Summarizing Numerical Data Sets

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Presentation transcript:

Summarizing Numerical Data Sets ~ adapted from Walch Education

Measures of Center The mean and median are two measures of center. The mean is the average value of the data. The median is the middle-most value in a data set. These measures are used to generalize data sets and identify common or expected values.

Measures of Center, continued… The mean is useful when data sets do not contain values that vary greatly.

Measures of Variability Interquartile range and mean absolute deviation describe variability of the data set. Interquartile range is the difference between the third and first quartiles. The first quartile is the median of the lower half of the data set. The third quartile is the median of the upper half of the data set.

Measures of Variability, continued… The mean absolute deviation is the average absolute value of the difference between each data point and the mean. Measures of spread describe the variance of data values (how spread out they are), and identify the diversity of values in a data set. Measures of spread are used to help explain whether data values are very similar or very different.

Measures of Variability, continued… The mean absolute deviation takes the average distance of the data points from the mean.

Measures of Variability, continued… The interquartile range finds the distance between the two data values that represent the middle 50% of the data.

Let’s Try a Problem! A website captures information about each customer’s order. The total dollar amounts of the last 8 orders are listed in the table to the right. What is the mean absolute deviation of the data?

To find the mean absolute deviation of the data, start by finding the mean of the data set. NEXT STEP….Find the absolute value of the difference between each data value and the mean: |data value – mean|

Here we go… Find the sum of the absolute values of the differences. |21 – 21| = 0 |15 – 21| = 6 |22 – 21| = 1 |26 – 21| = 5 |24 – 21| = 3 |17 – 21| = 4 Find the sum of the absolute values of the differences. 0 + 6 + 1 + 5 + 3 + 0 + 4 + 1 = 20

Divide the sum of the absolute values of the differences by the number of data values. THE MEAN ABSOLUTE DEVIATION OF THE DOLLAR AMOUNTS OF EACH ORDER SET IS 2.5. THIS SAYS THAT THE AVERAGE COST DIFFERENCE BETWEEN THE ORDERS AND THE MEAN ORDER IS $2.50

Thanks for Watching!!!!! ~Ms. Dambreville