Measures of Central Tendency Algebra A Unit 2, Lesson 1.

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Measures of Central Tendency Algebra A Unit 2, Lesson 1

Mean  The mean is the average value of a set of data.  How do we find the mean?  Add the values.  Divide by the number of values.

Sample Problem  Ramon received the following quiz scores: 36,92,86,88,79,94.  What was Ramon’s mean quiz score?  79

Median  The median is the middle value of a set of ordered data.  How do we find the median?  Put the values in order.  Locate the value in the middle.

Problem!  What if there are TWO middle values?  Add the two values and divide by two.

Sample Problem  Ramon received the following quiz scores: 36,92,86,88,79,94.  What was Ramon’s median quiz score? 36,79,86,88,92,94 87

Mode  The mode is the value that occurs most often.

Problem!  What if there are TWO modes?  What if there are NO modes?

Sample Problem  Ramon received the following quiz scores: 36,92,86,88,79,94.  What was Ramon’s mode quiz score? NO mode

Find the mean, median, and mode of the data below. Determine which measure of central tendency best describes the data Mode: none Median: List data in order. = 11.5For an even number of data items, find the mean of the two middle terms Mean: = total number of data items More Practice:

Vocabulary!  OUTLIER=A value in that is much higher or lower than the other numbers in the data set.  What effect does the outlier have on the MEAN?

Sample Problem  Ramon received the following quiz scores: 36,92,86,88,79,94.  Was there an outlier for Ramon’s quiz scores? YES, 36

Outlier effect: Ms. Sally teaches pre-school. Her students’ ages are: 3, 4, 4, 3, 2, 2, 3, 3, 3, 4, 5, 2, 3, and 4. Find the mean, median, and mode age of her students. 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 5 mean = 3.2 median = 3 mode = 3

Outlier effect: Ms. Sally is 52 years old. If her age is added to the data, find the new mean, median, and mode ages for her classroom. 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 5, 52 mean = 6.5 median = 3 mode = 3

Range: A Measure of Variation  The difference between the highest and lowest values.

Sample Problem  Ramon received the following quiz scores: 36,92,86,88,79,94.  What is the range of Ramon’s quiz scores? 94 – 36 = 58

Find the range and mean of each set of data. Use the range to compare the spread of the two sets of data Range: 47 – 34=13Range: 56 – 27=29 Both sets have a mean of 40. The first set of data has a lower range and is less spread out than the second data set. =40 Mean: Mean: