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Madison claims that two data sets with the same median will have the same variability. Which data set would provide good support for whether her claim is true or false?

Her claim is true and she should use these data sets to provide support:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11

Her claim is false and she should use these data sets to provide support:
0, 0, 0, 0, 0, 10, 10, 10, 10, 10
5, 5, 5, 5, 5, 5, 5, 5, 5, 5



Answer :

GinnyAnswer
To determine whether Madison's claim that two data sets with the same median have the same variability is true or false, we can analyze the given data sets.

Let's consider the following data sets provided:

1. Data set 1: [tex]\(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\)[/tex]
2. Data set 2: [tex]\(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\)[/tex]

First, we need to calculate the median of both data sets:

### Median Calculation:
The median is the middle number of a data set when it is ordered in ascending or descending order. If the data set has an even number of observations, the median is the average of the two middle numbers.

For both data set 1 and data set 2 (which are identical), there are 12 numbers. The middle numbers in the ordered list are the 6th and 7th numbers (considering 1-based indexing).

So, let's calculate the median:
[tex]\[ \text{Median} = \frac{\text{6th number} + \text{7th number}}{2} = \frac{5 + 6}{2} = 5.5 \][/tex]

### Variability Calculation (Standard Deviation):
Variability refers to how spread out the numbers in a data set are. A common measure of variability is the standard deviation. The standard deviation (σ) is calculated using the formula:
[tex]\[ \sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (x_i - \mu)^2} \][/tex]
where [tex]\( x_i \)[/tex] are the data points, [tex]\( \mu \)[/tex] is the mean of the data points, and [tex]\( N \)[/tex] is the number of data points.

For both data set 1 and data set 2:

1. Calculate the mean (μ):
[tex]\[ \mu = \frac{0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11}{12} = \frac{66}{12} = 5.5 \][/tex]

2. Calculate the variance:
[tex]\[ \text{Variance} = \frac{1}{12} \sum_{i=1}^{12} (x_i - 5.5)^2 \][/tex]

Using the variance calculation, we can then directly calculate the standard deviation as the square root of the variance.

After completing these steps for both data sets, we find that:
[tex]\[ \text{Standard Deviation of Data Set 1} = 3.452052529534663 \][/tex]
[tex]\[ \text{Standard Deviation of Data Set 2} = 3.452052529534663 \][/tex]

### Conclusion:
Since both data sets have the same median (5.5) and the same standard deviation (3.452052529534663), Madison's claim that two data sets with the same median will have the same variability can be supported with these specific data sets.

In this case, her claim holds true. These data sets provide good support for her claim.

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