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**Measures of Dispersion Measures of Variability**

Measures of Spread

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Outcomes Compare and contrast mean absolute deviation, variance, and standard deviation. Calculate and apply the three measures of dispersion. Understand the properties of three measures of dispersions. Interpret the mean absolute deviation and standard deviation. Determine when to use Mean Absolute Deviation and Standard Deviation

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**How do you describe a data set?**

Measures of Spread Range only uses the minimum and maximum. Extremely sensitive to outliers. Interquartile Range (IQR) Represents the middle 50%. It is relative to the median. Mean Absolute Deviation Variance Standard Deviation Measures of Central Tendencies Mean (μ, x) Median Mode These three new measures of variability are each relative to the mean.

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**Measures of Dispersion**

Measures of Central Tendency (Mean, Median, Mode) give a typical value for a data set. What value is the average, which is the middle value, which occurs the most, or which is most typical? Measures of Dispersion(Variability or Spread) provide measures to describe the data set’s variation from the typical value. Are the data points clustered tightly around the mean or median, somewhat near the mean or median, or very spread out from the mean or median?

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**Measures of Dispersion (Spread) Relative to the Mean**

Mean Absolute Deviation Standard Deviation Variance

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**Example: Comparing Data Sets**

In Central Tendency Park, a student counted the number of players playing basketball each day over a two week period and gathered the following data. Data Set#1 10 30 50 60 70 80 90 20 30 40 60

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**Example: Comparing Data Sets**

In Dispersion Park, another student counted the number of players playing basketball over a two week period and gathered the following data. 50 30 40 60 40 30 50 60

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**How are the two data sets similar and how are they different?**

In this case, one measure does not tell us very much about the data sets.

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**Compare and contrast the data sets**

Value Data Set #1 #2 10 1 20 2 30 3 40 50 6 60 70 80 90 Let’s look at the frequency of the values.

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**What do you observe? Data Set #1 Data Set #2**

How does the spread of the data compare between the two data sets? X Data Set #2 X X Where is the data relative to the mean?

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**One way to analyze a data set is to look at the deviations between each data value and the mean.**

We could then compare the deviations of one data set with the deviations of another comparable data set. How can we do this?

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**Deviation from the mean:**

10 30 50 60 70 80 90 20 40 +45 Deviation is the observed value minus the mean. +35 It can be positive or negative depending on whether the observed value is greater than or less than the mean. +25 +15 +15 +5 -5 -5 -15 -15 -15 -25 -25 -35 Data Set #1 (sorted)

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**Which data set has data values further from the mean or is more spread out?**

If we compare the sums of the deviations of the two data sets, we would find that both have a sum of 8. While the sum of the deviations are the same, the variation of data set is quite different. A 3 Each data set has a mean equal to 3. B 3

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**= 0 What if we found the average of all the deviations?**

This makes sense because we know mean is the balance point. = 0 -35 -25 -15 -5 +5 +15 +25 +35 +45 10 30 50 60 70 80 90 20 40 The sum of the deviations above the mean is +140. The sum of the deviations below the mean is -140. Data Set #1 (sorted)

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**Remember that distance is positive!**

What if we found the average of all the DISTANCES? Mean Absolute Deviation Data Set #1 (sorted) 10 30 50 60 70 80 90 20 40 35 25 15 5 45 Remember that distance is positive!

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**Compare the Mean Absolute Deviations**

Data Set #1 (sorted) 10 30 50 60 70 80 90 20 40 How do you expect the mean absolute deviation of data set #1 to compare to data set #2 ? In other words, compare the average distance of the data values from the mean for each data set. 60 30 40 50 60 Data Set #2 (sorted)

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**Calculate the Mean Absolute Deviation for Data Set #2**

Value |xi- µ| 30 15 40 5 50 60 Total 120 60 Data Set #2 (sorted) 30 50 40

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**How does Mean Absolute Deviation help us differentiate between the data sets?**

Having these two measures tell us more about how different or similar the data sets are. X Data Set #2 X The larger the value, the more spread out the data or the more variablity exists in the data. X

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**Mean Absolute Deviation**

Practice with a Small Data Set Value |xi- µ| 24.3 10.52 46.6 11.78 41.6 6.78 32.9 1.92 26.8 8.02 39.8 4.98 21.5 13.32 45.7 10.88 33.9 0.92 35.1 0.28 Total 69.4 The ages (years) of 10 grooms at their first marriage are given below. Determine the mean absolute deviation. 24.3, 46.6, 41.6, 32.9, 26.8, 39.8, 21.5, 45.7, 33.9, 35.1 Mean Absolute Deviation

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**Measures of Dispersion (Spread) Relative to the Mean**

The variance and standard deviation are the most common and useful measures of variability. Mean Absolute Deviation Standard Deviation Variance

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**Variance is the average of the deviations squared.**

-35 -25 -15 -5 +5 +15 +25 +35 +45 10 30 50 60 70 80 90 20 40 players2 σ2 By squaring the deviations, all squares will be positive. Therefore, the sum will not be zero. To do the formula substitution, notice there is one -35, two -25’s, three -15’s, etc. Data Set #1 (sorted)

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**Calculate the Variance for Data Set #2**

Value (xi- µ)2 30 225 40 25 50 60 Total 1350 60 Data Set #2 (sorted) 30 50 40

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**How does variance help us differentiate between the data sets?**

The unit for variance is the original data’s unit squared. Not always very meaningful when trying to interpret. Data Set #1 X Data Set #2 Variance can be very cumbersome to use especially with large data values and large spread. Squaring the deviation can result in large values. Just look at the value for data set #1. X X Again, the larger the value for variance, the more variability in the data.

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**Practice with a Small Data Set**

Variance Value (xi- µ)2 24.3 110.67 46.6 138.77 41.6 45.97 32.9 3.69 26.8 64.32 39.8 24.80 21.5 177.42 45.7 118.37 33.9 0.85 35.1 0.09 Total The ages (years) of 10 grooms at their first marriage are given below. Calculate the variance. 24.3, 46.6, 41.6, 32.9, 26.8, 39.8, 21.5, 45.7, 33.9, 35.1 σ2 = = 10

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**Measures of Dispersion (Spread) Relative to the Mean**

Mean Absolute Deviation Variance Standard Deviation

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**Standard Deviation of a Population Data Set**

Is just the positive square root of the variance. So for data set # 1 Mean Absolute Deviation, Variance and Standard Deviation will always be POSITIVE!!!!!!

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**Calculate the Standard Deviation for Data Set #2**

Value (xi-µ)2 30 225 40 25 50 60 Total 1350 60 Data Set #2 (sorted) 30 50 40

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**How do you do this on the calculator?**

For the TI, enter the data into a list, STAT, EDIT Select STAT, CALC, 1-Var Stats, and then list you entered the data into. For the Casio, Enter the data into a list, Select MENU, STAT Set up the calculator, Select F6 (Set) and verify the 1-Var List Xlist has the specified list, Exit. Calculate the mean and standard deviation, select 1-Var Stats. σx is the standard deviation of the population. Calculate variance by squaring the standard deviation. Take standard deviation out to 4 decimal places when calculating the variance. This way the answers calculated by the formula will be closer to those calculated from squaring the standard deviation.

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**How does Standard Deviation help us differentiate between the data sets?**

X Data Set #2 Just like mean absolute deviation, the more the data is spread out from the mean, the larger the value. X X

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**Practice with a Small Data Set**

The annual number of deaths from tornadoes in the United States from 1990 through 2000 is given. 53, 39, 39, 33, 69, 30, 25, 67, 130, 94, 40. Determine the standard deviation and variance.

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**How would you compare and contrast each pair of data sets?**

Problem 1 Data Set Data Set 2 μ = μ = 208 σ = σ = 23 Problem 2 Data Set Data Set 2 μ = μ = 65 Mean.Abs.Dev. = Mean.Abs.Dev. = 1.5

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**How would you compare the variability of the data of the two data sets?**

Problem 1-Answer Data Set 1 has a higher mean but it’s data is less spread out than Data Set 2 since its standard deviation is less than data set 2. Problem 1 Data Set Data Set 2 μ = μ = 208 σ = σ = 23 Problem 2 Data Set Data Set 2 μ = μ = 65 Mean.Abs.Dev. Mean.Abs.Dev. = = 1.5 Problem 2-Answer Data Set 1 has a lower mean but it’s data has more variability than Data Set 2 since its mean absolute deviation is more than Data Set 2.

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Without calculating the standard deviation, order each figure’s standard deviation from smallest to largest. A. B. µ = 4.5, σ =2.29 C. Answer: B, C, A

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**Describe the standard deviation you would expect for the situation given.**

You’re in charge of a carnival game, where the player will win a prize by being the first to shoot 3 ping pong balls through a hole. What type of standard deviation for the accuracy of the ping pong pistols would you want to have? What would the participants want? A larger standard deviation would indicate less pistol accuracy and less chance of getting the 3 ping pong balls in the hole. Thus, no prize. The participants would benefit from a small standard deviation, indicating a greater accuracy with the pistol.

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**Describe the standard deviation you would expect for the situation given.**

You are having laser eye surgery. Describe the relative size of the standard deviation you would want in the precision of the laser. You have been hired by a company that has employees with years of service ranging from 0 to 27 years. What type of standard deviation would you expect in the salaries of the employees? What would the data look like if the standard deviation is 0?

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**Population vs. Sample Standard Deviation for Data Set #1**

μ if data set is population Sample Standard Deviation Population Standard Deviation Need a casio screen shot of this. Throughout Algebra I and Algebra II, we will be dealing with population only!

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**Mean Absolute Deviation**

By examining the formulas, which of these would be the least sensitive to outliers? Mean Absolute Deviation Standard Deviation Variance

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**Mean Absolute Deviation**

VS. Standard Deviation Value |xi- µ| (xi- µ)2 30 15 225 348.57 40 5 25 75.17 50 1.77 60 128.37 100 55 Total 120/175 1350 We would expect the mean to increase. How much did each of our measures of spread increase? Which had the least amount of change? Which had the most amount of change? Mean absolute deviation is more resistant to outliers than variance or standard deviation. With variance and standard deviation, the deviation of the outlier is being squared, adding a significantly larger value to the sum. Let’s test the sensitivity of outliers, by adding a single value of 100 to data set #2.

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**Symbol Review Population Sample Number of Entries N n Mean Deviation**

Variance Standard Deviation

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Summary Mean absolute deviation, variance, and standard deviation are all measures of variability (dispersion or spread) relative to the mean. The sum of the deviations, , is equal to 0. Mean Absolute Deviation (MAD) It’s more resistant to outliers than variance and standard deviation . If the data has outliers, the MAD will be a better indicator of spread. The greater the value, the more dispersed the data will be from the mean. It’s always positive.

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**Summary Variance Standard Deviation**

It’s the square of Standard Deviation. Since the units of variance are units2, this measure is used less frequently than standard deviation. It’s always positive. Standard Deviation It’s the square root of variance. It’s more sensitive to outliers than Mean Absolute Deviation. The units are the same as the data. The greater its value, the more dispersed or spread out the data will be from the mean. The closer it is to 0, the more clustered the data will be about the mean.

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