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Mean Absolute Deviations. MAFS.6.SP.2.5C. Cluster 2: Summarize describe distributions. MAFS.6.SP.2.5(C) Summarize numerical data sets in relation to their context, such as by:
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Mean Absolute Deviations MAFS.6.SP.2.5C
Cluster 2: Summarize describe distributions • MAFS.6.SP.2.5(C) Summarize numerical data sets in relation to their context, such as by: • Giving quantitative measures of center (median and/or mean) and variability (interquartile range and/or mean absolute deviation), as well as describing any overall pattern and any striking deviations from the overall pattern with reference to the context in which the data were gathered.
Learning Objectives • Distinguish between measures of central tendency, measures of variability, and measures of shape • Compute mean and mean absolute deviation
Introduction to the lesson • Common mistakes students make when analyzing data are confusing the measures of center and spread of given data. • The center of the data set is the mean, median, and mode. • The spread of the data set is mean absolute deviation. • The mean absolute deviation of a set of data is the average distance between each data value and the mean.
Measures of Central Tendency • Common Measures of Location • Mode • Median • Mean • Students need to know these common measures of central tendency before doing Mean Absolute Deviation
Mean Absolute Deviation • The mean absolute deviation of a set of data is the average distance between each data value and the mean. • 1st you have to find the mean. • 2nd you have to find the distance between each data value from the mean. • 3rd you find the average of the absolute values of the differencesbetween each value in the data set.
Absolute Mean Deviation • When finding the absolute mean deviation you need to follow three steps: 1. Find the mean of all values 2. Find the distance of each value from that mean (subtract the mean from each value, ignore negative signs) To show "Absolute Value" we put "| |" marks either side like this: |-3| = 3 • 3. Then find the mean of those distances
Absolute Mean Deviation Step 1 Find the mean . Step 2 find the distance of each value from the mean Data set: 3,6,6,7,8,15,16,19 find the mean of the data set. (3+6+6+7+8+15+16+19) ÷8 = 10 the mean is 10
Absolute Mean Deviation Step 3 Find the mean of those distance (7+4+4+3+2+5+6+9) ÷ 8 = 5 the mean of this data is 5. In addition, this will be your absolute mean deviation. The absolute mean deviation is 5. It tells us how far, on average, all values are from the middle. In this example the values are, on average, 5 away from the middle.
Guided Practice Question 1 • The table shows the number of visitors at Carver Middle School on the last six days of school. Find the mean absolute deviation of the set of data.
Solutions • Range 11-0=11 • Mean=(0+1+3+5+10+11)/6=5 • Mean Absolute Deviation=3.67 • |0-5|= 5 • |1-5|= 4 • |3-5|= 2 • |5-5|= 0 • |10-5|= 5 • +|11-5|= 6 • 22
Guided Practice Question 2 • The table shows the weight of five babies on April 1, 2014. Find the mean and mean absolute deviation for the set of data.
Solutions • Mean=(10.8 + 7.5 + 9.7 + 6.8 + 5.2)/5 = 8 • Mean absolute deviation • |10.8 – 8| = 2.8 • |7.5 – 8| = 0.5 • |9.7 – 8| = 1.7 • |6.8 – 8| = 1.2 • + |5.2 – 8| = 2.8 9.0