 8.3006705 6.2502796 5.3124419 5.2274770 Mean: 4.55285 Median: (5.3124419+5.2274770)/2=5.269959 5 Number: 0.5188813 1.5681064 5.2699594 6.2502796 8.3100768 Standard deviation: 2.897736 Normal Distribution Lecture 8 Andrew L. Zaeske Review 6A: Characterizing Data 6B: Measures of Variation 6C: The Normal Distribution Normal: Symmetric, bell-shaped distribution with a single peak The peak is the mean, median and mode The standard deviation affects the height of the peak Rule: 66% of values are within 1 standard deviation of mean, 95% within 2, 99.7% within 3 Z-scores and Percentiles Lecture 8 Andrew L. Zaeske value−mean standarddeviation Review z-score = standard score = 6A: Characterizing Data Percentile of a value: Percentage of all values that are less than or equal to that value 6B: Measures of Variation 6C: The Normal Distribution Percentile and z-scores are the same: A z-score implies a specific percentile and each percentile has one z-score General procedure: Given a value, find a z-score and/or find what percentage will get above a particular score If given z-score, to solve for value rearrange: value = mean + z-score x standard deviation