Advanced Placement Statistics Ch 1.2: Describing Distributions

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Presentation transcript:

Advanced Placement Statistics Ch 1.2: Describing Distributions EQ: What are the measures of center and spread and how are they used to describe distributions?

Stem Plot Histogram Dot Plot  organize data in graphs Components of Data Analysis  organize data in graphs  identify patterns and/or departures in trend  analyze using numerical summaries Stem Plot Histogram Dot Plot

Measures of Center: Mode MEAN MEDIAN --- occurs the most Balancing Point in a data set MEAN “x-bar” numerical average MEDIAN “x-tilde” but just say “median” middle number in a chronological list of data Mode --- occurs the most

5 6 6 8.67 Find the mean and median for the data set: 1, 3, 6, 6, 9 Mean = __________ Median = _________ Now include the value 27 in your data set and recalculate the mean and median. 6 8.67 Mean = __________ Median = _________ What does that tell you about the resistance of these measures ? Not Resistant Resistant Mean Median

RECALL: Bell-Shaped Distributions Mean Median Mode

A C B B C A Locate the measures of center for each distribution. Mean: Median: Median: Mode: Mode:

Mean Median Mean Median If a distribution is skew right, then ______ > ______. If a distribution is skew left, then _____ < _______. Mean Median

AP STAT EXAM NOTE: Just because median = mean, you cannot assume the distribution is symmetric. You must also look at the histogram, boxplot, or stemplot for shape . Although the median of 129 and the mean of 130.2 are very close , this distribution is clearly skewed right.

Symmetric Distribution Mean Skewed Distribution Median

Not Resistant Mean Resistant Median

32. a) From 1986 to 2004 the mean number of home runs hit by Barry Bonds is 37 and the median is 37. b) Stem and Leaf of Home Runs Hit By Barry Bonds c) Barry Bonds typically hit around 37 home runs per season. He had an extremely unusual season in 2001 when he hit 73 homeruns.

Mean = Median = Mean = Median = Both plots have the same mean and median, but are clearly shaped differently.

IQR = 16 LF: 28 – (1.5 x 16) = 4 UF: 44 + (1.5 x 16) = 68 Based on the outlier formula, there are no outliers in this data set.

Compare the shapes. The shape of the distribution for Bond’s data appears to be skewed right while the shape of the distribution of both Ruth and Hank’s data appears to be skewed left.

Compare the centers. The median for Ruth’s data, 46, is larger than the median for both Bond and Hank’s data, 37 and 38 respectively.

OR Compare the distributions. SOCS or CUSS Compare the spreads. The spread of Bond’s data is from 16 to 73 while the spread of Ruth’s data is from 22 to 60. The smallest spread, 13 to 47, is in Hank’s data. OR The largest range, 57, is in Bond’s data. The next largest range, 38, is in Ruth’s data. The smallest range, 34, is in Hank’s data.

Since there are no data values below the lower fence or above the upper fence for any of these distributions, there are no outliers.

6 s = 3.07 s2= 9.43

6 s = 1.07 s2= 1.14

6 s = 3.07 6 s = 1.07 Larger standard deviation for Hybrid A implies more variation in bloom size than Hybrid B. More likely to have varying bloom sizes to choose from. Smaller standard deviation for Hybrid B implies less variation in bloom size than Hybrid A. More likely to have consistent bloom sizes to choose from.

Resistant Based on MEDIAN Not Resistant Based on MEAN

STANDARD DEVIATION VARIANCE SYMMETRIC

QUARTILES IQR SKEWED

Create a box plot and a histogram for PREZ. Describe the distribution. SOCS The shape of this distribution is skew right. The lower fence occurs at 51 – (1.5 x 6.5) = 41.25. The upper fence occurs at 57.5 + (1.5 x 6.5) = 67.25. Therefore 68 and 69 are outliers. An appropriate measure of center is the median which occurs at 54.5 . An appropriate measure of spread is IQR which is 6.5. 57.5 – 51 = 6.5

A D E B C F

Comparing Shapes of Histograms to the Corresponding Boxplot

4 2 1 3

Class Activity With Partner: Matching Histograms & Box Plots

KEY Class Activity With Partner: Matching Histograms & Box Plots Solutions

Assignment: p. 67 #19, 23, 25, 26 p. 74 #28 - 31 p. 82 #33, 34, 36, 38 Take Home Quiz #1 Due Fri, Jan 15

7 15.202 3.899 15.202 17 3.899

35 379.977 19.494 Variance Standard Deviation NONE!!